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
A normal space is a space (typically a topological space) which satisfies one of the stronger separation axioms.
the main separation axioms
number | name | statement | reformulation |
---|---|---|---|
$T_0$ | Kolmogorov | given two distinct points, at least one of them has an open neighbourhood not containing the other point | every 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 point | all points are closed | |
$T_2$ | Hausdorff | given two distinct points, they have disjoint open neighbourhoods | the 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 |
A topological space $X$ is normal if it satisfies:
By Urysohn's lemma this is equivalent to the condition
Often one adds the requirement
(Unlike with regular spaces, $T_0$ is not sufficient here.)
One may also see terminology where a normal space is any space that satisfies $T_4$, while a $T_4$-space must satisfy both $T_4$ and $T_1$. This has the benefit that a $T_4$-space is always also a $T_3$-space while still having a term available for the weaker notion. On the other hand, the reverse might make more sense, since you would expect any space that satisfies $T_4$ to be a $T_4$-space; this convention is also seen.
If instead of $T_1$, one requires
then the result may be called an $R_3$-space.
Any space that satisfies both $T_4$ and $T_1$ must be Hausdorff, and every Hausdorff space satisfies $T_1$, so one may call such a space a normal Hausdorff space; this terminology should be clear to any reader.
Any space that satisfies both $T_4$ and $R_0$ must be regular (in the weaker sense of that term), and every regular space satisfies $R_0$, so one may call such a space a normal regular space; however, those who interpret ‘normal’ to include $T_1$ usually also interpret ‘regular’ to include $T_1$, so this term can be ambiguous.
Every normal Hausdorff space is an Urysohn space, a fortiori regular and a fortiori Hausdorff.
It can be useful to rephrase $T_4$ in terms of only open sets instead of also closed ones:
This definition is suitable for generalisation to locales and also for use in constructive mathematics (where it is not equivalent to the usual one).
To spell out the localic case, a normal locale is a frame $L$ such that
Let $(X,d)$ be a metric space regarded as a topological space via its metric topology. Then this is a normal Hausdorff space.
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 $U_{C_2} \supset C_2$.
Consider the function
which computes distances from a subset $S \subset X$, by forming the infimum of the distances to all its points:
Then the unions of open balls
and
have the required properties.
Every regular second countable space is normal.
(Dieudonné‘s theorem)
Every paracompact Hausdorff space, in particular every compact Hausdorff space, is normal.
See at paracompact Hausdorff spaces are normal for details
The real numbers equipped with their K-topology $\mathbb{R}_K$ are a Hausdorff topological space which is not a regular Hausdorff space (hence in particular not a normal Hausdorff space).
By construction the K-topology is finer than the usual euclidean metric topology. Since the latter is Hausdorff, so is $\mathbb{R}_K$. It remains to see that $\mathbb{R}_K$ contains a point and a disjoint closed subset such that they do not have disjoint open neighbourhoods.
But this is the case essentially by construction: Observe that
is an open subset in $\mathbb{R}_K$, whence
is a closed subset of $\mathbb{R}_K$.
But every open neighbourhood of $\{0\}$ contains at least $(-\epsilon, \epsilon) \backslash K$ for some positive real number $\epsilon$. There exists then $n \in \mathbb{N}_{\geq 0}$ with $1/n \lt \epsilon$ and $1/n \in K$. An open neighbourhood of $K$ needs to contain an open interval around $1/n$, and hence will have non-trivial intersection with $(-\epsilon, \epsilon)$. Therefore $\{0\}$ and $K$ may not be separated by disjoint open neighbourhoods, and so $\mathbb{R}_K$ is not normal.
If $\omega_1$ is the first un-countable ordinal with the order topology, and $\widebar{\omega_1}$ its one-point compactification, then $X = \omega_1 \times \widebar{\omega_1}$ with the product topology is not normal.
Indeed, let $\infty \in \widebar{\omega_1}$ be the unique point in the complement of $\omega_1 \hookrightarrow \widebar{\omega_1}$; then it may be shown that every open set $U$ in $X$ that includes the closed set $A = \{(x, x): x \neq \infty\}$ in $X$ must somewhere intersect the closed set $\omega_1 \times \{\infty\}$ which is disjoint from $A$. For if that were false, then we could define an increasing sequence $x_n \in \omega_1$ by recursion, letting $x_0 = 0$ and letting $x_{n+1} \in \omega_1$ be the least element that is greater than $x_n$ and such that $(x_n, x_{n+1}) \notin U$. Then, letting $b \in \omega_1$ be the supremum of this increasing sequence, the sequence $(x_n, x_{n+1})$ converges to $(b, b)$, and yet the neighborhood $U$ of $(b, b)$ contains none of the points of this sequence, which is a contradiction.
This example also shows that general subspaces of normal spaces need not be normal, since $\omega_1 \times \widebar{\omega_1}$ is an open subspace of the compact Hausdorff space $\widebar{\omega_1} \times \widebar{\omega_1}$, which is itself normal.
A Dowker space is an example of a normal space which is not countably paracompact.
(normality in terms of topological closures)
A topological space $(X,\tau)$ is normal Hausdorff, 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 $U$:
In one direction, assume that $(X,\tau)$ is normal, and consider
It follows that the complement of the open subset $U$ is closed and disjoint from $C$:
Therefore by assumption of normality of $(X,\tau)$, there exist open neighbourhoods with
But this means that
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
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
Consider disjoint closed subsets
We need to produce disjoint open neighbourhoods for them.
From their disjointness it follows that
is an open neighbourhood. Hence by assumption there is an open neighbourhood $V$ with
Thus
are two disjoint open neighbourhoods, as required.
The Tietze extension theorem applies to normal spaces.
In fact the Tietze extension theorem can serve as a basis of a category theoretic characterization of normal spaces: a (Hausdorff) space $X$ is normal if and only if every function $f \colon A \to \mathbb{R}$ from a closed subspace $A \subset X$ admits an extension $\tilde{f}: X \to \mathbb{R}$, or what is the same, every regular monomorphism into $X$ in $Haus$ has the left lifting property with respect to the map $\mathbb{R} \to 1$. (See separation axioms in terms of lifting properties (Gavrilovich 14) for further categorical characterizations of various topological properties in terms of lifting problems.)
Although normal spaces are “nice topological spaces” (being for example Tychonoff spaces, by Urysohn's lemma), the category of normal topological spaces with continuous maps between them seems not to be very well-behaved. (Cf. the rule of thumb expressed in dichotomy between nice objects and nice categories.) It admits equalizers of pairs of maps $f, g: X \rightrightarrows Y$ (computed as in $Top$ or $Haus$; one uses the easy fact that closed subspaces of normal spaces are normal). However it curiously does not have products – or at least it is not closed under products in $Top$, as shown by Counter-Example 3. It follows that this category is not a reflective subcategory of $Top$, as $Haus$ is.
More at the page colimits of normal spaces.
The class of normal spaces was introduced by Tietze (1923) and Aleksandrov–Uryson (1924).
Ryszard Engelking, General topology, (Monographie Matematyczne, tom 60) Warszawa 1977; expanded Russian edition Mir 1986.
Misha Gavrilovich, Point set topology as diagram chasing computations, (arXiv:1408.6710).