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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
up vote 11 down vote accepted

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

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