Skip to the Main Content

Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser.

August 21, 2008

Polish Spaces

Posted by John Baez

Point-set 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 infinite-dimensional 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 second-countable 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 σ\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 countable-dimensional Hilbert space, and the closed unit interval… and they’re isomorphic! As measurable spaces with their Borel σ\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 countable-dimensional Hilbert space.

    Namely, every commutative von Neumann algebra of this sort is isomorphic to L (X)L^\infty(X), where the measure space XX is either:

    1. a countable set (equipped with counting measure),
    2. the closed unit interval (with Lebesgue measure),
    3. the disjoint union of a countable set and the closed unit interval.


  • Second-countable locally compact Hausdorff spaces show up in the theory of operator algebras. Thanks to the Gelfand-Naimark 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 second-countable. 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 second-countable locally compact Hausdorff space is automatically a Polish space.

This article suggests the answer is yes:

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 old-timers 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 second-countable.

which would imply:

Any second-countable locally compact Hausdorff space is homeomorphic to a metric space in which all bounded sets are relatively compact.

which should imply:

Any second-countable locally compact Hausdorff space is homeomorphic to a complete metric space.

or in other words:

Any second-countable locally compact Hausdorff space is Polish.

I should point out that Urysohn’s metrization theorem implies a weaker result:

Any second-countable 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”.

Posted at August 21, 2008 6:51 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1771

7 Comments & 0 Trackbacks

Re: Polish Spaces

It is true:

Let X be a second-countable compact Hausdorff space and let X+ be its one-point compactification.

Then X+ is compact and Hausdorff. Moreover, since X is sigma-compact, X+ is second-countable 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 second-countable if and only if the algebra is separable.

Posted by: otogo on August 21, 2008 11:04 PM | Permalink | Reply to this

Re: Polish Spaces

Your result is correct. Here’s a sketch: if XX is second countable locally compact Hausdorff, then the one-point compactification X +X^+ is metrizable and compact, hence complete (under any metric), and of course second countable, hence X +X^+ is Polish. An open subspace of a Polish space is Polish, hence XX is Polish.

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

Posted by: Todd Trimble on August 21, 2008 11:48 PM | Permalink | Reply to this

Re: Polish Spaces

Polish spaces are indeed awesome. I have a partially-written post explaining why Polish spaces are the natural setting for probability, but I never get around to finishing it. (They’re really the natural setting for those parts of analysis that do not rely directly on linearity.)

Posted by: Walt on August 22, 2008 12:53 AM | Permalink | Reply to this

Re: Polish Spaces

Thanks, otogo and Todd! I’d considered trying a proof using the one-point compactification, but 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 +X^+, remove the point at infinity, and ‘stretch’ the metric near the removed point to get a complete metric on XX. Alexandrov’s theorem does this kind of job in much more generality — I’m looking at the proof in Arveson’s An Invitation to C *C^*-Algebra, and it describes a very nice ‘stretching’ procedure.

otogo wrote:

P.S. The spectrum of an abelian C *C^*-algebra is second-countable if and only if the algebra is separable.

Great! So everything is maximally beautiful in this little circle of ideas.

You should finish that post on Polish space, Walt! I want to know more about ‘em.

Posted by: John Baez on August 22, 2008 2:10 AM | Permalink | Reply to this

Re: Polish Spaces

I learned about a remarkable property of Polish spaces the other day. This is Kuratowski’s theorem, see Parthasarathy Corollary I.3.3:

Let f : E -> F be bijective, with E subset X, F subset Y, X polish, E measurable, and Y separable.

Then F is measurable and f is an isomorphism of measurable spaces.

Posted by: Tom E on October 27, 2009 2:31 PM | Permalink | Reply to this

Re: Polish Spaces

I’m glad someone out there likes Polish spaces! They’re shockingly well-behaved.

If you haven’t read week272, you might like the discussion of Polish spaces there.

Posted by: John Baez on October 28, 2009 6:43 AM | Permalink | Reply to this

Re: Polish Spaces

Polish spaces are awesome, but more awesome are locally compact (LC) spaces. They have one point compactification, and other goodies.

And compact spaces are great. They have the Stone-Weierstrass theorem, which combined with the Riesz theorem make them very well suited for measure theory.

So it has long been believed that LC spaces are the “right” setting for measure thory, but the need for non-LC spaces (spaces of functions, of measures,…) has brought Polish spaces to the front of the scene.

LC spaces are not necessarily metric, and not necessarily separable. But metric spaces are so common, and measure theory is so fond of countability that sigma-compactness is a natural assumption. And it implies Polishness, as pointed out.

So settle for local compactness if you have it, because they’re very convenient, and use Polish spaces if you have no choice (or if you want to prove general theorems).

Posted by: Kebab on February 5, 2010 1:41 AM | Permalink | Reply to this

Post a New Comment