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Let $X$ be a topological space. The point-open game $G_{po}(X)$ is defined as folows. It is played by two players ONE and TWO. In the n'th step $(n \in \omega)$, ONE choose a finite subset $F$ of $X$, and TWO selects an open $G_n$ in $X$, $F_n \subset G_n$. ONE wins if $\bigcup \{ G_n : n \in \omega \} = X$, otherwise TWO wins.

Also:

Let $X$ be a topological space. The strict point-open game $G_{po}^s(X)$ is defined as folows. It is played by two players ONE and TWO. In the n'th step $(n \in \omega)$, ONE choose a finite subset $F$ of $X$, and TWO selects an open $G_n$ in $X$, $F_n \subset G_n$. ONE wins if $\underline {Lim} G_n = X$, otherwise TWO wins.

Theorem 1 here (page 154) states that ONE$\uparrow G_{po}^s(X)$ iff ONE$\uparrow G_{po}(X)$.

There is a notation in the proof which I am not familiar with and is not clear to me. In order to prove the non-trivial direction, the autors assume that S is a winning strategy for ONE in ONE$\uparrow G_{po}(X)$. They state then, that "a sequence $\langle \langle F_i,G_i \rangle : i < \omega \rangle$ is compatible with S, if $F_i$ is finite, $G_i$ is open, $F_i \subset G_i$ for $i < \omega$, and, for any $k < \omega$, $S(\langle G_i : i<k \rangle) \subset F_k$".

I don't understand the meaning of the notation $S(\langle G_i : i<k \rangle)$. It is, also, not explained previously in the article. Any help?

Thank you!

p.s. ONE$\uparrow G_{po}(X)$ means: ONE has a winning strategy in $G_{po}(X)$

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    $\begingroup$ It looks like $S(\langle G_i:i<k\rangle)$ is the finite set that ONE chooses in his winning strategy. By this I mean that $\langle\langle S(\langle G_i:i<k\rangle),G_k\rangle:\ k<\omega\rangle$ is a win for ONE. $\endgroup$ – A. Bellmunt May 1 '14 at 11:15
  • $\begingroup$ So, $S(\langle G_i:i<k\rangle)$ is $\bigcup_{i=1}^{k-1} F_k$? $\endgroup$ – topsi May 1 '14 at 12:36
  • $\begingroup$ I meant $\bigcup_{i=1}^{k-1} F_i$.. $\endgroup$ – topsi May 1 '14 at 12:45
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    $\begingroup$ Not necessarily, I'd say. If ONE has a winning strategy this means that he has a first play $S(\emptyset)$ which will eventually lead to a win, whatever TWO plays. Then TWO plays $G_1$ and ONE responds with $S(G_1)$ according to his winning strategy, which will lead him to victory whatever TWO does in the following play. Then TWO plays $G_2$ and ONE answers with $S(G_1,G_2)$ and so and so forth. $\endgroup$ – A. Bellmunt May 1 '14 at 13:45
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    $\begingroup$ @ShirSivroni I think it is better to give links to papers using DOI - in this case: dx.doi.org/10.1016/0166-8641(82)90065-7 In this way the link might be more stable. (DOI will be the same even if Elsevier changes the structure of their website.) $\endgroup$ – Martin Sleziak May 1 '14 at 15:09
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Let me start with two disclaimers. This is more or less what is written in A. Bellmunt's comments above. And I did not look at the rest of the proof, I only have tried to answer the meaning of the specific part the OP is asking about.


To understand the meaning of $S(\langle G_i; i<k \rangle)$ we only need to understand what is a strategy.

A strategy for the player I is a function $S$ which, depending on the moves players made so far, tells I what to play next.

So, in the first move, nothing was played so far and a strategy tells me what is $A_1$, the first set ONE should play. I.e. $S(\langle\rangle)=A_1$. (The $\langle\rangle$ there represents the list of all moves made so far.)

Now TWO makes some move $G_1$. The strategy tells ONE what to play next. I.e. it should be $S(\langle A_1,G_1\rangle)=A_2$. But since $A_1$ is uniquely determined by the chosen strategy, the authors simply omitted it and use $S(\langle G_1\rangle)=A_2$. (I am more used to the first notation, where strategy is something which assigns the next move based on the sequence of moves of both players. Clearly, some people use different convention.)

In the next step, if players so far played their moves $\langle A_1,G_1,A_2,G_2\rangle$, the next move of ONE is $S(\langle G_1,G_2\rangle)=A_3$. (Or, in the notation I am used to, $S(\langle A_1,G_1,A_2,G_2\rangle)=A_3$.) And so on.


Let us notice one more thing. If the winning strategy tells ONE to play a set $A_n$ in the $n$-th step, if he chooses to play some finite set $F_n\supseteq A_n$, this strategy works for him, too. (Since his goal is to make the set $\liminf G_n$ as big as possible.)


So now the sentence you have copied from the article simply says this: We will call a sequence $\langle F_i,G_i; i<\omega\rangle$ compatible with the strategy $S$, if this sequence represents a possible run of the game, where player ONE plays exactly according to the strategy $S$. (With the minor difference, he is also allowed to choose bigger finite sets. This is why they write $S(\langle G_i; i<k\rangle)\subset F_k$ and not $S(\langle G_i; i<k\rangle)=F_k$.)

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  • $\begingroup$ I see now. fully understood. Thank you for your detailed answer!! $\endgroup$ – topsi May 2 '14 at 9:22

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