nLab
normal space

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

Normal spaces

Idea

A normal space is a space (typically a topological space) which satisfies one of the stronger separation axioms.

the main separation axioms

numbernamestatementreformulation
T 0T_0Kolmogorovgiven 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 1T_1given two distinct points, both have an open neighbourhood not containing the other pointall points are closed
T 2T_2Hausdorffgiven two distinct points, they have disjoint open neighbourhoodsthe diagonal is a closed map
T >2T_{\gt 2}T 1T_1 and…all points are closed and…
T 3T_3regular 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 4T_4normal 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 XX is normal if it satisfies:

By Urysohn's lemma this is equivalent to the condition

Often one adds the requirement

  • T 1T_1: every point in XX is closed.

(Unlike with regular spaces, T 0T_0 is not sufficient here.)

One may also see terminology where a normal space is any space that satisfies T 4T_4, while a T 4T_4-space must satisfy both T 4T_4 and T 1T_1. This has the benefit that a T 4T_4-space is always also a T 3T_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 4T_4 to be a T 4T_4-space; this convention is also seen.

If instead of T 1T_1, one requires

  • R 0R_0: if xx is in the closure of {y}\{y\}, then yy is in the closure of {x}\{x\},

then the result may be called an R 3R_3-space.

Any space that satisfies both T 4T_4 and T 1T_1 must be Hausdorff, and every Hausdorff space satisfies T 1T_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 4T_4 and R 0R_0 must be regular (in the weaker sense of that term), and every regular space satisfies R 0R_0, so one may call such a space a normal regular space; however, those who interpret ‘normal’ to include T 1T_1 usually also interpret ‘regular’ to include T 1T_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 4T_4 in terms of only open sets instead of also closed ones:

  • T 4T_4: if G,HXG,H \subset X are open and GH=XG \cup H = X, then there exist open sets U,VU,V such that UGU \cup G and VHV \cup H are still XX but UVU \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 LL such that

  • T 4T_4: if G,HLG,H \in L are opens and GH=G \vee H = \top, then there exist opens U,VU,V such that UGU \vee G and VHV \vee H are still \top but UV=U \wedge V = \bot.

Examples

Example

Let (X,d)(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 is that given two disjoint closed subsets C 1,C 2XC_1, C_2 \subset X then their exists disjoint open neighbourhoods U C 1C 1U_{C_1} \subset C_1 and U C 2C 2U_{C_2} \supset C_2.

Consider the function

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

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

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

Then the unions of open balls

U C 1x 1C 1B x 1 (d(C 2,x 1)) U_{C_1} \coloneqq \underset{x_1 \in C_1}{\cup} B^\circ_{x_1}( d(C_2,x_1) )

and

U C 2x 2C 2B x 2 (d(C 1,x 2)). U_{C_2} \coloneqq \underset{x_2 \in C_2}{\cup} B^\circ_{x_2}( d(C_1,x_2) ) \,.

have the required properties.

Proposition

Every regular second countable space is normal.

Proposition

(Dieudonné’s theorem)

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

See at paracompact Hausdorff spaces are normal for details

Nonexample

The real numbers equipped with their K-topology K\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 K\mathbb{R}_K. It remains to see that K\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

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

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

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

is a closed subset of K\mathbb{R}_K.

But every open neighbourhood of {0}\{0\} contains at least (ϵ,ϵ)\K(-\epsilon, \epsilon) \backslash K for some positive real number ϵ\epsilon. There exists then n 0n \in \mathbb{N}_{\geq 0} with 1/n<ϵ1/n \lt \epsilon and 1/nK1/n \in K. An open neighbourhood of KK needs to contain an open interval around 1/n1/n, and hence will have non-trivial intersection with (ϵ,ϵ)(-\epsilon, \epsilon). Therefore {0}\{0\} and KK may not be separated by disjoint open neighbourhoods, and so K\mathbb{R}_K is not normal.

Counter-Example

If ω 1\omega_1 is the first un-countable ordinal with the order topology, and ω 1¯\widebar{\omega_1} its one-point compactification, then X=ω 1×ω 1¯X = \omega_1 \times \widebar{\omega_1} with the product topology is not normal.

Indeed, let ω 1¯\infty \in \widebar{\omega_1} be the unique point in the complement of ω 1ω 1¯\omega_1 \hookrightarrow \widebar{\omega_1}; then it may be shown that every open set UU in XX that includes the closed set A={(x,x):x}A = \{(x, x): x \neq \infty\} in XX must somewhere intersect the closed set ω 1×{}\omega_1 \times \{\infty\} which is disjoint from AA. For if that were false, then we could define an increasing sequence x nω 1x_n \in \omega_1 by recursion, letting x 0=0x_0 = 0 and letting x n+1ω 1x_{n+1} \in \omega_1 be the least element that is greater than x nx_n and such that (x n,x n+1)U(x_n, x_{n+1}) \notin U. Then, letting bω 1b \in \omega_1 be the supremum of this increasing sequence, the sequence (x n,x n+1)(x_n, x_{n+1}) converges to (b,b)(b, b), and yet the neighborhood UU of (b,b)(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 ω 1×ω 1¯\omega_1 \times \widebar{\omega_1} is an open subspace of the compact Hausdorff space ω 1¯×ω 1¯\widebar{\omega_1} \times \widebar{\omega_1}, which is itself normal.

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,τ)(X,\tau) is normal Hausdorff, precisely if all points are closed and for all closed subsets CXC \subset X with open neighbourhood UCU \supset C there exists a smaller open neighbourhood VCV \supset C whose topological closure Cl(V)Cl(V) is still contained in UU:

CVCl(V)U. C \subset V \subset Cl(V) \subset U \,.
Proof

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

CU. C \subset U \,.

It follows that the complement of the open subset UU is closed and disjoint from CC:

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

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

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

But this means that

VXW V \subset X \setminus W

and since the complement XWX \setminus W of the open set WW is closed, it still contains the closure of VV, so that we have

CVCl(V)XWU C \subset V \subset Cl(V) \subset X \setminus W \subset U

as required.

In the other direction, assume that for every open neighbourhood UCU \supset C of a closed subset CC there exists a smaller open neighbourhood VV with

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

Consider disjoint closed subsets

C 1,C 2X,AAAC 1C 2=. 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

XC 2C 1 X \setminus C_2 \supset C_1

is an open neighbourhood. Hence by assumption there is an open neighbourhood VV with

C 1VCl(V)XC 2. C_1 \subset V \subset Cl(V) \subset X \setminus C_2 \,.

Thus

VC 1,AAAAXCl(V)C 2 V \supset C_1 \,, \phantom{AAAA} X \setminus Cl(V) \supset C_2

are two disjoint open neighbourhoods, as required.

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 XX is normal if and only if every function f:Af \colon A \to \mathbb{R} from a closed subspace AXA \subset X admits an extension f˜:X\tilde{f}: X \to \mathbb{R}, or what is the same, every regular monomorphism into XX in HausHaus has the left lifting property with respect to the map 1\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 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:XYf, g: X \rightrightarrows Y (computed as in TopTop or HausHaus; 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 TopTop, as shown by Counter-Example 3. It follows that this category is not a reflective subcategory of TopTop, as HausHaus is.

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.

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

Revised on May 29, 2017 03:59:06 by Urs Schreiber (94.220.14.70)