\(\newcommand{\win}{\uparrow}
\newcommand{\mc}{\mathcal}
\newcommand{\markwin}{\win_{mark}}
\newcommand{\prewin}{\win_{pre}}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
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\)

Steven Clontz |
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University of South Alabama |

Many topological properties may be characterized in terms of who can guarantee victory (and how much memory they require) in certain two-player, infinite-length games. To illustrate this, we will explore a novel game-theoretic proof of the uncountability of the real numbers, and then discuss the class of "selection games" used throughout general topology.

Let \(W\subseteq\mathbb R\text{.}\) Then \(CG(W)\) denotes the following game played by players \(P1\) and \(P2\text{.}\)

During round \(0\text{,}\) \(P1\) chooses some real number \(a_0\text{,}\) followed by \(P2\) choosing \(b_0\) such that \(a_0\lt b_0\text{.}\)

During round \(n+1\text{,}\) \(P1\) chooses \(a_{n+1}\in(a_n,b_n)\text{,}\) followed by \(P2\) choosing \(b_{n+1}\in(a_{n+1},b_n)\text{.}\)

After \(\omega=\{0,1,2,\dots\}\)-many rounds, \(P1\) wins this game in the case that \(\displaystyle\lim_{n\to\infty}a_n\in W\text{.}\)

(This game seems to be introduced by Grossman and Turett in the problems section of Mathematics Magazine Vol. 71, with some further analysis done by Matt Baker in a later volume.)

It'd take some time to complete this kind of \(\omega\)-length game.

However, each player in such a game can define a strategy \(\sigma:M^{\lt\omega}\to M\) that designates their moves in response to their opponent's actions...

\begin{equation*}
\underbrace{\sigma}_{\text{player strategy}} :
\underbrace{M^{\lt\omega}}_{\text{finite sequence of opponent's moves}} \to
\underbrace{M}_{\text{player's next move}}
\end{equation*}

Typically, we want to prove something about the existence of winning strategies that defeat any counter-strategy chosen by the opponent.

If player \(P\) has a winning strategy for game \(G\text{,}\) we write \(P\win G\text{.}\)

\(P1\win CG(\mathbb R)\text{.}\)

If \(W\) is countable, then \(P2\win CG(W)\text{.}\)

\(\mathbb R\) is uncountable.

If \(W\) is countable, then \(P2\win CG(W)\text{.}\)

Let \(W=\{w_n:n\in\omega\}\text{.}\) Then if any element of \(W\) is legal to play during \(P2\)'s turn, their strategy is to play \(w_n\text{,}\) where \(n\) is the minimal ordinal such that \(w_n\) is legal to play. Otherwise, \(P2\) chooses arbitrarily.

At the end of the game, each element \(w_N\in W\) was either played before round \(N\text{,}\) or was illegal to be played in round \(N\text{.}\) Either way, \(w_N\not=\lim_{n\to\infty} a_n\text{,}\) so \(P2\)'s strategy is winning.

Given two sets \(\mc A,\mc B\text{,}\) the selection game \(G_1(\mc A,\mc B)\) is played as follows:

\(P2\) wins this game provided \(\{b_n:n\lt\omega\}\in\mc B\text{.}\)

\(G_{fin}(\mc A,\mc B)\) is similar, except \(P2\) can make finitely-many selections each round.

Let \(X\) be some topological space with a particular point \(x\text{.}\)

- \(G_*(\mc O,\mc O)\text{:}\) P1 chooses open covers of \(X\text{,}\) P2 tries to build their own open cover.
- \(G_*(\mc D,\Gamma_x)\text{:}\) P1 chooses dense subsets of \(X\text{,}\) P2 tries to build a sequence converging to \(x\text{.}\)

A strategy that can ignore all but the most recent move of the opponent (but might use the current round #) is said to be Markov.

A strategy that can ignore all moves of the opponent (but probably uses the current round #) is said to be predetermined.

For \(X=\mathbb R\) with the usual topology:

- \(P1\prewin G_1(\mc O,\mc O)\) but \(P2\markwin G_{fin}(\mc O,\mc O)\text{.}\)
- \(P2\markwin G_1(\mc D,\Gamma_0)\) (so \(P2\markwin G_{fin}(\mc D,\Gamma_0)\) also).

Notes:

- For second-countable spaces, \(P2\markwin G_*(\mc O,\mc O)\Leftrightarrow P2\win G_*(\mc O,\mc O)\text{.}\)
- Using the Continuum Hypothesis and the Axiom of Choice, one may construct Lusin sets \(X\subseteq\mathbb R\text{.}\) Any Lusin set satisfies both \(P1\not\win G_*(\mc O,\mc O)\) and \(P2\not\win G_*(\mc O,\mc O)\text{.}\)
- While \(P1\win G_*(\mc O,\mc O)\Leftrightarrow P1\prewin G_*(\mc O,\mc O)\) (Galvin 1979, Pawlikowski 1994), F. Jordan (2020) found a space where \(P1\win G_1(\mc K,\mc O)\) but \(P1\not\prewin G_1(\mc K,\mc O)\text{.}\)

Let \(\mc O\) be the set of open covers, and \(\mc P\) be the set of point-bases. Let \(\neg\) denote the complement of a set.

Then the point-open game \(G_1(\mc P,\neg\mc O)\) essentially has \(P1\) choosing points, and \(P2\) selecting a neighborhood for each point. Then \(P1\) wins if \(P2\)'s sets form a cover.

Compare with \(G_1(\mc O,\mc O)\text{:}\) \(P1\) chooses open covers, \(P2\) selects an open set from the cover. Here \(P2\) wants to build the open cover.

Galvin (1978) demonstrated the following:

- \(\displaystyle P1\win G_1(\mc O,\mc O)\Leftrightarrow P2\win G_1(\mc P,\neg\mc O)\)
- \(\displaystyle P2\win G_1(\mc O,\mc O)\Leftrightarrow P1\win G_1(\mc P,\neg\mc O)\)

The following also holds:

- \(\displaystyle P1\prewin G_1(\mc O,\mc O)\Leftrightarrow P2\markwin G_1(\mc P,\neg\mc O)\)
- \(\displaystyle P2\markwin G_1(\mc O,\mc O)\Leftrightarrow P1\prewin G_1(\mc P,\neg\mc O)\)

All together, we say that these games are dual.

It turns out that the key to this proof comes from a simple observation:

“For each \(x\in X\text{,}\) pick any neighborhood \(U_x\text{.}\) Then \(\{U_x:x\in X\}\) is an open cover.”

Put another way:

“For each \(B_x\in\mc P\text{,}\) pick any \(U_x\in B_x\text{.}\) Then \(\{U_x:B_x\in\mc P\}\in\mc O\text{.}\)”

So let's say \(\mc R\) reflects* \(\mc A\) given the following:

“For each \(R\in\mc R\text{,}\) pick any \(b_R\in R\text{.}\) Then \(\{b_R:R\in \mc R\}\in\mc A\text{.}\)” (*slight simplification)

So \(\mc P\) refelcts \(\mc O\text{,}\) and \(G_1(\mc O,\mc O)\) and \(G_1(\mc P,\neg\mc O)\) are dual.

Likewise, let \(\mc T\) be the topology of a space, and choose a point from each open set. That collection is a dense subspace, a member of \(\mc D\text{.}\)

What can we say about \(G_1(\mc D,\Gamma_x)\) and \(G_1(\mc T,\neg\Gamma_x)\text{?}\)

YES, they are dual!

If \(\mc R\) reflects \(\mc A\text{,}\) then \(G_1(\mc A,\mc B)\) and \(G_1(\mc R,\neg\mc B)\) are dual.