Can we construct from $[0,\omega_1)$ a space which is strictly-Frechet with no winning strategy in $G_{np}(q,E)$?

Question

I have asked in here a question which tured out to make no sense. I think I have found the confusion and would like to try and rephrase my question:

Let $E$ be a topological space, $q \in E$.

The neighbourhood point game $G_{np}(q,E)$, is defined as follows. It is played by two players, ONE and TWO.In the n's step $n \in \omega$, ONE chooses a neighbourhood $U_n$ of $q$, and TWO selects a point $q_n \in U_n$. ONE wins if the sequence $q_n$ converges to $q$ otherwise two wins.

Also:

E is strictly Frechet at $q$, if $A_n \subset E$, $q \in \overline {A_n}$,implies the existence of a sequence $q_n \in A_n$ converging to $q$.

I am trying to find an example of a topological space which is strictly-Frechet, but, in which, ONE does not have a winning strategy in the game $G_{np}(q,E)$.

I was thinking to start with the space $[0,\omega_1)$ (where $\omega_1$ is the first uncountable ordinal). As was explained to me. This space is indeed Frechet-Urysohn. I think it is also strictly Frechet-Urysohn because: If $A_n \subset E$, $q \in \overline {A_n}$, then $\overline {\bigcup A_n}$ is a countable space. So $E=[0,\omega_1)$ is strictly Frechet.

My question is: can I expand $[0,\omega_1)$ to a tree T, which will be strictly-Frechet and in which ONE will not have a winning strategy?

Thank you!

Remark: Feel free to remove any tag I have added which is not related to the subject. Thanks!

Answer 1

An example (due to A. Hajnal and I. Juhász) of a strictly Fréchet space $X$ for which ONE does not have a winning strategy in some game $G_{\text{np}} ( x,X )$ is described in

J. Gerlits & Zs. Nagy, Some properties of $C(X)$, I, Top. Appl. vol.14 (1982), no.2, pp.151–161, MR0667661, link (not free access)

It is the one-point compactification of an Aronszajn tree with the tree topology.

Recall that given a tree $\langle T , \leq_T \rangle$, the tree topology on $T$ is generated by the basis consisting of sets of the form

• $( s , t ] = \{ x \in Y : s <_T x \leq_T t \}$ for $s <_T t$ in $T$; and
• $( \leftarrow , t ] = \{ x \in T : x \leq_T t \}$ for $t \in T$.

Recall, also, that a tree $\langle T , \leq_T \rangle$ is Aronszajn if it has height $\omega_1$, no uncountable (cofinal) branches, and all levels are countable.

Note that if a tree $\langle T , \leq_T \rangle$ is given the tree-topology, then if $F \subseteq T$ is compact, then $F$ cannot include an infinite set of pairwise incomparable nodes of $T$. If follows that if $X = T \cup \{ \infty \}$ is the one-point compactification of $T$, then the sets of the form $$U ( s_1 , \ldots , s_n ) = \{ \infty \} \cup {\textstyle \bigcap_{i \leq n}} \{ x \in T : x \not\leq_T s_i \}$$ forms a neighbourhood basis at $\infty$.

So let $\langle T , \leq_T \rangle$ be an Aronszajn tree, and let $X = \{ \infty \} \cup T$ be the one-point compactification of $T$ with the tree topology.

Claim 1. ONE has no winning strategy for the game $G_{\text{np}} ( \infty , X )$.

proof. Let $\sigma$ be a strategy for ONE in this game. Without loss of generality we may assume that all of the moves suggested by $\sigma$ are basic open sets of the form above. Note that $\sigma$ then takes as input a finite sequence $\langle x_1 , \ldots , x_\ell \rangle$ from $X$, and outputs some basic open neighbourhood of $\infty$: $\sigma ( x_1 , \ldots , x_\ell ) = U ( s_1 , \ldots , s_k )$. We may then view $\sigma$ as a function $X^{<\omega} \to [ T ]^{<\omega}$ ($\hat{\sigma} : \langle x_1 , \ldots , x_\ell \rangle \mapsto \{ s_1 , \ldots , s_k \}$). Since the levels of $T$ are countable, it follows that there is a countable limit ordinal $\delta$ such that if $\langle x_1 , \ldots , x_\ell \rangle$ is a finite sequence of nodes of $T$ of height $< \delta$, then each node in $\hat{\sigma} ( x_1 , \ldots , x_\ell )$ is also of height $< \delta$.

Picking some $t \in T$ on the $\delta$th level, it follows that $t \in \sigma ( x_1 , \ldots , x_\ell )$ whenever $\langle x_1 , \ldots , x_\ell \rangle$ is a finite sequence of nodes of height $< \delta$. We then consider a strategy for TWO as follows: Given any play $U_\ell$ by ONE according to the strategy $\sigma$, TWO plays some some $s_\ell \in U_\ell$ which is strictly below $t$. It follows that $U(t)$ is a neighbourhood of $\infty$ which is disjoint from the sequence $\langle s_\ell \rangle_{\ell < \omega}$.$\;\;\Box$

Claim 2. $X$ is strictly Fréchet.

proof. Note that $\chi ( s , X ) \leq \aleph_0$ for each $s \in T$, and therefore $X$ is "strictly Fréchet at $s$" for each $s \in T$. So it remains to show that $X$ is "strictly Fréchet at $\infty$".

Note that if $A \subseteq T$ and $\infty \in \overline{A}$, then one of the following must hold:

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

Suppose that for each $n \in \omega$ we have an $A_n \subseteq T$ such that $\infty \in \overline{A_n}$. Without loss of generality, we may assume that each $A_n$ is either a countably infinite antichain in $T$, or is an "initial segment" of $T$ with no upper bound. Note that each $A_n$ is countable, and therefore so is $Y=\{ \infty \} \cup \bigcup_{n \in \omega} A_n$. Given any $s \in T$, note the following:

• If $t \in T \cap Y$ is such that $s \leq_T t$, then $U(t) \cap Y \subseteq U(s) \cap Y$.
• If there is no such $t \in T \cap Y$, then $U(s) \cap Y = U(u) \cap Y$ for all $u \geq_T t$.

It follows from this that $\chi ( Y , \infty ) \leq \aleph_0$, and so $Y$ is "strictly Fréchet at $\infty$". Thus there is a sequence $\langle s_n \rangle_{n \in \omega}$ such that $s_n \in A_n$ for each $n$, and the sequence converges (in $Y$) to $\infty$. But, of course, the sequence will also converge in $X$ to $\infty$.$\;\;\Box$

Answer 2

If $X$ is first countable at $q$, then ONE wins the neighbourhood point game $G_{np}(q,X)$. Just pick a decreasing countable neighbourhood base $U_n$ (so $U_{k+1} \subset U_k$ for all $k$) at $q$, and play these at the $n$-th move.

Also, first countable spaces are strictly Fréchet, quite similarly.

These notions like Fréchet, strictly Fréchet, sequential, ONE wins in $G_{np}(q,X)$, etc. are all meant to distinguish differences between non-first countable spaces, in a way.

So this space $[0,\omega_1)$ cannot be used as such: it is first countable, so ONE one wins all neighbourhood point games.