# A locally metrizable space

Is every locally metrizable space always first countable?

A locally metrizable space $X$ means for every point $x\in X$ has an open nbhd such that it is metrizable.

Thanks for help.

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If $x \in \rm X$ has an open metrizable nbhd $\rm U$, then the set of $\mathrm B(x, 1/n) \cap \mathrm U$ form a countable basis of open nbhd of $x$.

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This is badly stated, since the sets $B(x,1/n)$ make sense only within $U$ in the first place: they are necessarily subsets of $U$. What you mean is that there is a metric $d$ on $U$, and the sets $B_d(x,1/n)$ for $n\in\Bbb Z^+$ are a countable local base at $x$. –  Brian M. Scott Mar 9 '13 at 14:58
Fix $x\in X$ arbitrary. Fix $U\subseteq X$ a neighborhood of $x$ which is metrizable, say by $\rho:U\times U\to[0,+\infty]$ (or $[0,+\infty)$), so $(U,\rho)$ is a metric space. You don't even need to do things like $B(x,1/n)\cap U$ because the metric isn't defined off of $U\times U$.
Then as the original topology and the topology induced by $\rho$ are identical, what Damien L said is true. $\{B(x,1/n)\}_{n\in\mathbb{N}}$ is the countable neighborhood basis we need.