When $T_3$ implies first countable?

Question

In a $T_3$ space we can always define a chain $\{V_{j}\}_{j\in \omega}$ of neighborhoods of $x$ such that $\overline{V_{j+1}}\subset V_j$ for all $j$

Let $x\in X$, $x\in U$ open, assume $U\neq X$. $X\backslash U$ is closed and $x\notin X\backslash U,$ therefore, by $T_3$ there exist $V_0$ open and $W_0$ open disjoint such that $X\backslash U \subseteq W_0$ and $x\in V_0$. Therefore, $V_0\subseteq \overline{V_0}\subseteq X\backslash W_0 \subseteq U$. And define $V_{j+1}$ by uisng the above construction with $U=V_j$ and $V_0=V_{j+1}.$

Then, for every $U$ open neighborhood of $x$, there exist a $M>0$ such that $V_j\subset U$ and therefore, $\{V_{j}\}_{j\in \omega}$ is a local basis which is countable.

This is not true in general. For example, choose $X$ metrizable then, $V_{j+1}=B_{1+1/{(j+1)}}(x)$ is a chain with the desire property and $B_{1/2}(x)$ has not such $M$.

There are further conditions of the space such that $T_3+\text{Something}\implies \text{first countable}$?

There exists a construction of the $V_j$ such that the $V_j$ decrease as "small" as I want under stronger assumptions?

Answer 1

Being $T_3$ and being first countable are not very related. There are lots of first countable spaces that are not regular, and lots of regular spaces that are not first countable. Also, your claim that the $(V_j)_{j \in \omega}$ form a local base is not at all warranted: you start out with an open set $U$ in which you can construct such a sequence but you must give one sequence that works in for any open set, which is very much stronger. You don't come close to showing that.

The only thing you've shown is that in a regular space there is a decreasing sequence of sets around any point, inside a given open $U$. You cannot even show it gets arbitrarily small, or has intersection $\{x\}$ in general.

In short, there is no standard property $P$ (except trivially "first countable", or anything stronger, itself), weaker than first countability preferably, such that $P$ + $T_3$ implies first countable. At least not one that I can think of. With some strong $P$ you can do something: being compact $T_1$ and having $G_\delta$-points is one, but not very interesting.