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!



How to show that something is not completely metrizable
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.


