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Given $X$ a Hausdorff space, I have a hunch that

$X$ is locally compact $\iff X$ is metrizable.

I am not sure if it is true because I do not know how to prove that. To prove the implication (locally compact Hausdorff space is metrizable) I guess I need either $X$ is second countable or locally metrizable.

Could some one prove it or otherwise, give a counter example that the statement is not true. Thanks!

Chen M Ling
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    I don't think the equivalence is true, and you probably can find some answers here http://math.stackexchange.com/questions/74923/a-compact-hausdorff-space-that-is-not-metrizable – Darío G Apr 11 '16 at 14:05
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    Counterexample: $[0,1]^S$ for uncountable $S$. That's a compact Hausdorff space that's not metrizable. – David C. Ullrich Apr 11 '16 at 14:10
  • @Wore&DavidCUllrich,, yup. It's false statement. Thanks for the example. – Chen M Ling Apr 12 '16 at 09:38

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Both @Wore's and @David C Ullrich's comments above (and the linked question) give you examples of locally compact $T_2$ spaces which are not metrizable.

But the other direction doesn't hold either: an example of a metrizable space which is not locally compact is Example 30 from Steen and Seebach's Counterexamples in Topology (scroll quickly down to pg59, as it's a limited preview)

We consider the real line $\mathbb{R}$ with the Euclidean topology, and the irrationals $\mathbb{I}$ as a subspace with the inherited topology. The irrationals are metrizable (by the Euclidean metric) but not locally compact (can you prove this?).

'Counterexamples in Topology' is a very handy reference if you have a feeling that a property (or combination of) implies another, since it has several reference charts at the back which show you a multitude of properties which hold (or not) for the examples in the book.

Simon_Peterson
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    If $\mathbb{I}$ were locally compact, then for every irrational $r\in\mathbb{I}$ has compact neighbourhood, i.e. we can find a nbhd $U_r$ of $r$ where every sequence has a convergent subsequence. But since rationals are dense in $\mathbb{R}$ there always exists a rational $x \in U_r$, then the sequence approximating $x$ won't have a convergent subsequence. For e.g., the interval contains 1/2, since both rationals are not order-complete, there cannot be a sequence of irrationals converging to 1/2. So, the irrationals are not locally compact. – Chen M Ling Apr 12 '16 at 02:32
  • Thanks for the refference. That book seems very useful. I will get that book. Thanks again! So, my hunch is not true then. Now I will raise another question, haha. If $X$ is Haussdorf, second countable, and metrizable, is it always locally compact then? – Chen M Ling Apr 12 '16 at 02:38
  • The above is also a counterexample to this: the irrationals are Hausdorff, second countable and metrizable, but they are not locally compact. – Simon_Peterson Apr 14 '16 at 14:16