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This is from the paper: LEFT SEPARATED SPACES WITH POINT-COUNTABLE BASES by WILLIAM G. FLEISSNER.

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It is a little difficult for me to understand that what's the meaning of $I$ wins, and $II$ wins. Could sombody explaining more for me? Thanks for your help.

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$X$ is a topological space, and $\mathscr{B}$ is a specified base for the topology of $X$. $G(X,\mathscr{B})$ is a game played by two players, named I and II. Player I plays first, and after that they alternate.

Player I begins the game by choosing a set $B_0\in\mathscr{B}$. Then Player II takes his first turn: he chooses a countable set $A_0\subseteq X$. Then it’s Player I’s turn again, and he chooses another basic open set $B_1\in\mathscr{B}$. Then Player II chooses another countable set $A_1\subseteq X$, and they continue in this fashion for $\omega$ turns. At the end of the game Player I has chosen an infinite sequence $\langle B_n:n\in\omega\rangle$ of basic open sets in $\mathscr{B}$, and Player II has chosen an infinite sequence $\langle A_n:n\in\omega\rangle$ of countable subsets of the space $X$.

Let $\mathscr{U}=\{B_n:n\in\omega\}$ be the collection of sets chosen by Player I, and let $A=\bigcup_{n\in\omega}A_n$, the union of the countable subsets of $X$ chosen by Player II; note that $A$ is also a countable subset of $X$. Player I wins if there are a family $\mathscr{V}\subseteq\mathscr{U}$ and a point $x\in X\setminus A$ such that $\mathscr{V}$ is a nbhd base at the point $x$. Player II wins if Player I does not win.

To see just what it means for Player II to win, suppose that Player I does not win. Then for each $x\in X\setminus A$, $\mathscr{U}$ does not contain a nbhd base at $x$. This means that $x$ has an open nbhd $N_x$ such that if $x\in B_k\in\mathscr{U}$, then $U\nsubseteq N_x$. Thus, Player II wins if and only if for each $x\in X\setminus A$ there is an open nbhd $N_x$ of $x$ such that $x\in B_k\in\mathscr{U}$, then $U\nsubseteq N_x$.

The basic idea is that Player I is trying to choose a big enough family $\mathscr{U}$ of open sets to guarantee that $\mathscr{U}$ includes a nbhd base at at least one of the points that Player II does not pick. Player II is trying to choose all of the points at which Player I’s family $\mathscr{U}$ will turn out to be a base, so that $\mathscr{U}$ won’t be a base at any point of $X\setminus A$.

A strategy for Player I is a specific first move $B_0\in\mathscr{B}$ and a rule that tells him what basic open set to choose as $B_n$ based on the previous moves in the game. It’s a winning strategy if Player I is guaranteed to win when he plays according that rule. Similarly, a strategy for Player II is a rule that tells him what countable subset of $X$ to choose on turn $n$ based on the previous moves in the game, and it’s a winning strategy if Player II is guaranteed to win when he plays according to that rule.

Bill spells this out in detail what a strategy for Player II is. It’s a function $s$ that takes as input a finite sequence of members of $\mathscr{B}$ and produces as output a countable subset of $X$. The idea is that if Player I’s first $n+1$ moves are the sequence $\langle B_0,\dots,B_n\rangle$, then $s(B_0,\dots,B_n)$ is a countable subset of $X$, and if Player II follows the strategy $s$, he will choose $A_n=s(B_0,\dots,B_n)$ on his move $n$. Thus, $s$ is simply a function from ${}^{<\omega}\mathscr{B}$, the set of finite sequences of members of $\mathscr{B}$, to $[X]^{\le\omega}$, the family of all countable subsets of $X$.

The strategy $s$ is a winning strategy if it ensures that Player II chooses every point of $X$ at which Player I’s collection $\{B_n:n\in\omega\}$ contains a nbhd base. This is the case if and only if $s$ has the following property:

If $x\in X$, and $\{B_n:n\in\omega\}$ contains a nbhd base at $x$, then Player II has at some point chosen $x$, i.e., $$x\in\bigcup_{n\in\omega}A_n=\bigcup_{n\in\omega}s(B_0,\dots,B_n)\;.$$

Bill then gives the real line as an example of a space in which Player I has a winning strategy no matter what base $\mathscr{B}$ is used for the game: he simply makes sure to choose a family $\{B_n:n\in\omega\}$ that is itself a base for the topology of $\Bbb R$. Then $\{B_n:n\in\omega\}$ contains a nbhd base at every $x\in\Bbb R$. On the other hand, Player II can choose only countably many points altogether, so there is bound to be a point $x\in\Bbb R\setminus A$, and Player I’s collection $\{B_n:n\in\omega\}$ contains a nbhd base at $x$, so Player I wins.

On the other hand, Player II has a very simple winning strategy if $X$ is an uncountable discrete space. In that case $\{B_n:n\in\omega\}$ contains a nbhd base at a point $x\in X$ if and only if $\{x\}=B_n$ for some $n\in\omega$. Player II’s winning strategy is as follows: if Player I chooses $B_n=\{x\}$ at turn $n$, Player II chooses $A_n=\{x\}$, and otherwise Player II picks any countable subset of $X$. At the end of the game $\{B_n:n\in\omega\}$ contains a local base at a point $x$ if and only if some $B_n=\{x\}$, and in that case $x\in A_n\subseteq A$. Thus, Player I did not manage to build a nbhd base at any point that Player II didn’t pick, so Player I does loses, and Player II wins. (Note that this winning strategy for Player II is very simple: it depends only on Player I’s most recent move, not on the entire history of previous moves. It’s an example of what Bill calls a winning tactic.)

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Thanks, Brain. Your explanination is very clearly. –  Simple Mar 22 '13 at 8:31
    
@Simple: You’re welcome; I’m glad that it helped. –  Brian M. Scott Mar 22 '13 at 8:32
    
You are a hero in math. Could you give me a picture of yourself? –  Simple Mar 22 '13 at 8:34
    
@Simple: No hero: I just like to teach. I don’t have any very good pictures, but the casual shots here, here, and here are only two or three years old and maybe not too bad. –  Brian M. Scott Mar 22 '13 at 8:52
    
Thanks very much. –  Simple Mar 22 '13 at 8:56
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