In my post on the strong measure zero game, I mentioned a game $G_1(\mathcal {N}_H, \mathcal{N}_H)$, where player one, who I will call $I$, presents player two, who I will call $II$, a subset from a group $H$, where the normal closure of the set is the full group, and $II$, chooses one element from each set presented, trying to get a set whose normal closure is again the full group. The game lasts $\omega$ innings ($\mathbb{N}= \omega$ if you prefer that notation). Essentially the idea is that $I$ gives presentations of the trivial group, and $II$ is trying to end up with a presentation of the trivial group.

As a simple example, we can consider the group of integers $(\mathbb Z,+)$. Since the group is abelian any group that normally generates the group, also generates the group. This also means that there is some linear combination of elements from any presented set that has value $1$. What player $II$ can do to win, is choose a non identity element $x_0$, where $\vert x_0 \vert$ is the least among non identity elements. If $\vert x_0 \vert=1$, player $II$ has already chosen a set that normally generates and may choose arbitrarily. Suppose that $II$ has chosen $x_0,...,x_n$, if $d:=\gcd(x_0,...,x_n)=1$ we choose an arbitrary element. If we have $d>1$, then choose an element $x_{n+1}$, so that $\gcd(d,x_{n+1})<d$. Note that we can do that, since otherwise, every element in the presented set would have $d > 1$ dividing it, hence the greatest common divisor of the set, without the identity, would be greater than or equal to $d$, contradicting the set generates the group. So we can always decrease our greatest common divisor till we reach $1$. So we have a strategy where $II$ will win every time. A similar argument can be used to show that $II$ wins in groups where every strictly ascending chain of infinite index subgroups, is finite.

It is pretty easy to find groups where $I$ wins, by using large groups, like $(\mathbb{R},+)$, but what about finitely generated groups? We will show, in this post, that $I$ has a winning strategy to win $F_2$, in fact it will be a “fixed strategy” where $I$ will play the same no matter what $II$ picks on $II$’s turns.

Our proof will use facts from small cancellation theory, in particular small cancellation groups are not trivial. I will explain a bit of the idea of small cancellation conditions, and give key definitions. All the small cancellation that we discuss can be found in Combinatorial Group Theory Lyndon and Schupp, or Geometry of Defining Relations in Groups by Ol’shanskii, or Wikipedia.

The idea of small cancellation theory is that, if you have a set of relations, satisfying a “small cancellation condition”, that is when you multiply the two different words, the length of the reduced product (remove the $xx^{-1}$) is not much shorter than the the sum of the lengths of the word. This, in a sense, gives control over what words end up representing the identity, and how words can be reduced using the relations that satisfy the small cancellation conditions.

Basic definition for small cancellation theory

The information in this section is all fairly basic stuff, you may want to skip this section, or just read the $C'(1/6)$ definition and come back if there are some words you do not know.

Basic definitions for words in free groups. related to words and free groups. A word is a string of elements from some some set called an alphabet. For example $x, xyxx^{-1},$ and $xy$ are all (not equal) words from the alphabet $\{ x,y,x^{-1},y^{-1} \}$. A word is reduced if there is no occurrence of $tt^{-1}$ or $t^{-1}t$ in the word, and cyclically reduced provided it is not a nontrivial conjugate, so $w \neq t w' t^{-1}$ and $w \neq t^{-1} w' t$. The length of a word $w$, which we call $\vert w \vert$ is the number of elements in the reduced version of the word; this is well defined. So $\vert xyxx^{-1} \vert = \vert xy \vert =2$, for example.

Definition of symmetrized presentation. Suppose $G = \langle X \mid R \rangle$. We say the set of relations $R$ is symmetrized provided it is closed under taking inverses, and taking cyclic permutations of words up to being cyclically reduced, and every word in $R$ is cyclical reduced. $G$ has a symmetrized presentation provided $R$ is symmetrized. For example

$$\begin{align*} R = &\{ab^3ab^4,b^3ab^4a^{-1}, ab^4 a ^{-1} b^{-3}, \\ & b^4a^{-1} b^{-3} a ^{-1}, a^{-1}b^{-3}a^{-1}b^{-4},... \text{inverses} \}. \end{align*}$$

Definition of $C'(1/6)$: We say that $p$ is a piece relative to $R$ provided there exists two distinct elements $r_1,r_2 \in R$ with $r_1=pc_1$ and $r_2=pc_2$, where “$=$” is word equality, so no cancellation. We way $R$ has condition $C'(1/6)$ if $r \in R$ and $r=pc$ where $p$ is a piece, then $\vert p \vert < \frac{1}{6} \vert r \vert$. There is an entirely similar definition for $C'(\lambda).$

The reason I only mention $C'(1/6)$ is that it is a strong enough small cancellation condition to guarantee a group is infinite. Note that you can have a group that has a $C'(1/6)$ presentation, but may not be given to you that way. It is an interesting fact that for $\lambda > 1/5$, every group has a $C'(\lambda)$ presentation. There are many other small cancellation conditions, $C(p)$ and $T(p)$ for example, and you can mix-and-match different conditions too, but that is really beyond this post.

Player $I$ has a winning strategy in $G_1( \mathcal{N}_{F_2},\mathcal{N}_{F_2})$

Let $F_2$ be the free group of rank $2$, on the set $\{x,y\}$. We will show that $F_2$ has a winning strategy, by showing that there is a sequence of sets, $A_0,...,A_n,...$, that normally generate, even generates, the group $F_2$, and any one element choice from each of these sets gives a set $\{a_0,...,a_n,...\}$, which will correspond to a group, $G=\langle x,y \mid a_0,...,a_n,... \rangle$, that is $C'(1/6)$, which means that $\{a_0,...,a_n,...\}$ can not normally generate $F_2$. The strategy for player $I$ will be to play $A_n$ on the $n$th turn.

There will be a couple parts where I make some estimates on inequalities, and make no attempt to make these as good as possible(the estimates I make are good enough). If you are sensitive to such disregard of sharp bounds, you will need to have your eyes covered for the rest of this proof. You have been warned.

Proposition 1. Let

$$w_n=\prod_{i=1}^{9 \cdot 100^{n+1}} xy^{100^{n+1}+i}=xy^{100^{n+1}+1}...xy^{100^{n+2}}.$$

Then we have that $A_n=\{w_{n+1},xw_{n+1},w_{n+1}y\}$ generates the group $F_2$, but any one element choice from each $A_n$ results in a set of relations for a $C'(1/6)$ group. In particular this means that for any choice of $a_i$’s, where $a_i \in A_i$, that $\langle x ,y \mid a_0,...,a_n,... \rangle$ is an infinite group since it will be $C'(1/6)$.

proof of Lemma 1. First it is clear that each $A_n$ generates $F_2$ since we can multiply $w_{n+1}^{-1}$ to cancel $w_{n+1}$ on each term to end up with $x,y$, which generates the group. Let $R_w$ be the smallest symmetrized set of words containing the word $w$, which is all the cyclically reduced, cyclic permutations of the word $w$ and their inverses. For $R_{w_n}$ all pieces are of the form $b^i a^{\pm 1} b^j$, where $-100^{n+2} \leq i,j\leq 100^{n+2}$ (note that for many $i,j$ can not actually be obtained). So the length of a pieces $p$ are bounded by

$$2 \cdot 100^{n+2}+1<3\cdot 100^{n+2},$$

and the length of each $w_n$ is bounded from below by

$$9\cdot 100^{n+1} \cdot 100^{n+1}.$$

We get that

$$\frac{ \vert p \vert }{ \vert w_n \vert } \le \frac{ 3 \cdot 100^{n+2} }{9 \cdot 100^{2n+2}}= \frac{1}{3 \cdot 100^n}$$

which for $n>0$ we have the right hand side is less than $1/6$. A very similar argument can be carried out for $R_{w_ny}$ and $R_{xw_n}$, and in fact you can use the same bounds, since our bounds are so rough. So each $R_w$, for $w \in A_n$, gives a $C'(1/6)$ presentation. Say that $m>n$ and $w \in A_m$, and $u \in A_n$, and we consider $R_w,R_u$. Note, that at “worst” the amount of cancellation between words in $R_w$ and $R_u$ will be bounded by the cancellation bounds we established above for $R_u$, since elements in $R_u$ have small enough exponents on their $y$’s that they can not completely cancel out a $y^k$ term in $R_w$. This worked for arbitrary choices. So let $a_0,...,a_n,...$ be a sequence where $a_i \in A_i$, then $R=\bigcup_{i< \omega} R_{a_i}$ is a symmetrized set of relations, and we know that between two words, the cancellation is less than $1/6$ the length of either of the words, so it gives a $C'(1/6)$ presentation, hence $\langle x,y \mid a_0,...,a_n,... \rangle$ is a $C'(1/6)$ group, and is infinite. $\square$

I came up with this game in a class I took on selection principles, and studied it as a final project, and at the time I knew very little group theory (I still know very little), and even thought all finitely generated groups had a winning strategy for player $II$. At the end of the class I only really found some conditions for player $II$ winning in finitely generated groups. It had been at the back of my mind since the class, I would pick it up and work on it for a while before setting down, multiple times. In a way, thinking about this problem and closely related things really influenced my mathematical interests, I have a feeling it will show in future posts. In the end I am actually not to sure these ideas are of interest, at least in the finitely generated group case. I suspect it will be a while before I discuss these games on groups again, but I will end this with a couple questions to ponder:

• For finitely generated groups is there a nice demarcation line between player $I$ or $II$ having a winning strategy? For groups in general?
• Are there groups where there is no winning strategy for either player? Are there finitely generated groups where there is no winning strategy?
• There are a couple of other, similar ideas to consider, one where you replace normally generate with simply generate, or replace with generate finite index subgroup, or normally generate finite index subgroup, or mix-and-match the different sets. (I have not really thought about them much, but there are probably a lot of equivalences, because once you reach a finite index subgroup, in many cases, reach the full group) Are there any interesting connections and differences among them?
• For those who have looked up the connection of these sorts of games with selection principles and Ramsey statements(see the references in the strong measure zero game post): are there interesting connection between those and the games for finitely generated groups? For countable groups? For groups? In particular what would the Ramsey statements be?
• Can you think of any cool group theoretic consequences of a group that has a winning strategy for $I$, how about for $II$?

Background

Originally, my first post was going to be an introduction to selection principles, especially those in topology, and a few other related concepts. I decided I could not do justice to the subject in one post (especially with my minimal knowledge), and have opted to present some very cool results that will, hopefully, give a taste of the ideas that come up when studying the subject.

Definition of the game $G_1( \cal A, \cal B)$. Let $\mathcal{A}$ and $\mathcal{B}$ be collections of sets, we define the game $G_1(\cal{A}, \cal{B})$ to be a game with two players, that we will denote $I$ and $II$, and the game will have $\omega$ innings, where each inning consists of a turn by $I$ and then a turn by $II$. By $\omega$ innings we mean there is an inning for each $0,1,2,3, \dots$ (if you prefer $\omega = \mathbb N$). The innings follow this pattern: player $I$ starts off, in inning $n$, by presenting a nonempty set $A_n$ from $\cal A$ which marks the end of player $I$’s turn, and then player $II$ picks an $a_n \in A_n$. At the end of the $\omega$ innings we are left with a play

$$(A_0,a_0, \dots , A_i, a_i, \dots)$$

of what happened in that game. Player $II$ wins if $\{ a_i \mid i \in \omega \} \in \cal B$ and $I$ wins otherwise.

A couple of examples to consider:

• Let $D$ consist of all dense subsets of $\mathbb R$. Consider $G_1 (D,D)$.
• Let $X$ be a countable subset of some topological space $Y$ and ${\cal{O}}_Y$ is the set of all open covers of $Y$, and ${\cal {O}}_X$ the set of covers for $X$. Consider $G_1( {\cal{O}}_Y, {\cal{O}}_X)$.
• $V$ is a finite dimensional rational vector space, and $A$ is the set of infinite vector subspaces of $V$, and $B$ is the set of sets that contain infinite rational vector subspaces of $V$. Consider $G_1(A,B)$
• Let $U$ be an ultrafilter on some set. Consider $G_1(U,U)$.
• Let $H$ be a group and ${\mathcal N}_H$ the set of subsets of $H$ that normally generate the group. Consider $G_1({\mathcal N}_H, {\mathcal N}_H)$ . There are also variation of considering sets that generate the group $H$, ${\mathcal K}_H$ and looking at $G_1({\mathcal K}_H, {\mathcal N}_H)$, $G_1({\mathcal N}_H, {\mathcal K}_H)$, $G_1({\mathcal K}_H, {\mathcal K}_H)$ (there is a very good chance I am going to write about this bullet point, and related ideas, in future posts, since it is basically the reason I started the blog).
• etc. Think of some others!

A strategy for a player is a function that takes what has been played and outputs the next move. So $s_1,s_2$ will be strategies for players $I,II$ respectively, and $s_1: \varnothing \mapsto A_0$, and $s_1: (A_0,a_0,...,A_i,a_i) \mapsto A_{i+1}$. We get a similar description for $s_2$. Basically they are instructions on a game tree that tell the player what to do next given where they are in the game tree. We call $s_i$ a winning strategy provided the cooresponding player will always win following that strategy. It is interesting that in some games there is no winning strategy for either player, and in fact the two main theorems that will be presented will give such an example(assuming Borel’s conjecture is false)

Do any of the above (bulleted)examples have winning strategies for one of the players? How about the games you came up with?

Definition of strong measure zero. A set $X \subseteq \mathbb R$ is strong measure zero if, for all sequences $( \epsilon_n \mid n \in \omega)$ of positive real numbers, then there exists a sequence of intervals $I_n$, where the length of $I_n$ is less than $\epsilon_n$ for all $n$, and $X \subseteq \cup_{n \in \omega} I_n$.

Every strong measure zero set is Lebesgue measure zero, although not every measure zero set is strong measure zero, the Cantor set provides such an example. It is also easy to see that every countable set is strong measure zero. In fact Borel’s conjecture actually conjectures that every strong measure zero set is countable. That conjecture ended up being independent of the usual axioms, ZFC, so it is consistent for there to be uncountable strong measure zero sets, and consistent for there to be only countable strong measure zero sets. Luzin sets provide examples of strong measure zero sets that are not countable.

Let $J_{\epsilon}$, $\epsilon >0$, be the set of all intervals of length less than $\epsilon$, and say ${\mathcal J}:= \{J_\epsilon \mid \epsilon >0 \}$, and ${\mathcal O}_X$ the set of open covers of some set $X \subseteq \mathbb R$ (where the open sets come from $\mathbb R$ and not $X$ with the subspace topology).

Definition of the strong measure zero game. Let $X \subseteq \mathbb R$. Then the strong measure zero game is the game $G_1( {\mathcal J}, {\mathcal O}_X)$.

There are a couple of other definitions I could have chosen, but I want to stress that the game has a very strong connection to strong measure zero, since player $I$ can choose a sequence of $\epsilon_n$ and play $J_{\epsilon_n}$, in inning $n$, and then $II$ tries to cover $X$ by collecting one interval from each $J_{\epsilon_n}$.

A very natural question to ask is when does player $I$ have a winning strategy, and when does player $II$ have a winning strategy in $G_1( {\mathcal J}, {\mathcal O}_X)$? This brings us to “the main event”:

Theorem 1. Player $II$ has a winning strategy in the strong measure zero game, if, and only if, $X$ is countable.

Theorem 2. Player $I$ has a winning strategy in the strong measure zero game, if, and only if, $X$ is not strong measure zero.

One of the things that makes these results interesting is that this means it is consistent for winning strategies to always exist, that is when all strong measure zero sets are countable, and it is consistent for there to be sets where there is no winning strategy, which is when there are uncountable strong measure zero sets(as an example the Luzin sets mentioned before).

The Proofs of Theorem 1 and 2

The proofs of Theorem 1 and 2, that will be presented, is Galvin’s with some modification and can be found in his paper Indeterminacy of point-open games. I found out about this proof in a class, taught by Marion Scheepers, a few years ago.

First, some notation, $^\omega \omega$ will be the set of infinite sequences of order type $\omega$, $^{< \omega} \omega$ will be the set of finite sequences indexed by finite ordinals. And if $x,y$ are sequences, then $x \frown y$ will be the concatenation of the sequences. As examples

$$(1,2,3,...) \in {^\omega \omega},$$ $$(7,10,2) \in {^{<\omega}} \omega$$

and

$$(7,10,2) \frown (24,1)= (7,10,2,24,1).$$

Lemma 3. Let $X$ be a set and suppose that for each $x \in X$ we have that a set $B_x$, which is a set of positive integer, finite length, sequences such that:

1. For every infinite sequence $(n_1,...,n_i,...)$ there is a $k \in \omega$ such that $(n_1,...,n_k) \in B_x$.

2. For each finite sequence $(n_1,...,n_m)$,

   $$\vert \bigcap_{k \in \omega} \{ x \in X \mid (n_1,...,n_m,k) \in B_x \} \vert \leq \aleph_0,$$

then we have that $\vert X \vert \leq \aleph_0$ ($X$ is countable).

proof of Lemma 3. We will prove by contradiction, so suppose that $X$ is an uncountable set with the above properties. This means that

$$X \setminus \bigcup_{ \alpha \in { ^{< \omega} \omega} } \left( \bigcap_{n \in \omega} \{ x \in X \mid \alpha \frown (n) \in B_x \} \right) \neq \varnothing$$

since the part that is being removed from $X$ is countable. Let $x$ be an element from this nonempty set. By the definition of the above set, we know that $x$ has the property, that for any sequence $\alpha \in {^{<\omega} \omega}$, that there is an $n$ such that $\alpha \frown (n) \not\in B_x$. Starting with the empty sequence, $()$, we know there is a least $n_1$ such that $(n_1) \not\in B_x$, and we can continue doing this inductively, constructing sequences $(n_1,...,n_k) \not\in B_x$ of length $k$. This contradicts property 1. $\square$

Theorem 1. Player $II$ has a winning strategy in the strong measure zero game, if, and only if, $X$ is countable.

proof of Theorem 1. It is easy to see that $II$ has a winning strategy, when $X$ is countable, since we can enumerate the set $X$, and in turn $n$, $II$ covers the $n$-th element. For the other direction, we will need to define a different game. Let $(Y,d)$ be a separable metric space, so it also has a countable basis. (Separable means contains a countable dense set subset.) The game has these rules:

1. There is an inning for each $n \in \omega$.
2. In inning $n$ player $I$ chooses a set $\{S_1,S_2\}$, where $S_i \subseteq Y$ is open and the distance between $S_1$ and $S_2$ is positive, where distance between the two sets is defined to be $\inf \{ d(x,y) \mid (x,y) \in S_1 \times S_2 \}=d(S_1,S_2)$. We will call such pairs playable
3. Player $II$ chooses one of the sets, which we will call $T_n$.
4. Player $II$ wins if $\bigcap_{n \in \omega} T_n = \varnothing$.

Lets call this game $\Gamma(Y)$, or the disjoint open ball game on $Y$. We will now show that when $II$ has a winning strategy for $\Gamma(Y)$ then the set $Y$ is countably infinite. Let $\sigma$ be a winning strategy for $\Gamma(Y)$. Note that we may assume that $I$ only plays from sets from a basis, and since we have a countably basis, we may assume that $I$ plays from some countable basis. Let $W= \{ P_n \mid n \in \omega \}$ where $(P_n)_{n \in \omega}$ is an enumeration of all playable $\\{ S_1, S_2\\}$, where $S_i$ are from our chosen countable basis. For $x \in Y$ we define

$$\begin{align*} B_x := \{ (n_1,...,n_k ) \in {^{<\omega} \omega} \mid &x \in \sigma(P_{n_1}),...., \\ &x \in \sigma(P_{n_1},T_{n_1},...,P_{n_{k-1}}), \\ &x \not\in \sigma(P_{n_1},....,P_{n_{k-1}},T_{n_{k-1}},P_{n_k}) \} \end{align*}$$

Given that $\sigma$ is a winning strategy for $II$, we can see that for any $x \in B_x$ and sequence $(n_1,...) \in {^\omega \omega}$ that some segment $(n_1,...,n_k) \in B_x$, since otherwise $x \in \sigma(P_{n_1},...,P_{n_k})=T_{k-1}$ for all $k>0$, so $x \in \bigcap_{i>0} T_i$ contradicting that set being empty. Hence we have condition 1, from Lemma 3. We will now show condition 2 of Lemma 3. Suppose that we have distinct

$$x_1,x_2,x_3 \in \sigma( P_{n_1},....,P_{n_k} )$$

for $1 \leq k \leq m$. We can choose playable pairs

$$P_{n_{1,2}}=\{A_1,A_2\},P_{n_{1,3}}=\{B_1,B_3\},P_{n_{2,3}}= \{C_2,C_3\}$$

where the index $i$ on the open sets says that it contains $x_i$ but not the $x_j$ where $j \neq i$, since we are in a metric space. And note that only one of the $x_i=y_1$ can be in $\sigma(P_{n_1},...,P_{n_m},P_{n_{k,l} })$, hence other two, $y_2,y_3$, are not contained in that set. This is true for each combination, so we have

$$(n_1,...,n_m,n_{k,l}) \in \left(B_{y_2} \cap B_{y_3} \right)\setminus B_{y_1}.$$

So we get that $y_1 \not \in \{ x \mid (n_1,....,n_m,n_{k,l}) \in B_x$, and by shuffling things around, we get that each of the three points, $x_1,x_2,x_3$, have cooresponding sets that they are not in, hence they are not in the intersection of all those sets. This worked for an arbitrary choice of three points, so it must be

$$\bigcap_{n \in \omega} \{x \in Y \mid (n_1,...,n_m,n) \in B_x \}= \varnothing.$$

Using Lemma 3, we get that the space is countable.

Let $\Sigma$ be a winning strategy for $II$ in the strong measure zero game on $X$. We are going to show that the strategy $\Sigma$ will help us construct a winning strategy $\sigma$ in the game $\Gamma(X)$, hence $X$ will be countable.

Say that player $I$ plays $P_1=\{S_1,S_2\}$ in $\Gamma(X)$, and plays $J_{d(S_1,S_2)}=J_{\epsilon_1}$ in the strong measure zero game. We will have player $II$ play $A_1=\Sigma(J_{\epsilon_1})$, and then $\sigma(P_1)$ will be an element disjoint from $A_1$, which we can pick since the length of $A_1$ is less than $d(S_1,S_2)$. So we can define a strategy where $\sigma( \{S_1^n,S_2^n\} )= S_1^n$ if $\Sigma \left(J_{d(S_1^n,S_2^N)} \right) \cap S_1^n = \varnothing$ and $\sigma \left(\{S_1^n,S_2^n\} \right)= S_2^n$ otherwise. We will now show that this is in fact a winning strategy. Note that

$$\left(\bigcap_{n \in \omega} \sigma(P_1,...,P_n) \right) \cap \left( \bigcup_{n \in \omega} A_n \right) = \varnothing$$

but the union part covers the whole space so the intersection on the left must be empty. So player $II$ has a winning strategy for $\Gamma(X)$ when $II$ has a winning strategy for the strong measure zero game. Thus $X$ is countable.$\square$

Lemma 4 (Lebesgue covering lemma). If $X$ is a compact metric space, then for any open cover of $X$, there exists a postive real number $\delta$, the Lebesgue number, such that: for any $F \subseteq X$ of diameter less than $\delta$ there is a $U$ in the open cover such that $F \subseteq U$.

proof of Lemma 4. This is a well known theorem, so no proof will be given. $\square$

Theorem 2. Player $I$ has a winning strategy in the strong measure zero game, if, and only if, $X$ is not strong measure zero.

proof of Theorem 2. First we will prove ($\Leftarrow$), so suppose that $X$ is not strong measure zero. Hence there is a sequence $(\epsilon_n)$ for $n \in \omega$ such that, such that $X$ for all sequences of intervals $(A_n)$, for ${n \in \omega}$, length of $A_n$ is less than or equal too $\epsilon_n$. So have player $I$ plays $J_{\epsilon_n}$ in inning $n$, so player $II$ can not cover the space, since it will only pick intervals less than $\epsilon_n$ at inning $n$.

We will know prove ($\Rightarrow$), we will do the contrapositive, so we will show that if $X$ is strong measure zero then $I$ does not have a winning strategy. Let $\sigma$ be an arbitrary strategy for player $I$, and $X$ is strong measure zero. We will show that there is a play, by player $II$ that will beat this strategy. Here are some things that we are going to assume, to make the problem a bit easier:

1. That we are playing on strong measure zero sets, that are subsets of compact sets, and in fact we will assume to be working with $[0,1]$. This is enough since we can write $\mathbb {R}$ as a countable union of compact sets, for example $\bigcup_{i \in \mathbb Z} [i,i+1]$, and partition the innings, into countably infinite many infinite sets (think of sets of the form $\{p^n \mid n \in \mathbb{N} \}$, $p$ prime ), so we play the strong measure zero game on the $j$th compact set when the inning we are in is in the $j$th partition.
2. We will assume that the $J_\epsilon$’s that are played by player $I$ have $\epsilon$’s of the form $1/n$, since for some $\epsilon$ we can just move to a smaller number of the form $1/n$ without losing any generality.
3. Let $s=(J_{\epsilon_1},...,J_{\epsilon_n},K_n)$ be some sequence of turns, that is in the domain of player $I$’s strategy $\sigma$, then we define $\hat\sigma(s)= \epsilon \iff \sigma(s) = J_\epsilon$.

We will now construct a play in which $II$ which beats player $I$ strategy $\sigma$. Say that $\sigma( \varnothing)= J_{1/n_1}$. Lets say $U_1 \subset J_{1/n_1}$ is a finte open cover of $[0,1]$ (which exists since $[0,1]$ is compact). Consider $\min \{ \hat \sigma(K) \mid K \in U_1 \}$, and we will choose $m_1 \in \mathbb{N}$ large enough so that:

1. $n_1<m_1$,
2. $1/m_1 < \min \{ \hat \sigma(K) \mid K \in U_1 \}$,
3. and $1/m_1$ is a Lebesgue number of the cover $U_1$.

Basically we are going to go down, and play every possible outcome of the strategy $\sigma$ possible. Let $U_2 \subset J_{1/m_1}$, be a finite open cover, so in particular this $U_2$ works as a subcover for $\sigma(K)$ for $K \in U_2$ since $1/m_1 < \hat \sigma (K)$. Then we are going to choose an $m_2 \in \mathbb N$ large enough so that:

1. $m_1 < m_2$ (this also means that $\hat \sigma ( K ) < m_2$ for $K \in U_1$, which are the corresponding “n_i” terms ),
2. $1/m_2< \min\{ \hat \sigma (K_1,K_2) \mid K_1 \in U_1, K_2 \in U_2 \}$,
3. And $1/m_2$ is a Lebesgue number of the cover $U_2$.

Basically we continue in this fashion, and construct $m_t$, for arbitrary $t$ (which if that statement was good enough for you, I recommend skipping this paragraph). Suppose that we have $m_1,...,m_{t-1}$ constructed. Then we are going to construct $m_t$ by letting $U_t \subset J_{1/m_{t-1}}$ be a finite open cover. Then we choose $m_t \in \mathbb{N}$ large enough such that:

1. $m_{t-1} < m_t$,
2. $1/m_t < \min \{ \hat \sigma (K_1,...,K_t) \mid K_1 \in U_1,..., K_t \in U_t \}$,
3. And $1/m_t$ is a Lebesgue number of the cover $U_t$.

So know we have a sequence $(1/m_i)_{i \in \omega}$, where each $1/m_i$ is a Lebesgue number of the coverings $U_i$. Since $X$ is strong measure zero we have that there is a covering $\{L_i \mid i \in \omega\}$ where each $L_i$ has length less than $1/m_i$. Since $1/m_i$ is a Lebesgue number of the cooresponding coverings, we can find an $M_i \in U_i$ such that $L_i \subseteq M_i$. We can choose the $M_i$ as our sets, and so we found a play where $II$ wins, hence $\sigma$ is not a winning strategy, and since $\sigma$ was arbitrary we are done. $\square$

Resources

You can find out more about these sorts of things from this, non-exhaustive, list:

• F. Galvin, Indeterminacy of point-open games, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 26 :5 (1978), 445–449 (this is sort of difficult to find, so included all the information, for libraries)
• “The combinatorics of open covers (I): Ramsey theory” by Marion Scheepers,
• “The combinatorics of open covers (II)” by Winfried Just, Arnold W. Miller, Marion Scheepers, and Paul J. Szeptycki,
• …
• “The combinatorics of open covers ($N$)” by … (well there are at least 11 of these…),
• “Ramsey theory of open covers lecture 1” by Nadav Samet and Boaz Tsaban,
• …
• “Ramsey theory of open covers lecture 5” by Nadav Samet and Boaz Tsaban.
• You can browse around here

The “Ramsey theory of open covers” are more accessible and shorter than “The combinatorics of open covers…” and discuss some of the Ramsey theoretic aspects that are related to these games and to “selection principles” (which I did not discuss here, although are very related to games $G(\mathcal A,B)$).