What is the relation between Locally Compact Hausdorff Spaces and Complete Separable Metric Spaces?

Solution 1:

Both implications fail.

Product space $[0,1]^A$ with $A$ uncountable is compact Hausdorff but not separable and not metrizable.

Hilbert space $l_2$ is complete separable metric, but not locally compact.

Of course, many common spaces have both properties. Indeed, an open subset of $\mathbb R^n$ is completely metrizable separable locally compact Hausdorff.

Solution 2:

Let $I=[0,1]$ with the standard topology. Let $k$ be an infinite cardinal. By the Tychonoff Theorem ( a product of compact spaces is compact), the product-space $I^k$ is compact.

It is easy to show that a product of $T_2$ spaces is $T_2$ and it is easy to show that a compact $T_2$ space is $T_4$. So the "Tychonoff plank" $I^k$ is a compact normal space. It is also easy to show that any subspace of a normal space is a $T_{3\frac 1 2}$ space.

Theorem: If $S$ is a $T_{3\frac 1 2}$ space and if $S$ has a base (basis) $B$ with cardinal $|B|\le k$ then $S$ is homeomorphic to a subspace of $I^k.$

So the class of compact Hausdorff spaces and their subspaces is, in this sense, much bigger than the class of metrizable spaces.

In particular a separable metrizable space has a countable base so it is homeomorphic to a subspace of $I^{\aleph_0}.$

It is hard to define a useful countably-additive measure on the Borel sets of a space that is not locally compact. For example in an infinite-dimensional normed linear space (e.g. Hilbert space $\ell_2$ ) there exists $ r>0$ such that an open ball of radius $1$ contains an infinite pairwise-disjoint family of open balls, each of radius $r$.