Games Topologists Play

54th Spring Topology and Dynamics Conference Talk Math With Your Friends!

2020 March

Steven Clontz
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.

An Introductory \(\omega\)-length Game

The Cantor Game

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 \(P\win G\text{.}\)

A game-theoretic proof of a well-known result


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


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


\(\mathbb R\) is uncountable.

Proof of the theorem


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.

Selection Games


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.

Topological Examples

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 \(x\text{.}\)

Limited Information

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).

Spectrum of properties


  • 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 \(P1\not\prewin G_1(\mc K,\mc O)\text{.}\)

Dual Selection Games

The Point-Open Game

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)

Another example

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!

The Big Theorem

The importance of reflection

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