What can we say about the one point compactification of a Suslin tree?

Question

As a continuation to this question:

The Alexandroff one point compactification of $(X,\tau)$, is a space $X \cup \{a\}$, where open neighborhoods of $X$ are $\{ U : U \in \tau\} \cup \{ V \cup \{a\}: V \in \tau $ and $V^C$ is closed and compact in $ X \}$

A tree $T$ is a Suslin tree if: 1. The height of $T$ is $\omega_1$. 2. Every branch in $T$ is at most countable. 3. Every antichain in $T$ is at most countable.

Let $X=T \cup \{q\}$ be the Alexandroff one point compactification of a Suslin tree. I am trying to figure out whether $X$ is Frechet-Urysohn, and whether ONE wins in $G_{np}(q,X)$.

Any ideas or directions?

Thank you!

Answer 1

From this answer we have that if $X = \{ \infty \} \cup T$ is the one-point compactification of an Aronszajn tree $T$ with the tree topology, then the sets of the form $$U ( s_1 , \ldots , s_n ) = \{ \infty \} \cup {\textstyle \bigcap_{i \leq n}} \{ x \in T : x \not\leq_T s_i \}$$ where $s_1 , \ldots , s_n \in T$ form a neighbourhood basis at $\infty$.

The basic idea for show that $X$ is Fréchet-Urysohn is to notice first (and Henno Brandsma has) that for each $s \in T$ we have $\chi ( X , s ) \leq \aleph_0$, and so $X$ is "Fréchet-Urysohn at $s$."

Given $A \subseteq T$, it is relatively easy to show that $\infty \in \overline{A}$ iff either

1. $A$ includes an infinite family of pairwise incomparable nodes of $T$; or
2. $A$ includes an infinite chain with no upper bound in $T$.

(If $U = U(s_1 ,\ldots , s_n )$ is a basic open neighbourhood of $\infty$ and $A$ satisfies either (1) or (2), then there must be points of the antichain/chain which are not below any $s_i$. If $A$ fails to satisfy either of these conditions, then $A$ is the union of finitely many chains, each of which has an upper bound in $T$, and so $U(s_1 , \ldots , s_n)$ — where the $s_i$ are such chosen upper bounds — is an open neighbourhood of $\infty$ disjoint from $A$.)

In each of the above cases you basically do the obvious thing to construct a sequence in $A$ converging to $\infty$.

(That there is no winning strategy for ONE in $G_\text{np} ( \infty , X )$ is in the linked answer.)

Answer 2

The fact that $X$ is Fréchet-Urysohn is "folklore" according to this old paper, we only need $T$ to be Aronszajn for this. It's probably not too hard, as $T$ is already first countable.

And neither player has a winning strategy in $G_{np}(X,q)$ according to this nice survey paper by the author who introduced this game in the first place. It refers the proof to an older paper by Gerlits and Nagy, that might not be available online.

[added] Also see this answer which had exactly the same idea, but actually wrote out proofs!