# nLab Introduction to Topology -- 1

Point-set Topology

This page contains a detailed introduction to basic topology. Starting from scratch (required background is just a basic concept of sets), and amplifying motivation from analysis, it first develops standard point-set topology (topological spaces). In passing, some basics of category theory make an informal appearance, used to transparently summarize some conceptually important aspects of the theory, such as initial and final topologies and the reflection into Hausdorff and sober topological spaces. We close with discussion of the basics of topological manifolds and differentiable manifolds, hence of differential topology, laying the foundations for differential geometry.

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main page: Introduction to Topology

this chapter: Introduction to Topology 1 – Point-set topology

next chapter: Introduction to Topology 2 – Basic Homotopy Theory

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For introduction to more general and abstract homotopy theory see instead at Introduction to Homotopy Theory.

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# Point-set Topology

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The idea of topology is to study “spaces” with “continuous functions” between them. Specifically one considers functions between sets (whence “point-set topology”, see below) such that there is a concept for what it means that these functions depend continuously on their arguments, in that their values do not “jump”. Such a concept of continuity is familiar from analysis on metric spaces, (recalled below) but the definition in topology generalizes this analytic concept and renders it more foundational, generalizing the concept of metric spaces to that of topological spaces. (def. below).

Hence, topology is the study of the category whose objects are topological spaces, and whose morphisms are continuous functions (see also remark below). This category is much more flexible than that of metric spaces, for example it admits the construction of arbitrary quotients and intersections of spaces. Accordingly, topology underlies or informs many and diverse areas of mathematics, such as functional analysis, operator algebra, manifold/scheme theory, hence algebraic geometry and differential geometry, and the study of topological groups, topological vector spaces, local rings, etc. Not the least, it gives rise to the field of homotopy theory, where one considers also continuous deformations of continuous functions themselves (“homotopies”). Topology itself has many branches, such as low-dimensional topology or topological domain theory.

A popular imagery for the concept of a continuous function is provided by deformations of elastic physical bodies, which may be deformed by stretching them without tearing. The canonical illustration is a continuous bijective function from the torus to the surface of a coffee mug, which maps half of the torus to the handle of the coffee mug, and continuously deforms parts of the other half in order to form the actual cup. Since the inverse function to this function is itself continuous, the torus and the coffee mug, both regarded as topological spaces, are “the same” for the purposes of topology; one says they are homeomorphic. On the other hand, there is no homeomorphism from the torus to, for instance, the sphere, signifying that these represent two topologically distinct spaces. Part of topology is concerned with studying homeomorphism-invariants of topological spaces (“topological properties”) which allow to detect by means of algebraic manipulations whether two topological spaces are homeomorphic (or more generally homotopy equivalent) or not. This is called algebraic topology. A basic algebraic invariant is the fundamental group of a topological space (discussed below), which measures how many ways there are to wind loops inside a topological space.

Beware the popular imagery of “rubber-sheet geometry”, which only captures part of the full scope of topology, in that it invokes spaces that locally still look like metric spaces (called topological manifolds, see below). But the concept of topological spaces is a good bit more general. Notably, finite topological spaces are either discrete or very much unlike metric spaces (example below); the former play a role in categorical logic. Also, in geometry, exotic topological spaces frequently arise when forming non-free quotients. In order to gauge just how many of such “exotic” examples of topological spaces beyond locally metric spaces one wishes to admit in the theory, extra “separation axioms” are imposed on topological spaces (see below), and the flavour of topology as a field depends on this choice.

Among the separation axioms, the Hausdorff space axiom is the most popular (see below). But the weaker axiom of sobriety (see below) stands out, because on the one hand it is the weakest axiom that is still naturally satisfied in applications to algebraic geometry (schemes are sober) and computer science (Vickers 89), and on the other, it fully realizes the strong roots that topology has in formal logic: sober topological spaces are entirely characterized by the union-, intersection- and inclusion-relations (logical conjunction, disjunction and implication) among their open subsets (propositions). This leads to a natural and fruitful generalization of topology to more general “purely logic-determined spaces”, called locales, and in yet more generality, toposes and higher toposes. While the latter are beyond the scope of this introduction, their rich theory and relation to the foundations of mathematics and geometry provide an outlook on the relevance of the basic ideas of topology.

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In this first part we discuss the foundations of the concept of “sets equipped with topology” (topological spaces) and of continuous functions between them.

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(classical logic)

The proofs in the following freely use the principle of excluded middle, hence proof by contradiction, and in a few places they also use the axiom of choice/Zorn's lemma.

Hence we discuss topology in its traditional form with classical logic.

We do however highlight the role of frame homomorphisms (def. below) and that of sober topological spaces (def. below). These concepts pave the way to a constructive formulation of topology in terms not of topological spaces but in terms of locales (remark below). For further reading along these lines see Johnstone 83.

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(set theory)

Apart from classical logic, we assume the usual informal concept of sets. The reader (only) needs to know the concepts of

1. subsets$S \subset X$;

2. complements$X \setminus S$ of subsets;

3. image sets $f(X)$ and pre-image sets $f^{-1}(Y)$ under a function

$f \colon X \to Y$;

4. unions$\underset{i \in I}{\cup} S_i$ and intersections $\underset{i \in I}{\cap} S_i$ of indexed sets of subsets $\{S_i \subset X\}_{i \in I}$.

The only rules of set theory that we use are the

For reference, we recall these:

###### Proposition

(images preserve unions but not in general intersections)

Let $f \colon X \longrightarrow Y$ be a function between sets. Let $\{S_i \subset X\}_{i \in I}$ be a set of subsets of $X$. Then

1. $f\left( \underset{i \in I}{\cup} S_i\right) = \left(\underset{i \in I}{\cup} f(S_i)\right)$ (the image under $f$ of a union of subsets is the union of the images)

2. $f\left( \underset{i \in I}{\cap} S_i\right) \subset \left(\underset{i \in I}{\cap} f(S_i)\right)$ (the image under $f$ of the intersection of the subsets is contained in the intersection of the images).

The injection in the second item is in general proper. If $f$ is an injective function and if $I$ is non-empty, then this is a bijection:

• $(f\,\text{injective}) \Rightarrow \left(f\left( \underset{i \in I}{\cap} S_i\right) = \left(\underset{i \in I}{\cap} f(S_i)\right)\right)$
###### Proposition

(pre-images preserve unions and intersections)

Let $f \colon X \longrightarrow Y$ be a function between sets. Let $\{T_i \subset Y\}_{i \in I}$ be a set of subsets of $Y$. Then

1. $f^{-1}\left( \underset{i \in I}{\cup} T_i\right) = \left(\underset{i \in I}{\cup} f^{-1}(T_i)\right)$ (the pre-image under $f$ of a union of subsets is the union of the pre-images),

2. $f^{-1}\left( \underset{i \in I}{\cap} T_i\right) = \left(\underset{i \in I}{\cap} f^{-1}(T_i)\right)$ (the pre-image under $f$ of the intersection of the subsets is the intersection of the pre-images).

###### Proposition

(de Morgan's law)

Given a set $X$ and a set of subsets

$\{S_i \subset X\}_{i \in I}$

then the complement of their union is the intersection of their complements

$X \setminus \left( \underset{i \in I}{\cup} S_i \right) \;\;=\;\; \underset{i \in I}{\cap} \left( X \setminus S_i \right)$

and the complement of their intersection is the union of their complements

$X \setminus \left( \underset{i \in I}{\cap} S_i \right) \;\;=\;\; \underset{i \in I}{\cup} \left( X \setminus S_i \right) \,.$

Moreover, taking complements reverses inclusion relations:

$\left( S_1 \subset S_2 \right) \;\;\Leftrightarrow\;\, \left( X\setminus S_2 \subset X \setminus S_1 \right) \,.$

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## Metric spaces

The concept of continuity was first made precise in analysis, in terms of epsilontic analysis on metric spaces, recalled as def. below. Then it was realized that this has a more elegant formulation in terms of the more general concept of open sets, this is prop. below. Adopting the latter as the definition leads to a more abstract concept of “continuous space”, this is the concept of topological spaces, def. below.

Here we briefly recall the relevant basic concepts from analysis, as a motivation for various definitions in topology. The reader who either already recalls these concepts in analysis or is content with ignoring the motivation coming from analysis should skip right away to the section Topological spaces.

###### Definition

(metric space)

A metric space is

1. a set $X$ (the “underlying set”);

2. a function $d \;\colon\; X \times X \to [0,\infty)$ (the “distance function”) from the Cartesian product of the set with itself to the non-negative real numbers

such that for all $x,y,z \in X$:

1. (symmetry) $d(x,y) = d(y,x)$

2. (triangle inequality) $d(x,z) \leq d(x,y)+ d(y,z)$.

3. (non-degeneracy) $d(x,y) = 0 \;\;\Leftrightarrow\;\; x = y$

###### Definition

(open balls)

Let $(X,d)$, be a metric space. Then for every element $x \in X$ and every $\epsilon \in \mathbb{R}_+$ a positive real number, we write

$B^\circ_x(\epsilon) \;\coloneqq\; \left\{ y \in X \;\vert\; d(x,y) \lt \epsilon \right\}$

for the open ball of radius $\epsilon$ around $x$. Similarly we write

$B_x(\epsilon) \;\coloneqq\; \left\{ y \in X \;\vert\; d(x,y) \leq \epsilon \right\}$

for the closed ball of radius $\epsilon$ around $x$. Finally we write

$S_x(\epsilon) \;\coloneqq\; \left\{ y \in X \;\vert\; d(x,y) = \epsilon \right\}$

for the sphere of radius $\epsilon$ around $x$.

For $\epsilon = 1$ we also speak of the unit open/closed ball and the unit sphere.

###### Definition

For $(X,d)$ a metric space (def. ) then a subset $S \subset X$ is called a bounded subset if $S$ is contained in some open ball (def. )

$S \subset B^\circ_x(r)$

around some $x \in X$ of some radius $r \in \mathbb{R}$.

A key source of metric spaces are normed vector spaces:

###### Definition

(normed vector space)

1. a real vector space $V$;

2. a function (the norm)

${\Vert {-} \Vert} \;\colon\; V \longrightarrow \mathbb{R}_{\geq 0}$

from the underlying set of $V$ to the non-negative real numbers,

such that for all $c \in \mathbb{R}$ with absolute value ${\vert c \vert}$ and all $v , w \in V$ it holds true that

1. (linearity) ${\Vert c v \Vert} = {\vert c \vert} {\Vert v \Vert }$;

2. (triangle inequality) ${\Vert v+w \Vert} \leq {\Vert v \Vert } + {\Vert w \Vert}$;

3. (non-degeneracy) if ${\Vert v \Vert} = 0$ then $v = 0$.

###### Proposition

Every normed vector space $(V, {\Vert {-} \Vert})$ becomes a metric space according to def. by setting

$d(x,y) \coloneqq {\Vert x-y \Vert} \,.$

Examples of normed vector spaces (def. ) and hence, via prop. , of metric spaces include the following:

###### Example

(Euclidean space)

For $n \in \mathbb{N}$, the Cartesian space

$\mathbb{R}^n = \left\{ \vec x = (x_i)_{i = 1}^n \vert x_i \in \mathbb{R} \right\}$

carries a norm (the Euclidean norm ) given by the square root of the sum of the squares of the components:

${\Vert \vec x \Vert} \;\coloneqq\; \sqrt{ \underoverset{i = 1}{n}{\sum} (x_i)^2 } \,.$

Via prop. this gives $\mathbb{R}^n$ the structure of a metric space, and as such it is called the Euclidean space of dimension $n$.

###### Example More generally, for $n \in \mathbb{N}$, and $p \in \mathbb{R}$, $p \geq 1$, then the Cartesian space $\mathbb{R}^n$ carries the p-norm

${\Vert \vec x \Vert}_p \coloneqq \root p {\sum_i {|x_i|^p}}$

One also sets

${\Vert \vec x \Vert}_\infty \coloneqq \underset{i \in I}{max} \, {\vert x_i \vert}$

and calls this the supremum norm.

The graphics on the right (grabbed from Wikipedia) shows unit circles (def. ) in $\mathbb{R}^2$ with respect to various p-norms.

By the Minkowski inequality, the p-norm generalizes to non-finite dimensional vector spaces such as sequence spaces and Lebesgue spaces.

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### Continuity

The following is now the fairly obvious definition of continuity for functions between metric spaces.

###### Definition

(epsilontic definition of continuity) For $(X,d_X)$ and $(Y,d_Y)$ two metric spaces (def. ), then a function

$f \;\colon\; X \longrightarrow Y$

is said to be continuous at a point $x \in X$ if for every positive real number $\epsilon$ there exists a positive real number $\delta$ such that for all $x' \in X$ that are a distance smaller than $\delta$ from $x$ then their image $f(x')$ is a distance smaller than $\epsilon$ from $f(x)$:

$\left( f\,\, \text{continuous at}\, x \right) \;\coloneqq\; \underset{{\epsilon \in \mathbb{R}} \atop {\epsilon \gt 0}}{\forall} \left( \underset{{\delta \in \mathbb{R}} \atop {\delta \gt 0}}{\exists} \left( \left( d_X(x,x') \lt \delta \right) \;\;\Rightarrow\;\; \left( d_Y(\,f(x), f(x')\,) \lt \epsilon \right) \right) \right) \,.$

The function $f$ is said to be continuous if it is continuous at every point $x \in X$.

###### Example

(distance function from a subset is continuous)

Let $(X,d)$ be a metric space (def. ) and let $S \subset X$ be a subset of the underlying set. Define then the function

$d(S,-) \;\colon\; X \to \mathbb{R}$

from the underlying set $X$ to the real numbers by assigning to a point $x \in X$ the infimum of the distances from $x$ to $s$, as $s$ ranges over the elements of $S$:

$d(S,x) \coloneqq inf \left\{ d(s,x) \,\vert\, s\in S \right\} \,.$

This is a continuous function, with $\mathbb{R}$ regarded as a metric space via its Euclidean norm (example ).

In particular the original distance function $d(x,-) = d(\{x\},-)$ is continuous in both its arguments.

###### Proof

Let $x \in X$ and let $\epsilon$ be a positive real number. We need to find a positive real number $\delta$ such that for $y \in X$ with $d(x,y) \lt \delta$ then ${\vert d(S,x) - d(S,y) \vert} \lt \epsilon$.

For $s \in S$ and $y \in X$, consider the triangle inequalities

\begin{aligned} d(s,x) & \leq d(s,y) + d(y,x) \\ d(s,y) & \leq d(s,x) + d(x,y) \end{aligned} \,.

Forming the infimum over $s \in S$ of all terms appearing here yields

\begin{aligned} d(S,x) & \leq d(S,y) + d(y,x) \\ d(S,y) & \leq d(S,x) + d(x,y) \end{aligned}

which implies

${\vert d(S,x) - d(S,y) \vert} \leq d(x,y) \,.$

This means that we may take for instance $\delta \coloneqq \epsilon$.

###### Example

(rational functions are continuous)

Consider the real line $\mathbb{R}$ regarded as the 1-dimensional Euclidean space $\mathbb{R}$ from example .

For $P \in \mathbb{R}[X]$ a polynomial, then the function

$\array{ f_P &\colon& \mathbb{R} &\longrightarrow& \mathbb{R} \\ && x &\mapsto& P(x) }$

is a continuous function in the sense of def. . Hence polynomials are continuous functions.

Similarly rational functions are continuous on their domain of definition: for $P,Q \in \mathbb{R}[X]$ two polynomials, then $\frac{f_P}{f_Q} \colon \mathbb{R} \setminus \{x | f_Q(x) = 0\} \to \mathbb{R}$ is a continuous function.

Also for instance forming the square root is a continuous function $\sqrt(-) \colon \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$.

On the other hand, a step function is continuous everywhere except at the finite number of points at which it changes its value, see example below.

We now reformulate the analytic concept of continuity from def. in terms of the simple but important concept of open sets:

###### Definition

(neighbourhood and open set)

Let $(X,d)$ be a metric space (def. ). Say that:

1. A neighbourhood of a point $x \in X$ is a subset $U_x \subset X$ which contains some open ball $B_x^\circ(\epsilon) \subset U_x$ around $x$ (def. ).

2. An open subset of $X$ is a subset $U \subset X$ such that for every $x \in U$ it also contains an open ball $B^\circ_x(\epsilon)$ around $x$ (def. ).

3. An open neighbourhood of a point $x \in X$ is a neighbourhood $U_x$ of $x$ which is also an open subset, hence equivalently this is any open subset of $X$ that contains $x$.

The following picture shows a point $x$, some open balls $B_i$ containing it, and two of its neighbourhoods $U_i$: graphics grabbed from Munkres 75

###### Example

(the empty subset is open)

Notice that for $(X,d)$ a metric space, then the empty subset $\emptyset \subset X$ is always an open subset of $(X,d)$ according to def. . This is because the clause for open subsets $U \subset X$ says that “for every point $x \in U$ there exists…”, but since there is no $x$ in $U = \emptyset$, this clause is always satisfied in this case.

Conversely, the entire set $X$ is always an open subset of $(X,d)$.

###### Example

(open/closed intervals)

Regard the real numbers $\mathbb{R}$ as the 1-dimensional Euclidean space (example ).

For $a \lt b \in \mathbb{R}$ consider the following subsets:

1. $(a,b) \coloneqq \left\{ x \in \mathbb{R} \vert a \lt x \lt b \right\}$ $\phantom{AA}$ (open interval)

2. $(a,b] \coloneqq \left\{ x \in \mathbb{R} \vert a \lt x \leq b \right\}$ $\phantom{AA}$ (half-open interval)

3. $[a,b) \coloneqq \left\{ x \in \mathbb{R} \vert a \leq x \lt b \right\}$ $\phantom{AA}$ (half-open interval)

4. $[a,b] \coloneqq \left\{ x \in \mathbb{R} \vert a \leq x \leq b \right\}$ $\phantom{AA}$ (closed interval)

The first of these is an open subset according to def. , the other three are not. The first one is called an open interval, the last one a closed interval and the middle two are called half-open intervals.

Similarly for $a,b \in \mathbb{R}$ one considers

1. $(-\infty,b) \coloneqq \left\{ x \in \mathbb{R} \vert x \lt b \right\}$ $\phantom{AA}$ (unbounded open interval)

2. $(a,\infty) \coloneqq \left\{ x \in \mathbb{R} \vert a \lt x \right\}$ $\phantom{AA}$ $\,\,$ (unbounded open interval)

3. $(-\infty,b] \coloneqq \left\{ x \in \mathbb{R} \vert x \leq b \right\}$ $\phantom{AA}$ (unbounded half-open interval)

4. $[a,\infty) \coloneqq \left\{ x \in \mathbb{R} \vert a \leq x \right\}$ $\phantom{AA}$ $\,\,$ (unbounded half-open interval)

The first two of these are open subsets, the last two are not.

For completeness we may also consider

• $(-\infty , \infty) = \mathbb{R}$

• $(a,a) = \emptyset$

which are both open, according to def. .

We may now rephrase the analytic definition of continuity entirely in terms of open subsets (def. ):

###### Proposition

(rephrasing continuity in terms of open sets)

Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces (def. ). Then a function $f \colon X \to Y$ is continuous in the epsilontic sense of def. precisely if it has the property that its pre-images of open subsets of $Y$ (in the sense of def. ) are open subsets of $X$:

$\left( f \,\, \text{continuous} \right) \;\;\Leftrightarrow\;\; \left( \left( O_Y \subset Y \,\, \text{open} \right) \,\Rightarrow\, \left( f^{-1}(O_Y) \subset X \,\, \text{open} \right) \right) \,.$

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principle of continuity

$\,$ $\,$ Continuous pre-Images of open subsets are open.

###### Proof

Observe, by direct unwinding the definitions, that the epsilontic definition of continuity (def. ) says equivalently in terms of open balls (def. ) that $f$ is continuous at $x$ precisely if for every open ball $B^\circ_{f(x)}(\epsilon)$ around an image point, there exists an open ball $B^\circ_x(\delta)$ around the corresponding pre-image point which maps into it:

$\array{ \left( f \,\,\text{continuous at}\,\, x \right) & \Leftrightarrow \; \underset{\epsilon \gt 0}{\forall} \left( \underset{\delta \gt 0}{\exists} \left( f(\;B_x^\circ(\delta)\;) \;\subset\; B^\circ_{f(x)}(\epsilon) \right) \right) \\ & \Leftrightarrow \; \underset{\epsilon \gt 0}{\forall} \left( \underset{\delta \gt 0}{\exists} \left( B^\circ_x(\delta) \subset f^{-1}\left( B^\circ_{f(x)}(\epsilon) \right) \right) \right) } \,.$

With this observation the proof immediate. For the record, we spell it out:

First assume that $f$ is continuous in the epsilontic sense. Then for $O_Y \subset Y$ any open subset and $x \in f^{-1}(O_Y)$ any point in the pre-image, we need to show that there exists an open neighbourhood of $x$ in $f^{-1}(O_Y)$.

That $O_Y$ is open in $Y$ means by definition that there exists an open ball $B^\circ_{f(x)}(\epsilon)$ in $O_Y$ around $f(x)$ for some radius $\epsilon$. By the assumption that $f$ is continuous and using the above observation, this implies that there exists an open ball $B^\circ_x(\delta)$ in $X$ such that $f(B^\circ_x(\delta)) \subset B^\circ_{f(x)}(\epsilon) \subset Y$, hence such that $B^\circ_x(\delta) \subset f^{-1}\left(B^{\circ}_{f(x)}(\epsilon)\right) \subset f^{-1}(O_Y)$. Hence this is an open ball of the required kind.

Conversely, assume that the pre-image function $f^{-1}$ takes open subsets to open subsets. Then for every $x \in X$ and $B_{f(x)}^\circ(\epsilon) \subset Y$ an open ball around its image, we need to produce an open ball $B_x^\circ(\delta) \subset X$ around $x$ such that $f(B_x^\circ(\delta)) \subset B^\circ_{f(x)}(\epsilon)$.

But by definition of open subsets, $B^\circ_{f(x)}(\epsilon) \subset Y$ is open, and therefore by assumption on $f$ its pre-image $f^{-1}(B^\circ_{f(x)}(\epsilon)) \subset X$ is also an open subset of $X$. Again by definition of open subsets, this implies that it contains an open ball as required.

###### Example

(step function) Consider $\mathbb{R}$ as the 1-dimensional Euclidean space (example ) and consider the step function

$\array{ \mathbb{R} &\overset{H}{\longrightarrow}& \mathbb{R} \\ x &\mapsto& \left\{ \array{ 0 & \vert \, x \leq 0 \\ 1 & \vert \, x \gt 0 } \right. } \,.$

graphics grabbed from Vickers 89

Consider then for $a \lt b \in \mathbb{R}$ the open interval $(a,b) \subset \mathbb{R}$, an open subset according to example . The preimage $H^{-1}(a,b)$ of this open subset is

$H^{-1} \;\colon\; (a,b) \mapsto \left\{ \array{ \emptyset & \vert \, a \geq 1 \;\;\text{or} \;\; b \leq 0 \\ \mathbb{R} & \vert \, a \lt 0 \;\;\text{and}\;\; b \gt 1 \\ \emptyset & \vert \, a \geq 0 \;\;\text{and}\;\; b \leq 1 \\ (0,\infty) & \vert \, 0 \leq a \lt 1 \;\;\text{and}\;\; b \gt 1 \\ (-\infty, 0] & \vert \, a \lt 0 \;\;\text{and}\;\; b \leq 1 } \right. \,.$

By example , all except the last of these pre-images listed are open subsets.

The failure of the last of the pre-images to be open witnesses that the step function is not continuous at $x = 0$.

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### Compactness

A key application of metric spaces in analysis is that they allow a formalization of what it means for an infinite sequence of elements in the metric space (def. below) to converge to a limit of a sequence (def. below). Of particular interest are therefore those metric spaces for which each sequence has a converging subsequence: the sequentially compact metric spaces (def. ).

We now briefly recall these concepts from analysis. Then, in the above spirit, we reformulate their epsilontic definition in terms of open subsets. This gives a useful definition that generalizes to topological spaces, the compact topological spaces discussed further below.

###### Definition

(sequence)

Given a set $X$, then a sequence of elements in $X$ is a function

$x_{(-)} \;\colon\; \mathbb{N} \longrightarrow X$

from the natural numbers to $X$.

A sub-sequence of such a sequence is a sequence of the form

$x_{\iota(-)} \;\colon\; \mathbb{N} \overset{\iota}{\hookrightarrow} \mathbb{N} \overset{x_{(-)}}{\longrightarrow} X$

for some injection $\iota$.

###### Definition

(convergence to limit of a sequence)

Let $(X,d)$ be a metric space (def. ). Then a sequence

$x_{(-)} \;\colon\; \mathbb{N} \longrightarrow X$

in the underlying set $X$ (def. ) is said to converge to a point $x_\infty \in X$, denoted

$x_i \overset{i \to \infty}{\longrightarrow} x_\infty$

if for every positive real number $\epsilon$, there exists a natural number $n$, such that all elements in the sequence after the $n$th one have distance less than $\epsilon$ from $x_\infty$.

$\left( x_i \overset{i \to \infty}{\longrightarrow} x_\infty \right) \,\Leftrightarrow\, \left( \underset{ {\epsilon \in \mathbb{R}} \atop {\epsilon \gt 0} }{\forall} \left( \underset{n \in \mathbb{N}}{\exists} \left( \underset{ {i \in \mathbb{N}} \atop {i \gt n} }{\forall} \; d(x_i, x_\infty) \leq \epsilon \right) \right) \right) \,.$

Here the point $x_\infty$ is called the limit of the sequence. Often one writes $\underset{i \to \infty}{\lim}x_i$ for this point.

###### Definition

(Cauchy sequence)

Given a metric space $(X,d)$ (def. ), then a sequence of points in $X$ (def. )

$x_{(-)} \;\colon\; \mathbb{N} \longrightarrow X$

is called a Cauchy sequence if for every positive real number $\epsilon$ there exists a natural number $n \in \mathbb{N}$ such that the distance between any two elements of the sequence beyond the $n$th one is less than $\epsilon$

$\left( x_{(-)} \,\, \text{Cauchy} \right) \,\Leftrightarrow\, \left( \underset{{\epsilon \in \mathbb{R}} \atop {\epsilon \gt 0}}{\forall} \left( \underset{N \in \mathbb{N}}{\exists} \left( \underset{{i,j \in \mathbb{N}} \atop {i,j \gt N }}{\forall} \; d(x_i, x_j) \leq \epsilon \right) \right) \right) \,.$
###### Definition

(complete metric space)

A metric space $(X,d)$ (def. ), for which every Cauchy sequence (def. ) converges (def. ) is called a complete metric space.

A normed vector space, regarded as a metric space via prop. that is complete in this sense is called a Banach space.

Finally recall the concept of compactness of metric spaces via epsilontic analysis:

###### Definition

(sequentially compact metric space)

A metric space $(X,d)$ (def. ) is called sequentially compact if every sequence in $X$ has a subsequence (def. ) which converges (def. ).

The key fact to translate this epsilontic definition of compactness to a concept that makes sense for general topological spaces (below) is the following:

###### Proposition

(sequentially compact metric spaces are equivalently compact metric spaces)

For a metric space $(X,d)$ (def. ) the following are equivalent:

1. $X$ is sequentially compact;

2. for every set $\{U_i \subset X\}_{i \in I}$ of open subsets $U_i$ of $X$ (def. ) which cover $X$ in that $X = \underset{i \in I}{\cup} U_i$, then there exists a finite subset $J \subset I$ of these open subsets which still covers $X$ in that also $X = \underset{i \in J \subset I}{\cup} U_i$.

The proof of prop. is most conveniently formulated with some of the terminology of topology in hand, which we introduce now. Therefore we postpone the proof to below.

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In summary prop. and prop. show that the purely combinatorial and in particular non-epsilontic concept of open subsets captures a substantial part of the nature of metric spaces in analysis. This motivates to reverse the logic and consider more general “spaces” which are only characterized by what counts as their open subsets. These are the topological spaces which we turn to now in def. (or, more generally, these are the “locales”, which we briefly consider below in remark ).

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## Topological spaces

Due to prop. we should pay attention to open subsets in metric spaces. It turns out that the following closure property, which follow directly from the definitions, is at the heart of the concept:

###### Proposition

(closure properties of open sets in a metric space)

The collection of open subsets of a metric space $(X,d)$ as in def. has the following properties:

1. The union of any set of open subsets is again an open subset.

2. The intersection of any finite number of open subsets is again an open subset.

###### Remark

(empty union and empty intersection)

Notice the degenerate case of unions $\underset{i \in I}{\cup} U_i$ and intersections $\underset{i \in I}{\cap} U_i$ of subsets $U_i \subset X$ for the case that they are indexed by the empty set $I = \emptyset$:

1. the empty union is the empty set itself;

2. the empty intersection is all of $X$.

(The second of these may seem less obvious than the first. We discuss the general logic behind these kinds of phenomena below.)

This way prop. is indeed compatible with the degenerate cases of examples of open subsets in example .

Proposition motivates the following generalized definition, which abstracts away from the concept of metric space just its system of open subsets:

###### Definition

(topological spaces)

Given a set $X$, then a topology on $X$ is a collection $\tau$ of subsets of $X$ called the open subsets, hence a subset of the power set $P(X)$

$\tau \subset P(X)$

such that this is closed under forming

1. finite intersections;

2. arbitrary unions.

In particular (by remark ):

• the empty set $\emptyset \subset X$ is in $\tau$ (being the union of no subsets)

and

• the whole set $X \subset X$ itself is in $\tau$ (being the intersection of no subsets).

A set $X$ equipped with such a topology is called a topological space.

###### Remark

In the field of topology it is common to eventually simply say “space” as shorthand for “topological space”. This is especially so as further qualifiers are added, such as “Hausdorff space” (def. below). But beware that there are other kinds of spaces in mathematics.

In view of example below one generalizes the terminology from def. as follows:

###### Definition

(neighbourhood)

Let $(X,\tau)$ be a topological space and let $x \in X$ be a point. A neighbourhood of $x$ is a subset $U_x \subset X$ which contains an open subset that still contains $x$.

An open neighbourhood is a neighbourhood that is itself an open subset, hence an open neighbourhood of $x$ is the same as an open subset containing $x$.

###### Remark

The simple definition of open subsets in def. and the simple implementation of the principle of continuity below in def. gives the field of topology its fundamental and universal flavor. The combinatorial nature of these definitions makes topology be closely related to formal logic. This becomes more manifest still for the “sober topological space” discussed below. For more on this perspective see the remark on locales below, remark . An introductory textbook amplifying this perspective is (Vickers 89).

$\,$

Before we look at first examples below, here is some common further terminology regarding topological spaces:

There is an evident partial ordering on the set of topologies that a given set may carry:

###### Definition

(finer/coarser topologies)

Let $X$ be a set, and let $\tau_1, \tau_2 \in P(X)$ be two topologies on $X$, hence two choices of open subsets for $X$, making it a topological space. If

$\tau_1 \subset \tau_2$

hence if every open subset of $X$ with respect to $\tau_1$ is also regarded as open by $\tau_2$, then one says that

• the topology $\tau_2$ is finer than the topology $\tau_2$

• the topology $\tau_1$ is coarser than the topology $\tau_1$.

With any kind of structure on sets, it is of interest how to “generate” such structures from a small amount of data:

###### Definition

(basis for the topology)

Let $(X, \tau)$ be a topological space, def. , and let

$\beta \subset \tau$

be a subset of its set of open subsets. We say that

1. $\beta$ is a basis for the topology $\tau$ if every open subset $O \in \tau$ is a union of elements of $\beta$;

2. $\beta$ is a sub-basis for the topology if every open subset $O \in \tau$ is a union of finite intersections of elements of $\beta$.

Often it is convenient to define topologies by defining some (sub-)basis as in def. . Examples are the the metric topology below, example , the binary product topology in def. below, and the compact-open topology on mapping spaces below in def. . To make use of this, we need to recognize sets of open subsets that serve as the basis for some topology:

###### Lemma

(recognition of topological bases)

Let $X$ be a set.

1. A collection $\beta \subset P(X)$ of subsets of $X$ is a basis for some topology $\tau \subset P(X)$ (def. ) precisely if

1. every point of $X$ is contained in at least one element of $\beta$;

2. for every two subsets $B_1, B_2 \in \beta$ and for every point $x \in B_1 \cap B_2$ in their intersection, there exists a $B \in \beta$ that contains $x$ and is contained in the intersection: $x \in B \subset B_1 \cap B_2$.

2. A subset $B \subset \tau$ of open subsets is a sub-basis for a topology $\tau$ on $X$ precisely if $\tau$ is the coarsest topology (def. ) which contains $B$.

$\,$

### Examples

We discuss here some basic examples of topological spaces (def. ), to get a feeling for the scope of the concept. But topological spaces are ubiquituous in mathematics, so that there are many more examples and many more classes of examples than could be listed. As we further develop the theory below, we encounter more examples, and more classes of examples. Below in Universal constructions we discuss a very general construction principle of new topological space from given ones.

First of all, our motivating example from above now reads as follows:

###### Example

(metric topology)

Let $(X,d)$ be a metric space (def. ). Then the collection of its open subsets in def. constitutes a topology on the set $X$, making it a topological space in the sense of def. . This is called the metric topology.

The open balls in a metric space constitute a basis of a topology (def. ) for the metric topology.

While the example of metric space topologies (example ) is the motivating example for the concept of topological spaces, it is important to notice that the concept of topological spaces is considerably more general, as some of the following examples show.

The following simplistic example of a (metric) topological space is important for the theory (for instance in prop. ):

###### Example

(empty space and point space)

On the empty set there exists a unique topology $\tau$ making it a topological space according to def. . We write also

$\emptyset \;\coloneqq\; \left( \emptyset, \tau_{\emptyset} = \{ \emptyset \} \right)$

for the resulting topological space, which we call the empty topological space.

On a singleton set $\{1\}$ there exists a unique topology $\tau$ making it a topological space according to def. , namelyf

$\tau \coloneqq \left\{ \emptyset , \{1\} \right\} \,.$

We write

$\ast \coloneqq \left( \left\{1\right\}, \tau \coloneqq \left\{ \emptyset, \left\{1\right\}\right\} \right)$

for this topological space and call it the point topological space.

This is equivalently the metric topology (example ) on $\mathbb{R}^0$, regarded as the 0-dimensional Euclidean space (example ).

###### Example

On the 2-element set $\{0,1\}$ there are (up to permutation of elements) three distinct topologies:

1. the codiscrete topology (def. ) $\tau = \left\{ \emptyset, \{0,1\} \right\}$;

2. the discrete topology (def. ), $\tau = \left\{ \emptyset, \{0\}, \{1\}, \{0,1\} \right\}$;

3. the Sierpinski space topology $\tau = \left\{\emptyset, \{1\}, \{0,1\} \right\}$.

###### Example

The following shows all the topologies on the 3-element set (up to permutation of elements) graphics grabbed from Munkres 75

###### Example

(discrete and co-discrete topology)

Let $S$ be any set. Then there are always the following two extreme possibilities of equipping $X$ with a topology $\tau \subset P(X)$ in the sense of def. , and hence making it a topological space:

1. $\tau \coloneqq P(S)$ the set of all open subsets;

this is called the discrete topology on $S$, it is the finest topology (def. ) on $X$,

we write $Disc(S)$ for the resulting topological space;

2. $\tau \coloneqq \{ \emptyset, S \}$ the set containing only the empty subset of $S$ and all of $S$ itself;

this is called the codiscrete topology on $S$, it is the coarsest topology (def. ) on $X$,

we write $CoDisc(S)$ for the resulting topological space.

The reason for this terminology is best seen when considering continuous functions into or out of these (co-)discrete topological spaces, we come to this in example below.

###### Example

(cofinite topology)

Given a set $X$, then the cofinite topology or finite complement topology on $X$ is the topology (def. ) whose open subsets are precisely

1. all cofinite subsets $S \subset X$ (i.e. those such that the complement $X \setminus S$ is a finite set);

2. the empty set.

If $X$ is itself a finite set (but not otherwise) then the cofinite topology on $X$ coincides with the discrete topology on $X$ (example ).

$\,$

We now consider basic construction principles of new topological spaces from given ones:

1. disjoint union spaces (example )

2. subspaces (example ),

3. quotient spaces (example )

4. product spaces (example ).

Below in Universal constructions we will recognize these as simple special cases of a general construction principle.

###### Example

(disjoint union space)

For $\{(X_i, \tau_i)\}_{i \in I}$ a set of topological spaces, then their disjoint union

$\underset{i \in I}{\sqcup} (X_i, \tau_i)$

is the topological space whose underlying set is the disjoint union of the underlying sets of the summand spaces, and whose open subsets are precisely the disjoint unions of the open subsets of the summand spaces.

In particular, for $I$ any index set, then the disjoint union of $I$ copies of the point space (example ) is equivalently the discrete topological space (example ) on that index set:

$\underset{i \in I}{\sqcup} \ast \;=\; Disc(I) \,.$
###### Example

(subspace topology) Let $(X, \tau_X)$ be a topological space, and let $S \subset X$ be a subset of the underlying set. Then the corresponding topological subspace has $S$ as its underlying set, and its open subsets are those subsets of $S$ which arise as restrictions of open subsets of $X$.

$\left( U_S \subset S\,\,\text{open} \right) \,\Leftrightarrow\, \left( \underset{U_X \in \tau_X}{\exists} \left( U_S = U_X \cap S \right) \right) \,.$

(This is also called the initial topology of the inclusion map. We come back to this below in def. .)

The picture on the right shows two open subsets inside the square, regarded as a topological subspace of the plane $\mathbb{R}^2$:

graphics grabbed from Munkres 75

###### Example

(quotient topological space)

Let $(X,\tau_X)$ be a topological space (def. ) and let

$R_\sim \subset X \times X$

be an equivalence relation on its underlying set. Then the quotient topological space has

and

• a subset $O \subset X/\sim$ is declared to be an open subset precisely if its preimage $\pi^{-1}(O)$ under the canonical projection map

$\pi \;\colon\; X \to X/\sim$

is open in $X$.

(This is also called the final topology of the projection $\pi$. We come back to this below in def. . )

Often one considers this with input datum not the equivalence relation, but any surjection

$\pi \;\colon\; X \longrightarrow Y$

of sets. Of course this identifies $Y = X/\sim$ with $(x_1 \sim x_2) \Leftrightarrow (\pi(x_1) = \pi(x_2))$. Hence the quotient topology on the codomain set of a function out of any topological space has as open subsets those whose pre-images are open.

To see that this indeed does define a topology on $X/\sim$ it is sufficient to observe that taking pre-images commutes with taking unions and with taking intersections.

###### Example

(binary product topological space) For $(X_1,\tau_{X_1})$ and $(X_2, \tau_{X_2})$ two topological spaces, then their binary product topological space has as underlying set the Cartesian product $X_1 \times X_2$ of the corresponding two underlying sets, and its topology is generated from the basis (def. ) given by the Cartesian products $U_1 \times U_2$ of the opens $U_i \in \tau_i$.

graphics grabbed from Munkres 75

Beware for non-finite products, the descriptions of the product topology is not as simple. This we turn to below in example , after introducing the general concept of limits in the category of topological spaces.

$\,$

The following examples illustrate how all these ingredients and construction principles may be combined.

The following example examines in more detail below in example , after we have introduced the concept of homeomorphisms below.

###### Example

Consider the real numbers $\mathbb{R}$ as the 1-dimensional Euclidean space (example ) and hence as a topological space via the corresponding metric topology (example ). Moreover, consider the closed interval $[0,1] \subset \mathbb{R}$ from example , regarded as a subspace (def. ) of $\mathbb{R}$.

The product space (example ) of this interval with itself

$[0,1] \times [0,1]$

is a topological space modelling the closed square. The quotient space (example ) of that by the relation which identifies a pair of opposite sides is a model for the cylinder. The further quotient by the relation that identifies the remaining pair of sides yields a model for the torus. graphics grabbed from Munkres 75

###### Example

(spheres and disks)

For $n \in \mathbb{N}$ write

• $D^n$ for the n-disk, the closed unit ball (def. ) in the $n$-dimensional Euclidean space $\mathbb{R}^n$ (example ) and equipped with the induced subspace topology (example ) of the corresponding metric topology (example );

• $S^{n-1}$ for the (n-1)-sphere (def. ) also equipped with the corresponding subspace topology;

• $i_n \;\colon\; S^{n-1} \hookrightarrow D^n$ for the continuous function that exhibits this boundary inclusion.

Notice that

The following important class of topological spaces form the foundation of algebraic geometry:

###### Example

(Zariski topology on affine space)

Let $k$ be a field, let $n \in \mathbb{N}$, and write $k[X_1, \cdots, X_n]$ for the set of polynomials in $n$ variables over $k$.

For $\mathcal{F} \subset k[X_1, \cdots, X_n]$ a subset of polynomials, let the subset $V(\mathcal{F}) \subset k^n$ of the $n$-fold Cartesian product of the underlying set of $k$ (the vanishing set of $\mathcal{F}$) be the subset of points on which all these polynomials jointly vanish:

$V(\mathcal{F}) \coloneqq \left\{ (a_1, \cdots, a_n) \in k^n \,\vert\, \underset{f \in \mathcal{F}}{\forall} f(a_1, \cdots, a_n) = 0 \right\} \,.$

These subsets are called the Zariski closed subsets.

Write

$\tau_{\mathbb{A}^n_k} \;\coloneqq\; \left\{ k^n \setminus V(\mathcal{F}) \subset k^n \,\vert\, \mathcal{F} \subset k[X_1, \cdots, X_n] \right\}$

for the set of complements of the Zariski closed subsets. These are called the Zariski open subsets of $k^n$.

The Zariski open subsets of $k^n$ form a topology (def. ), called the Zariski topology. The resulting topological space

$\mathbb{A}^n_k \;\coloneqq\; \left( k^n, \tau_{\mathbb{A}^n_k} \right)$

is also called the $n$-dimensional affine space over $k$.

More generally:

###### Example

(Zariski topology on the prime spectrum of a commutative ring)

Let $R$ be a commutative ring. Write $PrimeIdl(R)$ for its set of prime ideals. For $\mathcal{F} \subset R$ any subset of elements of the ring, consider the subsets of those prime ideals that contain $\mathcal{F}$:

$V(\mathcal{F}) \;\coloneqq\; \left\{ p \in PrimeIdl(R) \,\vert\, \mathcal{F} \subset p \right\} \,.$

These are called the Zariski closed subsets of $PrimeIdl(R)$. Their complements are called the Zariski open subsets.

Then the collection of Zariski open subsets in its set of prime ideals

$\tau_{Spec(R)} \subset P(PrimeIdl(R))$

satisfies the axioms of a topology (def. ), the Zariski topology.

$Spec(R) \coloneqq (PrimeIdl(R), \tau_{Spec(R)})$

is called (the space underlying) the prime spectrum of the commutative ring.

$\,$

### Closed subsets

The complements of open subsets in a topological space are called closed subsets (def. below). This simple definition indeed captures the concept of closure in the analytic sense of convergence of sequences (prop. below). Of particular interest for the theory of topological spaces in the discussion of separation axioms below are those closed subsets which are “irreducible” (def. below). These happen to be equivalently the “frame homomorphisms” (def. ) to the frame of opens of the point (prop. below).

###### Definition

(closed subsets) Let $(X,\tau)$ be a topological space (def. ).

1. A subset $S \subset X$ is called a closed subset if its complement $X \setminus S$ is an open subset:

$\left( S \subset X\,\, \text{is closed} \right) \phantom{AA} \Leftrightarrow \phantom{AA} \left( X\setminus S \, \subset X \,\, \text{is open} \right) \,.$

graphics grabbed from Vickers 89

2. If a singleton subset $\{x\} \subset X$ is closed, one says that $x$ is a closed point of $X$.

3. Given any subset $S \subset X$, then its topological closure $Cl(S)$ is the smallest closed subset containing $S$:

$Cl(S) \;\coloneqq\; \underset{ {C \subset X\, \text{closed} } \atop {S \subset C } }{\cap} \left( C \right) \,.$
4. A subset $S \subset X$ such that $Cl(S) = X$ is called a dense subset of $(X,\tau)$.

Often it is useful to reformulate def. of closed subsets as follows:

###### Lemma

(alternative characterization of topological closure)

Let $(X,\tau)$ be a topological space and let $S \subset X$ be a subset of its underlying set. Then a point $x \in X$ is contained in the topological closure $Cl(S)$ (def. ) precisely if every open neighbourhood $U_x \subset X$ of $x$ (def. ) intersects $S$:

$\left( x \in Cl(S) \right) \phantom{AA} \Leftrightarrow \phantom{AA} \not\left( \underset{ {U \subset X \setminus S} \atop { U \subset X \, \text{open} } }{\exists} \left( x \in U \right) \right) \,.$
###### Proof

Due to de Morgan duality (prop. ) we may rephrase the definition of the topological closure as follows:

\begin{aligned} Cl(S) & \coloneqq \underset{ {S \subset C } \atop { C \subset X\,\text{closed} } }{\cap} \left(C \right) \\ & = \underset{ { U \subset X \setminus S } \atop {U \subset X\, \text{open}} }{\cap} \left( X \setminus U \right) \\ & = X \setminus \left( \underset{ {U \subset X \setminus S} \atop { U \subset X\, \text{open} }}{\cup} U \right) \end{aligned} \,.
###### Proposition

(closure of a finite union is the union of the closures)

For $I$ a finite set and $\{U_i \subset X\}_{i \in I}$ a finite set of subsets of a topological space, we have

$Cl(\underset{i \in I}{\cup}U_i) = \underset{i \in I}{\cup} Cl(U_i) \,.$
###### Proof

By lemma we use that a point is in the closure of a set precisely if every open neighbourhood (def. ) of the point intersects the set.

Hence in one direction

$\underset{i \in I}{\cup} Cl(U_i) \subset Cl(\underset{i \in I}{\cup}U_i)$

because if every neighbourhood of a point intersects some $U_i$, then every neighbourhood intersects their union.

The other direction

$Cl(\underset{i \in I}{\cup}U_i) \subset \underset{i \in I}{\cup} Cl(U_i)$

is equivalent by de Morgan duality to

$X \setminus \underset{i \in I}{\cup} Cl(U_i) \subset X \setminus Cl(\underset{i \in I}{\cup}U_i)$

On left now we have the point for which there exists for each $i \in I$ a neighbourhood $U_{x,i}$ which does not intersect $U_i$. Since $I$ is finite, the intersection $\underset{i \in I}{\cap} U_{x,i}$ is still an open neighbourhood of $x$, and such that it intersects none of the $U_i$, hence such that it does not intersect their union. This implis that the given point is contained in the set on the right.

###### Definition

(topological interior and boundary)

Let $(X,\tau)$ be a topological space (def. ) and let $S \subset X$ be a subset. Then the topological interior of $S$ is the largest open subset $Int(S) \in \tau$ still contained in $S$, $Int(S) \subset S \subset X$:

$Int(S) \coloneqq \underset{{O \subset S} \atop {O \subset X\, \text{open}}}{\cup} \left( U \right) \,.$

The boundary $\partial S$ of $S$ is the complement of its interior inside its topological closure (def. ):

$\partial S \;\coloneqq\; Cl(S) \setminus Int(S) \,.$
###### Lemma

(duality between closure and interior)

Let $(X,\tau)$ be a topological space and let $S \subset X$ be a subset. Then the topological interior of $S$ (def. ) is the same as the complement of the topological closure $Cl(X\setminus S)$ of the complement of $S$:

$X \setminus Int(S) \, = \, Cl(\, X \setminus S \,)$

and conversely

$X \setminus Cl(S) \, = \, Int(\, X \setminus S \,) \,.$
###### Proof

Using de Morgan duality (prop. ), we compute as follows:

\begin{aligned} X \setminus Int(S) & = X \setminus \left( \underset{ {U \subset S} \atop {U \subset X \, open} }{\cup}U \right) \\ & = \underset{ {U \subset S} \atop {U \subset X \, \text{open}} }{\cap} \left( X \setminus U \right) \\ & = \underset{ {C \supset X \setminus S} \atop {C\, closed} }{\cap} \left( C \right) \\ & = Cl(X \setminus S) \end{aligned}

Similarly for the other case.

###### Example

(topological closure and interior of closed and open intervals)

Regard the real numbers as the 1-dimensional Euclidean space (example ) and equipped with the corresponding metric topology (example ) . Let $a \lt b \in \mathbb{R}$. Then the topological interior (def. ) of the closed interval $[a,b] \subset \mathbb{R}$ (example ) is the open interval $(a,b) \subset \mathbb{R}$, moreover the closed interval is its own topological closure (def. ) and the converse holds (by lemma ):

$\array{ Cl\left( \,(a,b)\, \right) \,=\, [a,b] && Int\left( \, (a,b) \, \right) \,=\, (a,b) \\ \\ Cl\left( \,[a,b]\, \right) \,=\, [a,b] && Int\left(\,[a,b]\,\right) \,=\, (a,b) } \,.$

Hence the boundary of the closed interval is its endpoints, while the boundary of the open interval is empty

$\array{ \partial [a,b] = \{a\} \cup \{b\} && \partial(a,b) = \emptyset } \,.$

The terminology “closed” subspace for complements of opens is justified by the following statement, which is a further example of how the combinatorial concept of open subsets captures key phenomena in analysis:

###### Proposition

(convergence in closed subspaces)

Let $(X,d)$ be a metric space (def. ), regarded as a topological space via example , and let $V \subset X$ be a subset. Then the following are equivalent:

1. $V \subset X$ is a closed subspace according to def. .

2. For every sequence $x_i \in V \subset X$ (def. ) with elements in $V$, which converges as a sequence in $X$ (def. ) to some $x_\infty \in X$, we have $x_\infty \in V \subset X$.

###### Proof

First assume that $V \subset X$ is closed and that $x_i \overset{i \to \infty}{\longrightarrow} x_{\infty}$ for some $x_\infty \in X$. We need to show that then $x_\infty \in V$. Suppose it were not, hence that $x_\infty \in X\setminus V$. Since, by assumption on $V$, this complement $X \setminus V \subset X$ is an open subset, it would follow that there exists a real number $\epsilon \gt 0$ such that the open ball around $x$ of radius $\epsilon$ were still contained in the complement: $B^\circ_x(\epsilon) \subset X \setminus V$. But since the sequence is assumed to converge in $X$, this would mean that there exists $N_\epsilon$ such that all $x_{i \gt N_{\epsilon}}$ are in $B^\circ_x(\epsilon)$, hence in $X\setminus V$. This contradicts the assumption that all $x_i$ are in $V$, and hence we have proved by contradiction that $x_\infty \in V$.

Conversely, assume that for all sequences in $V$ that converge to some $x_\infty \in X$ then $x_\infty \in V \subset X$. We need to show that then $V$ is closed, hence that $X \setminus V \subset X$ is an open subset, hence that for every $x \in X \setminus V$ we may find a real number $\epsilon \gt 0$ such that the open ball $B^\circ_x(\epsilon)$ around $x$ of radius $\epsilon$ is still contained in $X \setminus V$. Suppose on the contrary that such $\epsilon$ did not exist. This would mean that for each $k \in \mathbb{N}$ with $k \geq 1$ then the intersection $B^\circ_x(1/k) \cap V$ were non-empty. Hence then we could choose points $x_k \in B^\circ_x(1/k) \cap V$ in these intersections. These would form a sequence which clearly converges to the original $x$, and so by assumption we would conclude that $x \in V$, which violates the assumption that $x \in X \setminus V$. Hence we proved by contradiction $X \setminus V$ is in fact open.

Often one considers closed subsets inside a closed subspace. The following is immediate, but useful.

###### Lemma

(subsets are closed in a closed subspace precisely if they are closed in the ambient space)

Let $(X,\tau)$ be a topological space (def. ), and let $C \subset X$ be a closed subset (def. ), regarded as a topological subspace $(C,\tau_{sub})$ (example ). Then a subset $S \subset C$ is a closed subset of $(C,\tau_{sub})$ precisely if it is closed as a subset of $(X,\tau)$.

###### Proof

If $S \subset C$ is closed in $(C,\tau_{sub})$ this means equivalently that there is an open subset $V \subset C$ in $(C, \tau_{sub})$ such that

$S = C \setminus V \,.$

But by the definition of the subspace topology, this means equivalently that there is a subset $U \subset X$ which is open in $(X,\tau)$ such that $V = U \cap C$. Hence the above is equivalent to the existence of an open subset $U \subset X$ such that

\begin{aligned} S & = C \setminus V \\ & = C \setminus (U \cap C) \\ & = C \setminus U \end{aligned} \,.

But now the condition that $C$ itself is a closed subset of $(X,\tau)$ means equivalently that there is an open subset $W \subset X$ with $C = X \setminus W$. Hence the above is equivalent to the existence of two open subsets $W,U \subset X$ such that

$S = (X \setminus W) \setminus U = X \setminus (W \cup U) \,.$

Since the union $W \cup U$ is again open, this implies that $S$ is closed in $(X,\tau)$.

Conversely, that $S \subset X$ is closed in $(X,\tau)$ means that there exists an open $T \subset X$ with $S = X \setminus T \subset X$. This means that $S = S \cap C = (X \setminus T) \cap C = C\setminus T = C \setminus (T \cap C)$, and since $T \cap C$ is open in $(C,\tau_{sub})$ by definition of the subspace topology, this means that $S \subset C$ is closed in $(C, \tau_{sub})$.

A special role in the theory is played by the “irreducible” closed subspaces:

###### Definition

(irreducible closed subspace)

A closed subset $S \subset X$ (def. ) of a topological space $X$ is called irreducible if it is non-empty and not the union of two closed proper (i.e. smaller) subsets. In other words, a non-empty closed subset $S \subset X$ is irreducible if whenever $S_1, S_2 \subset X$ are two closed subspace such that

$S = S_1 \cup S_2$

then $S_1 = S$ or $S_2 = S$.

###### Example

(closures of points are irreducible)

For $x \in X$ a point inside a topological space, then the closure $Cl(\{x\})$ of the singleton subset $\{x\} \subset X$ is irreducible (def. ).

###### Example

(no nontrivial closed irreducibles in metric spaces)

Let $(X,d)$ be a metric space, regarded as a topological space via its metric topology (example ). Then every point $x \in X$ is closed (def ), hence every singleton subset $\{x\} \subset X$ is irreducible according to def. .

Let $\mathbb{R}$ be the 1-dimensional Euclidean space (example ) with its metric topology (example ). Then for $a \lt c \subset \mathbb{R}$ the closed interval $[a,c] \subset \mathbb{R}$ (example ) is not irreducible, since for any $b \in \mathbb{R}$ with $a \lt b \lt c$ it is the union of two smaller closed subintervals:

$[a,c] \,=\, [a, b] \cup [b, c] \,.$

In fact we will see below (prop. ) that in a metric space the singleton subsets are precisely the only irreducible closed subsets.

Often it is useful to re-express the condition of irreducibility of closed subspaces in terms of complementary open subsets:

###### Proposition

(irreducible closed subsets in terms of prime open subsets)

Let $(X, \tau)$ be a topological space, and let $P \in \tau$ be a proper open subset of $X$, hence so that the complement $F \coloneqq X\setminus P$ is a non-empty closed subspace. Then $F$ is irreducible in the sense of def. precisely if whenever $U_1,U_2 \in \tau$ are open subsets with $U_1 \cap U_2 \subset P$ then $U_1 \subset P$ or $U_2 \subset P$:

$\left( X \setminus P \,\,\text{irreducible} \right) \;\Leftrightarrow\; \left( \underset{U_1, U_2 \in \tau}{\forall } \left( \left( U_1 \cap U_2 \subset P \right) \;\Rightarrow\; \left(U_1 \subset P \;\text{or}\; U_2 \subset P\right) \right) \right) \,.$

The open subsets $P \subset X$ with this property are also called the prime open subsets in $\tau_X$.

###### Proof

Observe that every closed subset $F_i \subset F$ may be exhibited as the complement

$F_i = F \setminus U_i$

of some open subset $U_i \in \tau$ with respect to $F$. Observe that under this identification the condition that $U_1 \cap U_2 \subset P$ is equivalent to the condition that $F_1 \cup F_2 = F$, because it is equivalent to the equation labeled $(\star)$ in the following sequence of equations:

\begin{aligned} F_1 \cup F_2 & = (F \setminus U_1) \cup (F \setminus U_2) \\ & = \left( X \setminus (P \cup U_1) \right) \cup \left( X \setminus P \cup U_2 \right) \\ & = X \setminus \left( \left( P \cup U_1 \right) \cap \left( P \cup U_2 \right) \right) \\ & = X \setminus ( P \cup (U_1 \cap U_2) ) \\ & \stackrel{(\star)}{=} X \setminus P \\ & = F \,. \end{aligned} \,.

Similarly, the condition that $U_i \subset P$ is equivalent to the condition that $F_i = F$, because it is equivalent to the equality $(\star)$ in the following sequence of equalities:

\begin{aligned} F_i &= F \setminus U_i \\ & = X \setminus ( P \cup U_i ) \\ & \stackrel{(\star)}{=} X \setminus P \\ & = F \end{aligned} \,.

Under these identifications, the two conditions are manifestly the same.

We consider yet another equivalent characterization of irreducible closed subsets, prop. below, which will be needed in the discussion of the separation axioms further below. Stating this requires the following concept of “framehomomorphism, the natural kind of homomorphisms between topological spaces if we were to forget the underlying set of points of a topological space, and only remember the set $\tau_X$ with its operations induced by taking finite intersections and arbitrary unions:

###### Definition

(frame homomorphisms)

Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces (def. ). Then a function

$\tau_X \longleftarrow \tau_Y \;\colon\; \phi$

between their sets of open subsets is called a frame homomorphism from $\tau_Y$ to $\tau_X$ if it preserves

1. arbitrary unions;

In other words, $\phi$ is a frame homomorphism precisely if

1. for every set $I$ and every $I$-indexed set $\{U_i \in \tau_Y\}_{i \in I}$ of elements of $\tau_Y$, then

$\phi\left(\underset{i \in I}{\cup} U_i\right) \;=\; \underset{i \in I}{\cup} \phi(U_i)\;\;\;\;\in \tau_X \,,$
2. for every finite set $J$ and every $J$-indexed set $\{U_j \in \tau_Y\}_{j \in J}$ of elements in $\tau_Y$, then

$\phi\left(\underset{j \in J}{\cap} U_j\right) \;=\; \underset{j \in J}{\cap} \phi(U_j) \;\;\;\;\in \tau_X \,.$
###### Remark

(frame homomorphisms preserve inclusions)

A frame homomorphism $\phi$ as in def. necessarily also preserves inclusions in that

• for every inclusion $U_1 \subset U_2$ with $U_1, U_2 \in \tau_Y \subset P(Y)$ then

$\phi(U_1) \subset \phi(U_2) \;\;\;\;\;\;\; \in \tau_X \,.$

This is because inclusions are witnessed by unions

$(U_1 \subset U_2) \;\Leftrightarrow\; \left( U_1 \cup U_2 = U_2 \right)$

or alternatively because inclusions are witnessed by finite intersections:

$(U_1 \subset U_2) \;\Leftrightarrow\; \left( U_1 \cap U_2 = U_1 \right) \,.$
###### Example

(pre-images of continuous functions are frame homomorphisms)

Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be two topological spaces. One way to obtain a function between their sets of open subsets

$\tau_X \longleftarrow \tau_Y \;\colon\; \phi$

is to specify a function

$f \colon X \longrightarrow Y$

of their underlying sets, and take $\phi \coloneqq f^{-1}$ to be the pre-image operation. A priori this is a function of the form

$P(X) \longleftarrow P(Y) \;\colon\; f^{-1}$

and hence in order for this to co-restrict to $\tau_X \subset P(X)$ when restricted to $\tau_Y \subset P(Y)$ we need to demand that, under $f$, pre-images of open subsets of $Y$ are open subsets of $Z$. Below in def. we highlight these as the continuous functions between topological spaces.

$f \;\colon\; (X,\tau_X) \longrightarrow (Y, \tau_Y)$

In this case then

$\tau_X \longleftarrow \tau_Y \;\colon\; f^{-1}$

is a frame homomorphism from $\tau_Y$ to $\tau_X$ in the sense of def. , by prop. .

For the following recall from example the point topological space $\ast = (\{1\}, \tau_\ast = \left\{\emptyset, \{1\}\right\})$.

###### Proposition

(irreducible closed subsets are equivalently frame homomorphisms to opens of the point)

For $(X,\tau)$ a topological space, then there is a natural bijection between the irreducible closed subspaces of $(X,\tau)$ (def. ) and the frame homomorphisms from $\tau_X$ to $\tau_\ast$, and this bijection is given by

$\array{ FrameHom(\tau_X, \tau_\ast) &\underoverset{}{\simeq}{\longrightarrow}& IrrClSub(X) \\ \phi &\mapsto& X \setminus \left( U_\emptyset(\phi) \right) }$

where $U_\emptyset(\phi)$ is the union of all elements $U \in \tau_x$ such that $\phi(U) = \emptyset$:

$U_{\emptyset}(\phi) \coloneqq \underset{{U \in \tau_X} \atop {\phi(U) = \emptyset} }{\cup} \left( U \right) \,.$

###### Proof

First we need to show that the function is well defined in that given a frame homomorphism $\phi \colon \tau_X \to \tau_\ast$ then $X \setminus U_\emptyset(\phi)$ is indeed an irreducible closed subspace.

To that end observe that:

$(\ast)$ If there are two elements $U_1, U_2 \in \tau_X$ with $U_1 \cap U_2 \subset U_{\emptyset}(\phi)$ then $U_1 \subset U_{\emptyset}(\phi)$ or $U_2 \subset U_{\emptyset}(\phi)$.

This is because

\begin{aligned} \phi(U_1) \cap \phi(U_2) & = \phi(U_1 \cap U_2) \\ & \subset \phi(U_{\emptyset}(\phi)) \\ & = \emptyset \end{aligned} \,,

where the first equality holds because $\phi$ preserves finite intersections by def. , the inclusion holds because $\phi$ respects inclusions by remark , and the second equality holds because $\phi$ preserves arbitrary unions by def. . But in $\tau_\ast = \{\emptyset, \{1\}\}$ the intersection of two open subsets is empty precisely if at least one of them is empty, hence $\phi(U_1) = \emptyset$ or $\phi(U_2) = \emptyset$. But this means that $U_1 \subset U_{\emptyset}(\phi)$ or $U_2 \subset U_{\emptyset}(\phi)$, as claimed.

Now according to prop. the condition $(\ast)$ identifies the complement $X \setminus U_{\emptyset}(\phi)$ as an irreducible closed subspace of $(X,\tau)$.

Conversely, given an irreducible closed subset $X \setminus U_0$, define $\phi$ by

$\phi \;\colon\; U \mapsto \left\{ \array{ \emptyset & \vert \, \text{if} \, U \subset U_0 \\ \{1\} & \vert \, \text{otherwise} } \right. \,.$

This does preserve

1. arbitrary unions

because $\phi(\underset{i}{\cup} U_i) = \{\emptyset\}$ precisely if $\underset{i}{\cup}U_i \subset U_0$ which is the case precisely if all $U_i \subset U_0$, which means that all $\phi(U_i) = \emptyset$ and because $\underset{i}{\cup}\emptyset = \emptyset$;

while $\phi(\underset{i}{\cup}U_1) = \{1\}$ as soon as one of the $U_i$ is not contained in $U_0$, which means that one of the $\phi(U_i) = \{1\}$ which means that $\underset{i}{\cup} \phi(U_i) = \{1\}$;

2. finite intersections

because if $U_1 \cap U_2 \subset U_0$, then by $(\ast)$ $U_1 \subset U_0$ or $U_2 \subset U_0$, whence $\phi(U_1) = \emptyset$ or $\phi(U_2) = \emptyset$, whence with $\phi(U_1 \cap U_2) = \emptyset$ also $\phi(U_1) \cap \phi(U_2) = \emptyset$;

while if $U_1 \cap U_2$ is not contained in $U_0$ then neither $U_1$ nor $U_2$ is contained in $U_0$ and hence with $\phi(U_1 \cap U_2) = \{1\}$ also $\phi(U_1) \cap \phi(U_2) = \{1\} \cap \{1\} = \{1\}$.

Hence this is indeed a frame homomorphism $\tau_X \to \tau_\ast$.

Finally, it is clear that these two operations are inverse to each other.

$\,$

## Continuous functions

With the concept of topological spaces in hand (def. ) it is now immediate to formally implement in abstract generality the statement of prop. :

principle of continuity

$\,$ $\,$ Continuous pre-Images of open subsets are open.

###### Definition

(continuous function)

A continuous function between topological spaces (def. )

$f \colon (X, \tau_X) \to (Y, \tau_Y)$

is a function between the underlying sets,

$f \colon X \longrightarrow Y$

such that pre-images under $f$ of open subsets of $Y$ are open subsets of $X$.

We may equivalently state this in terms of closed subsets:

###### Proposition

Let $(X, \tau_X)$ and $(Y,\tau_Y)$ be two topological spaces (def. ). Then a function

$f \;\colon\; X \longrightarrow Y$

between the underlying sets is continuous in the sense of def. precisely if pre-images under $f$ of closed subsets of $Y$ (def. ) are closed subsets of $X$.

###### Proof

This follows since taking pre-images commutes with taking complements.

$\,$

Before looking at first examples of continuous functions below we consider now an informal remark on the resulting global structure, the “category of topological spaces”, remark below. This is a language that serves to make transparent key phenomena in topology which we encounter further below, such as the Tn-reflection (remark below), and the universal constructions.

###### Remark

(concrete category of topological spaces)

For $X_1, X_2, X_3$ three topological spaces and for

$X_1 \overset{f}{\longrightarrow} X_2 \phantom{AA}\text{and}\phantom{AA} X_2 \overset{g}{\longrightarrow} X_3$

two continuous functions (def. ) then their composition

$f_2 \circ f_1 \;\colon\; X_1 \overset{f}{\longrightarrow} X_2 \overset{f_2}{\longrightarrow} X_3$

is clearly itself again a continuous function from $X_1$ to $X_3$.

Moreover, this composition operation is clearly associative, in that for

$X_1 \overset{f}{\longrightarrow} X_2 \phantom{AA}\text{and}\phantom{AA} X_2 \overset{g}{\longrightarrow} X_3 \phantom{AA}\text{and}\phantom{AA} X_3 \overset{h}{\longrightarrow} X_4$

three continuous functions, then

$f_3 \circ (f_2 \circ f_1) = (f_3 \circ f_2) \circ f_1 \;\colon\; X_1 \longrightarrow X_4 \,.$

Finally, the composition operation is also clearly unital, in that for each topological space $X$ there exists the identity function $id_X \colon X \to X$ and for $f \colon X_1 \to X_2$ any continuous function then

$id_{X_2} \circ f \;=\; f \;=\; f \circ id_{X_1} \,.$

One summarizes this situation by saying that:

1. topological spaces constitute the objects,

2. continuous functions constitute the morphisms (homomorphisms) of a category, called the category of topological spaces (“Top” for short).

It is useful to depict collections of objects with morphisms between them by diagrams, like this one:

$\,$

$\,$

$\,$

graphics grabbed from Lawvere-Schanuel 09.

There are other categories. For instance there is the category of sets (“Set” for short) whose

1. objects are sets,

2. morphisms are plain functions between these.

The two categories Top and Set are different, but related. After all,

1. an object of Top (hence a topological space) is an object of Set (hence a set) equipped with extra structure (namely with a topology);

2. a morphism in Top (hence a continuous function) is a morphism in Set (hence a plain function) with the extra property that it preserves this extra structure.

Hence we have the underlying set assigning function

$\array{ Top &\overset{U}{\longrightarrow}& Set \\ \\ (X,\tau) &\overset{\phantom{AAa}}{\mapsto}& X }$

from the class of topological spaces to the class of sets. But more is true: every continuous function between topological spaces is, by definition, in particular a function on underlying sets:

$\array{ Top &\overset{U}{\longrightarrow}& Set \\ \\ (X,\tau_X) &\overset{\phantom{AAA}}{\mapsto}& X \\ {}^{\mathllap{f}}\downarrow &\overset{}{\mapsto}& \downarrow^{\mathrlap{f}} \\ (Y, \tau_Y) &\underset{\phantom{AAA}}{\mapsto}& Y }$

and this assignment (trivially) respects the composition of morphisms and the identity morphisms.

Such a function between classes of objects of categories, which is extended to a function on the sets of homomorphisms between these objects in a way that respects composition and identity morphisms is called a functor. If we write an arrow between categories

$U \;\colon\; Top \longrightarrow Set$

then it is understood that we mean not just a function between their classes of objects, but a functor.

The functor $U$ at hand has the special property that it does not do much except forgetting extra structure, namely the extra structure on a set $X$ given by a choice of topology $\tau_X$. One also speaks of a forgetful functor.

This is intuitively clear, and we may easily formalize it: The functor $U$ has the special property that as a function between sets of homomorphisms (“hom sets”, for short) it is injective. More in detail, given topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$ then the component function of $U$ from the set of continuous function between these spaces to the set of plain functions between their underlying sets

$\array{ \left\{ (X,\tau_X) \underoverset{\text{function}}{\text{continuous}}{\longrightarrow} (Y,\tau_Y) \right\} &\;\overset{\phantom{AA}U\phantom{AA}}{\mapsto}\;& \left\{ X \underset{\text{function}}{\longrightarrow} Y \right\} }$

is an injective function, including the continuous functions among all functions of underlying sets.

A functor with this property, that its component functions between all hom-sets are injective, is called a faithful functor.

A category equipped with a faithful functor to Set is called a concrete category.

Hence Top is canonically a concrete category.

###### Example

(product topological space construction is functorial)

For $\mathcal{C}$ and $\mathcal{D}$ two categories as in remark (for instance Top or Set) then we obtain a new category denoted $\mathcal{C} \times \mathcal{D}$ and called their product category whose

1. objects are pairs $(c,d)$ with $c$ an object of $\mathcal{C}$ and $d$ an object of $\mathcal{D}$;
• morphisms are pairs $(f,g) \;\colon\; (c,d) \to (c', d')$ with $f \colon c \to c'$ a morphism of $\mathcal{C}$

and $g \colon d \to d'$ a morphism of $\mathcal{D}$,

• composition of morphisms is defined pairwise $(f', g') \circ (f,g) \coloneqq ( f' \circ f, g' \circ g )$.

This concept secretly underlies the construction of product topological spaces:

Let $(X_1,\tau_{X_1})$, $(X_2, \tau_{X_2})$, $(Y_1, \tau_{Y_1})$ and $(Y_2, \tau_{Y_2})$ be topological spaces. Then for all pairs of continuous functions

$f_1 \;\colon\; (X_1, \tau_{X_1}) \longrightarrow (Y_1, \tau_{Y_1})$

and

$f_2 \;\colon\; (X_2, \tau_{X_2}) \longrightarrow (Y_2, \tau_{Y_2})$

the canonically induced function on Cartesian products of sets

$\array{ X_1 \times X_2 & \overset{ f_1 \times f_2 }{\longrightarrow} & Y_1 \times Y_2 \\ (x_1, x_2) &\mapsto& ( f_1(x_1), f_2(x_2) ) }$

is clearly a continuous function with respect to the binary product space topologies (def. )

$f_1 \times f_2 \;\colon\; (X_1 \times X_2, \tau_{X_1 \times X_2}) \longrightarrow (Y_1 \times Y_2, \tau_{Y_1 \times Y_2}) \,.$

Moreover, this construction respects identity functions and composition of functions in both arguments.

In the language of category theory (remark ), this is summarized by saying that the product topological space construction $(-) \times (-)$ extends to a functor from the product category of the category Top with itself to itself:

$(-) \times (-) \;\colon\; Top \times Top \longrightarrow Top \,.$

$\,$

### Examples

We discuss here some basic examples of continuous functions (def. ) between topological spaces (def. ) to get a feeling for the nature of the concept. But as with topological spaces themselves, continuous functions between them are ubiquitous in mathematics, and no list will exhaust all classes of examples. Below in the section Universal constructions we discuss a general principle that serves to produce examples of continuous functions with prescribed “universal properties”.

###### Example

(point space is terminal)

For $(X,\tau)$ any topological space, then there is a unique continuous function

1. from the empty topological space (def. ) $X$

$\emptyset \overset{\phantom{AA} \exists ! \phantom{AA} }{\longrightarrow} X$
2. from $X$ to the point topological space (def. ).

$X \overset{\phantom{AA} \exists ! \phantom{AA}}{\longrightarrow} \ast$

In the language of category theory (remark ), this says that

1. the point space $\ast$ is the terminal object

in the category Top of topological spaces. We come back to this below in example .

###### Example

(constant continuous functions)

For $(X, \tau)$ a topological space then for $x \in X$ any element of the underlying set, there is a unique continuous function (which we denote by the same symbol)

$x \;\colon\; \ast \longrightarrow X$

from the point topological space (def. ), whose image in $X$ is that element. Hence there is a natural bijection

$\left\{ \ast \overset{f}{\to} X \,\vert\, f \,\,\text{continuous} \right\} \;\simeq\; X$

between the continuous functions from the point to any topological space, and the underlying set of that topological space.

More generally, for $(X,\tau_X)$ and $(Y,\tau_Y)$ two topological spaces, then a continuous function $X \to Y$ between them is called a constant function with value some point $y \in Y$ if it factors through the point spaces as

$const_y \;\colon\; X \overset{\exists !}{\longrightarrow} \ast \overset{y}{\longrightarrow} Y \,.$
###### Definition

(locally constant function)

For $(X,\tau_X)$, $(Y,\tau_Y)$ two topological spaces, then a continuous function $f \colon (X,\tau_X) \to (Y,\tau_Y)$ (def. ) is called locally constant if every point $x \in X$ has a neighbourhood (def. ) on which the function is constant.

###### Example

(continuous functions into and out of discrete and codiscrete spaces)

Let $S$ be a set and let $(X,\tau)$ be a topological space. Recall from example

1. the discrete topological space $Disc(S)$;

2. the co-discrete topological space $CoDisc(S)$

on the underlying set $S$. Then continuous functions (def. ) into/out of these satisfy:

1. every function (of sets) $Disc(S) \longrightarrow X$ out of a discrete space is continuous;

2. every function (of sets) $X \longrightarrow CoDisc(S)$ into a codiscrete space is continuous.

Also:

###### Example

(diagonal)

For $X$ a set, its diagonal $\Delta_X$ is the function from $X$ to the Cartesian product of $X$ with itself, given by

$\array{ X &\overset{\Delta_X}{\longrightarrow}& X \times X \\ x &\mapsto& (x,x) }$

For $(X,\tau)$ a topological space, then the diagonal is a continuous function to the product topological space (def. ) of $X$ with itself.

$\Delta_X \;\colon\; (X, \tau) \longrightarrow (X \times X, \tau_{X \times X}) \,.$

To see this, it is sufficient to see that the preimages of basic opens $U_1 \times U_2$ in $\tau_{X \times X}$ are in $\tau_X$. But these pre-images are the intersections $U_1 \cap U_2 \subset X$, which are open by the axioms on the topology $\tau_X$.

###### Example

(image factorization)

Let $f \;\colon\; (X, \tau_X) \longrightarrow (Y,\tau_Y)$ be a continuous function.

Write $f(X) \subset Y$ for the image of $f$ on underlying sets, and consider the resulting factorization of $f$ through $f(X)$ on underlying sets:

$f \;\colon\; X \overset{\text{surjective}}{\longrightarrow} f(X) \overset{\text{injective}}{\longrightarrow} Y \,.$

There are the following two ways to topologize the image $f(X)$ such as to make this a sequence of two continuous functions:

1. By example $f(X)$ inherits a subspace topology from $(Y,\tau_Y)$ which evidently makes the inclusion $f(X) \longrightarrow Y$ a continuous function.

Observe that this also makes $X \to f(X)$ a continuous function: An open subset of $f(X)$ in this case is of the form $U_Y \cap f(X)$ for $U_Y \in \tau_Y$, and $f^{-1}( U_Y \cap f(X) ) = f^{-1}(U_Y)$, which is open in $X$ since $f$ is continuous.

2. By example $f(X)$ inherits a quotient topology from $(X,\tau_X)$ which evidently makes the surjection $X \longrightarrow f(X)$ a continuous function.

Observe that this also makes $f(X) \longrightarrow Y$ a continuous function: The preimage under this map of an open subset $U_Y \in \tau_Y$ is the restriction $U_Y \cap f(X)$, and the pre-image of that under $X \to f(X)$ is $f^{-1}(U_Y)$, as before, which is open since $f$ is continuous, and therefore $U_Y \cap f(X)$ is open in the quotient topology.

$\,$

Beware, in general a continuous function itself (as opposed to its pre-image function) neither preserves open subsets, nor closed subsets, as the following examples show:

###### Example

Regard the real numbers $\mathbb{R}$ as the 1-dimensional Euclidean space (example ) equipped with the metric topology (example ). For $a \in \mathbb{R}$ the constant function (example )

$\array{ \mathbb{R} &\overset{const_a}{\longrightarrow}& \mathbb{R} \\ x &\mapsto& a }$

maps every open subset $U \subset \mathbb{R}$ to the singleton set $\{a\} \subset \mathbb{R}$, which is not open.

###### Example

Write $Disc(\mathbb{R})$ for the set of real numbers equipped with its discrete topology (def. ) and $\mathbb{R}$ for the set of real numbers equipped with its Euclidean metric topology (example , example ). Then the identity function on the underlying sets

$id_{\mathbb{R}} \;\colon\; Disc(\mathbb{R}) \longrightarrow \mathbb{R}$

is a continuous function (a special case of example ). A singleton subset $\{a\} \in Disc(\mathbb{R})$ is open, but regarded as a subset $\{a\} \in \mathbb{R}$ it is not open.

###### Example

Consider the set of real numbers $\mathbb{R}$ equipped with its Euclidean metric topology (example , example ). The exponential function

$\exp(-) \;\colon\; \mathbb{R} \longrightarrow \mathbb{R}$

maps all of $\mathbb{R}$ (which is a closed subset, since $\mathbb{R} = \mathbb{R} \setminus \emptyset$) to the open interval $(0,\infty) \subset \mathbb{R}$, which is not closed.

Those continuous functions that do happen to preserve open or closed subsets get a special name:

###### Definition

(open maps and closed maps)

A continuous function $f \colon (X,\tau_X) \to (Y, \tau_Y)$ (def. ) is called

• an open map if the image under $f$ of an open subset of $X$ is an open subset of $Y$;

• a closed map if the image under $f$ of a closed subset of $X$ (def. ) is a closed subset of $Y$.

###### Example

(image projections of open/closed maps are themselves open/closed)

If a continuous function $f \colon (X,\tau_X) \to (Y,\tau_Y)$ is an open map or closed map (def. ) then so its its image projection $X \to f(X) \subset Y$, respectively, for $f(X) \subset Y$ regarded with its subspace topology (example ).

###### Proof

If $f$ is an open map, and $O \subset X$ is an open subset, so that $f(O) \subset Y$ is also open in $Y$, then, since $f(O) = f(O) \cap f(X)$, it is also still open in the subspace topology, hence $X \to f(X)$ is an open map.

If $f$ is a closed map, and $C \subset X$ is a closed subset so that also $f(C) \subset Y$ is a closed subset, then the complement $Y \setminus f(C)$ is open in $Y$ and hence $(Y \setminus f(C)) \cap f(X) = f(X) \setminus f(C)$ is open in the subspace topology, which means that $f(C)$ is closed in the subspace topology.

###### Example

(projections are open continuous functions )

For $(X_1,\tau_{X_1})$ and $(X_2,\tau_{X_2})$ two topological spaces, then the projection maps

$pr_i \;\colon\; (X_1 \times X_2, \tau_{X_1 \times X_2}) \longrightarrow (X_i, \tau_{X_i})$

out of their product topological space (def. )

$\array{ X_1 \times X_2 &\overset{pr_1}{\longrightarrow}& X_1 \\ (x_1, x_2) &\overset{\phantom{AAA}}{\mapsto}& x_1 }$
$\array{ X_1 \times X_2 &\overset{pr_2}{\longrightarrow}& X_2 \\ (x_1, x_2) &\overset{\phantom{AAA}}{\mapsto}& x_2 }$

are open continuous functions (def. ).

This is because, by definition, every open subset $O \subset X_1 \times X_2$ in the product space topology is a union of products of open subsets $U_i \in X_1$ and $V_i \in X_2$ in the factor spaces

$O = \underset{i \in I}{\cup} \left( U_i \times V_i \right)$

and because taking the image of a function preserves unions of subsets

\begin{aligned} pr_1\left( \underset{i \in I}{\cup} \left( U_i \times V_i \right) \right) & = \underset{i \in I}{\cup} pr_1 \left( U_i \times V_i \right) \\ & = \underset{i \in I}{\cup} U_i \end{aligned} \,.

Below in prop. we find a large supply of closed maps.

$\,$

Sometimes it is useful to recognize quotient topological space projections via saturated subsets (essentially another term for pre-images of underlying sets):

###### Definition

(saturated subset)

Let $f \;\colon\; X \longrightarrow Y$ be a function of sets. Then a subset $S \subset X$ is called an $f$-saturated subset (or just saturated subset, if $f$ is understood) if $S$ is the pre-image of its image:

$\left(S \subset X \,\, f\text{-saturated} \right) \,\Leftrightarrow\, \left( S = f^{-1}(f(S)) \right) \,.$

Here $f^{-1}(f(S))$ is also called the $f$-saturation of $S$.

###### Example

(pre-images are saturated subsets)

For $f \;\colon\; X \to Y$ any function of sets, and $S_Y \subset Y$ any subset of $Y$, then the pre-image $f^{-1}(S_Y) \subset X$ is an $f$-saturated subset of $X$ (def. ).

Observe that:

###### Lemma

Let $f \colon X \longrightarrow Y$ be a function. Then a subset $S \subset X$ is $f$-saturated (def. ) precisely if its complement $X \setminus S$ is saturated.

###### Proposition

(recognition of quotient topologies)

A continuous function (def. )

$f \;\colon\; (X, \tau_X) \longrightarrow (Y,\tau_Y)$

whose underlying function $f \colon X \longrightarrow Y$ is surjective exhibits $\tau_Y$ as the corresponding quotient topology (def. ) precisely if $f$ sends open and $f$-saturated subsets in $X$ (def. ) to open subsets of $Y$. By lemma this is the case precisely if it sends closed and $f$-saturated subsets to closed subsets.

We record the following technical lemma about saturated subspaces, which we will need below to prove prop. .

###### Lemma

(saturated open neighbourhoods of saturated closed subsets under closed maps)

Let

1. $f \;\colon\; (X, \tau_X) \longrightarrow (Y, \tau_Y)$ be a closed map (def. );

2. $C \subset X$ be a closed subset of $X$ (def. ) which is $f$-saturated (def. );

3. $U \supset C$ be an open subset containing $C$;

then there exists a smaller open subset $V$ still containing $C$

$U \supset V \supset C$

and such that $V$ is still $f$-saturated.

###### Proof

We claim that the complement of $X$ by the $f$-saturation (def. ) of the complement of $X$ by $U$

$V \coloneqq X \setminus \left( f^{-1}\left( f\left( X \setminus U \right) \right) \right)$

has the desired properties. To see this, observe first that

1. the complement $X \setminus U$ is closed, since $U$ is assumed to be open;

2. hence the image $f(X\setminus U)$ is closed, since $f$ is assumed to be a closed map;

3. hence the pre-image $f^{-1}\left( f\left( X \setminus U \right)\right)$ is closed, since $f$ is continuous (using prop. ), therefore its complement $V$ is indeed open;

4. this pre-image $f^{-1}\left( f\left( X \setminus U \right) \right)$ is saturated (by example ) and hence also its complement $V$ is saturated (by lemma ).

Therefore it now only remains to see that $U \supset V \supset C$.

By de Morgan's law (prop. ) the inclusion $U \supset V$ is equivalent to the inclusion $f^{-1}\left( f\left( X \setminus U \right)\right) \supset X \setminus U$, which is clearly the case.

The inclusion $V \supset C$ is equivalent to $f^{-1}\left( f\left( X \setminus U \right) \right) \,\cap \, C = \emptyset$. Since $C$ is saturated by assumption, this is equivalent to $f^{-1}\left( f\left( X \setminus U \right)\right) \,\cap \, f^{-1}(f(C)) = \emptyset$. This in turn holds precisely if $f\left( X \setminus U \right) \,\cap \, f(C) = \emptyset$. Since $C$ is saturated, this holds precisely if $X \setminus U \cap C = \emptyset$, and this is true by the assumption that $U \supset C$.

$\,$

### Homeomorphisms

With the objects (topological spaces) and the morphisms (continuous functions) of the category Top thus defined (remark ), we obtain the concept of “sameness” in topology. To make this precise, one says that a morphism

$X \overset{f}{\to} Y$

in a category is an isomorphism if there exists a morphism going the other way around

$X \overset{g}{\longleftarrow} Y$

which is an inverse in the sense that both its compositions with $f$ yield an identity morphism:

$f \circ g = id_Y \;\;\;\;\; and \;\;\;\;\; g \circ f = id_X \,.$

Since such $g$ is unique if it exists, one often writes “$f^{-1}$” for this inverse morphism.

###### Definition

(homeomorphisms)

An isomorphism in the category Top (remark ) of topological spaces (def. ) with continuous functions between them (def. ) is called a homeomorphism.

Hence a homeomorphism is a continuous function

$f \;\colon\; (X, \tau_X) \longrightarrow (Y, \tau_Y)$

between two topological spaces $(X,\tau_X)$, $(Y,\tau_Y)$ such that there exists another continuous function the other way around

$(X, \tau_X) \longleftarrow (Y, \tau_Y) \;\colon\; g$

such that their composites are the identity functions on $X$ and $Y$, respectively:

$f \circ g = id_{Y} \;\;\;and\;\;\; g \circ f = id_{X} \,.$ graphics grabbed from Munkres 75

We notationally indicate that a continuous function is a homeomorphism by the symbol “$\simeq$”.

$f \;\colon\; (X,\tau_X) \overset{\simeq}{\longrightarrow} (Y,\tau_Y) \,.$

If there is some, possibly unspecified, homeomorphism between topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$, then we also write

$(X,\tau_X) \,\simeq\, (Y,\tau_Y)$

and say that the two topological spaces are homeomorphic.

A property/predicate $P$ of topological spaces which is invariant under homeomorphism in that

$\left( (X, \tau_X) \, \simeq \, (Y,\tau_Y) \right) \;\Rightarrow\; \left( P(X,\tau_X) \,\Leftrightarrow\, P(Y,\tau_Y) \right)$

is called a topological property or topological invariant.

###### Remark

(notation for homeomorphisms)

Beware the following notation:

1. In topology the notation $f^{-1}$ generally refers to the pre-image function of a given function $f$, while if $f$ is a homeomorphism (def. ), it is also used for the inverse function of $f$. This abuse of notation is convenient: If $f$ happens to be a homeomorphism, then the pre-image of a subsets under $f$ is its image under the inverse function $f^{-1}$.

2. Many authors strictly distinguish the symbols “$\cong$” and “$\simeq$” and use the former to denote homeomorphisms and the latter to refer to homotopy equivalences (which we consider in part 2). We use either symbol (but mostly “$\simeq$”) for “isomorphism” in whatever the ambient category may be and try to make that context always unambiguously explicit.

###### Remark

If $f \colon (X, \tau_X) \to (Y, \tau_Y)$ is a homeomorphism (def. ) with inverse continuous function $g$, then

1. also $g$ is a homeomophism, with inverse continuous function $f$;

2. the underlying function of sets $f \colon X \to Y$ of a homeomorphism $f$ is necessarily a bijection, with inverse bijection $g$.

But beware that not every continuous function which is bijective on underlying sets is a homeomorphism. While an inverse function $g$ will exists on the level of functions of sets, this inverse may fail to be continuous:

###### Counter Example

Consider the continuous function

$\array{ [0,2\pi) &\longrightarrow& S^1 \subset \mathbb{R}^2 \\ t &\mapsto& (cos(t), sin(t)) }$

from the half-open interval (def. ) to the unit circle $S^1 \coloneqq S_0(1) \subset \mathbb{R}^2$ (def. ), regarded as a topological subspace (example ) of the Euclidean plane (example ).

The underlying function of sets of $f$ is a bijection. The inverse function of sets however fails to be continuous at $(1,0) \in S^1 \subset \mathbb{R}^2$. Hence this $f$ is not a homeomorphism.

Indeed, below we see that the two topological spaces $[0,2\pi)$ and $S^1$ are distinguished by topological invariants, meaning that they cannot be homeomorphic via any (other) choice of homeomorphism. For example $S^1$ is a compact topological space (def. ) while $[0,2\pi)$ is not, and $S^1$ has a non-trivial fundamental group, while that of $[0,2\pi)$ is trivial (this prop.).

Below in example we discuss a practical criterion under which continuous bijections are homeomorphisms after all. But immediate from the definitions is the following characterization:

###### Proposition

(homeomorphisms are the continuous and open bijections)

Let $f \;\colon\; (X, \tau_X) \longrightarrow (Y,\tau_Y)$ be a continuous function between topological spaces (def. ). Then the following are equivalence:

1. $f$ is a homeomorphism;

2. $f$ is a bijection and an open map (def. );

3. $f$ is a bijection and a closed map (def. ).

###### Proof

It is clear from the definition that a homeomorphism in particular has to be a bijection. The condition that the inverse function $Y \leftarrow X \colon g$ be continuous means that the pre-image function of $g$ sends open subsets to open subsets. But by $g$ being the inverse to $f$, that pre-image function is equal to $f$, regarded as a function on subsets:

$g^{-1} = f \;\colon\; P(X) \to P(Y) \,.$

Hence $g^{-1}$ sends opens to opens precisely if $f$ does, which is the case precisely if $f$ is an open map, by definition. This shows the equivalence of the first two items. The equivalence between the first and the third follows similarly via prop. .

$\,$

Now we consider some actual examples of homeomorphisms:

###### Example

(concrete point homeomorphic to abstract point space)

Let $(X,\tau_X)$ be a non-empty topological space, and let $x \in X$ be any point. Regard the corresponding singleton subset $\{x\} \subset X$ as equipped with its subspace topology $\tau_{\{x\}}$ (example ). Then this is homeomorphic (def. ) to the abstract point space from example :

$(\{x\}, \tau_{\{x\}} ) \,\simeq\, \ast \,.$
###### Example

(open interval homeomorphic to the real line)

Regard the real line as the 1-dimensional Euclidean space (example ) with its metric topology (example ).

Then the open interval $(-1,1) \subset \mathbb{R}$ (def. ) regarded with its subspace topology (example ) is homeomorphic (def.) to all of the real line

$(-1,1) \,\simeq\, \mathbb{R}^1 \,.$

An inverse pair of continuous functions is for instance given (via example ) by

$\array{ f &\colon& \mathbb{R}^1 &\longrightarrow& (-1,+1) \\ && x &\mapsto& \frac{x}{\sqrt{1+ x^2}} }$

and

$\array{ g &\colon& (-1,+1) &\longrightarrow& \mathbb{R}^1 \\ && x &\mapsto& \frac{x}{\sqrt{1 - x^2}} } \,.$

But there are many other choices for $f$ and $g$ that yield a homeomorphism.

Similarly, for all $a \lt b \in \mathbb{R}$

1. the open intervals $(a,b) \subset \mathbb{R}$ (example ) equipped with their subspace topology are all homeomorphic to each other,

2. the closed intervals $[a,b]$ are all homeomorphic to each other,

3. the half-open intervals of the form $[a,b)$ are all homeomorphic to each other;

4. the half-open intervals of the form $(a,b]$ are all homeomorphic to each other.

Generally, every open ball in $\mathbb{R}^n$ (def. ) is homeomorphic to all of $\mathbb{R}^n$:

$\left( B^\circ_0(\epsilon) \subset \mathbb{R}^n \right) \,\simeq\, \mathbb{R}^n \,.$

While mostly the interest in a given homeomorphism is in it being non-obvious from the definitions, many homeomorphisms that appear in practice exhibit “obvious re-identifications” for which it is of interest to leave them consistently implicit:

###### Example

(homeomorphisms between iterated product spaces)

Let $(X,\tau_X)$, $(Y,\tau_Y)$ and $(Z, \tau_Z)$ be topological spaces.

Then:

1. There is an evident homeomorphism between the two ways of bracketing the three factors when forming their product topological space (def. ), called the associator:

$\alpha_{X,Y,Z} \;\colon\; \left( (X, \tau_X) \times (Y, \tau_Y) \right) \times (Z, \tau_Z) \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} (X,\tau_X) \times \left( (Y,\tau_Y) \times (Z, \tau_Z) \right) \,.$
2. There are evident homeomorphism between $(X,\tau)$ and its product topological space (def. ) with the point space $\ast$ (example ), called the left and right unitors:

$\lambda_X \;\colon\; \ast \times (X, \tau_X) \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} (X,\tau_X)$

and

$\rho_X \;\colon\; (X, \tau_X) \times \ast \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} (X, \tau_X) \,.$
3. There is an evident homeomorphism between the results of the two orders in which to form their product topological spaces (def. ), called the braiding:

$\beta_{X,Y} \;\colon\; (X,\tau_X) \times (Y,\tau_Y) \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} (Y,\tau_Y) \times (X, \tau_X) \,.$

Moreover, all these homeomorphisms are compatible with each other, in that they make the following diagrams commute (recall remark ):

1. (triangle identity)

$\array{ & (X \times \ast ) \times Y &\stackrel{\alpha_{X,\ast,Y}}{\longrightarrow} & X \times (\ast \times Y) \\ & {}_{\rho_x \times id_Y}\searrow && \swarrow_{id_X \times \lambda_Y} & \\ && X \times Y }$
2. $\array{ && (W \times X) \times (Y \times Z) \\ & {}^{\mathllap{\alpha_{W \times X, Y, Z}}}\nearrow && \searrow^{\mathrlap{\alpha_{W,X,Y \times Z}}} \\ ((W \times X ) \times Y) \times Z && && (W \times (X \times (Y \times Z))) \\ {}^{\mathllap{\alpha_{W,X,Y}} \times id_Z }\downarrow && && \uparrow^{\mathrlap{ id_W \times \alpha_{X,Y,Z} }} \\ (W \times (X \times Y)) \times Z && \underset{\alpha_{W,X \times Y, Z}}{\longrightarrow} && W \times ( (X \times Y) \times Z ) }$
3. (hexagon identities)

$\array{ (X \times Y) \times Z &\stackrel{\alpha_{X,Y,Z}}{\longrightarrow}& X \times (Y \times Z) &\stackrel{\beta_{X,Y \times Z}}{\longrightarrow}& (Y \times Z) \times X \\ \downarrow^{\beta_{X,Y} \times id_Z} &&&& \downarrow^{\alpha_{Y,Z,X}} \\ (Y \times X) \times Z &\stackrel{\alpha_{Y,X,Z}}{\longrightarrow}& Y \times (X \times Z) &\stackrel{id_Y \times \beta_{X,Y}}{\longrightarrow}& Y \times (Z \times X) }$

and

$\array{ X \times (Y \times Z) &\stackrel{\alpha^{inv}_{X,Y,Z}}{\longrightarrow}& (X \times Y) \times Z &\stackrel{\beta_{X \times Y, Z}}{\longrightarrow}& Z \times (X \times Y) \\ \downarrow^{id_X \times \beta_{Y,Z}} &&&& \downarrow^{\alpha^{inv}_{Z,X,Y}} \\ X \times (Z \times Y) &\stackrel{\alpha^{inv}_{X,Z,Y}}{\longrightarrow}& (X \times Z) \times Y &\stackrel{\beta_{X,Z} \times id}{\longrightarrow}& (Z \times X) \times Y } \,,$
4. (symmetry)

$\beta_{Y,X} \circ \beta_{X,Y} \;=\; id \;\colon\; (X_1 \times X_2 \tau_{X_1 \times X_2}) \to (X_1 \times X_2 \tau_{X_1 \times X_2}) \,.$

In the language of category theory (remark ), all this is summarized by saying that the the functorial construction $(-) \times (-)$ of product topological spaces (example ) gives the category Top of topological spaces the structure of a monoidal category which moreover is symmetrically braided.

From this, a basic result of category theory, the MacLane coherence theorem, guarantees that there is no essential ambiguity re-backeting arbitrary iterations of the binary product topological space construction, as long as the above homeomorphisms are understood.

Accordingly, we may write

$(X_1, \tau_1) \times (X_2, \tau_2) \times \cdots \times (X_n, \tau_n)$

for iterated product topological spaces without putting parenthesis.

$\,$

The following are a sequence of examples all of the form that an abstractly constructed topological space is homeomorphic to a certain subspace of a Euclidean space. These examples are going to be useful in further developments below, for example in the proof below of the Heine-Borel theorem (prop. ).

###### Example

(product of closed intervals homeomorphic to hypercubes)

Let $n \in \mathbb{N}$, and let $[a_i, b_i] \subset \mathbb{R}$ for $i \in \{1, \cdots, n\}$ be $n$ closed intervals in the real line (example ), regarded as topological subspaces of the 1-dimensional Euclidean space (example ) with its metric topology (example ). Then the product topological space (def. , example ) of all these intervals is homeomorphic (def. ) to the corresponding topological subspace of the $n$-dimensional Euclidean space (example ):

$[a_1, b_1] \times [a_2, b_2] \times \cdots \times [a_n, b_n] \;\simeq\; \left\{ \vec x \in \mathbb{R}^n \,\vert\, \underset{i}{\forall} (a_i \leq x_i \leq b_i) \right\} \subset \mathbb{R}^n \,.$

Similarly for open intervals:

$(a_1, b_1) \times (a_2, b_2) \times \cdots \times (a_n, b_n) \;\simeq\; \left\{ \vec x \in \mathbb{R}^n \,\vert\, \underset{i}{\forall} (a_i \lt x_i \lt b_i) \right\} \subset \mathbb{R}^n \,.$
###### Proof

There is a canonical bijection between the underlying sets. It remains to see that this, as well and its inverse, are continuous functions. For this it is sufficient to see that under this bijection the defining basis (def. ) for the product topology is also a basis for the subspace topology. But this is immediate from lemma .

###### Example

(closed interval glued at endpoints homeomorphic circle)

As topological spaces, the closed interval $[0,1]$ (def. ) with its two endpoints identified is homeomorphic (def. ) to the standard circle:

$[0,1]_{/(0 \sim 1)} \;\; \simeq \;\; S^1 \,.$

More in detail: let

$S^1 \hookrightarrow \mathbb{R}^2$

be the unit circle in the plane

$S^1 = \{(x,y) \in \mathbb{R}^2, x^2 + y^2 = 1\}$

equipped with the subspace topology (example ) of the plane $\mathbb{R}^2$, which is itself equipped with its standard metric topology (example ).

Moreover, let

$[0,1]_{/(0 \sim 1)}$

be the quotient topological space (example ) obtained from the interval $[0,1] \subset \mathbb{R}^1$ with its subspace topology by applying the equivalence relation which identifies the two endpoints (and nothing else).

Consider then the function

$f \;\colon\; [0,1] \longrightarrow S^1$

given by

$t \mapsto (cos(2\pi t), sin(2\pi t)) \,.$

This has the property that $f(0) = f(1)$, so that it descends to the quotient topological space

$\array{ [0,1] &\overset{}{\longrightarrow}& [0,1]_{/(0 \sim 1)} \\ & {}_{\mathllap{f}}\searrow & \downarrow^{\mathrlap{\tilde f}} \\ && S^1 } \,.$

We claim that $\tilde f$ is a homeomorphism (definition ).

First of all it is immediate that $\tilde f$ is a continuous function. This follows immediately from the fact that $f$ is a continuous function and by definition of the quotient topology (example ).

So we need to check that $\tilde f$ has a continuous inverse function. Clearly the restriction of $f$ itself to the open interval $(0,1)$ has a continuous inverse. It fails to have a continuous inverse on $[0,1)$ and on $(0,1]$ and fails to have an inverse at all on [0,1], due to the fact that $f(0) = f(1)$. But the relation quotiented out in $[0,1]_{/(0 \sim 1)}$ is exactly such as to fix this failure.

###### Example

(cylinder, Möbius strip and torus homeomorphic to quotients of the square)

The square $[0,1]^2$ with two of its sides identified is the cylinder, and with also the other two sides identified is the torus: If the sides are identified with opposite orientation, the result is the Möbius strip: graphics grabbed from Lawson 03

###### Example

(stereographic projection)

For $n \in \mathbb{N}$ then there is a homeomorphism (def. ) between the n-sphere $S^n$ (example ) with one point $p \in S^n$ removed and the $n$-dimensional Euclidean space $\mathbb{R}^n$ (example ) with its metric topology (example ):

$S^n \setminus \{p\} \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} \mathbb{R}^n \,.$ This homeomorphism is given by “stereographic projection”: One thinks of both the $n$-sphere as well as the Euclidean space $\mathbb{R}^n$ as topological subspaces (example ) of $\mathbb{R}^{n+1}$ in the standard way (example ), such that they intersect in the equator of the $n$-sphere. For $p \in S^n$ one of the corresponding poles, then the homeomorphism is the function which sends a point $x \in S^{n}\setminus \{p\}$ along the line connecting it with $p$ to the point $y$ where this line intersects the equatorial plane.

In the canonical ambient coordinates this stereographic projection is given as follows:

$\array{ \mathbb{R}^{n+1} \supset \;\;\; & S^n \setminus (1,0, \cdots, 0) &\overset{\phantom{AA} \simeq \phantom{AA}}{\longrightarrow}& \mathbb{R}^{n} & \;\;\; \subset \mathbb{R}^{n+1} \\ & (x_1, x_2, \cdots, x_{n+1}) &\overset{\phantom{AAAA}}{\mapsto}& \frac{1}{1 - x_1} \left( 0 , x_2, \cdots, x_{n+1} \right) } \,.$
###### Proof

First consider more generally the stereographic projection

$\sigma \;\colon\; \mathbb{R}^{n+1} \backslash (1,0,\cdots, 0) \longrightarrow \mathbb{R}^n = \{x \in \mathbb{R}^{n.1} \,\vert\, x_1 = 0 \}$

of the entire ambient space minus the point $p$ onto the equatorial plane, still given by mapping a point $x$ to the unique point $y$ on the equatorial hyperplane such that the points $p$, $x$ any $y$ sit on the same straight line.

This condition means that there exists $d \in \mathbb{R}$ such that

$p + d(x-p) = y \,.$

Since the only condition on $y$ is that $y_1 = 0$ this implies that

$p_1 + d(x_1-p_1) = 0 \,.$

This equation has a unique solution for $d$ given by

$d = \frac{1}{1 - x_1}$

$\sigma(x_1, x_2, \cdots, x_{n+1}) = \frac{1}{1-x_1}(0,x_2, \cdots, x_n) \,$

Since rational functions are continuous (example ), this function $\sigma$ is continuous and since the topology on $S^n\backslash p$ is the subspace topology under the canonical embedding $S^n \backslash p \subset \mathbb{R}^{n+1} \backslash p$ it follows that the restriction

$\sigma\vert_{S^n \backslash p} \;\colon\; S^n\backslash p \longrightarrow \mathbb{R}^n$

is itself a continuous function (because its pre-images are the restrictions of the pre-images of $\sigma$ to $S^n\backslash p$).

To see that $\sigma \vert_{S^n \backslash p}$ is a bijection of the underlying sets we need to show that for every

$(0, y_2, \cdots, y_{n+1})$

there is a unique $(x_1, \cdots , x_{n+1})$ satisfying

1. $(x_1, \cdots, x_{n+1}) \in S^{n} \backslash \{p\}$, hence

1. $x_1 \lt 1$;

2. $\underoverset{i = 1}{n+1}{\sum} (x_i)^2 = 1$;

2. $\underset{i \in \{2, \cdots, n+1\}}{\forall} \left(y_i = \frac{x_i}{1-x_1} \right)$.

The last condition uniquely fixes the $x_{i \geq 2}$ in terms of the given $y_{i \geq 2}$ and the remaining $x_1$, as

$x_{i \geq 2} = y_i \cdot (1-x_1) \,.$

With this, the second condition says that

$(x_1)^2 + (1-x_1)^2 \underset{r^2}{\underbrace{\underoverset{i = 2}{n+1}{\sum}(y_i)^2}} = 1$

hence equivalently that

$(r^2 + 1) (x_1)^2 - (2 r^2) x_1 + (r^2 - 1) = 0 \,.$

By the quadratic formula the solutions of this equation are

\begin{aligned} x_1 & = \frac { 2 r^2 \pm \sqrt{ 4 r^4 - 4 (r^4 - 1) } } { 2 (r^2 + 1) } \\ & = \frac { 2 r^2 \pm 2 } { 2 r^2 + 2 } \end{aligned} \,.

The solution $\frac{ 2 r^2 + 2 }{ 2 r^2 + 2 } = 1$ violates the first condition above, while the solution $\frac{ 2 r^2 - 2 }{ 2 r^2 + 2 } \lt 1$ satisfies it.

Therefore we have a unique solution, given by

$\left( \sigma\vert_{S^n \backslash \{p\}} \right)^{-1}(0,y_2, \cdots, y_{n+1}) \;=\; \left( \frac{2 r^2 - 2}{2 r^2 +2} , \left( 1- \frac{2 r^2 - 2}{2 r^2 +2} \right) y_2 , \cdots , \left( 1- \frac{2 r^2 - 2}{2 r^2 +2} \right) y_{n+1} \right)$

In particular therefore also an inverse function to the stereographic projection exists and is a rational function, hence continuous by example . So we have exhibited a homeomorphism as required.

$\,$

Important examples of pairs of spaces that are not homeomorphic include the following:

###### Theorem

(topological invariance of dimension)

For $n_1, n_2 \in \mathbb{N}$ but $n_1 \neq n_2$, then the Euclidean spaces $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ (example , example ) are not homeomorphic.

More generally, an open subset in $\mathbb{R}^{n_1}$ is never homeomorphic to an open subset in $\mathbb{R}^{n_2}$ if $n_1 \neq n_2$.

The proofs of theorem are not elementary, in contrast to how obvious the statement seems to be intuitively. One approach is to use tools from algebraic topology: One assigns topological invariants to topological spaces, notably classes in ordinary cohomology or in topological K-theory), quantities that are invariant under homeomorphism, and then shows that these classes coincide for $\mathbb{R}^{n_1} - \{0\}$ and for $\mathbb{R}^{n_2} - \{0\}$ precisely only if $n_1 = n_2$.

One indication that topological invariance of dimension is not an elementary consequence of the axioms of topological spaces is that a related “intuitively obvious” statement is in fact false: One might think that there is no surjective continuous function $\mathbb{R}^{n_1} \to \mathbb{R}^{n_2}$ if $n_1 \lt n_2$. But there are: these are called the Peano curves.

$\,$

## Separation axioms

The plain definition of topological space (above) happens to admit examples where distinct points or distinct subsets of the underlying set appear as more-or-less unseparable as seen by the topology on that set.

The extreme class of examples of topological spaces in which the open subsets do not distinguish distinct underlying points, or in fact any distinct subsets, are the codiscrete spaces (example ). This does occur in practice:

###### Example

(real numbers quotiented by rational numbers)

Consider the real line $\mathbb{R}$ regarded as the 1-dimensional Euclidean space (example ) with its metric topology (example ) and consider the equivalence relation $\sim$ on $\mathbb{R}$ which identifies two real numbers if they differ by a rational number:

$\left( x \sim y \right) \;\Leftrightarrow\; \left( \underset{p/q \in \mathbb{Q} \subset \mathbb{R}}{\exists} \left( x = y + p/q \right) \right) \,.$

Then the quotient topological space (def. )

$\mathbb{R}/\mathbb{Q} \;\coloneqq\; \mathbb{R}/\sim$

is a codiscrete topological space (def. ), hence its topology does not distinguish any distinct proper subsets.

Here are some less extreme examples:

###### Example

(open neighbourhoods in the Sierpinski space)

Consider the Sierpinski space from example , whose underlying set consists of two points $\{0,1\}$, and whose open subsets form the set $\tau = \{ \emptyset, \{1\}, \{0,1\} \}$. This means that the only (open) neighbourhood of the point $\{0\}$ is the entire space. Incidentally, also the topological closure of $\{0\}$ (def. ) is the entire space.

###### Example

(line with two origins)

Consider the disjoint union space $\mathbb{R} \sqcup \mathbb{R}$ (example ) of two copies of the real line $\mathbb{R}$ regarded as the 1-dimensional Euclidean space (example ) with its metric topology (example ), which is equivalently the product topological space (example ) of $\mathbb{R}$ with the discrete topological space on the 2-element set (example ):

$\mathbb{R} \sqcup \mathbb{R} \;\simeq\; \mathbb{R} \times Disc(\{0,1\})$

Moreover, consider the equivalence relation on the underlying set which identifies every point $x_i$ in the $i$th copy of $\mathbb{R}$ with the corresponding point in the other, the $(1-i)$th copy, except when $x = 0$:

$\left( x_i \sim y_j \right) \;\Leftrightarrow\; \left( \left( x = y \right) \,\text{and}\, \left( \left( x \neq 0 \right) \,\text{or}\, \left( i = j \right) \right) \right) \,.$ The quotient topological space by this equivalence relation (def. )

$\left( \mathbb{R} \sqcup \mathbb{R} \right)/\sim$

is called the line with two origins. These “two origins” are the points $0_0$ and $0_1$.

We claim that in this space every neighbourhood of $0_0$ intersects every neighbourhood of $0_1$.

Because, by definition of the quotient space topology, the open neighbourhoods of $0_i \in \left( \mathbb{R} \sqcup \mathbb{R} \right)/\sim$ are precisely those that contain subsets of the form

$(-\epsilon, \epsilon)_i \;\coloneqq\; (-\epsilon,0) \cup \{0_i\} \cup (0,\epsilon) \,.$

But this means that the “two origins” $0_0$ and $0_1$ may not be separated by neighbourhoods, since the intersection of $(-\epsilon, \epsilon)_0$ with $(-\epsilon, \epsilon)_i$ is always non-empty:

$(-\epsilon, \epsilon)_0 \cap (-\epsilon, \epsilon)_1 \;=\; (-\epsilon, 0) \cup (0, \epsilon) \,.$

In many applications one wants to exclude at least some such exotic examples of topological spaces from the discussion and instead concentrate on those examples for which the topology recognizes the separation of distinct points, or of more general disjoint subsets. The relevant conditions to be imposed on top of the plain axioms of a topological space are hence known as separation axioms which we discuss in the following.

These axioms are all of the form of saying that two subsets (of certain kinds) in the topological space are ‘separated’ from each other in one sense if they are ‘separated’ in a (generally) weaker sense. For example the weakest axiom (called $T_0$) demands that if two points are distinct as elements of the underlying set of points, then there exists at least one open subset that contains one but not the other.

In this fashion one may impose a hierarchy of stronger axioms. For example demanding that given two distinct points, then each of them is contained in some open subset not containing the other ($T_1$) or that such a pair of open subsets around two distinct points may in addition be chosen to be disjoint ($T_2$). Below in Tn-spaces we discuss the following hierarchy:

the main separation axioms

numbernamestatementreformulation
$T_0$Kolmogorovgiven two distinct points, at least one of them has an open neighbourhood not containing the other pointevery irreducible closed subset is the closure of at most one point
$T_1$given two distinct points, both have an open neighbourhood not containing the other pointall points are closed
$T_2$Hausdorffgiven two distinct points, they have disjoint open neighbourhoodsthe diagonal is a closed map
$T_{\gt 2}$$T_1$ and…all points are closed and…
$T_3$regular Hausdorff…given a point and a closed subset not containing it, they have disjoint open neighbourhoods…every neighbourhood of a point contains the closure of an open neighbourhood
$T_4$normal Hausdorff…given two disjoint closed subsets, they have disjoint open neighbourhoods…every neighbourhood of a closed set also contains the closure of an open neighbourhood
… every pair of disjoint closed subsets is separated by an Urysohn function

The condition, $T_2$, also called the Hausdorff condition is the most common among all separation axioms. Historically this axiom was originally taken as part of the definition of topological spaces, and it is still often (but by no means always) considered by default.

However, there are respectable areas of mathematics that involve topological spaces where the Hausdorff axiom fails, but a weaker axiom is still satisfied, called sobriety. This is the case notably in algebraic geometry (schemes are sober) and in computer science (Vickers 89). These sober topological spaces are singled out by the fact that they are entirely characterized by their sets of open subsets with their union and intersection structure (as in def. ) and may hence be understood independently from their underlying sets of points. This we discuss further below.

hierarchy of separation axioms
$\array{ &&&\text{metric space} \\ &&& \Downarrow \\ &&& \vdots \\ &&& \Downarrow \\ &&& T_4 = \text{normal Hausdorff} \\ &&& \Downarrow \\ &&& T_3 = \text{regular Hausdorff} \\ &&& \Downarrow \\ &&& T_2 = \text{Hausdorff} \\ && \swArrow && \seArrow \\ \, & T_1 && && \text{sober} & \, \\ && \seArrow && \swArrow \\ &&& T_0 = \text{Kolmogorov} \\ }$

All separation axioms are satisfied by metric spaces (example , example below), from whom the concept of topological space was originally abstracted above. Hence imposing some of them may also be understood as gauging just how far one allows topological spaces to generalize away from metric spaces

### $T_n$ spaces

There are many variants of separation axioms. The classical ones are labeled $T_n$ (for German “Trennungsaxiom”) with $n \in \{0,1,2,3,4,5\}$ or higher. These we now introduce in def. and def. .

###### Definition

(the first three separation axioms)

Let $(X,\tau)$ be a topological space (def. ).

For $x \neq y \in X$ any two points in the underlying set of $X$ which are not equal as elements of this set, consider the following propositions: • (T0) There exists a neighbourhood of one of the two points which does not contain the other point.

• (T1) There exist neighbourhoods of both points which do not contain the other point.

• (T2) There exists neighbourhoods_ of both points which do not intersect each other.

graphics grabbed from Vickers 89

The topological space $X$ is called a $T_n$-topological space or just $T_n$-space, for short, if it satisfies condition $T_n$ above for all pairs of distinct points.

A $T_0$-topological space is also called a Kolmogorov space.

A $T_2$-topological space is also called a Hausdorff topological space.

For definiteness, we re-state these conditions formally. Write $x,y \in X$ for points in $X$, write $U_x, U_y \in \tau$ for open neighbourhoods of these points. Then:

• (T0) $\underset{x \neq y}{\forall} \left( \left( \underset{U_y}{\exists} \left( \{x\} \cap U_y = \emptyset \right) \right) \,\text{or}\, \left( \underset{U_x}{\exists} \left( U_x \cap \{y\} = \emptyset \right) \right) \right)$

• ((T1) $\underset{x \neq y}{\forall} \left(\underset{U_x,U_y}{\exists} \left(\left( \{x\} \cap U_y = \emptyset\right) \, \text{and} \, \left( U_x \cap \{y\} = \emptyset \right)\right) \right)$

• (T2) $\underset{x \neq y}{\forall} \left( \underset{U_x, U_y}{\exists} \left( U_x \cap U_y = \emptyset\right) \right)$

The following is evident but important:

###### Proposition

($T_n$ are topological properties of increasing strength)

The separation properties $T_n$ from def. are topological properties in that if two topological spaces are homeomorphic (def. ) then one of them satisfies $T_n$ precisely if the other does.

Moreover, these properties imply each other as

$T2 \Rightarrow T1 \Rightarrow T0 \,.$
###### Example

Examples of topological spaces that are not Hausdorff (def. ) include

1. the Sierpinski space (example ),

2. the line with two origins (example ),

3. the quotient topological space $\mathbb{R}/\mathbb{Q}$ (example ).

###### Example

(finite $T_1$-spaces are discrete)

For a finite topological space $(X,\tau)$, hence one for which the underlying set $X$ is a finite set, the following are equivalent:

1. $(X,\tau)$ is $T_1$ (def. );

2. $(X,\tau)$ is a discrete topological space (def. ).

###### Example

(metric spaces are Hausdorff)

Every metric space (def ), regarded as a topological space via its metric topology (example ) is a Hausdorff topological space (def. ).

Because for $x \neq y \in X$ two distinct points, then the distance $d(x,y)$ between them is positive number, by the non-degeneracy axiom in def. . Accordingly the open balls (def. )

$B^\circ_x(d(x,y)) \supset \{x\} \phantom{AA} \text{and} \phantom{AA} B^\circ_y(d(x,y)) \supset \{y\}$

are disjoint open neighbourhoods.

###### Example

(subspace of $T_n$-space is $T_n$)

Let $(X,\tau)$ be a topological space satisfying the $T_n$ separation axiom for some $n \in \{0,1,2\}$ according to def. . Then also every topological subspace $S \subset X$ (example ) satisfies $T_n$.

(Beware that this fails for some higher $n$ discussed below in def. . Open subspaces of normal spaces need not be normal.)

$\,$

Separation in terms of topological closures

The conditions $T_0$, $T_1$ and $T_2$ have the following equivalent formulation in terms of topological closures (def. ).

###### Proposition

($T_0$ in terms of topological closures)

A topological space $(X,\tau)$ is $T_0$ (def. ) precisely if the function $Cl(\{-\})$ that forms topological closures (def. ) of singleton subsets from the underlying set of $X$ to the set of irreducible closed subsets of $X$ (def. , which is well defined according to example ), is injective:

$Cl(\{-\}) \;\colon\; X \overset{\phantom{AAA}}{\hookrightarrow} IrrClSub(X)$
###### Proof

Assume first that $X$ is $T_0$. Then we need to show that if $x,y \in X$ are such that $Cl(\{x\}) = Cl(\{y\})$ then $x = y$. Hence assume that $Cl(\{x\}) = Cl(\{y\})$. Since the closure of a point is the complement of the union of the open subsets not containing the point (lemma ), this means that the union of open subsets that do not contain $x$ is the same as the union of open subsets that do not contain $y$:

$\underset{ {U \subset X \, \text{open}} \atop { U \subset X\setminus \{x\} } }{\cup} \left( U \right) \;=\; \underset{ {U \subset X \, \text{open}} \atop { U \subset X\setminus \{y\} } }{\cup} \left( U \right)$

But if the two points were distinct, $x \neq y$, then by $T_0$ one of the above unions would contain $x$ or $y$, while the other would not, in contradiction to the above equality. Hence we have a proof by contradiction.

Conversely, assume that $\left( Cl\{x\} = Cl\{y\}\right) \Rightarrow \left( x = y\right)$, and assume that $x \neq y$. Hence by contraposition $\mathrm{Cl}(\{x\}) \neq \mathrm{Cl}(\{y\})$. We need to show that there exists an open set which contains one of the two points, but not the other.

Assume there were no such open subset, hence that every open subset containing one of the two points would also contain then other. Then by lemma this would mean that $x \in \mathrm{Cl}(\{y\})$ and that $y \in \mathrm{Cl}(\{x\})$. But this would imply that $Cl(\{x\}) \subset \mathrm{Cl}(\{y\})$ and that $\mathrm{Cl}(\{y\}) \subset \mathrm{Cl}(\{x\})$, hence that $\mathrm{Cl}(\{x\}) = \mathrm{Cl}(\{y\})$. This is a proof by contradiction.

###### Proposition

($T_1$ in terms of topological closures)

A topological space $(X,\tau)$ is $T_1$ (def. ) precisely if all its points are closed points (def. ).

###### Proof

We have

\begin{aligned} \text{all points in}\, (X, \tau)\, \text{are closed} &\coloneqq\, \underset{x \in X}{\forall} \left( Cl(\{x\}) = \{x\} \right) \\ & \Leftrightarrow\, X \setminus \left( \underset{ { U \subset X\, \text{open} } \atop { x \notin U } }{\cup} \left( U \right) \right) \;=\; \{x\} \\ & \Leftrightarrow\, \left( \underset{ { U \subset X\, \text{open} } \atop { x \notin U } }{\cup} \left( U \right) \right) \;=\; X \setminus \{x\} \\ & \Leftrightarrow \underset{y \in Y}{\forall} \left( \left( \underset{ { U \subset X \, \text{open} } \atop { x \notin U } }{\exists} \left( y \in U \right) \right) \Leftrightarrow (y \neq x) \right) \\ & \Leftrightarrow\, (X,\tau)\, \text{is}\, T_1 \end{aligned} \,.

Here the first step is the reformulation of closure from lemma , the second is another application of the de Morgan law (prop. ), the third is the definition of union and complement, and the last one is manifestly by definition of $T_1$.

###### Proposition

($T_2$ in terms of topological closures)

A topological space $(X,\tau_X)$ is $T_2$=Hausdorff precisely if the image of the diagonal

$\array{ X &\overset{\Delta_X}{\longrightarrow}& X \times X \\ x &\overset{\phantom{AAA}}{\mapsto}& (x,x) }$

is a closed subset in the product topological space $(X \times X, \tau_{X \times X})$.

###### Proof

Observe that the Hausdorff condition is equivalently rephrased in terms of the product topology as: Every point $(x,y) \in X$ which is not on the diagonal has an open neighbourhood $U_{(x,y)} \times U_{(x,y)}$ which still does not intersect the diagonal, hence:

\begin{aligned} & (X,\tau)\,\text{Hausdorff} \\ \Leftrightarrow & \underset{(x,y) \in (X \times X) \setminus \Delta_X(X) }{\forall} \left( \underset{ { U_{(x,y)} \times V_{(x,y)} \in \tau_{X \times Y} } \atop { (x,y) \in U_{(x,y)} \times V_{(x,y)} } }{\exists} \left( U_{(x,y)} \times V_{(x,y)} \cap \Delta_X(X) = \emptyset \right) \right) \end{aligned}

Therefore if $X$ is Hausdorff, then the diagonal $\Delta_X(X) \subset X \times X$ is the complement of a union of such open sets, and hence is closed:

$(X, \tau)\, \text{Hausdorff} \;\;\;\Rightarrow \;\;\; \Delta_X(X) = X \setminus \left( \underset{(x,y) \in (X \times X) \setminus \Delta_X(X)}{\cup} U_{(x,y)} \times V_{(x,y)} \right) \,.$

Conversely, if the diagonal is closed, then (by lemma ) every point $(x,y) \in X \times X$ not on the diagonal, hence with $x \neq y$, has an open neighbourhood $U_{(x,y)} \times V_{(x,y)}$ still not intersecting the diagonal, hence so that $U_{(x,y)} \cap V_{(x,y)} = \emptyset$. Thus $(X,\tau)$ is Hausdorff.

$\,$

Further separation axioms

Clearly one may and does consider further variants of the separation axioms $T_0$, $T_1$ and $T_2$ from def. . Here we discuss two more:

###### Definition

Let $(X,\tau)$ be topological space (def. ).

Consider the following conditions

• (T3) The space $(X,\tau)$ is $T_1$ (def. ) and for $x \in X$ a point and $C \subset X$ a closed subset (def. ) not containing $x$, then there exist disjoint open neighbourhoods $U_x \supset \{x\}$ and $U_C \supset C$.

• (T4) The space $(X,\tau)$ is $T_1$ (def. ) and for $C_1, C_2 \subset X$ two disjoint closed subsets (def. ) then there exist disjoint open neighbourhoods $U_{C_i} \supset C_i$.

If $(X,\tau)$ satisfies $T_3$ it is said to be a $T_3$-space also called a regular Hausdorff topological space.

If $(X,\tau)$ satisfies $T_4$ it is to be a $T_4$-space also called a normal Hausdorff topological space.

###### Example

(metric spaces are normal Hausdorff)

Let $(X,d)$ be a metric space (def. ) regarded as a topological space via its metric topology (example ). Then this is a normal Hausdorff space (def. ).

###### Proof

By example metric spaces are $T_2$, hence in particular $T_1$. What we need to show is that given two disjoint closed subsets $C_1, C_2 \subset X$ then their exists disjoint open neighbourhoods $U_{C_1} \subset C_1$ and $U_{C_2} \supset C_2$.

Recall the function

$d(S,-) \colon X \to \mathbb{R}$

computing distances from a subset $S \subset X$ (example ). Then the unions of open balls (def. )

$U_{C_1} \coloneqq \underset{x_1 \in C_1}{\cup} B^\circ_{x_1}( d(C_2,x_1)/2 )$

and

$U_{C_2} \coloneqq \underset{x_2 \in C_2}{\cup} B^\circ_{x_2}( d(C_1,x_2)/2 ) \,.$

have the required properties.

Observe that:

###### Proposition

($T_n$ are topological properties of increasing strength)

The separation axioms from def. , def. are topological properties (def. ) which imply each other as

$T_4 \Rightarrow T_3 \Rightarrow T_2 \Rightarrow T_1 \Rightarrow T_0 \,.$
###### Proof

The implications

$T_2 \Rightarrow T_1 \Rightarrow T_0$

and

$T_4 \Rightarrow T_3$

are immediate from the definitions. The remaining implication $T_3 \Rightarrow T_2$ follows with prop. : This says that by assumption of $T_1$ then all points in $(X,\tau)$ are closed, and with this the condition $T_2$ is manifestly a special case of the condition for $T_3$.

Hence instead of saying “$X$ is $T_1$ and …” one could just as well phrase the conditions $T_3$ and $T_4$ as “$X$ is $T_2$ and …”, which would render the proof of prop. even more trivial.

The following shows that not every $T_2$-space/Hausdorff space is $T_3$/regular

###### Example

(K-topology)

Write

$K \coloneqq \{1/n \,\vert\, n \in \mathbb{N}_{\geq 1}\} \subset \mathbb{R}$

for the subset of natural fractions inside the real numbers.

Define a topological basis $\beta \subset P(\mathbb{R})$ on $\mathbb{R}$ consisting of all the open intervals as well as the complements of $K$ inside them:

$\beta \;\coloneqq\; \left\{ (a,b), \,\vert\, a\lt b \in \mathbb{R} \right\} \,\cup\, \left\{ (a,b) \setminus K, \,\vert\, a\lt b \in \mathbb{R} \right\} \,.$

The topology $\tau_{\beta} \subset P(\mathbb{R})$ which is generated from this topological basis is called the K-topology.

We may denote the resulting topological space by

$\mathbb{R}_K \;\coloneqq\; ( \mathbb{R}, \tau_{\beta}\} \,.$

This is a Hausdorff topological space (def. ) which is not a regular Hausdorff space, hence (by prop. ) in particular not a normal Hausdorff space (def. ).

$\,$

Further separation axioms in terms of topological closures

As before we have equivalent reformulations of the further separation axioms.

###### Proposition

($T_3$ in terms of topological closures)

A topological space $(X,\tau)$ is a regular Hausdorff space (def. ), precisely if all points are closed and for all points $x \in X$ with open neighbourhood $U \supset \{x\}$ there exists a smaller open neighbourhood $V \supset \{x\}$ whose topological closure $Cl(V)$ is still contained in $U$:

$\{x\} \subset V \subset Cl(V) \subset U \,.$

The proof of prop. is the direct specialization of the following proof for prop. to the case that $C = \{x\}$ (using that by $T_1$, which is part of the definition of $T_3$, the singleton subset is indeed closed, by prop. ).

###### Proposition

($T_4$ in terms of topological closures)

A topological space $(X,\tau)$ is normal Hausdorff space (def. ), precisely if all points are closed and for all closed subsets $C \subset X$ with open neighbourhood $U \supset C$ there exists a smaller open neighbourhood $V \supset C$ whose topological closure $Cl(V)$ is still contained in $U$:

$C \subset V \subset Cl(V) \subset U \,.$
###### Proof

In one direction, assume that $(X,\tau)$ is normal, and consider

$C \subset U \,.$

It follows that the complement of the open subset $U$ is closed and disjoint from $C$:

$C \cap X \setminus U = \emptyset \,.$

Therefore by assumption of normality of $(X,\tau)$, there exist open neighbourhoods with

$V \supset C \,, \phantom{AA} W \supset X \setminus U \phantom{AA} \text{with} \phantom{AA} V \cap W = \emptyset \,.$

But this means that

$V \subset X \setminus W$

and since the complement $X \setminus W$ of the open set $W$ is closed, it still contains the closure of $V$, so that we have

$C \subset V \subset Cl(V) \subset X \setminus W \subset U$

as required.

In the other direction, assume that for every open neighbourhood $U \supset C$ of a closed subset $C$ there exists a smaller open neighbourhood $V$ with

$C \subset V \subset Cl(V) \subset U \,.$

Consider disjoint closed subsets

$C_1, C_2 \subset X \,, \phantom{AAA} C_1 \cap C_2 = \emptyset \,.$

We need to produce disjoint open neighbourhoods for them.

From their disjointness it follows that

$X \setminus C_2 \supset C_1$

is an open neighbourhood. Hence by assumption there is an open neighbourhood $V$ with

$C_1 \subset V \subset Cl(V) \subset X \setminus C_2 \,.$

Thus

$V \supset C_1 \,, \phantom{AAAA} X \setminus Cl(V) \supset C_2$

are two disjoint open neighbourhoods, as required.

But the $T_4$/normality axiom has yet another equivalent reformulation, which is of a different nature, and will be important when we discuss paracompact topological spaces below:

The following concept of Urysohn functions is another approach of thinking about separation of subsets in a topological space, not in terms of their neighbourhoods, but in terms of continuous real-valued “indicator functions” that take different values on the subsets. This perspective will be useful when we consider paracompact topological spaces below.

But the Urysohn lemma (prop. below) implies that this concept of separation is in fact equivalent to that of normality of Hausdorff spaces.

###### Definition

(Urysohn function)

Let $(X,\tau)$ be a topological space, and let $A,B \subset X$ be disjoint closed subsets. Then an Urysohn function separating $A$ from $B$ is

to the closed interval equipped with its Euclidean metric topology (example , example ), such that

• it takes the value 0 on $A$ and the value 1 on $B$:

$f(A) = \{0\} \phantom{AAA} \text{and} \phantom{AAA} f(B) = \{1\} \,.$
###### Proposition

(Urysohn's lemma)

Let $X$ be a normal Hausdorff topological space (def. ), and let $A,B \subset X$ be two disjoint closed subsets of $X$. Then there exists an Urysohn function separating $A$ from $B$ (def. ).

###### Remark

Beware, the Urysohn function in prop. may take the values 0 or 1 even outside of the two subsets. The condition that the function takes value 0 or 1, respectively, precisely on the two subsets corresponds to “perfectly normal spaces”.

###### Proof

of Urysohn's lemma, prop.

Set

$C_0 \coloneqq A \phantom{AAA} U_1 \coloneqq X \setminus B \,.$

Since by assumption

$A \cap B = \emptyset \,.$

we have

$C_0 \subset U_1 \,.$

That $(X,\tau)$ is normal implies, by lemma , that every open neighbourhood $U \supset C$ of a closed subset $C$ contains a smaller neighbourhood $V$ together with its topological closure $Cl(V)$

$U \subset V \subset Cl(V) \subset C \,.$

Apply this fact successively to the above situation to obtain the following infinite sequence of nested open subsets $U_r$ and closed subsets $C_r$

$\array{ C_0 && && && &\subset& && && && U_1 \\ C_0 && &\subset& && U_{1/2} &\subset& C_{1/2} && &\subset& && U_1 \\ C_0 &\subset& U_{1/4} &\subset& C_{1/4} &\subset& U_{1/2} &\subset& C_{1/2} &\subset& U_{3/4} &\subset& C_{3/4} &\subset& U_1 }$

and so on, labeled by the dyadic rational numbers $\mathbb{Q}_{dy} \subset \mathbb{Q}$ within $(0,1]$ $\{ U_{r} \subset X \}_{r \in (0,1] \cap \mathbb{Q}_{dy}}$

with the property

$\underset{r_1 \lt r_2 \, \in (0,1] \cap \mathbb{Q}_{dy}}{\forall} \left( U_{r_1} \subset Cl(U_{r_1}) \subset U_{r_2} \right) \,.$

Define then the function

$f \;\colon\; X \longrightarrow [0,1]$

to assign to a point $x \in X$ the infimum of the labels of those open subsets in this sequence that contain $x$:

$f(x) \coloneqq \underset{U_r \supset \{x\}}{\lim} r$

Here the limit is over the directed set of those $U_r$ that contain $x$, ordered by reverse inclusion.

This function clearly has the property that $f(A) = \{0\}$ and $f(B) = \{1\}$. It only remains to see that it is continuous.

To this end, first observe that

$\array{ (\star) && \left( x \in Cl(U_r) \right) &\Rightarrow& \left( f(x) \leq r \right) \\ (\star\star) && \left( x \in U_r \right) &\Leftarrow& \left( f(x) \lt r \right) } \,.$

Here it is immediate from the definition that $(x \in U_r) \Rightarrow (f(x) \leq r)$ and that $(f(x) \lt r) \Rightarrow (x \in U_r \subset Cl(U_r))$. For the remaining implication, it is sufficient to observe that

$(x \in \partial U_r) \Rightarrow (f(x) = r) \,,$

where $\partial U_r \coloneqq Cl(U_r) \setminus U_r$ is the boundary of $U_r$.

This holds because the dyadic numbers are dense in $\mathbb{R}$. (And this would fail if we stopped the above decomposition into $U_{a/2^n}$-s at some finite $n$.) Namely, in one direction, if $x \in \partial U_r$ then for every small positive real number $\epsilon$ there exists a dyadic rational number $r'$ with $r \lt r' \lt r + \epsilon$, and by construction $U_{r'} \supset Cl(U_r)$ hence $x \in U_{r'}$. This implies that $\underset{U_r \supset \{x\}}{\lim} = r$.

Now we claim that for all $\alpha \in [0,1]$ then

1. $f^{-1}(\,(\alpha, 1]\,) = \underset{r \gt \alpha}{\cup} \left( X \setminus Cl(U_r) \right)$

2. $f^{-1}(\,[0,\alpha)\,) = \underset{r \lt \alpha}{\cup} U_r$

Thereby $f^{-1}(\,(\alpha, 1]\,)$ and $f^{-1}(\,[0,\alpha)\,)$ are exhibited as unions of open subsets, and hence they are open.

Regarding the first point:

\begin{aligned} & x \in f^{-1}( \,(\alpha,1]\, ) \\ \Leftrightarrow\, & f(x) \gt \alpha \\ \Leftrightarrow\, & \underset{r \gt \alpha}{\exists} (f(x) \gt r) \\ \overset{(\star)}{\Rightarrow}\, & \underset{r \gt \alpha}{\exists} \left( x \notin Cl(U_r) \right) \\ \Leftrightarrow\, & x \in \underset{r \gt \alpha}{\cup} \left(X \setminus Cl(U_r)\right) \end{aligned}

and

\begin{aligned} & x \in \underset{r \gt \alpha}{\cup} \left(X \setminus Cl(U_r)\right) \\ \Leftrightarrow\, & \underset{r \gt \alpha}{\exists} \left( x \notin Cl(U_r) \right) \\ \Rightarrow\, & \underset{r \gt \alpha}{\exists} \left( x \notin U_r \right) \\ \overset{(\star \star)}{\Rightarrow}\, & \underset{r \gt \alpha}{\exists} \left( f(x) \geq r \right) \\ \Leftrightarrow\, & f(x) \gt \alpha \\ \Leftrightarrow\, & x \in f^{-1}(\, (\alpha,1] \,) \end{aligned} \,.

Regarding the second point:

\begin{aligned} & x \in f^{-1}(\, [0,\alpha) \,) \\ \Leftrightarrow\, & f(x) \lt \alpha \\ \Leftrightarrow\, & \underset{r \lt \alpha}{\exists}( f(x) \lt r ) \\ \overset{(\star \star)}{\Rightarrow}\, & \underset{r \lt \alpha}{\exists }( x \in U_r ) \\ \Leftrightarrow\, & x \in \underset{r \lt \alpha}{\cup} U_r \end{aligned}

and

\begin{aligned} & x \in \underset{r \lt \alpha}{\cup} U_r \\ \Leftrightarrow\, & \underset{r \lt \alpha}{\exists }( x \in U_r ) \\ \overset{}{\Rightarrow}\, & \underset{r \lt \alpha}{\exists }( x \in Cl(U_r) ) \\ \overset{(\star)}{\Rightarrow}\, & \underset{r \lt \alpha}{\exists }( f(x) \leq r ) \\ \Leftrightarrow\, & f(x) \lt \alpha \\ \Leftrightarrow\, & x \in f^{-1}(\, [0,\alpha) \,) \end{aligned} \,.

(In these derivations we repeatedly use that $(0,1] \cap \mathbb{Q}_{dy}$ is dense in $[0,1]$ (def. ), and we use the contrapositions of $(\star)$ and $(\star \star)$.)

Now since the subsets $\{ [0,\alpha), (\alpha,1]\}_{\alpha \in [0,1]}$ form a sub-base (def. ) for the Euclidean metric topology on $[0,1]$, it follows that all pre-images of $f$ are open, hence that $f$ is continuous.

As a corollary of Urysohn's lemma we obtain yet another equivalent reformulation of the normality of topological spaces, this one now of a rather different character than the re-formulations in terms of explicit topological closures considered above:

###### Proposition

(normality equivalent to existence of Urysohn functions)

A $T_1$-space (def. ) is normal (def. ) precisely if it admits Urysohn functions (def ) separating every pair of disjoint closed subsets.

###### Proof

In one direction this is the statement of the Urysohn lemma, prop. .

In the other direction, assume the existence of Urysohn functions (def. ) separating all disjoint closed subsets. Let $A, B \subset X$ be disjoint closed subsets, then we need to show that these have disjoint open neighbourhoods.

But let $f \colon X \to [0,1]$ be an Urysohn function with $f(A) = \{0\}$ and $f(B) = \{1\}$ then the pre-images

$U_A \coloneqq f^{-1}([0,1/3) \phantom{AAA} U_B \coloneqq f^{-1}((2/3,1])$

are disjoint open neighbourhoods as required.

$\,$

### $T_n$ reflection

While the topological subspace construction preserves the $T_n$-property for $n \in \{0,1,2\}$ (example ) the construction of quotient topological spaces in general does not, as shown by examples and .

Further below we will see that, generally, among all universal constructions in the category Top of all topological spaces those that are limits preserve the $T_n$ property, while those that are colimits in general do not.

But at least for $T_0$, $T_1$ and $T_2$ there is a universal way, called reflection (prop. below), to approximate any topological space “from the left” by a $T_n$ topological spaces

Hence if one wishes to work within the full subcategory of the $T_n$-spaces among all topological space, then the correct way to construct quotients and other colimits (see below) is to first construct them as usual quotient topological spaces (example ), and then apply the $T_n$-reflection to the result.

###### Proposition

($T_n$-reflection)

Let $n \in \{0,1,2\}$. Then for every topological space $X$ there exists

1. a $T_n$-topological space $T_n X$

2. $t_n(X) \;\colon\; X \longrightarrow T_n X$

called the $T_n$-reflection of $X$,

which is the “closest approximation from the left” to $X$ by a $T_n$-topological space, in that for $Y$ any $T_n$-space, then continuous functions of the form

$f \;\colon\; X \longrightarrow Y$

are in bijection with continuous function of the form

$\tilde f \;\colon\; T_n X \longrightarrow Y$

and such that the bijection is constituted by

$f = \tilde f \circ t_n(X) \;\colon\; X \overset{ t_n(X)}{\longrightarrow} T_n X \overset{\tilde f}{\longrightarrow} Y \phantom{AAAA}i.e.: \phantom{AAA} \array{ X && \overset{f}{\longrightarrow} && Y \\ & {}_{\mathllap{t_n(X)}}\searrow && \nearrow_{\mathrlap{\tilde f}} \\ && T_n X } \,.$
• For $n = 0$ this is known as the Kolmogorov quotient construction (see prop. below).

• For $n = 2$ this is known as Hausdorff reflection or Hausdorffication or similar.

Moreover, the operation $T_n(-)$ extends to continuous functions $f \colon X \to Y$

$(X \overset{f}{\to} Y) \;\mapsto\; (T_n X \overset{T_n f}{\to} T_n Y)$

such as to preserve composition of functions as well as identity functions:

$T_n g \circ T_n f = T_n(g \circ f) \phantom{AA} \,, \phantom{AA} T_n id_X = id_{T_n X} \,$

Finally, the comparison map is compatible with this in that

$t_n(Y) \circ f = T_n(f) \circ t_n(X) \phantom{AAAA} i.e.: \phantom{AAAA} \array{ X &\overset{f}{\longrightarrow}& Y \\ {}^{\mathllap{t_n(X)}}\downarrow && \downarrow^{\mathrlap{t_n(Y)}} \\ T_n X &\underset{T_n(f)}{\longrightarrow}& T_n Y } \,.$

We prove this via a concrete construction of $T_n$-reflection in prop. below. But first we pause to comment on the bigger picture of the $T_n$-reflection:

###### Remark

(reflective subcategories)

In the language of category theory (remark ) the $T_n$-reflection of prop. says that

1. $T_n(-)$ is a functor $T_n \;\colon\; Top \longrightarrow Top_{T_n}$ from the category Top of topological spaces to the full subcategory $Top_{T_n} \overset{\iota}{\hookrightarrow} Top$ of Hausdorff topological spaces;

2. $t_n(X) \colon X \to T_n X$ is a natural transformation from the identity functor on Top to the functor $\iota \circ T_n$

3. $T_n$-topological spaces form a reflective subcategory of all topological spaces in that $T_n$ is left adjoint to the inclusion functor $\iota$; this situation is denoted as follows:

$Top_{T_n} \underoverset{\underset{\phantom{A}\iota\phantom{A}}{\hookrightarrow}}{\overset{\phantom{A}H\phantom{A}}{\longleftarrow}}{\bot} Top \,.$

Generally, an adjunction between two functors

$L \;\colon\; \mathcal{C} \leftrightarrow \mathcal{D} \;\colon\; R$

is for all pairs of objects $c \in \mathcal{C}$, $d \in \mathcal{D}$ a bijection between sets of morphisms of the form

$\left\{ L(c) \longrightarrow d \right\} \phantom{A} \leftrightarrow \phantom{A} \left\{ c \longrightarrow R(d) \right\} \,.$

i.e.

$Hom_{\mathcal{D}}(L(c), d) \underoverset{\phantom{AA}\simeq \phantom{AA}}{\phi_{c,d}}{\longrightarrow} Hom_{\mathcal{C}}(c, R(d))$

and such that these bijections are “natural” in that they for all pairs of morphisms $f \colon c' \to c$ and $g \colon d \to d'$ then the folowing diagram commutes:

$\array{ Hom_{\mathcal{D}}(L(c), d) &\underoverset{\phantom{AA}\simeq\phantom{AA}}{\phi_{c,d}}{\longrightarrow}& Hom_{\mathcal{C}}(c, R(d)) \\ {\mathllap{g \circ (-) \circ L(f)}}\downarrow && \downarrow{\mathrlap{ R(g) \circ (-) \circ f }} \\ Hom_{\mathcal{C}}(L(c'), d') &\underoverset{\phantom{AA}\simeq\phantom{AA}}{\phi_{c',d'}}{\longrightarrow}& Hom_{\mathcal{D}}(c', R(d')) } \,.$

One calls the image under $\phi_{c,L(c)}$ of the identity morphism $id_{L(x)}$ the unit of the adjunction, written

$\eta_x \;\colon\; c \longrightarrow R(L(c)) \,.$

One may show that it follows that the image $\tilde f$ under $\phi_{c,d}$ of a general morphism $f \colon c \to d$ (called the adjunct of $f$) is given by this composite:

$\tilde f \;\colon\; c \overset{\eta_c}{\longrightarrow} R(L(c)) \overset{R(f)}{\longrightarrow} R(d) \,.$

In the case of the reflective subcategory inclusion $(T_n \dashv \iota)$ of the category of $T_n$-spaces into the category Top of all topological spaces this adjunction unit is precisely the $T_n$-reflection $t_n(X) \colon X \to \iota( T_n(X))$ (only that we originally left the re-embedding $\iota$ notationally implicit).

$\,$

There are various ways to see the existence and to construct the $T_n$-reflections. The following is the quickest way to see the existence, even though it leaves the actual construction rather implicit.

###### Proposition

($T_n$-reflection via explicit quotients)

Let $n \in \{0,1,2\}$. Let $(X,\tau)$ be a topological space and consider the equivalence relation $\sim$ on the underlying set $X$ for which $x_1 \sim x_2$ precisely if for every surjective continuous function $f \colon X \to Y$ into any $T_n$-topological space $Y$ (def. ) we have $f(x_1) = f(x_2)$:

$(x_1 \sim x_2) \;\coloneqq\; \underset{ { Y \in Top_{T_n} } \atop { X \underoverset{\text{surjective}}{f}{\to} Y } }{\forall} \left( f(x) = f(y) \right) \,.$

Then

1. the set of equivalence classes

$T_n X \coloneqq X /{\sim}$

equipped with the quotient topology (example ) is a $T_n$-topological space,

2. the quotient projection

$\array{ X & \overset{t_n(X)}{\longrightarrow} & X/{\sim} \\ x &\overset{\phantom{AAA}}{\mapsto}& [x] }$

exhibits the $T_n$-reflection of $X$, according to prop. .

###### Proof

First we observe that every continuous function $f \colon X \longrightarrow Y$ into a $T_n$-topological space $Y$ factors uniquely, via $t_n(X)$ through a continuous function $\tilde f$ (this makes use of the “universal property” of the quotient topology, which we dwell on a bit more below in example ):

$f = \tilde f \circ t_n(X)$

Clearly this continuous function $\tilde f$ is unique if it exists, because its underlying function of sets must be given by

$\tilde f \colon [x] \mapsto f(x) \,.$

First observe that this is indeed well defined as a function of underlying sets. To that end, factor $f$ through its image $f(X)$

$f \;\colon\; X \longrightarrow f(X) \hookrightarrow Y$

equipped with its subspace topology as a subspace of $Y$ (example ). By prop. also the image $f(X)$ is a $T_n$-topological space, since $Y$ is. This means that if two elements $x_1, x_2 \in X$ have the same equivalence class, then, by definition of the equivalence relation, they have the same image under all continuous surjective functions into a $T_n$-space, hence in particular they have the same image under $f \colon X \overset{\text{surjective}}{\longrightarrow} f(X) \hookrightarrow Y$:

\begin{aligned} ( [x_1] = [x_2]) & \Leftrightarrow\, (x_1 \sim x_2) \\ & \Rightarrow\, ( f(x_1) = f(x_2) ) \,. \end{aligned}

This shows that $\tilde f$ is well defined as a function between sets.

To see that $\tilde f$ is also continuous, consider $U \in Y$ an open subset. We need to show that the pre-image $\tilde f^{-1}(U)$ is open in $X/\sim$. But by definition of the quotient topology (example ), this is open precisely if its pre-image under the quotient projection $t_n(X)$ is open, hence precisely if

\begin{aligned} (t_n(X))^{-1} \left( \tilde f^{-1}\left(U\right) \right) & = \left( \tilde f \circ t_n(X) \right)^{-1}(U) \\ & = f^{-1}(U) \end{aligned}

is open in $X$. But this is the case by the assumption that $f$ is continuous. Hence $\tilde f$ is indeed the unique continuous function as required.

What remains to be seen is that $T_n X$ as constructed is indeed a $T_n$-topological space. Hence assume that $[x] \neq [y] \in T_n X$ are two distinct points. Depending on the value of $n$, need to produce open neighbourhoods around one or both of these points not containing the other point and possibly disjoint to each other.

Now by definition of $T_n X$ the assumption $[x] \neq [y]$ means that there exists a $T_n$-topological space $Y$ and a surjective continuous function $f \colon X \overset{surjective}{\longrightarrow} Y$ such that $f(x) \neq f(y) \in Y$:

$( [x_1] \neq [x_2] ) \;\Leftrightarrow\; \underset{ { Y \in Top_{T_m} \atop { X \underoverset{\text{surjective}}{f}{\longrightarrow} Y } } }{\exists} \left( f(x_1) \neq f(x_2) \right) \,.$

Accordingly, since $Y$ is $T_n$, there exist the respective kinds of neighbourhoods around $f(x_1)$ and $f(x_2)$ in $Y$. Moreover, by the previous statement there exists the continuous function $\tilde f \colon T_n X \to Y$ with $\tilde f([x_1]) = f(x_1)$ and $\tilde f([x_2]) = f(x_2)$. By the nature of continuous functions, the pre-images of these open neighbourhoods in $Y$ are still open in $X$ and still satisfy the required disjunction properties. Therefore $T_n X$ is a $T_n$-space.

Here are alternative constructions of the reflections:

###### Proposition

(Kolmogorov quotient)

Let $(X,\tau)$ be a topological space. Consider the relation on the underlying set by which $x_1 \sim x_2$ precisely if neither $x_i$ has an open neighbourhood not containing the other. This is an equivalence relation. The quotient topological space $X \to X/\sim$ by this equivalence relation (def. ) exhibits the $T_0$-reflection of $X$ according to prop. .

A more explicit construction of the Hausdorff quotient than given by prop. is rather more involved. The issue is that the relation “$x$ and $y$ are not separated by disjoint open neighbourhoods” is not transitive;

###### Proposition

(more explicit Hausdorff reflection)

For $(Y,\tau_Y)$ a topological space, write $r_Y \subset Y \times Y$ for the transitive closure of the relation given by the topological closure $Cl(\Delta_Y)$ of the image of the diagonal $\Delta_Y \colon Y \hookrightarrow Y \times Y$.

$r_Y \coloneqq Trans(Cl(Delta_Y)) \,.$

Now for $(X,\tau_X)$ a topological space, define by induction for each ordinal number $\alpha$ an equivalence relation $r^\alpha$ on $X$ as follows, where we write $q^\alpha \colon X \to H^\alpha(X)$ for the corresponding quotient topological space projection:

We start the induction with the trivial equivalence relation:

• $r^0_X \coloneqq \Delta_X$;

For a successor ordinal we set

• $r_X^{\alpha+1} \coloneqq \left\{ (a,b) \in X \times X \,\vert\, (q^\alpha(a), q^\alpha(b)) \in r_{H^\alpha(X)} \right\}$

and for a limit ordinal $\alpha$ we set

• $r_X^\alpha \coloneqq \underset{\beta \lt \alpha}{\cup} r_X^\beta$.

Then:

1. there exists an ordinal $\alpha$ such that $r_X^\alpha = r_X^{\alpha+1}$

2. for this $\alpha$ then $H^\alpha(X) = H(X)$ is the Hausdorff reflection from prop. .

A detailed proof is spelled out in (vanMunster 14, section 4).

###### Example

(Hausdorff reflection of the line with two origins)

The Hausdorff reflection ($T_2$-reflection, prop. )

$T_2 \;\colon\; Top \longrightarrow Top_{Haus}$

of the line with two origins from example is the real line itself:

$T_2\left( \left( \mathbb{R} \sqcup \mathbb{R} \right)/\sim \right) \;\simeq\; \mathbb{R} \,.$

$\,$

## Sober spaces

While the original formulation of the separation axioms $T_n$ from def. and def. clearly does follow some kind of pattern, its equivalent reformulation in terms of closure conditions in prop. , prop. , prop , prop. and prop. suggests rather different patterns. Therefore it is worthwhile to also consider separation-like axioms that are not among the original list.

In particular, the alternative characterization of the $T_0$-condition in prop. immediately suggests the following strengthening, different from the $T_1$-condition (see example below):

###### Definition

(sober topological space)

A topological space $(X,\tau)$ is called a sober topological space precisely if every irreducible closed subspace (def. ) is the topological closure (def. ) of a unique point, hence precisely if the function

$Cl(\{-\}) \;\colon\; X \longrightarrow IrrClSub(X)$

from the underlying set of $X$ to the set of irreducible closed subsets of $X$ (def. , well defined according to example ) is bijective.

###### Proposition

(sober implies $T_0$)

Every sober topological space (def. ) is $T_0$ (def. ).

By prop. .

###### Proposition

(Hausdorff spaces are sober)

Every Hausdorff topological space (def. ) is a sober topological space (def. ).

More specifically, in a Hausdorff topological space the irreducible closed subspaces (def. ) are precisely the singleton subspaces (def. ).

Hence, by example , in particular every metric space with its metric topology (example ) is sober.

###### Proof

The second statement clearly implies the first. To see the second statement, suppose that $F$ is an irreducible closed subspace which contained two distinct points $x \neq y$. Then by the Hausdorff property there would be disjoint neighbourhoods $U_x, U_y$, and hence it would follow that the relative complements $F \setminus U_x$ and $F \setminus U_y$ were distinct closed proper subsets of $F$ with

$F = (F \setminus U_x) \cup (F \setminus U_y)$

in contradiction to the assumption that $F$ is irreducible.

This proves by contradiction that every irreducible closed subset is a singleton. Conversely, generally the topological closure of every singleton is irreducible closed, by example .

By prop. and prop. we have the implications on the right of the following diagram:

separation axioms
$\array{\\ &&& T_2 = \text{Hausdorff} \\ && \swArrow && \seArrow \\ \, & T_1 && && \text{sober} & \, \\ && \seArrow && \swArrow \\ &&& T_0 = \text{Kolmogorov} \\ }$

But there there is no implication betwee $T_1$ and sobriety:

###### Proposition

The intersection of the classes of sober topological spaces (def. ) and $T_1$-topological spaces (def. ) is not empty, but neither class is contained within the other.

That the intersection is not empty follows from prop. . That neither class is contained in the other is shown by the following counter-examples:

###### Example

($T_1$ neither implies nor is implied by sobriety)

Finally, sobriety is indeed strictly weaker that Hausdorffness:

###### Example

(schemes are sober but in general not Hausdorff)

The Zariski topology on an affine space (example ) or more generally on the prime spectrum of a commutative ring (example ) is

1. sober (def );

2. in general not Hausdorff (def. ).

For details see at Zariski topology this prop and this example.

### Frames of opens

What makes the concept of sober topological spaces special is that for them the concept of continuous functions may be expressed entirely in terms of the relations between their open subsets, disregarding the underlying set of points of which these opens are in fact subsets.

Recall from example that for every continuous function $f \colon (X, \tau_X) \to (Y, \tau_Y)$ the pre-image function $f^{-1} \colon \tau_Y \to \tau_X$ is a frame homomorphism (def. ).

For sober topological spaces the converse holds:

###### Proposition

If $(X,\tau_X)$ and $(Y,\tau_Y)$ are sober topological spaces (def. ), then for every frame homomorphism (def. )

$\tau_X \longleftarrow \tau_Y \;\colon\; \phi$

there is a unique continuous function $f \colon X \to Y$ such that $\phi$ is the function of forming pre-images under $f$:

$\phi = f^{-1} \,.$
###### Proof

We first consider the special case of frame homomorphisms of the form

$\tau_\ast \longleftarrow \tau_X \;\colon\; \phi$

and show that these are in bijection to the underlying set $X$, identified with the continuous functions $\ast \to (X,\tau)$ via example .

By prop. , the frame homomorphisms $\phi \colon \tau_X \to \tau_\ast$ are identified with the irreducible closed subspaces $X \setminus U_\emptyset(\phi)$ of $(X,\tau_X)$. Therefore by assumption of sobriety of $(X,\tau)$ there is a unique point $x \in X$ with $X \setminus U_{\emptyset} = Cl(\{x\})$. In particular this means that for $U_x$ an open neighbourhood of $x$, then $U_x$ is not a subset of $U_\emptyset(\phi)$, and so it follows that $\phi(U_x) = \{1\}$. In conclusion we have found a unique $x \in X$ such that

$\phi \;\colon\; U \mapsto \left\{ \array{ \{1\} & \vert \,\text{if}\, x \in U \\ \emptyset & \vert \, \text{otherwise} } \right. \,.$

This is precisely the inverse image function of the continuous function $\ast \to X$ which sends $1 \mapsto x$.

Hence this establishes the bijection between frame homomorphisms of the form $\tau_\ast \longleftarrow \tau_X$ and continuous functions of the form $\ast \to (X,\tau)$.

With this it follows that a general frame homomorphism of the form $\tau_X \overset{\phi}{\longleftarrow} \tau_Y$ defines a function of sets $X \overset{f}{\longrightarrow} Y$ by composition:

$\array{ X &\overset{f}{\longrightarrow}& Y \\ (\tau_\ast \leftarrow \tau_X) &\mapsto& (\tau_\ast \leftarrow \tau_X \overset{\phi}{\longleftarrow} \tau_Y) } \,.$

By the previous analysis, an element $U_Y \in \tau_Y$ is sent to $\{1\}$ under this composite precisely if the corresponding point $\ast \to X \overset{f}{\longrightarrow} Y$ is in $U_Y$, and similarly for an element $U_X \in \tau_X$. It follows that $\phi(U_Y) \in \tau_X$ is precisely that subset of points in $X$ which are sent by $f$ to elements of $U_Y$, hence that $\phi = f^{-1}$ is the pre-image function of $f$. Since $\phi$ by definition sends open subsets of $Y$ to open subsets of $X$, it follows that $f$ is indeed a continuous function. This proves the claim in generality.

###### Remark

(locales)

Proposition is often stated as saying that sober topological spaces are equivalently the “locales with enough points” (Johnstone 82, II 1.). Here “locale” refers to a concept akin to topological spaces where one considers just a “frame of open subsets$\tau_X$, without requiring that its elements be actual subsets of some ambient set. The natural notion of homomorphism between such generalized topological spaces are clearly the frame homomorphisms $\tau_X \leftarrow \tau_Y$ from def. .

From this perspective, prop. says that sober topological spaces $(X, \tau_X)$ are entirely characterized by their frames of opens $\tau_X$ and just so happen to “have enough points” such that these are actual open subsets of some ambient set, namely of $X$.

### Sober reflection

We saw above in prop. that every $T_n$-topological space for $n \in \{0,1,2\}$ has a “best approximation from the left” by a $T_n$-topological space (for $n = 2$: “Hausdorff reflection”). We now discuss the analogous statement for sober topological spaces.

Recall again the point topological space $\ast \coloneqq ( \{1\}, \tau_\ast = \left\{ \emptyset, \{1\}\right\} )$ (example ).

###### Definition

(sober reflection)

Let $(X,\tau)$ be a topological space.

Define $S X$ to be the set

$S X \coloneqq FrameHom( \tau_X, \tau_\ast )$

of frame homomorphisms (def. ) from the frame of opens of $X$ to that of the point. Define a topology $\tau_{S X} \subset P(S X)$ on this set by declaring it to have one element $\tilde U$ for each element $U \in \tau_X$ and given by

$\tilde U \;\coloneqq\; \left\{ \phi \in S X \,\vert\, \phi(U) = \{1\} \right\} \,.$

Consider the function

$\array{ X &\overset{s_X}{\longrightarrow}& S X \\ x &\mapsto& (const_x)^{-1} }$

which sends an element $x \in X$ to the function which assigns inverse images of the constant function $const_x \;\colon\; \{1\} \to X$ on that element.

We are going to call this function the sober reflection of $X$.

###### Lemma

(sober reflection is well defined)

The construction $(S X, \tau_{S X})$ in def. is a topological space, and the function $s_X \colon X \to S X$ is a continuous function

$s_X \colon (X, \tau_X) \longrightarrow (S X, \tau_{S X})$
###### Proof

To see that $\tau_{S X} \subset P(S X)$ is closed under arbitrary unions and finite intersections, observe that the function

$\array{ \tau_X &\overset{\widetilde{(-)}}{\longrightarrow}& \tau_{S X} \\ U &\mapsto& \tilde U }$

in fact preserves arbitrary unions and finite intersections. With this the statement follows by the fact that $\tau_X$ is closed under these operations.

To see that $\widetilde{(-)}$ indeed preserves unions, observe that (e.g. Johnstone 82, II 1.3 Lemma)

\begin{aligned} p \in \underset{i \in I}{\cup} \widetilde{U_i} \; & \Leftrightarrow \underset{i \in I}{\exists} p(U_i) = \{1\} \\ & \Leftrightarrow \underset{i \in I}{\cup} p(U_i) = \{1\} \\ & \Leftrightarrow p\left(\underset{i \in I}{\cup} U_i\right) = \{1\} \\ & \Leftrightarrow p \in \widetilde{ \underset{i \in I}{\cup} U_i } \end{aligned} \,,

where we used that the frame homomorphism $p \colon \tau_X \to \tau_\ast$ preserves unions. Similarly for intersections, now with $I$ a finite set:

\begin{aligned} p \in \underset{i \in I}{\cap} \widetilde{U_i} \; & \Leftrightarrow \underset{i \in I}{\forall} p(U_i) = \{1\} \\ & \Leftrightarrow \underset{i \in I}{\cap} p(U_i) = \{1\} \\ & \Leftrightarrow p\left(\underset{i \in I}{\cap} U_i\right) = \{1\} \\ & \Leftrightarrow p \in \widetilde{ \underset{i \in I}{\cap} U_i } \end{aligned} \,,

where we used that the frame homomorphism $p$ preserves finite intersections.

To see that $s_X$ is continuous, observe that $s_X^{-1}(\tilde U) = U$, by construction.

###### Lemma

(sober reflection detects $T_0$ and sobriety)

For $(X, \tau_X)$ a topological space, the function $s_X \colon X \to S X$ from def. is

1. an injection precisely if $(X,\tau_X)$ is $T_0$ (def. );

2. a bijection precisely if $(X,\tau_Y)$ is sober (def. ), in which case $s_X$ is in fact a homeomorphism (def. ).

###### Proof

By lemma there is an identification $S X \simeq IrrClSub(X)$ and via this $s_X$ is identified with the map $x \mapsto Cl(\{x\})$.

Hence the second statement follows by definition, and the first statement by prop. .

That in the second case $s_X$ is in fact a homeomorphism follows from the definition of the opens $\tilde U$: they are identified with the opens $U$ in this case (…expand…).

###### Lemma

(soberification lands in sober spaces, e.g. Johnstone 82, lemma II 1.7)

For $(X,\tau)$ a topological space, then the topological space $(S X, \tau_{S X})$ from def. , lemma is sober.

###### Proof

Let $S X \setminus \tilde U$ be an irreducible closed subspace of $(S X, \tau_{S X})$. We need to show that it is the topological closure of a unique element $\phi \in S X$.

Observe first that also $X \setminus U$ is irreducible.

To see this use prop. , saying that irreducibility of $X \setminus U$ is equivalent to $U_1 \cap U_2 \subset U \Rightarrow (U_1 \subset U) or (U_2 \subset U)$. But if $U_1 \cap U_2 \subset U$ then also $\tilde U_1 \cap \tilde U_2 \subset \tilde U$ (as in the proof of lemma ) and hence by assumption on $\tilde U$ it follows that $\tilde U_1 \subset \tilde U$ or $\tilde U_2 \subset \tilde U$. By lemma this in turn implies $U_1 \subset U$ or $U_2 \subset U$. In conclusion, this shows that also $X \setminus U$ is irreducible .

By lemma this irreducible closed subspace corresponds to a point $p \in S X$. By that same lemma, this frame homomorphism $p \colon \tau_X \to \tau_\ast$ takes the value $\emptyset$ on all those opens which are inside $U$. This means that the topological closure of this point is just $S X \setminus \tilde U$.

This shows that there exists at least one point of which $X \setminus \tilde U$ is the topological closure. It remains to see that there is no other such point.

So let $p_1 \neq p_2 \in S X$ be two distinct points. This means that there exists $U \in \tau_X$ with $p_1(U) \neq p_2(U)$. Equivalently this says that $\tilde U$ contains one of the two points, but not the other. This means that $(S X, \tau_{S X})$ is T0. By prop. this is equivalent to there being no two points with the same topological closure.

###### Proposition

(unique factorization through soberification)

For $(X, \tau_X)$ any topological space, for $(Y,\tau_Y^{sob})$ a sober topological space, and for $f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y)$ a continuous function, then it factors uniquely through the soberification $s_X \colon (X, \tau_X) \longrightarrow (S X, \tau_{S X})$ from def. , lemma

$\array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow_{\exists !} \\ (S X , \tau_{S X}) } \,.$
###### Proof

By the construction in def. , we find that the outer part of the following square commutes:

$\array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau^{sob}_Y) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow& \downarrow^{\mathrlap{s_{S X}}} \\ (S X, \tau_{S X}) &\underset{S f}{\longrightarrow}& (S S X, \tau_{S S X}) } \,.$

By lemma and lemma , the right vertical morphism $s_{S X}$ is an isomorphism (a homeomorphism), hence has an inverse morphism. This defines the diagonal morphism, which is the desired factorization.

To see that this factorization is unique, consider two factorizations $\tilde f, \overline{f} \colon \colon (S X, \tau_{S X}) \to (Y, \tau_Y^{sob})$ and apply the soberification construction once more to the triangles

$\array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow_{ \tilde f, \overline{f}} \\ (S X , \tau_{S X}) } \phantom{AAA} \mapsto \phantom{AAA} \array{ (S X, \tau_{S X}) &\overset{S f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\simeq}\downarrow & \nearrow_{ \tilde f, \overline{f}} \\ (S X , \tau_{S X}) } \,.$

Here on the right we used again lemma to find that the vertical morphism is an isomorphism, and that $\tilde f$ and $\overline{f}$ do not change under soberification, as they already map between sober spaces. But now that the left vertical morphism is an isomorphism, the commutativity of this triangle for both $\tilde f$ and $\overline{f}$ implies that $\tilde f = \overline{f}$.

In summary we have found

###### Proposition

(sober reflection)

For every topological space $X$ there exists

1. a sober topological spaces $S X$;

2. a continuous function $s_X \colon X \longrightarrow S X$

such that …

As before for the $T_n$-reflection in remark , the statement of prop. may neatly be re-packaged:

###### Remark

(sober topological spaces are a reflective subcategory)

In the language of category theory (remark ) and in terms of the concept of adjoint functors (remark ), proposition simply says that sober topological spaces form a reflective subcategory $Top_{sob}$ of the category Top of all topological spaces

$Top_{sob} \underoverset {\underset{}{\hookrightarrow}} {\overset{s}{\longleftarrow}} {\bot} Top \,.$

$\,$

## Universal constructions

We have seen above various construction principles for topological spaces above, such as topological subspaces and topological quotient spaces. It turns out that these constructions enjoy certain “universal properties” which allow us to find continuous functions into or out of these spaces, respectively (examples , example and below).

Since this is useful for handling topological spaces (we secretly used the universal property of the quotient space construction already in the proof of prop. ), we next consider, in def. below, more general “universal constructions” of topological spaces, called limits and colimits of topological spaces (and to be distinguished from limits in topological spaces, in the sense of convergence of sequences as in def. ).

Moreover, we have seen above that the quotient space construction in general does not preserve the $T_n$-separation property or sobriety property of topological spaces, while the topological subspace construction does. The same turns out to be true for the more general “colimiting” and “limiting” universal constructions. But we have also seen that we may universally “reflect” any topological space to becomes a $T_n$-space or sober space. The remaining question then is whether this reflection breaks the desired universal property. We discuss that this is not the case, that instead the universal construction in all topological spaces followed by these reflections gives the correct universal constructions in $T_n$-separated and sober topological spaces, respectively (remark below).

After these general considerations, we finally discuss a list of examples of universal constructions in topological spaces.

$\,$

To motivate the following generalizations, first observe the universal properties enjoyed by the basic construction principles of topological spaces from above

###### Example

(universal property of binary product topological space)

Let $X_1, X_2$ be topological spaces. Consider their product topological space $X_1 \times X_2$ from example . By example the two projections out of the product space are continuous functions

$\array{ X_1 &\overset{pr_1}{\longleftarrow}& X_1 \times X_2 &\overset{pr_2}{\longrightarrow}& X_2 } \,.$

Now let $Y$ be any other topological space. Then, by composition, every continuous function $Y \to X_1 \times X_2$ into the product space yields two continuous component functions $f_1$ and $f_2$:

$\array{ && Y \\ & {}^{\mathllap{f_1}}\swarrow & \downarrow & \searrow^{\mathrlap{f_2}} \\ X_1 &\underset{pr_1}{\longleftarrow}& X_1 \times X_2 &\underset{pr_2}{\longrightarrow}& X_2 } \,.$

But in fact these two components completely characterize the function into the product: There is a (natural) bijection between continuous functions into the product space and pairs of continuous functions into the two factor spaces:

$\array{ & \left\{ Y \longrightarrow X_1 \times X_2 \right\} &\simeq& \left\{ \left( \array{ Y \longrightarrow X_1, \\ Y \longrightarrow X_2 } \right) \right\} \\ \text{i.e.:} & \\ & Hom(Y, X_1 \times X_2) &\simeq& Hom(Y,X_1) \times Hom(Y, X_2) } \,.$
###### Example

(universal property of disjoint union spaces)

Let $X_1, X_2$ be topological spaces. Consider their disjoint union space $X_1 \sqcup X_2$ from example . By definition, the two inclusions into the disjoint union space are clearly continuous functions

$\array{ X_1 &\overset{i_1}{\longrightarrow}& X_1 \sqcup X_2 &\overset{i_2}{\longleftarrow}& X_2 } \,.$

Now let $Y$ be any other topological space. Then by composition a continuous function $X_1 \sqcup X_2 \longrightarrow Y$ out of the disjoint union space yields two continuous component functions $f_1$ and $f_2$:

$\array{ X_1 &\overset{i_1}{\longleftarrow}& X_1 \sqcup X_2 &\overset{i_2}{\longrightarrow}& X_2 \\ & {}_{\mathllap{f_1}}\searrow & \downarrow & \swarrow_{\mathrlap{f_2}} \\ && Y } \,.$

But in fact these two components completely characterize the function out of the disjoint union: There is a (natural) bijection between continuous functions out of disjoint union spaces and pairs of continuous functions out of the two summand spaces:

(1)$\array{ & \left\{ X_1 \sqcup X_2 \longrightarrow Y \right\} &\simeq& \left\{ \left( \array{ X_1 \longrightarrow Y, \\ X_2 \longrightarrow Y } \right) \right\} \\ \text{i.e.:} \\ & Hom(X_1 \sqcup X_2, Y) &\simeq& Hom(X_1, Y) \times Hom(X_2, Y) } \,.$
###### Example

(universal property of quotient topological spaces)

Let $X$ be a topological space, and let $\sim$ be an equivalence relation on its underlying set. Then the corresponding quotient topological space $X/\sim$ together with the corresponding quotient continuous function $p \colon X \to X/\sim$ has the following universal property:

Given $f \colon X \longrightarrow Y$ any continuous function out of $X$ with the property that it respects the given equivalence relation, in that

$(x_1 \sim x_2) \;\Rightarrow\; \left( f(x_1) = f(x_2) \right)$

then there is a unique continuous function $\tilde f \colon X/\sim \longrightarrow Y$ such that

$f = \tilde f\circ p \phantom{AAAA} i.e. \phantom{Aaaa} \array{ X &\overset{f}{\longrightarrow}& Y \\ {}^{\mathllap{p}}\downarrow & \nearrow_{\exists ! \tilde f} \\ X/\sim } \,.$

(We already made use of this universal property in the construction of the $T_n$-reflection in the proof of prop. .)

###### Proof

First observe that there is a unique function $\tilde f$ as claimed on the level of functions of the underlying sets: In order for $f = \tilde f \circ p$ to hold, $\tilde f$ must send an equivalence class in $X/\sim$ to one of its members

$\tilde f \;\colon\; [x] \mapsto x$

and that this is well defined and independent of the choice of representative $x$ is guaranteed by the condition on $f$ above.

Hence it only remains to see that $\tilde f$ defined this way is continuous, hence that for $U \subset Y$ an open subset, then its pre-image $\tilde f^{-1}(U) \subset X/\sim$ is open in the quotient topology. By definition of the quotient topology (example ), this is the case precisely if its further pre-image under $p$ is open in $X$. But by the fact that $f = \tilde f \circ p$, this is the case by the continuity of $f$:

\begin{aligned} p^{-1} \left( \tilde f^{-1} \left( U \right) \right) & = \left( \tilde f \circ p \right)^{-1}(U) \\ & = f^{-1}(U) \end{aligned} \,.

This kind of example we now generalize.

$\,$

### Limits and colimits

We consider now the general definition of free diagrams of topological spaces (def. below), their cones and co-cones (def. ) as well as limiting cones and colimiting cocones (def. below).

Then we use these concepts to see generally (remark below) why limits (such as product spaces and subspaces) of $T_{n \leq 2}$-spaces and of sober spaces are again $T_n$ or sober, respectively, and to see that the correct colimits (such as disjoint union spaces and quotient spaces) of $T_n$- or sober spaces are instead the $T_n$-reflection (prop. ) or sober reflection (prop. ), respectively, of these colimit constructions performed in the context of unconstrained topological spaces.

$\,$

###### Definition

(free diagram of sets/topological spaces)

A free diagram $X_\bullet$ of sets or of topological spaces is

1. an indexed set $\{ X_i \}_{i \in I}$ of sets or of topological spaces, respectively;

2. for every pair $(i,j) \in I \times I$ of labels, a set $\{ X_i \overset{ f_\alpha }{\longrightarrow} X_j\}_{\alpha \in I_{i,j}}$ of functions of continuous functions, respectively, between these.

$\,$

Here is a list of basic and important examples of free diagrams

• discrete diagrams and the empty diagram (example );

• pairs of parallel morphisms (example );

• span and cospan diagram (example );

• tower and cotower diagram (example ).

###### Example

(discrete diagram and empty diagram)

Let $I$ be any set, and for each $(i,j) \in I \times I$ let $I_{i,j} = \emptyset$ be the empty set.

The corresponding free diagrams (def. ) are simply a set of sets/topological spaces with no specified (continuous) functions between them. This is called a discrete diagram.

For example for $I = \{1,2,3\}$ the set with 3-elements, then such a diagram looks like this:

$X_1 \phantom{AAA} X_2 \phantom{AAA} X_3 \,.$

Notice that here the index set may be empty set, $I = \emptyset$, in which case the corresponding diagram consists of no data. This is also called the empty diagram.

###### Definition

(parallel morphisms diagram)

Let $I = \{a, b\}$ be the set with two elements, and consider the sets

$I_{i,j} \;\coloneqq\; \left\{ \array{ \{ 1,2 \} & \vert & (i = a) \,\text{and}\, (j = b) \\ \emptyset & \vert & \text{otherwise} } \right\} \,.$

The corresponding free diagrams (def. ) are called pairs of parallel morphisms. They may be depicted like so:

$X_a \underoverset {\underset{f_2}{\longrightarrow}} {\overset{f_1}{\longrightarrow}} {\phantom{AAAAA}} X_b \,.$
###### Example

(span and cospan diagram)

Let $I = \{a,b,c\}$ the set with three elements, and set

$I_{i ,j} = \left\{ \array{ \{f_1\} & \vert & (i = c) \,\text{and}\, (j = a) \\ \{f_2\} & \vert & (i = c) \,\text{and}\, (j = b) \\ \emptyset & \vert & \text{otherwise} } \right.$

The corresponding free diagrams (def. ) look like so:

$\array{ && X_c \\ & {}^{\mathllap{f_1}}\swarrow && \searrow^{\mathrlap{f_2}} \\ X_a && && X_b } \,.$

These are called span diagrams.

Similarly, there is the cospan diagram of the form

$\array{ && X_c \\ & {}^{\mathllap{f_1}}\nearrow && \nwarrow^{\mathrlap{f_2}} \\ X_a && && X_b } \,.$
###### Example

(tower diagram)

Let $I = \mathbb{N}$ be the set of natural numbers and consider

$I_{i,j} \;\coloneqq\; \left\{ \array{ \{f_{i,j}\} & \vert & j = i+1 \\ \emptyset & \vert & \text{otherwise} } \right.$

The corresponding free diagrams (def. ) are called tower diagrams. They look as follows:

$X_0 \overset{\phantom{A}f_{0,1} \phantom{A} }{\longrightarrow} X_1 \overset{\phantom{A} f_{1,2} \phantom{A} }{\longrightarrow} X_2 \overset{\phantom{A} f_{2,3} \phantom{A} }{\longrightarrow} X_3 \overset{}{\longrightarrow} \cdots \,.$

Similarly there are co-tower diagram

$X_0 \overset{\phantom{A} f_{0,1} \phantom{A} }{\longleftarrow} X_1 \overset{\phantom{A} f_{1,2} \phantom{A}}{\longleftarrow} X_2 \overset{\phantom{A} f_{2,3} \phantom{A}}{\longleftarrow} X_3 \overset{}{\longleftarrow} \cdots \,.$

$\,$

###### Definition

(cone over a free diagram)

Consider a free diagram of sets or of topological spaces (def. )

$X_\bullet \,=\, \left\{ X_i \overset{f_\alpha}{\longrightarrow} X_j \right\}_{i,j \in I, \alpha \in I_{i,j}} \,.$

Then

1. a cone over this diagram is

1. a set or topological space $\tilde X$ (called the tip of the cone);

2. for each $i \in I$ a function or continuous function $\tilde X \overset{p_i}{\longrightarrow} X_i$

such that

• for all $(i,j) \in I \times I$ and all $\alpha \in I_{i,j}$ then the condition

$f_{\alpha} \circ p_i = p_j$

holds, which we depict as follows:

$\array{ && \tilde X \\ & {}^{\mathllap{p_i}}\swarrow && \searrow^{\mathrlap{p_j}} \\ X_i && \underset{f_\alpha}{\longrightarrow} && X_j }$
2. a co-cone over this diagram is

1. a set or topological space $\tilde X$ (called the tip of the co-cone);

2. for each $i \in I$ a function or continuous function $q_i \colon X_i \longrightarrow \tilde X$;

such that

• for all $(i,j) \in I \times I$ and all $\alpha \in I_{i,j}$ then the condition

$q_j \circ f_{\alpha} = q_i$

holds, which we depict as follows:

$\array{ X_i && \overset{f_\alpha}{\longrightarrow} && X_j \\ & {}_{\mathllap{q_i}}\searrow && \swarrow_{\mathrlap{q_j}} \\ && \tilde X } \,.$
###### Example

(solutions to equations are cones)

Let $f,g \colon \mathbb{R} \to \mathbb{R}$ be two functions from the real numbers to themselves, and consider the corresponding parallel morphism diagram of sets (example ):

$\mathbb{R} \underoverset {\underset{f_2}{\longrightarrow}} {\overset{f_1}{\longrightarrow}} {\phantom{AAAAA}} \mathbb{R} \,.$

Then a cone (def. ) over this free diagram with tip the singleton set $\ast$ is a solution to the equation $f(x) = g(x)$

$\array{ && \ast \\ & {}^{\mathllap{const_x}}\swarrow && \searrow^{\mathrlap{const_y}} \\ \mathbb{R} && \underoverset {\underset{f_2}{\longrightarrow}} {\overset{f_1}{\longrightarrow}} {\phantom{AAAAA}} && \mathbb{R} } \,.$

Namely the components of the cone are two functions of the form

$cont_x, const_y \;\colon\; \ast \to \mathbb{R}$

hence equivalently two real numbers, and the conditions on these are

$f_1 \circ const_x = const_y \phantom{AAAA} f_2 \circ const_x = const_y \,.$
###### Definition

(limiting cone over a diagram)

Consider a free diagram of sets or of topological spaces (def. ):

$\left\{ X_i \overset{f_\alpha}{\longrightarrow} X_j \right\}_{i,j \in I, \alpha \in I_{i,j}} \,.$

Then

1. its limiting cone (or just limit for short, also “inverse limit”, for historical reasons) is the cone

$\left\{ \array{ && \underset{\longleftarrow}{\lim}_k X_k \\ & {}^{\mathllap{p_i}}\swarrow && \searrow^{\mathrlap{p_j}} \\ X_i && \underset{f_\alpha}{\longrightarrow} && X_j } \right\}$

over this diagram (def. ) which is universal among all possible cones, in that for

$\left\{ \array{ && \tilde X \\ & {}^{\mathllap{p'_i}}\swarrow && \searrow^{\mathrlap{p'_j}} \\ X_i && \underset{f_\alpha}{\longrightarrow} && X_j } \right\}$

any other cone, then there is a unique function or continuous function, respectively

$\phi \;\colon\; \tilde X \overset{}{\longrightarrow} \underset{\longrightarrow}{\lim}_i X_i$

that factors the given cone through the limiting cone, in that for all $i \in I$ then

$p'_i = p_i \circ \phi$

which we depict as follows:

$\array{ \tilde X \\ {}^{\mathllap{ \exists !\, \phi}}\downarrow & \searrow^{\mathrlap{p'_i}} \\ \underset{\longrightarrow}{\lim}_i X_i &\underset{p_i}{\longrightarrow}& X_i }$
2. its colimiting cocone (or just colimit for short, also “direct limit”, for historical reasons) is the cocone

$\left\{ \array{ X_i && \overset{f_\alpha}{\longrightarrow} && X_j \\ & {}^{\mathllap{q_i}}\searrow && \swarrow^{\mathrlap{q_j}} \\ \\ && \underset{\longrightarrow}{\lim}_i X_i } \right\}$

under this diagram (def. ) which is universal among all possible co-cones, in that it has the property that for

$\left\{ \array{ X_i && \overset{f_\alpha}{\longrightarrow} && X_j \\ & {}^{\mathllap{q'_i}}\searrow && \swarrow_{\mathrlap{q'_j}} \\ && \tilde X } \right\}$

any other cocone, then there is a unique function or continuous function, respectively

$\phi \;\colon\; \underset{\longrightarrow}{\lim}_i X_i \overset{}{\longrightarrow} \tilde X$

that factors the given co-cone through the co-limiting cocone, in that for all $i \in I$ then

$q'_i = \phi \circ q_i$

which we depict as follows:

$\array{ X_i &\overset{q_i}{\longrightarrow}& \underset{\longrightarrow}{\lim}_i X_i \\ & {}_{q'_i}\searrow & \downarrow^{\mathrlap{\exists ! \phi}} \\ && \tilde X }$

$\,$

We now briefly mention the names and comment on the general nature of the limits and colimits over the free diagrams from the list of examples above. Further below we discuss examples in more detail.

shapes of free diagrams and the names of their limits/colimits