Lemma 1 - Banach-Mazur Game - Dan Ma's topology Blog

Question

I have been given a presentation for a course based on this particular post in Dan Ma's topology blog. The post itself is very clear and educational, but there is one proof that I'm not sure why it is complete.

Lemma 1 has a rather standard application of Zorn's lemma, but I feel that I'm missing something in the second paragraph of the proof.

Lemma 1: Let $X$ be a space, Let $O \subset X$ be a nonempty open set. Let $\tau$ be the set of all nonempty open subsets of $O$. Let $f: \tau \longrightarrow \tau$ be a function such that for each $V \in \tau$, $f(V) \subset V$. Then, there exists a disjoint collection $\mathcal{U}$ consisting of elements of $f(\tau)$ such that $\bigcup \mathcal{U}$ is dense in $O$.

The proof begins by explaining the trivial case of $O$ having just one point, then if I understand correctly everything holds in a rather obvious way, namely: the property of $f$ gives me the same set as an image and clearly this is dense in $O$.

Then, we assume $O$ has two points and the author shows that $\mathcal{P}$ (the set of all collections $\mathcal{F}$ such that $\mathcal{F}$ is a disjoint collection of elements of $f(\tau)$) is non-empty. For this, two disjoint open subsets of $O$ are taken based on the assumption that $O$ has at least two points.

Why must these subsets be disjoint? Isn't the author assuming anything in addition?

I underestand that by applying $f$ later those sets become exactly of the form of an element of $\mathcal{P}$ which is the goal of that particular part of the proof. I thought we could possibly assume they are disjoint and otherwise apply $f$ again and again until they are but no assumption really guarantees this process would stop.

Answer 1

You are correct: unless the author is assuming that $X$ is Hausdorff, $O$ need not contain two disjoint non-empty open sets. However, the argument can easily be patched. If $O$ does not contain two disjoint non-empty open sets, let $\mathscr{U}=\{\tau(O)\}$; every non-empty open subset of $O$ meets $\tau(O)$, so $\bigcup\mathscr{U}=\tau(O)$ is dense in $O$, and we’re done right away.

Answer 2

In the introduction, before the heading Banach-Mazur Game, he says that all spaces are assumed to be Hausdorff.

And then if $O$ has at least two points, say $p \neq q$, we can find disjoint open sets $U_p$ and $U_q$ around them and then $\{U_p \cap O, U_q \cap O\}$ is a starting point, one of the members of your poset of disjoint open subfamilies of $O$ etc.