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Let $X$ be a countably infinite set. While investigasting the literature on Polish spaces, I met so far only examples for compact or locally compact Polish topologies on $X$:

  • the order topology on $[0, \Gamma)$ for a countable limit ordinal $\Gamma$ (see here) - this one is locally compact but not compact (for $\Gamma = \omega$ we have the discrete topology on $\mathbb{N} = [0, \omega)$)
  • one-point-compactifications of $X \setminus \{ x \}$ for some $x \in X$ are compact Polish, e.g. $[0, \Gamma]$ (but see also the example of the countable Fort space. Wikipedia also mentions that the Fort space arises as a one-point-compactification of some discrete space).

Question 1: If $X$ carries a Polish topology is $X$ necessarily locally compact?

Probably we can do more:

Question 2: Are the above examples some kind of "prototypes" for any (locally) compact Polish topology on $X$? By prototype I mean something like any Polish topology arises from these examples by those operations for which $X$ is Polish and $X$ remains countably infinite, e.g. finite disjoint unions, finite products, countably infinite $G_\delta$-subsets.

I somehow doubt that such a simple classification holds. There are for sure examples that I have overlooked.

Steven Clontz
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yada
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1 Answers1

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If $X$ is countable and Polish, it means that every closed subset of it has an isolated point (or it would not be Baire!), and so the space is scattered, and has some countable Cantor-Bendixson rank. One can show that locally compact countable metric spaces (which are Polish, but are a strict subclass) are homeomorphic to some countable ordinal.

(This covers your first class of examples; Countable Fort space is just $\omega+1$ of course).

Q1 is no: Let $X$ be the space $(\mathbb{N} \times \mathbb{N}) \cup \{0\}$, where the metric is defined as follows: $d((m,n),0) = 2^{-m}$, $d((m,n),(m',n')) = 2^{-m} + 2^{-m'}$, except for $\forall x: d(x,x) = 0$ of course. This is Polish, but not locally compact at $0$.

Q2 is yes, as said. For locally compact such spaces we can take the one-point compactification and apply this question, or look it up in books (I'm not sure of a reference in a paper, some say it's folklore), I know it is in Semadeni's book "Banach spaces of continuous functions", which might be obscure. It's probably in other books as well (Engelking, as an exercise?), but this is where I learnt it from. Locally compact, countable metric implies homeomorphic to an ordinal, and no extra operations are required.

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Henno Brandsma
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  • What do you mean by "essentially"? Are you saying that every countable Polish space is homeomorphic to a countable ordinal (with the order topology)? I can't quite see how to get there from the Cantor-Bendixson fact. – Nate Eldredge Nov 06 '15 at 16:25
  • @NateEldredge All metric scattered spaces are homeomorphic to some countable ordinal. This is a classical result. It does require a bit more argumentation. The ordinal is determined by scattering height, and the number of points in the next to last stage, IIRC. – Henno Brandsma Nov 06 '15 at 16:37
  • I see. Do you happen to know a reference for this? – Nate Eldredge Nov 06 '15 at 16:42
  • @NateEldredge I misremembered. It's only true for locally compact countable metric spaces, which is a strictly smaller class. The proof of this is in the book "Banach spaces of continuous functions" by Semadeni (which is where I first saw its proof). – Henno Brandsma Nov 06 '15 at 18:52
  • Very nice example! It should be added to $\pi$-base or Steen and Seebach. I don't think you need the $n$-component: $\mathbb{N}\cup{\infty}$ is enough. The space is Polish since every Cauchy sequence that is not eventually constant converges to $\infty$. It is also a nice example of a metric with balls of the form ${ m } \cup M$ where $M = [n, \infty]$, $n \leq \infty$ or $M = \emptyset$. It looks like the compactification of $\mathbb{N}$ by $\infty$ extended with additional open sets of the form ${ m, \infty }$ in order to make the topology not locally compact at $\infty$. – yada Nov 06 '15 at 23:33