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 $p_n \in U_n$. 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 am thinking that maybe the space $[0,\omega_1]$ (where $\omega_1$ is the first uncountable ordinal) might have this property.
Proof: 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. On the other hand, taking $q=\omega_1$, any sequence in $[0,\omega_1)$ is converging to a point $\alpha < \omega_1$. So, ONE cannot have a winning strategy.
What do you think? Am I right and is my proof correct?
Thank you!