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 $T_0$:

A space is $T_0$ if, 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 $U, V$, then $V \subset\!\!\!\!\subset U$ 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 Tychonoff space for more.

For locales, there is also a weaker notion called weak regularity?, which uses the notion of fiberwise closed sublocale? instead of ordinary closed sublocales.

A uniform space is automatically regular and even completely regular, at least in classical mathematics. In constructive mathematics this may not be true, and there is an intermediate notion of interest called uniform regularity.

Every regular space comes with a naturally defined (point-point) apartness relation: we say $x # y$ if there is an open set containing $x$ but not $y$. This can be defined for any topological space and is obviously irreflexive, but in a regular space it is symmetric and a comparison, hence an apartness. For symmetry, if $x\in U$ and $y\notin U$, let $V$ be an open set containing $x$ and $G$ an open set such that $V\cap G = \emptyset$ and $G\cup U = X$; then $y\in G$ (since $y\notin U$) while $x\notin G$ (since $x\in V$). With the same notation, to prove comparison, for any $z$ we have either $z\in G$, in which case $z # x$, or $z\in U$, in which case $z # y$. Note that this argument is valid constructively; indeed, classically, the much weaker $R_0$ separation axiom is enough to make this relation symmetric, and it is a comparison on any topological space whatsoever.

Note that if a space is localically strongly Hausdorff (a weaker condition than regularity), then it has an apartness relation defined by $x \# y$ if there are disjoint open sets containing $x$ and $y$. If $X$ is regular, then this coincides with the above-defined apartness.

Revised on February 3, 2017 07:04:47
by Mike Shulman
(76.167.222.204)