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 08 '12 at 23:55
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1Probably Locally Compact with a Countable Base – ncmathsadist Jul 08 '12 at 23:56
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yes, sorry just added it. – percojazz Jul 08 '12 at 23:57
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Did you mean to ask if a locally compact second-countable metrizable space is neccessarily Polish? – tomasz Jul 09 '12 at 02:36
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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 09 '12 at 03:23
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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:
Jisang Yoo
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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
Martin Sleziak
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percojazz
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2The 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
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This is true if the space is Hausdorff. There are also LCCB spaces which are not Hausdorff, and hence not metrizable. But the "nice enough" ones are still quasi-Polish, which is a natural generalization of the Polish property for non-Hausdorff spaces. – Robin Saunders Feb 27 '24 at 03:44
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To extend Jisang Yoo's answer, every sober LCCB space is quasi-Polish. Sober and quasi-Polish are natural generalizations of Hausdorff and Polish, and this result was noted in the paper which introduced quasi-Polish spaces (section 8).
In fact, being sober is not much of a restriction. If we define spaces in terms of the relationships between open (or closed) sets, rather than focusing on points, then every LCCB space is, essentially, sober.
Robin Saunders
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