A non-first-countable topological space in which ONE has a winning strategy in $G_{np}(q,E)$ .

Question

It is mentioned in this article, that, the one point compactification of an uncountable discrete space, is A non-first-countable topological space in which ONE has a winning strategy in $G_{np}(q,E)$. I have tried to use the Alexandroff one point compactification of the Mrowka space:

Mrówka’s space $\Psi$: Subsets of $\omega$ are said to be almost disjoint if their intersection is finite. Let $\mathscr{A}$ be a maximal almost disjoint family of subsets of $\omega$, and let $\Psi=\omega\cup\mathscr{A}$. Points of $\omega$ are isolated. Basic open nbhds of $A\in\mathscr{A}$ are sets of the form $\{A\}\cup(A\setminus F)$, where $F$ is any finite subset of $A$. $\Psi$ is not even countably compact, since $\mathscr{A}$ is an infinite (indeed uncountable) closed, discrete set in $\Psi$.

Claim: $X=\Psi \cup \{ \infty \}$ is a non-first-countable topological space in which ONE has a winning strategy in $G_{np}(q,E)$ .

Proof:It is obvious that $X$ is not first countabe, since every point in $\mathscr{A}$, has an uncountable local base. We will show now, that, it satisfies $G_{np}(q,E)$.

Let $q \in \overline A$. If $q \in \omega$, then it is a discrete point. So, suppose $q$ is not in $\omega$. If $q \neq \infty$, then, every open neighbothood of $q$, is a cofinite subset of $q$. Since $q \in \mathscr{A}$ contains only points from $\omega$, That means that, every infinite sequence of points $\{q_n\} \subset q$ from $\omega$ converges to $q$.

If $q = \infty$, Then, every open neighbourhood of $q$, is a complement of a compact set in $X$. Again, sinse an infinite set of points from $\omega$ or from $\mathscr{A}$ is not compact, every sequence of points, that will be picked by TWO will converge to $p$.

What do you think? Is my proof ok?

Thank you!

Answer 1

Suppose that $\mathscr{B} \subseteq \mathscr{A}$, and $B \subseteq \omega$. It is not too difficult to show that $\mathscr{B} \cup B$ is a compact subset of $\Psi$ iff $\mathscr{B}$ is finite, and $B \setminus \bigcup \mathscr{B}$ is finite. From this it follows that if we make the very mild assumption that $\omega = \bigcup \mathscr{A}$, then the sets of the form $$\{ \infty \} \cup ( \Psi \setminus {\textstyle \bigcup_{i \leq n}} ( \{ A_i \} \cup A_i )) = X \setminus ( {\textstyle \bigcup_{i \leq n}} ( \{ A_i \} \cup A_i ) ), \tag{$\star$}$$ where the $A_i$ belong to $\mathscr{A}$, form a neighbourhood basis for $\infty$ in the one-point compactification $X = \Psi \cup \{ \infty \}$. In the sequel I will denote by $U(A_0 , \ldots , A_n)$ the set described in ($\star$).

ONE may not have a winning strategy for $G_{\text{np}} ( \infty , X )$. In fact, I'll describe a mad family $\mathscr{A}$ in which TWO has a winning strategy in this game. We will assume that ONE always plays a basic open neighbourhood of $\infty$ of the kind described in ($\star$).

Consider the full binary tree of height $\omega$: $C = 2^{<\omega}$, and let $\mathscr{A}$ be a mad family of subsets of $C$ which includes every branch through $C$. It follows that every set in $\mathscr{A}$ is either branch through $C$, or includes at most finitely many nodes from any branch.

Before play begins, TWO takes $x_{-1}$ to be the root of $C$. TWO's basic strategy is to play $x_{n+1}$ which properly extends $x_n$. (In this way, TWO's moves will constitute an increasing chain in $C$, which will have the unique branch containing each $x_i$ as its limit in $X$.)

Suppose that ONE's $n$th move is $U(B_1, \ldots , B_k, A_1 , \ldots , A_\ell)$ where each each $B_i$ is a branch through $C$, and each $A_i$ is a non-branch set in $\mathscr{A}$. Since there are uncountably many branches in $C$ through $x_n$, TWO picks some branch $B \in \mathscr{A}$ through $x_n$ which is not among the $B_i$. Then $B \cap A_i$ is finite for each $i \leq \ell$, and so TWO can pick some $x_{n+1}$ in $B$ extending $x_n$ which is not in any $A_i$.

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Addendum

In fact, given any mad family $\mathscr{A}$ on $\omega$ TWO has a winning strategy for the game $G_{\text{np}} ( \infty , X )$. This is because no sequence of points in $\omega$ can converge to $\infty$ in $X$.

Suppose that $\langle n_i \rangle_{i \in \omega}$ is a one-to-one sequence in $\omega$. This means that $B = \{ n_i : i \in \omega \}$ is infinite, and so there is an $A \in \mathscr{A}$ which has infinite intersection with $B$. But then the open neighbourhood $U(A)$ of $\infty$ cannot include a tail of the sequence, and so the sequence cannot converge to $\infty$.

Since open neighbourhoods of $\infty$ must include infinitely many natural numbers, it follows that as long as TWO plays natural numbers, the sequence constructed cannot converge to $\infty$.

Answer 2

I think that if we take, $\mathbb R$ with the discrete topology and $X = \mathbb R \{ \infty \}$ to be the Alexandroff one point compactification of $\mathbb R$, we get a space which is not first counable in which ONE has a winning strategy in the game: $G_{np}(\infty,X)$.

Proof: Suppose we take $\mathbb R$ with the discrete topology and add the point $\infty$. Then, every open neighborhood of $\infty$ is the complement of a finite set. So, take $\mathcal U$ a countable set of open sets $U_n=\mathbb R \setminus A_n$ where $A_n$ is finite for each $n$. So, for every $x \notin \bigcup A_n$, $(\mathbb R \setminus \{ x \})$ is an open neignborhood of $\infty$, and for each $n \in \mathbb N$ $(\mathbb R \setminus \{ x \}) \nsubseteq U_n$. So,to me, it seems that $\mathbb R \cup \{ \infty \}$ is not first countable.

To show that ONE does not have a winning strategy, I think that it is enough to point out that, every infinite sequence of points from $\mathbb R$ is converging to $\infty$.

What do you think? am I right?

Thank you!