a question about first countable dense subset If $X$ is first countable and dense subset of  regular space $Y$, can we say $Y$ has a countable local base for all $x\in X$? I think, we need construct a new local base in $Y$?
thanks,
 A: $\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$Yes, the conjecture is true.
Let $X$ be a dense subset of a $T_3$-space $Y$, let $x\in X$, and let $\mathscr{B}$ be a local base at $x$ in the space $X$, not $Y$. If $\mathscr{B}$ is countable, we may assume that $\mathscr{B}=\{B_n:n\in\Bbb N\}$, and we may use the regularity of $Y$ to assume further that $\cl_X B_{n+1}\subseteq B_n$ for each $n\in\Bbb N$. For $n\in\Bbb N$ let $V_n=\int_Y\cl_YB_n$, and let $\mathscr{V}=\{V_n:n\in\Bbb N\}$. Clearly $\mathscr{V}$ is a family of open neighborhoods of $x$ in $Y$, and I claim that it’s a local base at $x$.
To see this, let $U$ be any open nbhd of $x$ in $Y$. $Y$ is regular, so there is an open $W$ in $Y$ such that $x\in W\subseteq\cl_YW\subseteq U$. $W\cap X$ is an open nbhd of $x$ in $X$, so there is an $n\in\Bbb N$ such that $x\in B_n\subseteq W\cap X$. Then $$x\in V_n=\int_Y\cl_YB_n\subseteq\int_Y\cl_Y(W\cap X)\subseteq\int_Y\cl_YW\subseteq\int_YU=U\;,$$ i.e., $x\in V_n\subseteq U$. $U$ was an arbitrary open nbhd of $x$ in $Y$, so $\mathscr{V}$ is a local base at $x$ in $Y$.
It follows that if $X$ is first countable as a space, each point of $X$ is a point of first countability of $Y$.
