# Polish space in which the interior of each compact set is empty

If anyone could give me an example of Polish space, in which the interior of each compact set is empty? I guess it is trivial, but can't find such an example.

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Did you try any countably infinite product of noncompact spaces? – user83827 Aug 14 '11 at 23:55
Try the irrational numbers (that's of course a special case of @ccc 's suggestion) – t.b. Aug 15 '11 at 0:04
@Nate Yes, thanks, it was so easy, I was looking for this in the midst of a very specific cases...I did not think that it is so large class of spaces. – dawid Aug 15 '11 at 0:14
@Theo I will sure check this. – dawid Aug 15 '11 at 0:14

Consider the irrational numbers with the standard topology. However we consider a space homeomorphic to them:

The Baire space, namely $\omega^\omega$ - all the infinite sequences of natural numbers. It is a metric space endowed with the metric:

$$d(x,y) = \begin{cases} 2^{-n} & n=\min\{k\in\omega\mid x(k)\neq y(k)\}\\ 0 & x=y\end{cases}$$

That is, we count how long is the initial segment which is the same for two sequences. If they differ there must be a least integer as above, if they do not differ they are the same point.

The basic open sets are going to be defined by finite sequences. For each $s=\langle x_1,\ldots,x_n\rangle$, a finite sequence of natural numbers, consider $O_s$ as the set of all sequences $x$ such that $x(k)=s(k)$ for all $k<n+1$, that is $s$ is a finite initial segment of $x$.

This is in fact a clopen set, and the collection of all $O_s$ form a countable base of clopen sets.

Now suppose $X\subseteq\omega^\omega$ was compact, for every $n$ let $U_n$ be the finite subcover of $X$ taken from $\{O_s\mid \operatorname{Length}(s)=n\}$.

Assume by contradiction that $X$ contains some open set, without loss of generality it is some $O_s$ which is a basic open set. Suppose $k$ is the length of $s$. Since the $k+1$ level can be covered by finitely many basic open sets, there exists some $m$ so $t=\langle s\rangle^\smallfrown\langle m\rangle$ (that is an end extension of $s$ by the value $m$) is not such that $O_t\in U_{k+1}$.

In particular for any $x$ that has $t$ as a proper initial segment, then $x\notin X$, however $s$ is an initial segment of $t$, therefore of $x$. This in contradiction that $O_s\subseteq X$.

Therefore if $X$ is compact, it does not contain any open set.

Note that this does not imply that compact sets in the Baire space are finite, but rather that for every $n\in\omega$, $\{x(n)\mid x\in X\}$ is a finite set. A very good example is the Cantor space, which is a subspace of the Baire space given by $\{0,1\}^\omega$.

A very nice theorem states that if $X$ is a Polish space, zero dimensional and every compact set has an empty interior, then $X\cong\omega^\omega$. This means that in the class of zero dimensional Polish spaces, there is only one representative (up to homeomorphism, of course), which is quite the property if you ask me.

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Thanks you for this example, I'm very grateful. – dawid Aug 15 '11 at 13:18
@dawid: No problems. – Asaf Karagila Aug 15 '11 at 13:19

Any infinite-dimensional separable Banach space.

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Indeed. Proving this boils down to showing that the closed unit ball in such a space is non-compact. This is essentially Riesz's lemma. – Mark Aug 14 '11 at 23:56