nLab normal space

Idea

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

Definitions

A topological space $X$ is normal if it satisfies:

• $T_4$: for every two closed disjoint subsets $A,B \subset X$ there are (optionally open) neighborhoods $U \supset A$, $V \supset B$ such that their intersection $U \cap V$ is empty.

By Urysohn's lemma this is equivalent to the condition

• for every two closed disjoint subsets $A,B \subset X$ there exists an Urysohn function that separates them.

Often one adds the requirement

• $T_1$: every point in $X$ is closed.

(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

• $R_0$: if $x$ is in the closure of $\{y\}$, then $y$ is in the closure of $\{x\}$,

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:

• $T_4$: if $G,H \subset X$ are open and $G \cup H = X$, then there exist open sets $U,V$ such that $U \cup G$ and $V \cup H$ are still $X$ but $U \cap V$ is empty.

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

• $T_4$: if $G,H \in L$ are opens and $G \vee H = \top$, then there exist opens $U,V$ such that $U \vee G$ and $V \vee H$ are still $\top$ but $U \wedge V = \bot$.

Examples

Example

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

Proof

We need to show that given two disjoint closed subsets $C_1, C_2 \subset X$, there exist disjoint open neighbourhoods $U_{C_1} \supset C_1$ and $U_{C_2} \supset C_2$.

Consider the function

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

which computes distances from a subset $S \subset X$, by forming the infimum of the distances to all its points:

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

If $S$ is closed and $x \notin S$, then $d(S, x) \gt 0$. Then the unions of open balls

$$U_{C_1} \coloneqq \underset{x_1 \in C_1}{\bigcup} B^\circ_{x_1}( \frac1{2}d(C_2,x_1) )$$

and

$$U_{C_2} \coloneqq \underset{x_2 \in C_2}{\bigcup} B^\circ_{x_2}( \frac1{2}d(C_1,x_2) ) \,.$$

have the required properties. For if there exist $x_1 \in C_1, x_2 \in C_2$ and $y \in B^\circ_{x_1}( \frac1{2}d(C_2,x_1) ) \cap B^\circ_{x_2}( \frac1{2}d(C_1,x_2) )$, then

$$d(x_1, x_2) \leq d(x_1, y) + d(y, x_2) \lt \frac1{2} (d(C_2, x_1) + d(C_1, x_2)) \leq \max\{d(C_2, x_1), d(C_1, x_2)\}$$

and if $d(C_1, x_2) \leq d(C_2, x_1)$ say, then $d(x_2, x_1) = d(x_1, x_2) \lt d(C_2, x_1)$, contradicting the definition of $d(C_2, x_1)$.

Proposition

Every regular second countable space is normal.

See Urysohn metrization theorem for details.

Proposition

(Dieudonné‘s theorem)

Every paracompact Hausdorff space, in particular every compact Hausdorff space, is normal.

See paracompact Hausdorff spaces are normal for details.

Nonexample

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).

Proof

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

$$\mathbb{R} \backslash K \;=\; (-\infty,-1/2) \cup \left( (-1,1) \backslash K \right) \cup (1/2, \infty)$$

is an open subset in $\mathbb{R}_K$, whence

$$K = \mathbb{R} \backslash ( \mathbb{R} \backslash K )$$

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.

Counter-Example

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.

Counter-Example

An uncountable product of infinite discrete spaces $X$ is not normal. More generally, a product of $T_1$ spaces $X_i$ uncountably many of which are not limit point compact is not normal.

Indeed, by a simple application of Remark below and the fact that closed subspaces of normal Hausdorff spaces are normal Hausdorff, it suffices to see that the archetypal example $\mathbb{N}^{\omega_1}$ is not normal. For a readable and not overly long account of this result, see Dan Ma’s blog.

Example

A Dowker space is an example of a normal space which is not countably paracompact.

Properties

Basic properties

Proposition

(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 $U$:

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

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.

In terms of lifting properties

The separation conditions $T_0$ to $T_4$ may equivalently be understood as lifting properties against certain maps of finite topological spaces, among others.

This is discussed at separation axioms in terms of lifting properties, to which we refer for further details. Here we just briefly indicate the corresponding lifting diagrams.

In the following diagrams, the relevant finite topological spaces are indicated explicitly by illustration of their underlying point set and their open subsets:

• points (elements) are denoted by $\bullet$ with subscripts indicating where the points map to;

• boxes are put around open subsets,

• an arrow $\bullet_u \to \bullet_c$ means that $\bullet_c$ is in the topological closure of $\bullet_u$.

In the lifting diagrams for $T_2-T_4$ below, an arrow out of the given topological space $X$ is a map that determines (classifies) a decomposition of $X$ into a union of subsets with properties indicated by the picture of the finite space.

Notice that the diagrams for $T_2$-$T_4$ below do not in themselves imply $T_1$.

Proposition

(Lifting property encoding $T_0$)
The following lifting property in Top equivalently encodes the separation axiom $T_0$:

Proposition

(Lifting property encoding $T_1$)
The following lifting property in Top equivalently encodes the separation axiom $T_1$:

Proposition

(Lifting property encoding $T_2$)
The following lifting property in Top equivalently encodes the separation axiom $T_2$:

Proposition

(Lifting property encoding $T_3$)
The following lifting property in Top equivalently encodes the separation axiom $T_3$:

Proposition

(Lifting property encoding $T_4$)
The following lifting property in Top equivalently encodes the separation axiom $T_4$:

Tietze extension and lifting property

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.)

The category of normal spaces

Although normal (Hausdorff) 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 . It follows that this category is not a reflective subcategory of $Top$, as $Haus$ is.

Remark

A small saving grace is that the category of normal spaces is Cauchy complete, in fact is closed under retracts in $Top$. This is so whether or not the Hausdorff condition is included. (If $r: Y \to X$ retracts $i: X \to Y$, then $r$ is a quotient map and $i$ is a subspace inclusion. If $A, B$ are closed and disjoint in $X$, then $r^{-1}(A), r^{-1}(B)$ are closed and disjoint in $Y$. Separate them by disjoint open sets $U \supseteq r^{-1}(A), V \supseteq r^{-1}(B)$ in $Y$; then $i^{-1}(U), i^{-1}(V)$ are disjoint open sets separating $i^{-1} r^{-1}(A) = A, i^{-1} r^{-1}(B) = B$ in $X$.)

More at the page colimits of normal spaces.

References

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.

Discussion of separation axioms in terms of lifting properties is due to:

• Misha Gavrilovich, Point set topology as diagram chasing computations, (arXiv:1408.6710).