> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $X$ is **regular** if it is $T_1$ and that points and [[Closed Set|closed sets]] can be [[Separation Axioms|separated]] by disjoint open [[Neighbourhood|neighbourhoods]]: > $ > \forall A \text{ closed}, x \not\in A, \exists U \in \cn(A)^o, V \in \cn(x)^o: U \cap V = \emptyset > $ > [!theorem] Alternate Characterisation > > A $T_1$ space $X$ is regular ($T_3$) if and only if for any $x \in X$ and $V \in \cn(x)^o$, there exists $U \in \cn(x)^o$ such that $x \in U \subset \ol{U} \subset V$. > > *Proof*. Let $x \in X$ and $V \in \cn(x)^o$. > > Suppose that $X$ is regular. Since $V^c$ is closed, there exists open neighbourhoods $U \in \cn(x)^o$ and $V' \in \cn(V^c)^o$ such that $U \cap V' = \emptyset$. As $(V')^c \supset U$ is a closed set, $(V')^c \supset \ol{U}$, and we have found $U \in \cn(x)^o$ such that > $ > x \in U \subset \ol{U} \subset (V')^c \subset V > $ > > Let $A \subset X$ be a closed set such that $x \not\in A$. Then $A^c$ is a neighbourhood of $x$, and there exists $U \in \cn(x)^o$ such that > $ > x \in U \subset \ol{U} \subset A^c > $ > and we have found $U \in \cn(x)$ and $V = \ol{U}^c \in \cn(A)$ such that $U \cap V = \emptyset$.