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Now I would like to change question in If $X$ is locally compact, second countable and Hausdorff, then $X^*$ is metrizable and hence $X$ is metrizable.

Is it true that

If $X$ is Hausdorff, second countable, and metrizable, then $X$ is locally compact.

To show that $X$ is locally compact, need to show that for every $x\in X$ has compact neighbourhood, i.e. we can find a nbhd $U_x$ where every sequence has a convergent subsequence. I raise this question because my hunch at Locally compact Hausdorff space is metrizable is not true.

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Chen M Ling
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  • This is not true. Take $C([0,1],\mathbb{R})$. It is separable and metrizable, and hence Hausdorff and secound countable, but it can't be locally compact, since it is a infinite dimensional topological vector space. – Aloizio Macedo Apr 12 '16 at 03:11

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The irrationals are certainly metrizable (implying Hausdorff also), second countable, but not locally compact. In fact, they are nowhere locally compact.

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