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A Polish space ​is a separable completely metrizable topological space. On the wikipidia article and in the book measure theory from Bauer (§26, Example 4) is stated that any open set of a polish space is again a polish space.

In the book from Bauer one can find a proof that shows that any open set of a Polish space is homeomorphic to another Polish space. But how is it possible that a open set is again completely metrizable? For instance, the set $[0,1]$ is Polish and $(0,1)$ is an open subset which is not completely metrizable, right?

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    $\begingroup$ $(0,1)$ is homeomorphic to $\mathbb R$, which is completely metrizable. $\endgroup$ – user2345215 Nov 15 '14 at 10:29
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    $\begingroup$ A space homeomorphic to a Polish space is Polish. It's a topological definition, not a metric one. $\endgroup$ – Henno Brandsma Nov 15 '14 at 10:49
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Note the subtle difference between definitions:

The difference between completely metrizable space and complete metric space is in the words there exists at least one metric in the definition of completely metrizable space, which is not the same as there is given a metric (the latter would yield the definition of complete metric space)

(From wikipedia)

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In fact, every $G_\delta$ subset of a Polish space is Polish; and if $X$ is a Polish space, and $Y\subseteq X$ is a Polish space (with the induced topology from $X$), then we can prove that $Y$ is a $G_\delta$ subset of $X$.

You need to remember that being metrizable means that you can choose a different metric than the one you started with. It just has to induce the same topology.

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    $\begingroup$ I believe separability has nothing to do with it; i.e., a subset of a completely metrizable space is completely metrizable if and only if it's a $G_\delta$. $\endgroup$ – bof Nov 15 '14 at 10:39
  • $\begingroup$ You're right; but since the question is essentially about Polish spaces, I stayed in the confines of Polish spaces. $\endgroup$ – Asaf Karagila Nov 15 '14 at 10:44

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