# How to show that something is not completely metrizable

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!

• 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). – Nate Eldredge Nov 7 '16 at 13:11
• 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... – Tom Nov 17 '16 at 15:06

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.
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).