Explanation for Oxtoby's proof: a nonempty topological space $X$ is Baire iff player (I) has no winning strategy in the Choquet game

Question

A nonempty topological space $X$ is a Baire space iff player I has no winning strategy in the Choquet game $G_X$.

Oxtoby's proof

I have several questions about this proof.

$(\Leftarrow)$

1. How can he assume the existence of a sequence $(G_n)$ of dense open sets?

2. Moreover, how can he assume that $\bigcap_n G_n \cap U_0 = \emptyset,$ if they're dense?

3. He takes a $V_0 \subseteq U_0$ and claims that $V_0 \cap G_0 \neq \emptyset$. But I thought we only knew that $\bigcap_n G_n \cap U_0 = \emptyset$? How can we know for sure that $G_0$ kills it?

$(\Rightarrow)$

1. What is it mean to show that $U_0$ is not "Baire"? I thought we had to show that $X$ is not Baire. I am confused about the logical implication too. If assuming player I has a winning strategy implies that $X$ is not Baire, then player I not having a winning strategy doesn't necessarily have to imply that $X$ is Baire?

2. Where he says that $\cup \mathcal {U}_p$ is dense in $U_n$, does that mean that the union of all possible responses $U_{n+1}$ is dense in $U_n$?

3. If $W_n = \cup \{U_n: (U_0, V_0, \ldots, U_n) \in S \}$ then isn't that the same as $\cup \mathcal{U}_p$ for $p = (U_0, V_0, \ldots, U_n)$? Then wouldn't we only have that $W_n$ is dense in $U_{n-1}$? How do we get that it is dense in $U_0$?

4. We claim that $\bigcap_n W_n = \emptyset.$ Otherwise, if $x \in \bigcap_n W_n$, there is unique $(U_0, V_0, U_1, V_1, \ldots ) \in [S]$ with $x \in U_n$ for each $n$, so $\bigcap_n U_n \neq \emptyset,$ a contradiction.

   My question is how do we know that $(U_0, V_0, U_1, V_1, \ldots )$ is unique? I see how the $U_i$ are unique since they are disjoint, but what about the $V_i$?

5. Let, by an application of Zorn's Lemma, $\mathcal{V}_p$ be a maximal collection of nonempty open subsets $V_n \subseteq U_n$ such that $\{V_n^*: V_n \in \mathcal{V}_p \}$ is pairwise disjoint. Put in $S$ all $ (U_0, V_0, \ldots U_,n, V_n, U_{n+1})$ with $V_n \in \mathcal {V}_p$. Then $\mathcal{U}_p = \{U_{n+1}:(V_0, \ldots, U_n, V_n, U_{n+1}) \in S \} = \{V_n^*: V_n \in \mathcal{V}_p \}$ is a family of pairwise disjoint sets and $\bigcup \mathcal {U}_p$ is dense in $U_n$ by the maximality of $\mathcal {V}_p,$ since if $\bar{V_n} \subseteq U_n$ is nonempty, open and disjoint from $\bigcup \mathcal{U}_p$, then $\mathcal {V}_p \cup \{\bar{V}_n \}$ violates the maximality of $\mathcal {V}_p$.

   I honestly have no idea what's going on here. Why distinguish between $\mathcal {V}_p$ and $\mathcal {U}_p$? What is $\bigcup \mathcal {U}_p$? Why do we need the sets to be pairwise disjoint? Why does the maximality of $\mathcal {V}_p$ imply that $\bigcup \mathcal {U}_p$ is dense and why do we need that this is dense?

Answer 1

$\Leftarrow$:

1. Rearrange the statement as follows: since $X$ is not a Baire space, there exists a sequence $(G_n)$ of dense open sets in $X$ such that $\bigcap_n G_n$ is not dense in $X$ (we are simply denying one of the equivalent definitions of Baire Spaces). Hence we have a non-empty open set $U_0$ such that $\bigcap_n G_n \cap U_0 = \emptyset$.
2. see above
3. $G_0$ is a dense open set, as for all other $G_n$.

$\Rightarrow$:

1. Every open subset of a Baire space, is a Baire space. So if we prove that a non-empty open set $U_0 \subseteq X$ is not Baire, then we are actually proving that $X$ is not a Baire space.
2. What Oxtoby is trying to do is to provide a specific play of the game in which player I plays by $\sigma$ and such that player I winning ($\sigma$ is a winning strategy by assumption, so player I wins no matter what) implies that $U_0$ is not Baire. So the tree $S\subseteq \sigma$ he's building is a proper subset of all possible plays. So, no, $\bigcup \mathcal{U}_p$ being dense in $U_n$ means that the union of all responses $U_{n+1}$ *contained in $S$ * is dense in $U_n$, and those responses are a proper subset of all possible responses $U_{n+1}$ of player I.
3. No. $W_n$ is the union of all player 1 responses $U_n$ in the n-th stage of the plays. You can see it as $$W_n = \bigcup[\bigcup_{p = (U_0,V_0,\dots,U_{n-1}) \in S} \mathcal{U}_p]$$ It is easy to show that therefore $W_n$ is dense in $W_{n-1}$, and by a simple induction, proving that therefore $W_n$ is dense in $U_0 = W_0$ for all n.
4. The sequence is unique since not only $\mathcal{U}_p$ is a disjoint collection of open sets, but, again we can show this by a simple induction, that $$\bigcup_{p = (U_0,V_0,\dots,U_{n-1}) \in S} \mathcal{U}_p$$ is still a disjoint collection of open sets in $U_0$. Intuitively player 1's response in the subsequent stages of the games contained in $S$, are smaller and smaller disjoint open "refinements" of $U_0$. Keeping this in mind, it is easy to see that for all $n$, $\exists! s_n = (U_0,V_0,\dots,U_n) \in S$ such that $x \in U_n$ and that $s_{n} \subseteq s_{n+1}$ for all $n$ (hence we are defining a unique sequence by Dependent Choice).
5. Here we are simply constructing the subtree $S\subseteq \sigma$ we are working in from the start. What do we need? A subtree $S \subseteq \sigma$ such that $\mathcal{U}_p$ with $p = (U_0,\dots,U_n)$ is a disjoint collection of open sets such that $\bigcup \mathcal{U}_p$ is dense in $U_n$ for all $p \in S$. This kind of tree exists by Zorn's Lemma. See this, in this page the author explicitate this passage by a lemma. I find it quite clearer.