nLab
regular space

Regular spaces

Idea

A regular space is a topological space (or variation) that has, in a certain sense, enough regular open subsets. The condition of regularity is one the separation axioms satsified by every metric space (in this case, by every pseudometric space).

Definitions

Fix a topological space XX.

The classical definition is this: that if a point and a closed set are disjoint, then they are separated by neighbourhoods. In detail, this means:

Definition

Given any point aa and closed set FF, if aFa \notin F, then there exist a neighbourhood VV of aa and a neighbourhood GG of FF such that VGV \cap G is empty.

In many contexts, it is more helpful to change perspective, from a closed set that aa does not belong to, to an open set that aa does belong to. Then the definition reads:

Definition

Given any point aa and neighbourhood UU of aa, there exist a neighbourhood VV of aa and an open set GG such that VG=V \cap G = \empty but UG=XU \cup G = X.

You can think of VV as being half the size of UU, with GG the exterior of VV. (In a metric space, or even in a uniform space, this can be made into a proof.)

If we apply the regularity condition twice, then we get what at first might appear to be a stronger result:

Definition

Given any point aa and neighbourhood UU of aa, there exist a neighbourhood WW of aa and an open set GG such that Cl(W)Cl(G)=Cl(W) \cap Cl(G) = \empty but UG=XU \cup G = X (where ClCl indicates topological closure).

Proof of equivalence

Find VV and GG as above. Now apply the regularity axiom to aa and the interior Int(V)Int(V) of VV to get WW (and HH).

In terms of the classical language of separation axioms, this says that aa and FF are separated by closed neighbourhoods.

Sometimes one includes in the definition that a regular space must be T 0T_0:

Definition (of T 0T_0)

Given any two points, if each neighbourhood of either is a neighbourhood of the other, then they are equal.

Other authors use the weaker definition above but call a regular T 0T_0 space a T 3T_3 space, but then that term is also used for a (merely) regular space. An unambiguous term for the weaker condition is an R 2R_2 space, but hardly anybody uses that.

We have

Theorem

Every T 3T_3 space is Hausdorff.

Proof

Suppose every neighbourhood of aa meets every neighbourhood of bb; by T 0T_0 (and symmetry), it's enough to show that each neighbourhood UU of aa is a neighbourhood of bb. Use regularity to get VV and GG. Then GG cannot be a neighbourhood of bb, so UU is.

Since every Hausdorff space is T 0T_0, a less ambiguous term for a T 3T_3 space is a regular Hausdorff space.

It is possible to describe the regularity condition fairly simply entirely in terms of the algebra of open sets. First notice the relevance above of the condition that Cl(V)UCl(V) \subset U; we write VUV \subset\subset U in that case and say that VV is well inside UU. We now rewrite this condition in terms of open sets and regularity in terms of this condition.

Definition

Given sets U,VU, V, VUV \subset\subset U iff there exists an open set GG such that VG=V \cap G = \empty but UG=XU \cup G = X. Then XX is regular iff, given any open set UU, UU is the union of all of the open sets that are well inside UU.

This definition is suitable for locales. As the definition of a Hausdroff locale is rather more complicated, one often speaks of compact regular locales where classically one would have spoken of compact Hausdorff spaces. (The theorem that compact regular T 0T_0 spaces and compact Hausdorff spaces are the same works also for locales, and every locale is T 0T_0, so compact regular locales and compact Hausdorff locales are the same.)

The condition that a space XX be regular is related to the regular open sets in XX, that is those open sets GG such that GG is the interior of its own closure. (In the Heyting algebra of open subsets of XX, this means precisely that GG is its own double negation; this immediately generalises the concept to locales.) Basically, we start with a neighbourhood UU of xx and reduce that to a closed neighbourhood Cl(V)Cl(V) of xx; then Int(Cl(V))Int(Cl(V)) is a regular open set.

This is enough to characterise regular spaces, as follows:

Definition

Given a neighbourhood UU of xx, there is a closed neighbourhood of xx that is contained in UU. Equivalently, xx has a regular open neighbourhood contained in UU. In other words, the closed neighbourhoods of xx, or equivalently the regular open neighbourhoods of xx, form a local base (a base of the neighbourhood filter) at xx.

In constructive mathematics, Definition 2 is good; then everything else follows without change, except for the equivalence with 1. Even then, the classical separation axioms hold for a regular space; they just are not sufficient.

Definition 5 suggests a slightly weaker condition, that of a semiregular space:

Definition (of semiregular)

The regular open sets form a basis for the topology of XX.

As we've seen above, a regular T 0T_0 space (T 3T_3) is Hausdorff (T 2T_2); we can also remove the T 0T_0 condition from the latter to get R 1R_1:

Definition (of R 1R_1)

Given points aa and bb, if every neighbourhood of aa meets every neighbourhood of bb, then every neighbourhood of aa is a neighbourhood of bb.

It is immediate that T 2R 1T 0T_2 \equiv R_1 \wedge T_0, and the proof above that T 3T 2T_3 \Rightarrow T_2 becomes a proof that R 2R 1R_2 \Rightarrow R_1; that is, every regular space is R 1R_1. An R 1R_1 space is also called preregular (in HAF) or reciprocal (in convergence space theory).

A bit stronger than regularity is complete regularity; a bit stronger than T 3T_3 is T 312T_{3\frac{1}{2}}. The difference here is that we require that aa and FF be separated by a function, that is by a continuous real-valued function. See (or write) Tychonoff space for more.

Revised on August 5, 2011 20:21:49 by Toby Bartels (64.89.48.241)