topology (point-set topology, point-free topology)
see also differential topology, algebraic topology, functional analysis and topological homotopy theory
Basic concepts
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
continuous metric space valued function on compact metric space is uniformly continuous
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Analysis Theorems
t
A topological space is connected if it can not be split up into two independent parts.
Every topological space may be decomposed into disjoint maximal connected subspaces, called its connected components. The underlying set of a topological space is the disjoint union of the underlying sets of its connected components, but the space itself is not necessarily the coproduct of its connected components in the category of spaces.
One often studies topological ideas first for connected spaces and then generalises to general spaces. This is especially true if one is studying such nice topological spaces that every space is a coproduct of connected components (such as for example locally connected spaces; see below).
(connected topological space)
A topological space $(X, \tau)$ is connected if the following equivalent conditions hold:
For all pairs of topological spaces $(X_1, \tau_1), (X_2, \tau_2)$ such that $(X, \tau)$ is homeomorphic to their disjoint union space
then exactly one of the two spaces is the empty space.
For all pairs of open subsets $U_1, U_2 \subset X$ if
then exactly one of the two subsets is the empty set
if a subset $CO \subseteq X$ is clopen (both closed and open), then $CO = X$ if and only if $CO$ is inhabited.
According to def. 1 the empty topological space is not regarded as connected. Some authors do want the empty space to count as a connected space. This means to change in the first item of def. 1 the “exactly one” to “at least one” and in the second item “if and only if” to “if”.
The conditions in def. 1 are indeed equivalent.
First consider the equivalence of the first two statements:
Suppose that in every disjoint union decomposition of $(X,\tau)$ then exactly one summand is empty. Now consider two disjoint open subsets $U_1, U_2 \subset X$ whose union is $X$ and whose intersection is empty. We need to show that exactly one of the two subsets is empty.
Write $(U_1, \tau_{1})$ and $(U_2, \tau_2)$ for the corresponding topological subspaces. Then observe that from the definition of subspace topology and the disjoint union space we have a homeomorphism
because by assumption every open subset $U \subset X$ is the disjoint union of open subsets of $U_1$ and $U_2$, respectively:
which is the definition of the disjoint union topology.
Hence by assumption exactly one of the two summand spaces is the empty space and hence the underlying set is the empty set.
Conversely, suppose that for every pair of open subsets $U_1, U_2 \subset U$ with $U_1 \cup U_2 = X$ and $U_1 \cap U_2 = \emptyset$ then exactly one of the two is empty. Now consider a homeomorphism of the form $(X,\tau) \simeq (X_1, \tau_1) \sqcup (X_2,\tau_2)$. By the nature of the disjoint union space this means that $X_1, X_2 \subset X$ are disjoint open subsets of $X$ which cover $X$. So by asumption precisely one of the two subsets is the empty set and hence precisely one of the two topological spaces is the empty space.
Now regarding the equivalence to the third statement:
If a subset $CO \subset X$ is both closed and open, this means equivalently that it is open and that its complement $X \setminus CO$ is also open, hence equivalently that there are two open subsets $CO, X \backslash CO \subset X$ whose union is $X$ and whose intersection is empty. This way the third condition is equivalent to the second.
In the language of category theory def. 1 may be rephrased as follows:
Write Top for the category of all topological space.
Then a topological space $X$ is connected precisely if the representable functor
preserves coproducts.
It is equivalent to just require that it preserves binary coproducts (a detailed proof in a more general setting is given at connected object in this proposition. In that case, notice that we always have a map
so $X$ is connected if this is always a bijection. This definition generalises to the notion of connected object in an extensive category.
The variant of the definition according to remark 1, which regards the empty space means in terms of category theoretic language that one requires only that the maps
be surjections.
However, many results come out more cleanly by disqualifying the empty space (much as one disqualifies $1$ when one defines the notion of prime number). See also the discussion at empty space and too simple to be simple.
Every topological space $X$ admits an equivalence relation $\sim$ where $x \sim y$ means that $x$ and $y$ belong to some subspace which is connected. The equivalence class $Conn(x)$ of an element $x$ is thus the union of all connected subspaces containing $x$; it follows readily from Result 4 that $Conn(x)$ is itself connected. It is called the connected component of $x$. It is closed, by Result 5. A space is connected if and only if it has exactly one connected component (or at most one, if you allow the empty space to be connected).
(connected components)
For $(X,\tau)$ a topological space, then its connected components are the equivalence classes under the equivalence relation on $X$ which regards two points as equivalent if they both sit in some open subset which, as a topological subspace (example \ref{SubspaceTopology}), is connected (def. \ref{ConnectedTopologicalSpace}):
There is another equivalence relation $\sim_q$ where $x \sim_q y$ if $f(x) = f(y)$ for every continuous $f: X \to D$ mapping to a discrete space $D$. The equivalence class of $x$ may be alternatively described as the intersection of all clopens that contain $x$. This is called the quasi-component of $x$, denoted here as $QConn(x)$. It is easy to prove that
and that equality holds if $X$ is compact Hausdorff or is locally connected (see below), but also in other circumstances (such as the space of rational numbers as a topological subspace of the real line).
For an example where $Conn(x) \neq QConn(x)$, take $X$ to be the following subspace of $[0, 1] \times [0, 1]$:
In this example, $Conn((0, 1)) = \{(0, 1)\}$, but $QConn((0, 1)) = \{(0, 0), (0, 1)\}$.
(locally constant function on connected topological spaces are constant functions)
If $X$ is a connected topological space and $f \colon X \to Y$ is a locally constant continuous function, then $f$ is in fact a constant function.
By definition of locally constant functions, every point $x \in X$ has an open neighbourhood $U_x$ such that the restriction $f\vert_{U_x}$ is a constant function. The unions of these neighbourhoods for a fixed constant value hence are disjoint open subsets that constitute a cover of $X$. By connectedness this cover must consist of a single non-empty element. But by construction this mans that $f$ is constant.
(continuous images of connected spaces are connected)
The regular image of a connected space $X$ under a continuous map $f: X \to Y$ (i.e., the set-theoretic image with the subspace topology inherited from $Y$) is connected. Or, what is essentially the same: if $X$ is connected and $f: X \to Y$ is epic, then $Y$ is connected:
Let $X$ be a connected topological space, let $Y$ be any topological space, and let
be a continuous function. This factors via continuous functions through the image
for $f(X)$ equipped either with he subspace topology relative to $Y$ or the quotient topology relative to $X$. In either case:
If $X$ is a connected topological space, then so is $f(X)$.
Let $U_1,U_2 \subset f(X)$ be two open subsets such that $U_1 \cup U_2 = f(X)$ and $U_1 \cap U_2 = \emptyset$. We need to show that precisely one of them is the empty set.
Since $p$ is a continuous function, also the pre-images $p^{-1}(U_1), p^{-1}(U_2) \subset X$ are open subsets and are still disjoint. Since $p$ is surjective it also follows that $p^{-1}(U_1) \cup p^{-1}(U_2) = X$. Since $X$ is connected, it follows that one of these two pre-images is the empty set. But again sicne $p$ is surjective, this implies that precisely one of $U_1, U_2$ is empty, which means that $f(X)$ is connected.
Wide pushouts of connected spaces are connected. (This would of course be false if the empty space were considered to be connected.) This follows from the hom-functor definition of connectedness, plus the fact that coproducts in $Set$ commute with wide pullbacks. More memorably: connected colimits of connected spaces are connected.
If $S \subseteq X$ is a connected subspace and $S \subseteq T \subseteq \overline{S}$ (i.e. if $T$ is between $S$ and its closure), then $T$ is connected. Or, what is essentially the same: if $T$ has a dense connected subspace $S$, then $T$ is connected.
(see also prop. 3 below)
(product space of connected spaces is connected)
Let $\{X_i\}_{i \in I}$ be a set of connected spaces. Then also their product topological space $\underset{i \in I}{\prod}X_i$ (with the Tychonoff topology) is connected.
This relies on some special features of Top An general abstract proof is given at connected object in this theorem and this remark.
Here is an alternative elementary proof in point-set topology:
Let $U_1, U_2 \subset \underset{i \in I}{\prod}X_i$ be an open cover of the product space by two disjoint open subsets. We need to show that precisely one of the two is empty. Since each $X_i$ is connected and hence non-empty, the product space is not empty, and hence it is sufficient to show that at lest one of the two is empty.
Assume on the contrary that both $U_1$ and $U_2$ are non-empty.
Observe first that if so, then we could find $x_1 \in U_1$ and $x_2 \in U_2$ whose coordinates differed only a a finite subset of $I$. This is since by the nature of the Tychonoff topology $\pi_i(U_1) = X_i$ and $\pi_i(U_2) = X_i$ for all but a finite number of $i \in iI$.
Next observe that we then could even find $x'_1 \in U_1$ that differed only in a single coordinate from $x_2$: Because pick one coordinate in which $x_1$ differs from $x_2$ and change it to the corresponding coordinate of $x_2$. Since $U_1$ and $U_2$ are a cover, the resulting point is either in $U_1$ or in $U_2$. If it is in $U_2$, then $x_1$ already differed in only one coordinate from $x_2$ and we may take $x'_1 \coloneqq x_1$. If instead the new point is in $U_1$, then rename it to $x_1$ and repeat the argument. By induction this finally yields an $x'_1$ as claimed.
Therefore it is now sufficient to see that it leads to a contradiction to assume that there are points $x_1 \in U_1$ and $x_2 \in U_2$ that differ in only the $i_0$th coordinate, for some $i_0 \in I$ then $x_1 = x_2$.
Observe that the inclusion
which is the identity on the $i_0$th component and is otherwise constant on the $i$th component of $x_1$ or equivalently of $x_2$ is a continuous function, by the nature of the Tychonoff topology.
Therefore also the restrictions $\iota^{-1}(U_1)$ and $\iota^{-1}(U_2)$ are open subsets. Moreover they are still disjoint and cover $X_i$. Hence by the connectedness of $X_i$, precisely one of them is empty. This means that the $i_0$-component of both $x_1$ and $x_2$ must be in the other subset of $X_i$, and hence that $x_1$ and $x_2$ must both be in $U_1$ or both in $U_2$, contrary to the assumption.
(connected subspaces of the real line are the intervals)
Regard the real line with its Euclidean metric topology. Then a subspace $S \subset \mathbb{R}$ is connected (def. 1) precisely if it is an interval, hence precisely if
In particular for $\{ I_i \subset \mathbb{R} \}_{i \in I}$ a set of disjoint intervals, then $I$ is the set of connected components of the union $\underset{i \in I}{\cup} I_i$.
Suppose on the contrary that we have $x \lt r \lt y$ but $r \notin S$. Then by the nature of the subspace topology there would be a decomposition of $S$ as a disjoint union of disjoint open subsets:
But since $x \lt r$ and $r \lt y$ both these open subsets were non-empty, thus contradicting the assumption that $S$ is connected. This yields a proof by contradiction.
The basic results above give a plethora of ways to construct connected spaces. More exotic examples are sometimes useful, especially for constructing counterexamples.
The following, due to Bing, is a countable connected Hausdorff space. Let $Q = \{(x, y) \in \mathbb{Q} \times \mathbb{Q}: y \geq 0\}$, topologized by defining a basis of neighborhoods $N_{\epsilon, a, b}$ for each point $(a, b) \in Q$ and $\epsilon \gt 0$:
where $\theta \lt 0$ is some chosen fixed irrational number. It is easy to see this space is Hausdorff (using the fact that $\theta$ is irrational). However, the closure of $N_{\epsilon, a, b}$ consists of points $(x, y)$ of $Q \times Q$ with either $(x-a) - \epsilon \leq (y-b)/\theta \leq (x-a) + \epsilon$ or $(x-a) - \epsilon \leq -(y-b)/\theta \leq (x-a) + \epsilon$, in other words, the union of two infinitely long strips of width $2\epsilon$ and slopes $\theta$, $-\theta$. Clearly any two such closures intersect, and therefore the space is connected.
This example is due to Golomb. Topologize the set of natural numbers $\mathbb{N}$ by taking a basis to consist of sets $A_{a,b} \coloneqq \{a k + b | k = 1,2, \ldots\}$, where $a, b \in \mathbb{N}$ are relatively prime. The space is Hausdorff, but the intersection of the closures of two non-empty open sets is never empty, so this space is connected.
Warning
It is not generally true that a topological space is the disjoint union space (coproduct in Top) of its connected components, nor of its quasi-components.
The spaces such that all their open subspaces are the disjoint union of their connected components are the locally connected topological spaces.
The connected components in Cantor space $2^{\mathbb{N}}$ (with its topology as a product of 2-point discrete spaces) are just the singletons, but the coproduct of the singleton subspaces carries the discrete topology, which differs from that of Cantor space.
Similarly for set of rational numbers with its absolute-value topology (the one induced as a topological subspace of the real line).
An important variation on the theme of connectedness is path-connectedness.
(continuous path in a topological space)
Let $(X,\tau)$ be a topological space, Then a continuous path (or just path, for short) in $(X,\tau)$ is a continuous function of the form
where the domain is the closed interval equipped with its Euclidean subspace topology.
One says that the path connects the point $\gamma(0) \in X$ with the point $\gamma(1) \in X$.
For $x \in X$ a fixed point, then the subset
is called the path-connected component of $x$.
The set of path connected components of $X$ is denoted
The set $\pi_0(X)$ of path components (the 0th “homotopy group”) is thus the coequalizer in
Observe that this is a reflexive coequalizer, as witnessed by the mutual right inverse $\hom(!, X): \hom(1, X) \to \hom([0, 1], X)$.
(We can even topologize $\pi_0(X)$ by taking the coequalizer in $Top$ of
taking advantage of the fact that the locally compact Hausdorff space $[0, 1]$ is exponentiable. The resulting quotient space will be discrete if $X$ is locally path-connected.)
We say $X$ is path-connected if it has exactly one path component.
It follows easily from the basic results above that:
A path connected space $X$ is connected.
Assume it were not, then it would be covered by two disjoint inhabited open subsets $U_1, U_2 \subset X$. But by path connectedness there were a continuous path $\gamma \colon [0,1] \to X$ from a point in one of the open subsets to a point in the other. The continuity would imply that $\gamma^{-1}(U_1), \gamma^{-1}(U_2) \subset [0,1]$ were a disjoint open cover of the interval. This would be in contradiction to the fact that intervals are connected. Hence we have a proof by contradiction.
However, it need not be closed (and therefore need not be the connected component of $x$); see the following example. The path components and connected components do coincide if $X$ is locally path-connected.
The topologist’s sine curve
provides a classic example where the path component of a point need not be closed. (Specifically, consider a point on the locus of $y = \sin(1/x)$.)
The basic categorical Results 3, 4, and 6 above carry over upon replacing “connected” by “path-connected”. (As of course does example 7, trivially.)
Finally, as a contrast to a path-connected space, a totally path-disconnected space is a space such that its set of path components is equal to the underlying set of the space. Equivalently, that there are no non-constant paths. This by far does not mean that the space is discrete!
A refinement of the notion of path-connected space is that of arc-connected (or arcwise-connected) space:
A space $X$ is arc-connected if for any two distinct $x, y \in X$ there exists an injective continuous map $\alpha: I \to X$ such that $\alpha(0) = x$ and $\alpha(1) = y$.
Arc-connected spaces are of course path-connected, but there are trivial examples (using an indiscrete topology) that the converse fails to hold. A rather nontrivial theorem is the following:
A path-connected Hausdorff space $X$ is arc-connected.
This immediately generalizes to the statement that in a Hausdorff space $X$, any two points that can be connected by a path $\alpha: I \to X$ can be connected by an arc: just apply the theorem to the image $\alpha(I)$.
For a proof of this theorem, see Willard, theorem 31.2. More precisely, that result states that a Peano space, i.e., a compact, connected, locally connected, and metrizable space, is arc-connected if it is path-connected. It then suffices to observe that the continuous image $\alpha(I) \subseteq X$ of a path is in fact a Peano space, so that the path $\alpha: I \to \alpha(I)$ can be replaced by an arc.
If $X$ is Hausdorff and there is a continuous surjection $f: I \to X$, then $X$ is a Peano space.
Obviously $X$ is compact (Hausdorff) and connected. $X$ is a quotient space of $I$, since $f$ is a closed surjection (using compactness of $I$ and Hausdorffness of $X$), and therefore $X$ is locally connected by this lemma. Being compact Hausdorff, $X$ is regular, so to show metrizability it suffices by the Urysohn metrization theorem to show $X$ is second-countable.
Let $\mathcal{B}$ be a countable base for $I$ and let $\mathcal{C}$ be the collection consisting of finite unions of elements of $\mathcal{B}$. We claim $\forall_f(\mathcal{C}) \coloneqq \{\forall_f(C) = \neg f(\neg C): C \in \mathcal{C}\}$ is an (evidently countable) base for $X$. Indeed, suppose $U \subseteq X$ is open and $p \in U$; then $f^{-1}(p)$ is compact, so there exist finitely many $B_1, \ldots, B_n \in \mathcal{B}$ with
Put $C = B_1 \cup \ldots \cup B_n$. The first inclusion is equivalent to $p \in \forall_f(C)$ by the adjunction $f^{-1} \dashv \forall_f$. The second inclusion implies $\forall_f(C) \subseteq \forall_f f^{-1}(U) = U$, where the equality $\forall_f f^{-1} = id$, equivalent to $\exists_f f^{-1} = id$, follows from surjectivity of $f$. Thus we have shown $\forall_f(\mathcal{C})$ is a base.
The converse of this lemma is the celebrated Hahn-Mazurkiewicz theorem:
Let $X$ be a nonempty Hausdorff space. Then there exists a continuous surjection $\alpha: [0, 1] \to X$ if $X$ is a Peano space. In particular, a nonempty Peano space is path-connected.
(The terminology “Peano space” is given in recognition of Peano’s discovery of space-filling curves, as for example the unit square.)
As above, let $\pi_0 \colon Top \to Set$ be the functor which assigns to each space $X$ its set of path components $\pi_0(X)$.
The functor $\pi_0 \colon Top \to Set$ preserves arbitrary products.
Let $X_i$ be a family of spaces; we must show that the comparison map
is invertible. Injectivity: suppose $(x_i), (y_i) \in \prod_i X_i$ are tuples that map to the same tuple of path-components $(c_i)$; we must show that $(x_i)$ and $(y_i)$ belong to the same path component. For each $i$, both $x_i$ and $y_i$ belong to $c_i$, so we may choose a path $\alpha_i: I \to X_i$ connecting $x_i$ to $y_i$. Then $\langle \alpha_i \rangle \colon I \to \prod_i X_i$ connects $(x_i)$ to $(y_i)$. (Note this uses the axiom of choice.) Surjectivity: for any tuple $(c_i) \in \prod_i \pi_0(X_i)$, the component $c_i$ is nonempty for each $i$, so we may choose an element $x_i$ therein. Then $(x_i)$ maps to $(c_i)$. Again this uses the axiom of choice.
An elegant proof of the previous proposition but for preservation of finite products is as follows: both $\hom(I, -)$ and $\hom(1, -)$ preserve products, and a reflexive coequalizer of product-preserving functors $C \to Set$, being a sifted colimit, is also product-preserving.
The functor $\pi_0 \colon Top \to Set$ preserves arbitrary coproducts.
####### Proof
The functor $\hom(I, -) \colon Top \to Set$ preserves coproducts since $I$ is connected, and similarly for $\hom(1, -)$. The coequalizer of a pair of natural transformations between coproduct-preserving functors is also a coproduct-preserving functor.
Point-set topology is filled with counterexamples. An unusual type of example is that of pseudo-arc:
A pseudo-arc is a metric continuum with more than one point such that every subcontinuum (a subspace that is a continuum) cannot be expressed as a union of two proper subcontinua.
A pseudo-arc $X$ is necessarily totally path-disconnected: two distinct points $x, y$ of $X$ cannot be connected by a path in $X$. Indeed, the image of such a# path would be a path-connected Hausdorff space, hence arc-connected by Theorem 1. Letting $\alpha: [0, 1] \to X$ be an arc from $x$ to $y$, we have that the continuum $\alpha([0, 1])$ is a union of proper subcontinua $\alpha([0, 1/2])$ and $\alpha([1/2, 1])$, a contradiction. Thus, a pseudo-arc is an example of a compact connected metrizable space that is totally path-disconnected.
Remarkably, all pseudo-arcs are homeomorphic, and a pseudo-arc is a homogeneous space. Perhaps also remarkable is the fact that the collection of pseudo-arcs in the Hilbert cube $Q$ (or in any Euclidean space) is a dense $G_\delta$ set (see G-delta set) in the Polish space of all nonempty compact subsets of $Q$ under the Hausdorff metric; see Bing2, theorem 2.
A typical way in which pseudo-arcs arise is through inverse limits of dynamical systems. One of the original constructions is due to Henderson:
There is a $C^\infty$ function $f: I \to I$ such that the limit of the diagram
is a pseudo-arc.
Roughly speaking, Henderson’s $f$ is a small “notched” perturbation of the squaring function $[0, 1] \to [0, 1]: x \mapsto x^2$, as illustrated on page 38 (of 58) here.
Regard the real numbers $\mathbb{R}$ with their Euclidean metric topology, and consider a closed interval $[a,b] \subset \mathbb{R}$ equipped with its subspace topology.
Then a continuous function
takes every value in between $f(a)$ and f(b).
By example 7 the interval $[a,b]$ is connected. By example 3 also its image $f([a,b]) \subset \mathbb{R}$ is connected. By example 7 that image is hence itself an interval. This implies the claim.
(topological closure of connected subspace is connected)
Let $(X,\tau)$ be a topological space and let $S \subset X$ be a subset which, as a subspace, is connected. Then also the topological closure $Cl(S) \subset X$ is connected
Suppose that $Cl(S) = A \sqcup B$ with $A,B \subset X$ disjoint open subsets. We need to show that one of the two is empty.
But also the intersections $A \cap S\,,B \cap S \subset S$ are disjoint subsets, open as subsets of the subspace $S$ with $S = (A \cap S) \sqcup (B \cap S)$. Hence by the connectedness of $S$, one of $A \cap S$ or $B \cap S$ is empty. Assume $B \cap S$ is empty, otherwise rename. Hence $A \cap S = S$, or equivalently: $S \subset A$. By disjointness of $A$ and $B$ this means that $S \subset Cl(S) \setminus B$. But since $B$ is open, $Cl(S) \setminus B$ is still closed, so that
This means that $B = \emptyset$, and hence that $Cl(S)$ is connected.
(connected components are closed)
Let $(X,\tau)$ be a topological space. Then its connected components (def. 2) are closed subsets.
By definition, the connected components are maximal elements in the set of connected subspaces pre-ordered by inclusion. By prop. 3 this means that they must contain their closures, hence they must equal their closures.
Prop. 4 implies that when a space has a finite set of connected components, then they are not just closed but also open, hence clopen subsets (because then each is the complement of a finite union of closed subspaces).
For a non-finite set of connected components this remains true if the space is locally connected. See this prop.
A space in which the only connected subspaces are the singletons and the empty set is called totally disconnected space.
The connected subsets of a space form a connectology.
Examples of countable connected Hausdorff spaces were give in
R.H. Bing, A connected countable Hausdorff space, Proc. Amer. Math. Soc. 4 (1953), 474.
Solomon W. Golomb, A Connected Topology for the Integers, Amer. Math. Monthly, Vol. 66 No. 8 (Oct. 1959), 663-665.
Material on arc-connected spaces and the Hahn-Mazurkiewicz theorem can be found in Chapter 31 of
Material on pseudo-arcs can be found in