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\begin{document}

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\section*{regular space}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{regular_spaces}{}\section*{{Regular spaces}}\label{regular_spaces}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak
\noindent\hyperlink{variations}{Variations}\dotfill \pageref*{variations} \linebreak
\noindent\hyperlink{Examples}{Examples}\dotfill \pageref*{Examples} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A \emph{regular space} is a [[topological space]] (or variation, such as a [[locale]]) 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).

[[!include main separation axioms -- table]]

\hypertarget{definitions}{}\subsection*{{Definitions}}\label{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:

\begin{defn}
\label{classical}\hypertarget{classical}{}
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.

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

\begin{defn}
\label{constructive}\hypertarget{constructive}{}
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 $U \cup G = X$.

\end{defn}
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:

\begin{defn}
\label{closed}\hypertarget{closed}{}
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).

\end{defn}
\begin{proof}
Find $V$ and $G$ as above. Now apply the regularity axiom to $a$ and the interior $Int(V)$ of $V$ to get $W$ (and $H$).

\end{proof}
In terms of the classical language of separation axioms, this says that $a$ and $F$ are separated by \emph{closed} neighbourhoods.

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

\begin{udefn}
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.

\end{udefn}
Other authors use the weaker definition above but call a regular $T_0$ space a \textbf{$T_3$ space}, but then that term is also used for a (merely) regular space. An unambiguous term for the weaker condition is an \textbf{$R_2$ space}, but hardly anybody uses that.

We have

\begin{utheorem}
Every $T_3$ space is [[Hausdorff space|Hausdorff]].

\end{utheorem}
\begin{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.

\end{proof}
Since every Hausdorff space is $T_0$, a less ambiguous term for a $T_3$ space is a \textbf{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 \textbf{well inside} $U$. We now rewrite this condition in terms of open sets and regularity in terms of this condition.

\begin{defn}
\label{locale}\hypertarget{locale}{}
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 $U$.

\end{defn}
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 space|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.)

\begin{defn}
\label{closed_basis}\hypertarget{closed_basis}{}
Given a neighbourhood $U$ of $x$, there is a closed neighbourhood of $x$ that is contained in $U$. Equivalently, $x$ has an open neighbourhood well inside $U$. In other words, the closed neighbourhoods of $x$ form a local base (a base of the [[neighbourhood filter]]) at $x$.

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

\hypertarget{variations}{}\subsection*{{Variations}}\label{variations}

The condition that a space $X$ be regular is related to the \textbf{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.)

\begin{cor}
\label{regular_basis}\hypertarget{regular_basis}{}
For any regular space $X$, the regular open sets form a basis for the topology of $X$.

\end{cor}
\begin{proof}
For any closed neighbourhood $V$ of $x \in X$, the interior $Int(V)$ is a regular open neighbourhood of $x$. Using Definition \ref{closed_basis} finishes the proof.

\end{proof}
Corollary \ref{regular_basis} suggests a slightly weaker condition, that of a \textbf{semiregular space}:

\begin{udefn}
The regular open sets form a basis for the topology of $X$.

\end{udefn}
As we've seen above, a regular $T_0$ space ($T_3$) is [[Hausdorff space|Hausdorff]] ($T_2$); we can also remove the $T_0$ condition from the latter to get $R_1$:

\begin{udefn}
\textbf{(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 $b$.

\end{udefn}
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 \emph{preregular} (in \emph{[[HAF]]}) or \emph{reciprocal} (in [[convergence space]] theory).

A bit stronger than regularity is \emph{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 \emph{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 space]]s.

For locales, there is also a weaker notion called [[weakly regular locale|weak regularity]], which uses the notion of [[fiberwise closed sublocale]] instead of ordinary closed [[sublocales]].

\hypertarget{Examples}{}\subsection*{{Examples}}\label{Examples}

\begin{example}
\label{}\hypertarget{}{}
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.

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

Consider the function

\begin{displaymath}
d(S,-) \colon X \to \mathbb{R}
\end{displaymath}
which computes [[distances]] from a subset $S \subset X$, by forming the [[infimum]] of the distances to all its points:

\begin{displaymath}
d(S,x) \coloneqq inf\left\{ d(s,x) \vert s \in S \right\}
  \,.
\end{displaymath}
Then the [[unions]] of [[open balls]]

\begin{displaymath}
U_{C_1}
    \coloneqq
  \underset{x_1 \in C_1}{\cup} B^\circ_{x_1}( d(C_2,x_1) )
\end{displaymath}
and

\begin{displaymath}
U_{C_2}
    \coloneqq
  \underset{x_2 \in C_2}{\cup} B^\circ_{x_2}( d(C_1,x_2) )
  \,.
\end{displaymath}
have the required properties.

\end{proof}
\begin{example}
\label{}\hypertarget{}{}
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]]).

\end{example}
\begin{proof}
By construction the K-topology is [[finer topology|finer]] than the usual [[Euclidean space|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 subset|disjoint]] closed subset such that they do not have disjoint [[open neighbourhoods]].

But this is the case essentially by construction: Observe that

\begin{displaymath}
\mathbb{R} \backslash K
  \;=\;
  (-\infty,-1/2)
    \cup
  \left( 
    (-1,1) \backslash K
  \right)
    \cup
  (1/2, \infty)
\end{displaymath}
is an open subset in $\mathbb{R}_K$, whence

\begin{displaymath}
K = \mathbb{R} \backslash ( \mathbb{R} \backslash K  )
\end{displaymath}
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.

\end{proof}
\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\begin{itemize}%
\item [[second-countable regular spaces are paracompact]]


\item [[paracompact Hausdorff spaces are normal|paracompact Hausdorff spaces are regular]]



\end{itemize}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

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 [[Hausdorff space|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.

[[!redirects completely regular]] [[!redirects completely regular space]]

[[!redirects regular space]] [[!redirects regular spaces]] [[!redirects R2 space]] [[!redirects R2 spaces]] [[!redirects T3 space]] [[!redirects T3 spaces]]

[[!redirects regular Hausdorff space]] [[!redirects regular Hausdorff spaces]] [[!redirects regular Hausdorff topological space]] [[!redirects regular Hausdorff topological spaces]]

[[!redirects regular topological space]] [[!redirects regular topological spaces]] [[!redirects regular topology]] [[!redirects regular topologies]] [[!redirects R2 topology]] [[!redirects R2 topologies]] [[!redirects T3 topology]] [[!redirects T3 topologies]] [[!redirects T3 axiom]]

[[!redirects regular locale]] [[!redirects regular locales]]

[[!redirects semiregular space]] [[!redirects semiregular spaces]]



\end{document}