Are 2nd-countable locally compact Hausdorff spaces Polish?

[John Baez]

Pondering operator algebras and their relation to measure theory,
I've blundered into this question:

Are second-countable locally compact Hausdorff spaces Polish?

A Polish space is a topological space that's homeomorphic to
a separable complete metric space.

This article suggests the answer is YES:

H. E. Vaughan
On locally compact metrisable spaces
Bull. Amer. Math. Soc. Volume 43, Number 8 (1937), 532-535.
http://www.projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bams/1183499919

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.

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

Any second-countable locally compact Hausdorff space is Polish.

But, I'm wondering if this is really true. One reason I'm
worried is that the rather erudite-looking Wikipedia article on
Polish spaces:

http://en.wikipedia.org/wiki/Polish_space

doesn't mention this result.

[G. A. Edgar]

I did a Google search, found this

http://books.google.com/books?id=g64TCEiYULAC&pg=PA260&lpg=PA260&dq=poli
sh+%22locally+compact%22&source=web&ots=5w8B80MAUV&sig=WxKOW1q3mtPjXuRQU
1Cmv4_l2H0&hl=en&sa=X&oi=book_result&resnum=1&ct=result

with the quote

One reason for the abundance of Polish spaces is that if X is a second
countable locally compact Hausdorff space, then X is Polish. To verify
this fact, ...

It's page 260 of Descriptive Set Theory and Dynamical Systems By M.
Foreman, A. S. Kechris, A. Louveau, B. Weiss .

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

[William Elliot]

On Fri, 22 Aug 2008, John Baez wrote:

> Are second-countable locally compact Hausdorff spaces Polish?
>
> A Polish space is a topological space that's homeomorphic to
> a separable complete metric space.
>

> 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.
>

That property is called Borel compact in honor of Borel's theorem.

The following definitions and observations are useful for
this discussion.

S Borel compact when for all bounded K, cl K compact

Borel compact S ==> S locally compact, sigma compact

S regularly bounded when for all closed bounded K, K totally bounded

S Borel compact iff S regularly bounded and complete

Thus the theorem can be stated as

S homeomorphic Borel compact metric space iff
S 2nd countable locally compact Hausdorff

Since Borel compact spaces are sigma compact and because
compact metric spaces are separable, Borel compact spaces
are separable, hence 2nd countable.

Conversely
2nd countable locally compact Hausdorff spaces are metrizable
They are also sigma compact. Why are they Borel compact?

Anyway, the theorem shows 2nd countable locally compact Hausdorff
spaces are Polish. Is the converse false?

> 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
>
> Any second-countable locally compact Hausdorff space is Polish.

----

[Henno Brandsma]

On Aug 23, 7:00 pm, William Elliot <ma...@hevanet.remove.com> wrote:
> On Fri, 22 Aug 2008, John Baez wrote:

[snip]

>
> Thus the theorem can be stated as
>
> S homeomorphic Borel compact metric space iff
>         S 2nd countable locally compact Hausdorff
>
> Since Borel compact spaces are sigma compact and because
> compact metric spaces are separable, Borel compact spaces
> are separable, hence 2nd countable.
>
> Conversely
> 2nd countable locally compact Hausdorff spaces are metrizable
> They are also sigma compact.  Why are they Borel compact?
>
> Anyway, the theorem shows 2nd countable locally compact Hausdorff
> spaces are Polish.  Is the converse false?

A Polish space need not be locally compact, or Borel compact.
E.g. the irrationals are Polish (being a G-delta in a complete metric
separable space)
but not locally compact.

Henno