# 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 $X$.

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 $a$ and closed set $F$, if $a \notin F$, then there exist a neighbourhood $V$ of $a$ and a neighbourhood $G$ of $F$ such that $V \cap G$ is empty.

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

###### Definition

Given any point $a$ and neighbourhood $U$ of $a$, there exist a neighbourhood $V$ of $a$ and an open set $G$ such that $V \cap G = \empty$ but $U \cup G = X$.

You can think of $V$ as being half the size of $U$, with $G$ the exterior of $V$. (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 $a$ and neighbourhood $U$ of $a$, there exist a neighbourhood $W$ of $a$ and an open set $G$ such that $Cl(W) \cap Cl(G) = \empty$ but $U \cup G = X$ (where $Cl$ indicates topological closure).

###### Proof of equivalence

Find $V$ and $G$ as above. Now apply the regularity axiom to $a$ and the interior $Int(V)$ of $V$ to get $W$ (and $H$).

In terms of the classical language of separation axioms, this says that $a$ and $F$ are separated by closed neighbourhoods.

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

###### Definition (of $T_0$)

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 $T_0$ space a $T_3$ space, but then that term is also used for a (merely) regular space. An unambiguous term for the weaker condition is an $R_2$ space, but hardly anybody uses that.

We have

###### Theorem

Every $T_3$ space is Hausdorff.

###### Proof

Suppose every neighbourhood of $a$ meets every neighbourhood of $b$; by $T_0$ (and symmetry), it's enough to show that each neighbourhood $U$ of $a$ is a neighbourhood of $b$. Use regularity to get $V$ and $G$. Then $G$ cannot be a neighbourhood of $b$, so $U$ is.

Since every Hausdorff space is $T_0$, a less ambiguous term for a $T_3$ 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 $Cl(V) \subset U$; we write $V \subset\subset U$ in that case and say that $V$ is well inside $U$. We now rewrite this condition in terms of open sets and regularity in terms of this condition.

###### Definition

Given sets $U, V$, $V \subset\subset U$ iff there exists an open set $G$ such that $V \cap G = \empty$ but $U \cup G = X$. Then $X$ is regular iff, given any open set $U$, $U$ is the union of all of the open sets that are well inside $U$.

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 $T_0$ spaces and compact Hausdorff spaces are the same works also for locales, and every locale is $T_0$, so compact regular locales and compact Hausdorff locales are the same.)

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

This is enough to characterise regular spaces, as follows:

###### Definition

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

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 $X$.

As we've seen above, a regular $T_0$ space ($T_3$) is Hausdorff ($T_2$); we can also remove the $T_0$ condition from the latter to get $R_1$:

###### Definition (of $R_1$)

Given points $a$ and $b$, if every neighbourhood of $a$ meets every neighbourhood of $b$, then every neighbourhood of $a$ is a neighbourhood of $b$.

It is immediate that $T_2 \equiv R_1 \wedge T_0$, and the proof above that $T_3 \Rightarrow T_2$ becomes a proof that $R_2 \Rightarrow R_1$; that is, every regular space is $R_1$. An $R_1$ 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 $T_3$ is $T_{3\frac{1}{2}}$. The difference here is that we require that $a$ and $F$ 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)