Polish Spaces
Posted by John Baez
Pointset topology has a bit of a dry reputation. I guess this is because a lot is known about it, and people who need it can usually find what they want in textbooks. But it keeps coming up…
For example, Yetter’s paper on infinitedimensional categorified Hilbert spaces has forced me to learn about Polish spaces. A Polish space is a topological space that’s homeomorphic to a separable complete metric space.
And now I’m dying to know the answer to this:
Is every secondcountable locally compact Hausdorff space a Polish space?
Can you help? I have some evidence that the answer is yes, but I don’t completely trust it.
First I should say a bit about why I care about this question.

Polish spaces are a nice framework for doing measure theory. We can take any Polish space and regard it as a measurable space with its $\sigma$algebra of Borel sets. And then, remarkably, there’s a complete classification of Polish spaces up to measurable bijection: there’s one for each cardinality that’s countable, and one whose cardinality is that of the continuum, and that’s all!
So, for example, you can take a countabledimensional Hilbert space, and the closed unit interval… and they’re isomorphic! As measurable spaces with their Borel $\sigma$algebras, that is.
Why are Polish spaces ‘not very big’? In other words, why are there none with cardinality exceeding the continuum? It’s because any Polish space has a countable dense subset — it’s separable — and you can write any point as a limit of a sequence of points in this subset. So, you only need a sequence of integers to specify any point in a Polish space.
As a close relative of this remarkable classification of Polish spaces, there’s a nice classification of all abelian von Neumann algebras that are ‘not too big’: that are subalgebras of the algebra of bounded operators on a countabledimensional Hilbert space.
Namely, every commutative von Neumann algebra of this sort is isomorphic to $L^\infty(X)$, where the measure space $X$ is either:
 a countable set (equipped with counting measure),
 the closed unit interval (with Lebesgue measure),
 the disjoint union of a countable set and the closed unit interval.
 Secondcountable locally compact Hausdorff spaces show up in the theory of operator algebras. Thanks to the GelfandNaimark theorem, the spectrum of any commutative C*algebra is a locally compact Hausdorff space. If our C*algebra is ‘not too big’, its spectrum will also be secondcountable. I’m afraid I don’t know a precise theorem along these lines — will it suffice for our C*algebra to be separable?
So, it’s natural to wonder if a secondcountable locally compact Hausdorff space is automatically a Polish space.
This article suggests the answer is yes:
 H. E. Vaughan On locally compact metrisable spaces, Bull. Amer. Math. Soc. 43 (1937), 532535.
The author claims to show:
Theorem 2: In order that a Hausdorff space be homeomorphic to a totally complete metric space it is necessary and sufficient that it be locally compact and perfectly separable.
He says “a totally complete metric space is a metrisable space in which the metric is chosen such that every bounded set is compact.” Hmm? Every bounded set is compact? That seems rather drastic… maybe he means they’re all relatively compact? In another paper he cites Kuratowski’s Topologie I, page 196 for this definition. I seem to recall that oldtimers sometimes used ‘compact’ to mean relatively compact.
He doesn’t define ‘perfectly separable’, but my readings suggest that this is often used as a synonym for ‘second countable’.
So, rather optimistically, I can hope the author has proved:
In order that a Hausdorff space be homeomorphic to a metric space in which all bounded sets are relatively compact, it is necessary and sufficient that it be locally compact and secondcountable.
which would imply:
Any secondcountable locally compact Hausdorff space is homeomorphic to a metric space in which all bounded sets are relatively compact.
which should imply:
Any secondcountable locally compact Hausdorff space is homeomorphic to a complete metric space.
or in other words:
Any secondcountable locally compact Hausdorff space is Polish.
I should point out that Urysohn’s metrization theorem implies a weaker result:
Any secondcountable locally compact Hausdorff space is homeomorphic to a metric space.
But as the Wikipedia article on Polish spaces notes, “The problem of determining whether a metrizable space is completely metrizable is more difficult”.
Re: Polish Spaces
It is true:
Let X be a secondcountable compact Hausdorff space and let X^{+} be its onepoint compactification.
Then X^{+} is compact and Hausdorff. Moreover, since X is sigmacompact, X^{+} is secondcountable and thus homeomorphic to a closed (due to its compactness) subset of the Tychonoff cube (see the proof of Urysohn’s metrization theorem). Hence X^{+} is complete in the subspace metric, i.e. Polish.
On the other hand, an open subset of a Polish space is Polish by Alexandrov’s theorem and X is open in X^{+}.
P.S. The spectrum of an abelian C^{*}algebra is secondcountable if and only if the algebra is separable.