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!
Best Answer
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.