regular space

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).

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:

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:

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:

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).

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 $T_0$:

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

Every $T_3$ space is Hausdorff.

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.

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:

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**:

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 $R_1$:

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)