Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 6 down vote accepted

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 $x$.

share|improve this answer
2  
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

Yes.

A topological space is called first countable if every point has a countable neighborhood basis.

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.