Regular spaces

# Regular spaces

## Idea

A regular space is a topological space (or variation, such as a locale) that has, in a certain sense, enough regular open subspaces. The condition of regularity is one of the separation axioms satsified by every metric space (and 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 T₀)

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

###### 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 $U, V$, 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 neighbourhood of $x$.

This gives us another way 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, or indeed any neighbourhood, well inside $U$.) In other words, the closed neighbourhoods of $x$ form a local base (a base of the neighbourhood filter) at $x$.

###### Remark

(Warning)
It is not sufficient that the regular open neighbourhoods themselves form a local base of each point; see Counterexample . It's the closures of the regular open neighbourhoods (which are arbitrary closed neighbourhoods) that form the basis. But compare semiregular spaces below.

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

## Variations

Following up on Definition , we have:

###### Corollary

For any regular space $X$, the regular open sets form a basis for the topology of $X$.

###### Proof

For any closed neighbourhood $V$ of $x \in X$, the interior $Int(V)$ is a regular open neighbourhood of $x$. Using Definition finishes the proof.

This 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 R₁)

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 for a completely regular space we require that $a$ and $F$ be separated by a function, that is by a continuous real-valued function. See Tychonoff space for more. This strengthening implies (Tychonoff Embedding Theorem) that the space embeds into a product of metric spaces.

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

## Examples

###### Example

Let $(X,d)$ be a metric space regarded as a topological space via its metric topology. Then this is a normal Hausdorff space, in particular hence a regular Hausdorff space.

###### Proof

We need to show is that given two disjoint closed subsets $C_1, C_2 \subset X$ then their exists disjoint open neighbourhoods $U_{C_1} \subset C_1$ and $U_{C_2} \supset C_2$.

Consider the function

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

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

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

Then the unions of open balls

$U_{C_1} \coloneqq \underset{x_1 \in C_1}{\cup} B^\circ_{x_1}( d(C_2,x_1) )$

and

$U_{C_2} \coloneqq \underset{x_2 \in C_2}{\cup} B^\circ_{x_2}( d(C_1,x_2) ) \,.$

have the required properties.

###### Example

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

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

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

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

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

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

###### Example

(counter example)
Let $\bigl((0, 1)\times(0, 1)\bigr)\cup\{0\}$ be equipped with the Euclidean topology on $(0, 1)\times(0,1)$ and have the sets of the form $(0, 1/2)\times(0, \varepsilon)\cup\{0\}$ (for $\varepsilon \in (0, 1)$) as a basis of open neighbourhoods for the point $0$.

• This space is not regular since we cannot separate $0$ from $[1/2, 1)\times(0,1)$
• Every point $p = (p_1, p_2) \neq 0$ has the euclidean balls of centre $p$ and radius $\varepsilon \in (0, p_1)$ as regular neighbourhood basis.
• The provided basis for the neighbourhoods of $0$ already is a system of regular open sets.

Therefore, this space has the property that every point has a neighbourhood basis of regular open sets (and consequently, the space is semiregular, an even weaker property), but $0$ does not have a neighbourhood basis of closed sets (and consequently, the space is not regular). The problem is that, while every basic neighbourhood of $0$ (and therefore every neighbourhood of $0$ whatsoever) contains a regular open neighbourhood of $0$, none of these basic neighbourhoods contains the closure of any of these regular open neighbourhoods (or any other closed neighbourhood of $0$).

###### Example

Locally compact Hausdorff spaces are completely regular.

(e.g. Engelking 1989, Thm. 3.3.1)
###### Remark

Example plays a key role in discussion of slice theorems, see there for more.

## Properties

the main separation axioms

numbernamestatementreformulation
$T_0$Kolmogorovgiven 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
$T_1$given two distinct points, both have an open neighbourhood not containing the other pointall points are closed
$T_2$Hausdorffgiven two distinct points, they have disjoint open neighbourhoodsthe diagonal is a closed map
$T_{\gt 2}$$T_1$ and…all points are closed and…
$T_3$regular 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
$T_4$normal 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

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.

## References

Textbook accounts:

Last revised on September 22, 2021 at 14:48:33. See the history of this page for a list of all contributions to it.