What is the relation between Locally Compact Hausdorff Spaces and Complete Separable Metric Spaces? In the book "Real and Complex Analysis" by Rudin, he often uses the condition that a space is locally compact Hausdorff in order to present results in a general manner. The thing is, I'm not very used to such condition. Most books of analysis/measure-theory that I've read present results in terms of metric spaces/separable/complete.
Thus, I was wondering if there is a precise relation between such notions. For example, does locally compact Hausdorff implies completeness or separability? Is the opposite implication true?
Take for example the following theorem by Rudin:
If $X$ is locally compact Hausdorff, and $\mu$ a measure on the borelians of $X$. Then for $1\leq p < \infty$, $C_c(X)$ is dense in $L^p(\mu)$.
Now, I was wondering if this theorem could be somehow stated, but something like, if $X$ is Polish, then this is true. Hence, what I'm really interested in is to know if there is a way to somehow relate this type of spaces. If the implications are not true, is there an extra condition that tie them together?
 A: 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.
A: 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 $r$.
