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I have a Polish space $X$ and a subset $A \subset X$.

I know that $A$ is completely metrizable (in its induced topology) if and only if $A$ is a $G_\delta$-set in $X$.

This means: If I want to show that $A$ is completly metrizable then it suffices to find a sequence $U_1,U_2,\ldots$ of open sets such that $A=\bigcap_j U_j$.

But what if I want to show that $A$ is not completely metrizable? Are there any criteria I can use? The only criterium I know is that if $A$ is not Baire then it is not completely metrizable (This shows that $\mathbb Q$ is not $G_\delta$ in $\mathbb R$) but this does not help if my set $A$ is a Baire space. Also there may be cases where it is not easy to decide wether $A$ is Baire.

I hope that something can be said in this generality... Thanks for your help in advance!

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  • $\begingroup$ Is $A$ at least Borel? If you think it might not be Borel, then there is a whole different set of possible techniques to show that (non-measurable, has the wrong cardinality, etc). $\endgroup$
    – Nate Eldredge
    Commented Nov 7, 2016 at 13:11
  • $\begingroup$ I know that my set has full cardinality and I would be very surprised if it was not Borel, but to be honest, I don't know that yet... $\endgroup$
    – Tom
    Commented Nov 17, 2016 at 15:06

2 Answers 2

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There is an equivalent condition for complete metrizability in terms of a game known as the Choquet game. The game is described, for example, in the book Classical Descriptive Set Theory by Kechris, who calls it the strong Choquet game.

The game goes like this. The rounds are labeled with natural numbers. On round $i$, the first player chooses an open set $U_i$ and a point $x_i \in U_i$. The second player then chooses an open set $V_i$ with $x_i \in V_i \subseteq U_i$. At the second and subsequent rounds, the first player must additionally ensure that $U_{i+1} \subseteq V_i$, so the open sets that are chosen are nested $U_1 \supseteq V_1 \supseteq U_2 \supseteq V_2 \supseteq \cdots$. The second player wins if $\bigcap_{i\in\mathbb{N}} U_{i} \not = \emptyset$, which is equivalent to $\bigcap_{i\in\mathbb{N}} V_{i} \not = \emptyset$.

It is an easy exercise to prove that if $(A,d)$ is completely metrizable then the second player has a winning strategy. Choquet proved the converse: a space is completely metrizable if and only if the second player has a winning strategy in the game for that space. Only the easy half of the equivalence is needed for our current purpose.

So, if you can prove the second player does not have a winning strategy for a space, that is sufficient to disprove complete metrizability. In particular, one possible method in the current situation is to attempt to prove that the first player has a winning strategy in the game.

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  • $\begingroup$ I liked both answers but this one seems more promising. Plus I like infinite games :-). Therefore I accepted it. $\endgroup$
    – Tom
    Commented Nov 17, 2016 at 15:05
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A classical trick is to find a closed subset of $A$ which is not a Baire space: then $A$ is not completely metrizable. Note that this is not an equivalent condition, but a Borel set which is such that all closed subsets are Baire must actually be $G_\delta$ (see Hurewicz's theorem in the book of Kechris on descriptive set theory).

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