A type of $T_2$ topological space such that for every point there's a $T_3$ subspace then the whole space is $T_3$

Question

Given a $T_2$ topological space $(X,\tau)$ such that $\forall x\in X, \exists V\in \tau$ such that $x \in V$ and the subspace $(cl(V),\tau_{cl(V)})$ is $T_3$ then $(X,\tau)$ is $T_3$.

I really don't know how to solve this, any tips would be of help.

Answer 1

Suppose $x\in V\subseteq Y\subseteq X$, with $V$ open in $X$ and $Y$ a closed subspace of $X$ which is regular in the subspace topology. Note that such a subspace $Y$ will exist if and only if $cl_X(V)$ is regular, since regularity is an hereditary property and $cl_X(V)\subseteq Y$ (we can always take $Y=cl_X(V)$). I only want to introduce $Y$ as I will need closure operators in both $X$ and its subspace $Y$, and the notation below becomes somewhat more approachable.

To proceed let $A\subseteq X$ be a closed set not containing $x$. Then $Y\cap A$ is a closed subset of $Y$ which does not contain $x$. Since $Y$ is regular, there is a relatively open set $W'\subset Y$ such that $x\in W'\subseteq cl_Y(W')\subseteq Y\setminus A$.

Letting $W=W'\cap V$ we obtain a relatively open subset of $V$ containing $x$. Since $V$ is open in $X$, and $W$ is open in $V$, the set $W$ is open in $X$. Since $W\subseteq Y$ and $Y$ is closed we have $cl_X(W)\subseteq Y$. From this we obtain $$cl_X(W)=cl_X(W)\cap Y=cl_Y(W)\subseteq cl_Y(W')\subseteq Y\setminus A\subseteq X\setminus A,$$ which is what we needed to show. We conclude that $x$ and any disjoint closed set have disjoint neighbourhoods. In particular, if each point of $X$ has a closed regular neighbourhood, then $X$ is regular.

So far we haven't used the $T_2$ axiom. However this is hereditary, so if $X$ is $T_2$, then each regular subspace of $X$ is $T_3$.