Are Polish space and LCCB space related?

It seems these spaces are the most useful ones for doing probabilities. Are LCCB (locally compact with countable basis) somewhat more general spaces that when endowed with a metric become Polish? I think I once knew the answer to this question. Thanks

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What is a LCCB? – William Jul 8 '12 at 23:55
Probably Locally Compact with a Countable Base – ncmathsadist Jul 8 '12 at 23:56
yes, sorry just added it. – Antoine P Jul 8 '12 at 23:57
Did you mean to ask if a locally compact second-countable metrizable space is neccessarily Polish? – tomasz Jul 9 '12 at 2:36
By locally compact with countable basis, do you mean sigma-locally-compact ? if yes, the space would have lindelöf property, and if it is endowed with a metric, it would be separable. – saposcat Jul 9 '12 at 3:23

2 Answers

Theorem. Every locally-compact second-countable Hausdorff space is a Polish space.

I thought I'd quote a sketch of proof of above fact from somewhere else.

The following sketch is from https://golem.ph.utexas.edu/category/2008/08/polish_spaces.html

if X is second countable locally compact Hausdorff, then the one-point compactification X+ is metrizable and compact, hence complete (under any metric), and of course second countable, hence X+ is Polish. An open subspace of a Polish space is Polish, hence X is Polish.

Bourbaki (which I found as a Google Book result) provided invaluable assistance.

As for why an open subspace of a Polish space is Polish, again from the same link:

I had no idea that an open set of a Polish space was Polish — it seemed pretty tough removing that extra point and finding a complete metric on what’s left.

I now see how easy this is: we start with a metric on the one-point compactification X+, remove the point at infinity, and ‘stretch’ the metric near the removed point to get a complete metric on X.

As for why the one-point compactification is metrizable, the selected answer from the following thread explains how to build a local countable basis at infinity:

If $X$ is locally compact, second countable and Hausdorff, then $X^*$ is metrizable and hence $X$ is metrizable

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After a bit of research I found that A locally compact space that is Hausdorff (LCH) will be sigma-locally-compact. Also that a LCCB will be metrizable (with a complete metric) and separable thus Polish too. thanks

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The converse doesn't hold, though. Take, for example, $L^1(\mathbb R)$, the set of absolutely integrable Lebesgue-measurable functions on $\mathbb R$ (with any two functions equal almost everywhere being identified). It is a separable Banach space, so it is Polish (and second countable). However, it is not locally compact, because no infinite-dimensional normed vector space is. – triple_sec Jan 29 '15 at 10:24