3.1 Setup and inductive hypothesis

Let $\beta \in (0,\tfrac 13]$ . We will supply ${\mathsf {Alice}}$ with a winning strategy for each system.

First, we define the following terms.

• • A point $z\in [0,1]$ is called an nth-order vertex (also denoted by vertex $^{(n)}$ ) if $z\in T^{-n}(0)$ .

• • An interval is called an nth-order cylinder (also denoted by cylinder $^{(n)}$ ) if its endpoints are two consecutive (ordered by the natural order of real numbers) vertices $^{(n)}$ .

Denote the balls chosen by ${\mathsf {Bob}}$ by

$$ \begin{align*}\underbrace{B_1^{(1)}}_{\subset [0,1]}\to \underbrace{B_2^{(1)}}_{\subset B_1^{(1)}\backslash A_1^{(1)}}\to B_3^{(1)}\to \cdots \to B_1^{(n)}\to B_2^{(n)}\to \cdots \to B_{k_n}^{(n)}=B_1^{(n+1)}\end{align*} $$

and those chosen by ${\mathsf {Alice}}$ by

$$ \begin{align*}\underbrace{A_1^{(1)}}_{\subset B_1^{(1)}}\to A_2^{(1)}\to \cdots \to \cdots \to B_1^{(n)}\to B_2^{(n)}\to \cdots \to B_{k_n}^{(n)}=B_1^{(n+1)},\end{align*} $$

where the natural number $k_n$ will be defined over the course of the proof. We will call the superscript ‘ $(n)$ ’ the ‘levels’ n, and we call $B_k^{(n)}$ the kth round in the nth level, with this level starting from $B_1^{(n)}$ . The idea of levels is that for a $\mathsf {Bob}$ ’s ball $B_\ell $ , $\mathsf {Alice}$ responds to $B_\ell $ on the nth level if $T^{n-1}$ is injective on $B_\ell $ and $T^{n-1}B_\ell $ is not contained in any first-order cylinder. Solely for notational purposes, we will allow zero and negative indices for these balls; for example, $B_{0}^{(n)}$ is the ball one round before $B_{1}^{(n)}$ and $B_{-5}^{(n)}$ is the ball six rounds before $B_{1}^{(n)}$ .

• • For all $n\geq 1$ , a point $z\in [0,1]$ is called an nth-order *-vertex (denoted by *-vertex $^{(n)}$ ) if $z\in T^{-(n-1)}(f_{n-1}(c^{(n-1)}))$ where $c^{(n-1)}$ is the center of $B_1^{(n-1)}$ .

• • Denote the collection of the *-vertices of order less than or equal to n by $\mathcal {S}^{(n)}$ .

• • Denote the diameter of $B_k^{(n)}$ by $\tau _k^{(n)}$ , denote the radius of $B_k^{(n)}$ by $r_k^{(n)}$ , and denote the diameter of $T^{n-1}B_k^{(n)}$ by $\rho _k^{(n)}$ .

In both proofs, we will verify that the following hypotheses hold for all $n\in \mathbb {N}$ . Let $\rho>0$ and $\unicode{x3bb}>0$ be constants that will be specified in the strategies.

Denote by $J^{(n)}$ the last appearance of a Type III level before level n; that is,

$$ \begin{align*}& J^{(n)} \\ & =\begin{cases} \max \{k:\, k<n\text{ and }\text{level}^{(k)}\text{ is Type III}\} & \text{if }\{k:\, k<n\text{ and }\text{level}^{(k)}\text{ is Type III}\}\neq \varnothing,\\ 0& \text{otherwise}. \end{cases}\end{align*} $$

To prove Statement 3 holds for all $m,n$ , we will need the following lemma.

We claim that under the given strategy, the following property holds.

Proposition 5. There exists some $\delta>0$ such that for all $n>1$ and for all $x\in B_{1}^{(n+1)}$ , one has that

$$ \begin{align*}{\mathrm{dist}}(T^{n} x,f_n(c^{(n)}))\geq \delta.\end{align*} $$

3.2 Construction of strategy, I

Let $\rho =\beta \cdot \min \{\rho _1^{(1)}, \rho _\#\}$ , where $\rho _\#$ is a positive solution to

(2) $$ \begin{align} \frac{-\log\rho_\#-\log\beta}{\log\gamma}=-4, \end{align} $$

and $\unicode{x3bb} =(({-\log \rho -\log \beta })/{\log \gamma })+1$ .

The base case for Lemma 1 easily holds.

Let $M\in \mathbb {N}\backslash \{0\}$ . We make the following assumption.

Next, we state the strategy for ${\mathsf {Alice}}$ .

Case I: If the nth level is of Type I, then ${\mathsf {Alice}}$ relabels $B_1^{(n)}$ by $B_1^{(n+1)}$ . Note that this ‘action’ of relabeling by ${\mathsf {Alice}}$ does not alter the state of the game.

Case II: If the nth level is of Type II, then ${\mathsf {Alice}}$ takes $A_1^{(n)}=B(v^{(n)},\beta \cdot \tau _1^{(n)})$ , where $v^{(n)}$ is the vertex $^{(n)}$ in $B_1^{(n)}$ , and then ${\mathsf {Alice}}$ relabels $B_2^{(n)}$ by $B_1^{(n+1)}$ .

Case III: Suppose that the nth level is of Type III. Denote the collection of the *-vertices of order less than or equal to n by $\mathcal {S}^{(n)}$ . We will see that by induction $|\mathcal {S}^{(n-1)}\cap B_1^{(n)}|\leq \unicode{x3bb} $ . For all $k\geq 1$ , we ask ${\mathsf {Alice}}$ to take the center of $A_k^{(n)}$ to be

$$ \begin{align*}\begin{cases} \textit{an element of }\mathcal{S}^{(n-1)}\cap B_k^{(n)}& \textit{if }B_k^{(n)}\cap \mathcal{S}^{(n-1)}\neq\varnothing ,\\[6pt] \textit{the center of }B_k^{(n)} & \textit{if }B_k^{(n)}\cap \mathcal{S}^{(n-1)}=\varnothing, \end{cases}\end{align*} $$

with the maximal allowed radii $\beta \cdot r_k^{(n)}$ , until the time

$$ \begin{align*}H^{(n)}:= \min \bigg\{k:\, B_k^{(n)}\cap \mathcal{S}^{(n-1)}=\varnothing \text{ and } {\mathrm{diam}}(B_k^{(n)})<\frac{\rho}{\gamma} \bigg\};\end{align*} $$

we denote the smallest k such that ${\mathrm {diam}}(B_k^{(n)})<({\rho }/{\gamma })$ by $L^{(1)}$ . Note that it takes at most $\lceil {\log \unicode{x3bb} }/{\log 2}\rceil $ rounds for $\mathsf {Alice}$ to remove all the points in $\mathcal {S}^{(n-1)}\cap B_1^{(n)}$ , so

(3) $$ \begin{align} H^{(n)}\leq L^{(n)}+\bigg\lceil \frac{\log \unicode{x3bb}}{\log 2}\bigg\rceil. \end{align} $$

Note that all cylinders contained $T^{n-1}B_1^{(n)}\subset [0,1]$ have diameter ${1}/{\gamma }$ , so $B_{H^{(n)}}^{(n)}$ contains at most one vertex $^{(n)}$ and two *-vertices $^{(n)}$ . $\mathsf {Alice}$ removes these in two turns. Then

$$ \begin{align*} {\mathrm{dist}}(T^{n-1}(B_{H^{(n)}+2}^{(n)}\backslash A_{H^{(n)}+2}^{(n)}),\mathcal{S}^{(n)})& \geq {\mathrm{radius}} (T^{n-1}A_{H^{(n)}+2})\\[6pt] & \geq \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+3}\cdot {\mathrm{radius}}(T^{n-1}B_{L^{(n)}-1}^{(n)})\\[6pt] & \geq \frac{1}{2} \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+3}\cdot \frac{\beta}{2\gamma}, \end{align*} $$

so as a consequence, for all $x\in B_1^{(n+1)}\subset B_{H^{(n)}+2}^{(n)}\backslash A_{H^{(n)}+2}^{(n)})$ , one has

(4) $$ \begin{align} {\mathrm{dist}}(T^k x,f(c^{(k)}))\geq \gamma^{-\unicode{x3bb}}\, \frac{1}{2}\, \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+3}\cdot \frac{\beta}{2\gamma}\quad \text{for all }J^{(n)}<k\leq n. \end{align} $$

In addition, we have

(5) $$ \begin{align} \rho_1^{(n+1)}&= {\mathrm{diam}} (T^nB_1^{(n+1)})\notag \\[6pt] &= \gamma^n \cdot {\mathrm{diam}} (B_1^{(n+1)})\notag \\[6pt] & \geq \gamma^n \cdot \beta \cdot {\mathrm{diam}} (B_{H^{(n)}+2}^{(n)})\notag \\[6pt] & \overset{(3)}{\geq} \gamma^n \cdot \beta \cdot {\mathrm{diam}} (B_{(L^{(n)}-1)+\lceil{\log \unicode{x3bb}}/{\log 2}\rceil+3})\notag \\[6pt] & \geq \gamma^n \cdot \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot {\mathrm{diam}}(B_{L^{(n)}-1}^{(n)})\notag \\[6pt] &= \gamma \cdot \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot {\mathrm{diam}}(T^{n-1}B_{L^{(n)}-1}^{(n)})\notag \\[6pt] & \geq \gamma \cdot \beta^{\lceil \log_2({-\log\rho-\log\beta})/{\log\gamma}\rceil+4}\cdot \frac{\rho}{\gamma}\notag \\[6pt] & \overset{(2)}{=} \rho. \end{align} $$

Remark (Levels versus rounds). We make the correspondence of levels and rounds explicit by counting the number of rounds in each type of levels in the following table. Denote by $k_n$ the number of moves $\mathsf {Alice}$ makes while the strategy is in level n (or equivalently, the number of rounds in level n). Then we have:

$$ \begin{align*} \begin{array}{llll} \text{Level type} & k_n & B^{(n+1)}_1 \\[2pt] \hline \text{Type I} & 0 & B^{(n)}_1 \\ \text{Type II} & 1 & B^{(n)}_2 \\ \text{Type III } & H^{(n)}+2 & B^{(n)}_{H^{(n)}+3} \end{array} \end{align*} $$

A Type I level n performs a pure relabeling process and does not advance the game, so $k_n=0$ there. In the notation of §2, we have that

$$ \begin{align*} B_k^{(n)}=B_{\sum_{m=1}^n k_m}. \end{align*} $$

For example, suppose that the strategy produces, in order, a Type I level, a Type III level with $H^{(2)}=3$ , a Type II level, and a Type I level. The bookkeeping unfolds as follows:

$$ \begin{align*} B_1&=B_1^{(1)}=B_1^{(2)}\to B_2=B_2^{(2)} \to B_3=B_3^{(2)}\to B_4=B_4^{(2)} \\ \to B_5&=B_5^{(2)}\to B_6=B_6^{(2)}=B_1^{(3)}\to B_7=B_2^{(3)}=B_1^{(4)}=B_1^{(5)}. \end{align*} $$

Now we verify our hypotheses hold in the progression.

Proof of Lemma 1 for System I.

For Statement 1, note that if a ball B contains two vertices $^{(n)}$ , then the diameter of $T^{n-1}B$ is at least ${1}/{\gamma }$ , so if $B_1^{(n)}$ contains a vertex $^{(n)}$ (not of Type I), then either it contains exactly one vertex $^{(n)}$ (of Type II) or $T^{n-1}B_1^{(n)}\geq ({1}/{\gamma })$ (of Type III).

By assuming Statements 2 and 3 for $n<M$ , if level $(M-1)$ is Type III, then Statement 2 for $n=M$ follows immediately from (5). Recall that

$$ \begin{align*}&J^{(n)}\\ & =\begin{cases} \max \{k:\, k<n\text{ and }\text{level}^{(k)}\text{ is Type III}\} & \text{if }\{k:\, k<n\text{ and }\text{level}^{(k)}\text{ is Type III}\}\neq \varnothing,\\ 0& \text{otherwise}. \end{cases}\end{align*} $$

We now prove Lemma 2 for this system and then use it to show Statement 3 holds.

Proof of Lemma 2 for System I.

Assume the contrary; suppose there are two consecutive Type II levels $n,m$ with $n<m$ without any intermediate Type III level. Then $B_1^{(n)}$ contains a vertex $^{(n)}\, v_1$ and a vertex $^{(m)}\, v_2$ , which are distinct. Then

$$ \begin{align*}{\mathrm{diam}}(B_1^{(m)})=\beta\cdot {\mathrm{diam}}(B_1^{(n)})\geq \beta\cdot {\mathrm{dist}}(v_1,v_2)\geq \beta\cdot \frac{1}{\gamma^m},\end{align*} $$

contradicting that level m is Type II.

Then by Lemma 2, there is at most one Type II level between level $J^{(M)}$ and level M. Then

$$ \begin{align*} \rho_1^{(M)}& = {\mathrm{diam}}(T^{M-1}B_1^{(M)})\geq \beta\cdot {\mathrm{diam}}(T^{M-1}B_1^{(J^{(M)}+1)})\\ & \geq \beta\cdot {\mathrm{diam}}(T^{J^{(M)}}B_1^{(J^{(M)}+1)})\overset{(5)}{\geq} \beta\cdot \rho. \end{align*} $$

Since Statement 3 is assumed for all $0<n<M$ , by the expanding property of T, note that for all $0<n<M$ ,

$$ \begin{align*} {\mathrm{diam}}(T^{n+\unicode{x3bb}-1}B_1^{(n+\unicode{x3bb})})\geq \beta \cdot \gamma^{\unicode{x3bb}-1} {\mathrm{diam}}(T^n B_1^{(n)})=\beta\cdot \frac{1}{\gamma}\cdot \frac{\gamma}{\rho\cdot \beta}\cdot \rho=1, \end{align*} $$

then $T^{\lceil \unicode{x3bb} \rceil } B_1^{(n)}=[0,1]$ since its diameter exceeds $1$ , meaning that within $\unicode{x3bb} $ levels any Type III level $(n)$ , there must be another Type III level and proving Statement 3 for all n.

From (4), we finally obtain the most crucial result: Proposition 5. We will finish our proof of Theorem 4 in §3.4.

3.3 Construction of strategy, II

Before we start our statement of the strategy, we state two well-known properties of System II.

Lemma 3. (Expanding property for System II)

There exists a constant $R>0$ such that for all positive $n\in \mathbb {N}$ and for all $x\in (0,1)$ ,

$$ \begin{align*} |D_xT^n|\geq R \cdot \bigg[\bigg(\frac{1+\sqrt{5}}{2}\bigg)^2\bigg]^n. \end{align*} $$

For the remainder of the paper, we denote $\psi :=(({1+\sqrt {5}})/{2})^2$ .

Proof. From [Reference Khinchin8, Ch. 1, §2], in view of the notation in this book, the denominators of the convergents satisfy the recurrence relation

$$ \begin{align*} \begin{cases} q_0=1,\quad q_1=a_1,&\\ q_n=a_nq_{n-1}+q_{n-2}& \text{ for all } \text{ integers } n>1, \end{cases} \end{align*} $$

together with the facts that $a_n\geq 1$ for all n, so $q_n$ is greater than or equal to the nth Fibonacci number, and hence $q_n\geq \psi ^{n/2}$ . By the mean value theorem, the diameter of a cylinder $^{(n)}$ cannot be less than ${1}/{|D_xT^n|}$ . The diameter of a cylinder $^{(n)}$ is ${1}/({q_n(q_n+q_{n-1})})$ , so the lemma is proved.

Lemma 4. (Bounded distortion for System II)

For all positive $n\in \mathbb {N}$ and for all $x,y$ in an nth-order cylinder,

$$ \begin{align*} \frac{1}{4}\leq \frac{|D_xT^n|}{|D_yT^n|}\leq 4. \end{align*} $$

Lemma 4 is also well known (see, for example, [Reference Cornfeld, Fomin and Sinăi3, §7.4, Lemma 2]).

Let $\rho _\#,\unicode{x3bb} $ be positive solutions to the system

(6) $$ \begin{align} \begin{cases} (\beta/4)^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+6}\geq 50\rho_\#,\\[6pt] \dfrac{R}{4}\beta \psi^{\unicode{x3bb}-1}\geq \dfrac{1}{\rho_\#}. \end{cases} \end{align} $$

One can simply take $\unicode{x3bb} $ large enough such that

$$ \begin{align*} \frac{4}{R\beta}\cdot \bigg(\frac{1}{\psi}\bigg)^{\unicode{x3bb}-1}\leq \frac{1}{50}\cdot \bigg(\frac{\beta}{4}\bigg)^{({\log\unicode{x3bb}}/{\log2})+6}, \end{align*} $$

and $\rho _\#$ to be any number in $(({4}/{R\beta })\cdot \psi ^{1-\unicode{x3bb} },({1}/{50})\cdot (\beta /4)^{({\log \unicode{x3bb} }/{\log 2})+6})$ . Let $\rho =\min \{\rho _1^{(1)},\rho _\#\}$ .

The base case for Lemma 1 easily holds.

Let $M\in \mathbb {N}\backslash \{0\}$ . We make the following assumption.

Next, we state the strategy for ${\mathsf {Alice}}$ .Case I: If the nth level is of Type I, then ${\mathsf {Alice}}$ relabels $B_1^{(n)}$ by $B_1^{(n+1)}$ .

Case II: If the nth level is of Type II, then ${\mathsf {Alice}}$ takes $A_1^{(n)}=B(v^{(n)},\beta \cdot r_1^{(n)})$ , where $v^{(n)}$ is the vertex $^{(n)}$ in $B_1^{(n)}$ and $r_1^{(n)}$ is the radius of $B_1^{(n)}$ , and then ${\mathsf {Alice}}$ relabels $B_2^{(n)}$ by $B_1^{(n+1)}$ .

Case III: Suppose that the nth level is of Type III. ${\mathsf {Alice}}$ first selects $A_1^{(n)}$ with radius $\beta \cdot {\mathrm {radius}}(B_1^{(n)})$ so that $T^{n-1}A_1^{(n)}$ is placed as far to the left as possible in $T^{n-1}B_1^{(n)}$ . Note that

$$ \begin{align*}\rho_1^{(n)} \begin{cases} <({1}/{N^{(n)}-1}) & \text{if }N^{(1)}\neq 1,\\[6pt] \leq ({1}/{N^{(n)}}) & \text{if }N^{(1)}=1; \end{cases}\end{align*} $$

in particular, we can conclude that

(7) $$ \begin{align} \rho_1^{(n)}<\frac{1}{N^{(n)}}. \end{align} $$

In Case III, we need to consider the following subcases.

• • Subcase 1: Suppose $\underline {B_1^{(n)}\backslash A_1^{(n)}}$ contains zero or one vertex $\underline {{}^{(n)}}$ , say $\underline { v^{(n)}}$ . By induction, $|\mathcal {S}^{(n-1)}\cap B_1^{(n)}|\leq \unicode{x3bb} $ . Then $B_2^{(n)}$ contains at most $\unicode{x3bb} $ *-vertices from $\mathcal {S}^{(n-1)}$ , where we denote the collection of the *-vertices of order less than or equal to n by $\mathcal {S}^{(n)}$ . ${\mathsf {Alice}}$ then selects each $A_k^{(n)}$ in $A_2^{(n)},\ldots ,A_{1+\lceil \log _2(\unicode{x3bb} +2)\rceil }^{(n)}$ with the maximal radii allowed and first centered at $v^{(n)}$ and then centered at the *-vertices in $\mathcal {S}^{(n-1)}\cap B_2^{(n)}$ if $\mathcal {S}^{(n-1)}\cap B_k^{(n)}\neq \varnothing $ , and centered at the center of $B_k^{(n)}$ if $\mathcal {S}^{(n-1)}\cap B_k^{(n)}=\varnothing $ . Consequently, for all $S^{(n)}<k<n$ where $S^{(n)}$ is the last appeared Type III level, we have

  (8) $$ \begin{align} {\mathrm{dist}}&\ (T^{k-1}B_{2+\lceil\log_2(\unicode{x3bb}+2)\rceil}^{(n)},T^{-1}(f_k(c^{(k)})))\nonumber\\[6pt] & \geq \tfrac{1}{4}\cdot {\mathrm{radius}}(T^{k-1}A_{1+\lceil\log_2(\unicode{x3bb}+2)\rceil}^{(n)}) \nonumber\\[6pt] & \geq \tfrac{1}{4}\cdot \beta^{1+\lceil\log_2(\unicode{x3bb}+2)\rceil}\cdot \rho_1^{(k)} \nonumber\\[6pt] & \geq \tfrac{1}{4}\cdot \beta^{1+\lceil\log_2(\unicode{x3bb}+2)\rceil}\cdot \rho. \end{align} $$
  Consequently, for all $x\in B_1^{(n)}=B_{2+\lceil \log _2(\unicode{x3bb} +2)\rceil }^{(n)}$ , one has
  (9) $$ \begin{align} {\mathrm{dist}}\ (T^k x, f_k(c^{(k)}))\geq \tfrac{1}{4}\cdot \beta^{1+\lceil\log_2(\unicode{x3bb}+2)\rceil}\cdot \rho.\\[-22pt]\nonumber \end{align} $$
  Note that $B_2^{(n)}$ does not contain any vertices $^{(n)}$ , so it contains at most one *-vertex $^{(n)}$ ; ${\mathsf {Alice}}$ centers $A_{2+\lceil \log _2(\unicode{x3bb} +2)\rceil }$ at this *-vertex $^{(n)}$ with maximal radius. Consequently, we obtain
  (10) $$ \begin{align} {\mathrm{dist}}\ (T^n x, f_n(c^{(n)}))\geq \tfrac{1}{4}\cdot \beta^{2+\lceil\log_2(\unicode{x3bb}+2)\rceil}\cdot \rho.\\[-16pt]\nonumber \end{align} $$
  Then ${\mathsf {Alice}}$ relabels $B_{3+\lceil \log _2(\unicode{x3bb} +2)\rceil }^{(n)}$ by $B_1^{(n+1)}$ . Note that
  (11) $$ \begin{align} {\mathrm{diam}}\ (T^{n-1}B_1^{(n+1)})&= {\mathrm{diam}}\ (T^{n-1}B_{3+\lceil\log_2(\unicode{x3bb}+2)\rceil}^{(n)})\notag \\[6pt] & \geq {\mathrm{diam}}\ (T^{n-1}B_1^{(n)})\cdot \beta^{3+\lceil\log_2(\unicode{x3bb}+2)\rceil}\notag \\[6pt] & \geq \frac{1}{4(N^{(n)})^2}\cdot \beta^{3+\lceil\log_2(\unicode{x3bb}+2)\rceil}. \end{align} $$
  On the other hand, since the ball $T^{n-1}B_1^{(n+1)}$ does not contain any vertices $^{(1)}$ , $T^{n-1}B_1^{(n+1)}$ is included in a cylinder $J^{(n)}$ of order $1$ . Since $J^{(n)}$ is to the left of the cylinder $^{(1)}\, [{1}/{N^{(n)}},{1}/({N^{(n)}-1})]$ , its diameter will not exceed that of $[{1}/{N^{(n)}},{1}/({N^{(n)}-1})]$ ; hence ${\mathrm {diam}}(J^{(n)})\leq ({1}/({N^{(n)}(N^{(n)}-1)}))\leq ({2}/{(N^{(n)})^2})$ , so together with the bounded distortion property (Lemma 4) and (11), we have that
  (12) $$ \begin{align} \rho_1^{(n+1)}=\frac{{\mathrm{diam}}\ (T^{n}B_1^{(n+1)})}{{\mathrm{diam}}([0,1])}& \geq \bigg[\frac{{\mathrm{diam}} (T^{n-1}B_1^{(n+1)})}{{\mathrm{diam}}(J^{(n)})}\bigg]\cdot \frac{1}{4}\notag \\[6pt]&\hspace{-1.5pt} \overset{(11)}{\geq} \bigg[\frac{1}{4(N^{(n)})^2}\cdot \beta^{3+\lceil\log_2(\unicode{x3bb}+2)\rceil} \bigg/ {\mathrm{diam}}(J^{(n)})\bigg]\cdot \frac{1}{4}\notag \\[6pt]& \geq \bigg[\frac{1}{4(N^{(n)})^2}\cdot \beta^{3+\lceil\log_2(\unicode{x3bb}+2)\rceil} \bigg/ \frac{2}{(N^{(n)})^2}\bigg]\cdot \frac{1}{4}\notag\\[6pt]&= \frac{\beta^{3+\lceil\log_2(\unicode{x3bb}+2)\rceil}}{32} \end{align} $$
  (13) $$ \begin{align} &\overset{(6)}{\geq} \rho. \end{align} $$

• • Subcase 2: Suppose ${B_1^{(n)}\backslash A_1^{(1)}}$ contains more than one vertex ${{}^{(n)}.}$ Let ${1}/{K^{(n)}}$ be the leftmost vertex $^{(1)}$ in $T^{n-1}(B_1^{(n)}\backslash A_1^{(n)})$ . Since $T^{n-1}(A_1^{(n)})$ is the leftmost subinterval of $T^{n-1}(B_1^{(n)})$ and the length of $T^{n-1}(A_1^{(n)})$ is at least $({\beta }/{4})\cdot \rho _1^{(n)}$ by Lemma 4, so $({1}/{K^{(n)}})>({\beta }/{4})\cdot \rho _1^{(n)}$ . Consequently, we have

  $$ \begin{align*} \frac{1}{N^{(n)}}-\frac{1}{K^{(n)}}< {\mathrm{diam}}(T^{n-1}B_1^{(n)})=\rho_1^{(n)}<\frac{4}{\beta}\cdot \frac{1}{K^{(n)}}, \end{align*} $$
  so
  (14) $$ \begin{align} \frac{1}{N^{(n)}}<\bigg(1+\frac{4}{\beta}\bigg)\cdot \frac{1}{K^{(n)}}. \end{align} $$
  Since $B_1^{(n)}$ does not contain any vertex $^{(n-1)}$ , it contains at most $\unicode{x3bb} $ *-vertices of order strictly less than n. Then ${\mathsf {Alice}}$ takes the radius of $A_k^{(n)}$ to be $\beta \cdot r_k^{(n)}$ (maximal allowed radius) and its center to be
  $$ \begin{align*} \begin{cases} \text{an element of }\mathcal{S}^{(n-1)}\cap B_k^{(n)}& \text{if }B_k^{(n)}\cap \mathcal{S}^{(n-1)}\neq\varnothing ,\\ \text{the center of }B_k^{(n)} & \text{if }B_k^{(n)}\cap \mathcal{S}^{(n-1)}=\varnothing, \end{cases} \end{align*} $$
  until the time (until ${\mathsf {Alice}}$ ’s turn $A_{H^{(n)}}^{(n)}$ )
  (15) $$ \begin{align} H^{(n)}:= \min \bigg\{k:\, k> |\mathcal{S}^{(n)}|+1 \text{ and } {\mathrm{diam}}(T^{n-1}B_k^{(n)})<\frac{\beta^2}{25(N^{(n)})^2} \bigg\}; \end{align} $$
  we denote the smallest k for which ${\mathrm {diam}}(T^{n-1}B_k^{(n)})<({\beta ^2}/({25(N^{(n)})^2}))$ by $L^{(n)}$ . It is easy to observe that $H^{(n)}\leq L^{(n)}+\lceil {\log \unicode{x3bb} }/{\log 2}\rceil $ . In addition, for all cylinders $^{(1)}\, J$ contained in $T^{n-1}(B_1^{(n)}\backslash A_1^{(n)})$ , the cylinder J is contained in $[{1}/{K^{(n)}},{1}/{N^{(n)}}]$ . By (14), in particular, $({1}/{N^{(n)}})<(1+4/\beta )\cdot ({1}/{K^{(n)}})<(5/\beta )\cdot ({1}/{K^{(n)}})$ so $[{1}/{K^{(n)}},{1}/{N^{(n)}}]\subset [{\beta /5}/{N^{(n)}},{1}/{N^{(n)}}]$ , and hence
  $$ \begin{align*} \frac{\beta}{5N^{(n)}}\cdot \frac{\beta}{5N^{(n)}}<{\mathrm{diam}}(J)<\frac{1}{N^{(n)}}\cdot \frac{1}{N^{(n)}}, \end{align*} $$
  and hence $B_{L^{(n)}}^{(n)}$ does not contain any cylinder $^{(n)}$ . Therefore, $B_{L^{(n)}+\lceil {\log \unicode{x3bb} }/{\log 2}\rceil }^{(n)}$ contains at most one vertex $^{(n)}$ , say $v^{(n)}$ , and at most two *-vertices $^{(n)}$ , say $s_1^{(n)},s_2^{(n)}$ , which $\mathsf {Alice}$ will remove from the game as centers of $A_{H^{(n)}+1}$ and $A_{H^{(n)}+2}$ ; that is, $A_{H^{(n)}+1}=B(v^{(n)},\beta \rho _{H^{(n)}+1}^{(n)})$ and $A_{H^{(n)}+2}=B(s_i^{(n)},\beta \rho _{H^{(n)}+2}^{(n)})$ where $s_i^{(n)}\in B_{H^{(n)}+1}\backslash A_{H^{(n)}+1}$ .

  Then ${\mathsf {Alice}}$ relabels $B_{H^{(n)}+3}^{(n)}$ by $B_1^{(n+1)}$ . We find

  $$ \begin{align*} {\mathrm{dist}}(T^{n-1}B_1^{(n+1)},f_n(c^{(n)}))& \geq {\mathrm{dist}}(T^{n-1}(B_{H^{(n)}+2}^{(n)}\backslash A_{H^{(n)}+2}^{(n)}),f_n(c^{(n)}))\\[6pt] & \geq \frac{1}{4}\cdot {\mathrm{radius}}(T^{n-1}A_{L^{(n)}+\lceil{\log\unicode{x3bb}}/{\log 2}\rceil+2})\\[6pt] & \geq \beta^{\lceil{\log\unicode{x3bb}}/{\log 2}\rceil+4}\cdot {\mathrm{radius}}(T^{n-1}B_{L^{(n)}-1}^{(n)})\\ & \geq \beta^{\lceil{\log\unicode{x3bb}}/{\log 2}\rceil+4}\cdot \frac{\beta^2}{50(N^{(n)})^2}\\ &\overset{(7)}{\geq} \frac{\beta^{\lceil{\log\unicode{x3bb}}/{\log 2}\rceil+6}}{50}\cdot (\rho_1^{(n)}/2)^2\\[6pt] &\overset{\text{Statement 2}}{\geq } \frac{\beta^{\lceil{\log\unicode{x3bb}}/{\log 2}\rceil+6}}{50}\cdot (\rho\beta/2)^2. \end{align*} $$
  In addition, for all $S^{(n)}<k< n$ , we have that
  $$ \begin{align*} {\mathrm{dist}}(T^{k-1}B_1^{(n+1)},f_k(c^{(k)})) & \geq {\mathrm{dist}}(T^{k-1} B_{\lceil\log\unicode{x3bb}/\log 2\rceil}^{(J^{(n)}+1)},f_k(c^{(k)}))\\[6pt] & \geq {\mathrm{radius}}(T^{k-1}A_{\lceil\log\unicode{x3bb}/\log 2\rceil+1}^{(J^{(n)}+1)})\\[6pt] & \geq \tfrac{1}{4}\cdot {\mathrm{radius}} (T^{k-1}A_1^{(J^{(n)}+1)})\cdot \beta^{\lceil\log\unicode{x3bb}/\log 2\rceil}\\[6pt] & \geq \tfrac{1}{4}\cdot {\mathrm{radius}} (T^{J^{(n)}}A_1^{(J^{(n)}+1)})\cdot \beta^{\lceil\log\unicode{x3bb}/\log 2\rceil}\\[6pt] & \geq \tfrac{1}{16}\cdot {\mathrm{radius}} (T^{(J^{(n)})}B_1^{(J^{(n)}+1)})\cdot \beta^{\lceil\log\unicode{x3bb}/\log 2\rceil+1}\\[6pt] & \overset{\text{Statement 2}}{\geq} \tfrac{1}{16}\cdot \beta\rho\cdot \beta^{\lceil\log\unicode{x3bb}/\log 2\rceil+1}. \end{align*} $$
  Consequently, for all $x\in B_1^{(n+1)}$ and for all $S^{(n)}<k\leq n$ , we have
  (16) $$ \begin{align} {\mathrm{dist}}(T^{k-1}x,f_{k-1}(c^{(k)}))&\geq \min \bigg\{\frac{\beta^{\lceil{\log\unicode{x3bb}}/{\log 2}\rceil+6}}{50}\cdot (\rho\beta/2)^2,\frac{1}{16}\notag\\[6pt] &\qquad\qquad \cdot \beta\rho\cdot \beta^{\lceil\log\unicode{x3bb}/\log 2\rceil+1}\bigg\}.\\[-20pt]\nonumber \end{align} $$
  Finally, note that
  $$ \begin{align*} \rho_1^{(n+1)}& = {\text{diam}}(T^nB_1^{(n+1)})\\[3pt] & \geq \dfrac{1}{4} \cdot \beta\cdot {\mathrm{diam}}(T^{n}B_{H^{(n)}+2})\\[3pt] & = \dfrac{1}{4} \cdot \beta \cdot {\mathrm{diam}} (T^{n}B_{(L^{(n)}-1)+\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+3})\\ & \geq \dfrac{1}{4} \cdot \beta \cdot {\mathrm{diam}}(B_{(L^{(n)}-1)+\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+3})\cdot \dfrac{1}{R}\cdot \psi^{n-1}\\ & \geq 4^{-\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+3}\cdot \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot {\mathrm{diam}}(T^{n}B_{L^{(n)}-1}^{(n)}). \end{align*} $$
  Note that $T^{n-1}B_{L^{(n)}-1}^{(n)}$ is included in a cylinder $J^{(1)}$ of order $1$ with diameter not exceeding that of $[{1}/{N^{(n)}},{1}/({N^{(n)}-1})]$ and ${\mathrm {diam}}(J^{(1)})=({1}/({N^{(n)}(N^{(n)}-1)}))\leq ({2}/({(N^{(n)})^2}))$ . Hence, by Lemma 4, we have
  (17) $$ \begin{align} \rho_1^{(n+1)} &= \frac{{\mathrm{diam}}(T^n B_1^{(n+1)})}{{\mathrm{diam}}([0,1])}\notag \\ & \geq [{\mathrm{diam}}(T^{n-1}B_1^{(n+1)})/{\mathrm{diam}}(J^{(1)})]\cdot \frac{1}{4}\notag \\ & \geq [(1/4)^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot {\mathrm{diam}}(T^{n-1}B_{L^{(n)}-1}^{(n)})/{\mathrm{diam}}(J^{(1)})]\cdot \frac{1}{4}\notag \\ & \geq \bigg[\frac{\beta^2}{25(N^{(n)})^2}\bigg/{\mathrm{diam}}(J^{(1)})\bigg]\cdot (1/4)^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot \frac{1}{4}\notag \\[6pt] & \geq \bigg[\frac{\beta^2}{25(N^{(n)})^2}\bigg/\frac{2}{(N^{(n)})^2}\bigg]\cdot (1/4)^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+4}\cdot \frac{1}{4} \end{align} $$
  (18) $$ \begin{align} &= \frac{1}{50}\cdot 4^{-\lceil {\log \unicode{x3bb}}/{\log 2}\rceil-5}\cdot \beta^{\lceil {\log \unicode{x3bb}}/{\log 2}\rceil+6}\notag \\[6pt] & \overset{(6)}{\geq} \rho. \end{align} $$

Now we verify our hypotheses hold in the progression.

We first prove Lemma 2 for this system and then use it to show Statement 3 holds.

Proof of Lemma 1 for System II.

For Statement 1, note that if a ball B contains two vertices $^{(n)}$ , $T^{n-1}B$ contains a cylinder of order n, so ${\mathrm {diam}}(T^{n-1}B)\geq ({1}/({N^{(n)}(N^{(n)}+1)}))$ where ${1}/{N^{(n)}}$ is the rightmost vertex $^{(1)}$ in $T^{n-1}B$ . Hence, if $B_1^{(n)}$ is not of Type I, then either it is Type II, or ${\mathrm {diam}}\ (T^{n-1}B_1^{(n)})\geq ({1}/({N^{(n)}(N^{(n)}+1)}))\geq ({\beta ^2}/({4(N^{(n)})^2}))$ (of Type III).

By assuming Statements 2 and 3 for $n<M$ , if level $(M-1)$ is Type III, then Statement 2 for $n=M$ follows immediately from (13) and (18). Next, we show that Lemma 2 also holds for System II.

Proof of Lemma 2 for II.

Assume the contrary; suppose there are two consecutive Type II levels, say n and m with $n<m$ , without any intermediate Type III level. Since $n,m$ are consecutive Type II levels without any intermediate Type III level, we know that $B_2^{(n)}=B_1^{(n+1)}=B_1^{(m)}$ . Let $v_1$ be the vertex $^{(n)}$ in $B_1^{(n)}$ and let $v_2$ be the vertex $^{(m)}$ in $B_1^{(m)}$ ; denote $N_1:=({1}/({T^{n-1}v_1}))$ and $N_2:=({1}/({T^{m-1}v_2}))$ . Then we have two cases to consider, depending on which connected component of $B_1^{(n)}\backslash A_1^{(n)}=B_1^{(n)}\backslash B(v_1,\beta \cdot ({\rho _1^{(n)}}/{2}))\, \mathsf {Bob}$ picks to place $B_2^{(n)}$ .

• • Case 1: Suppose that $T^{n-1}B_2^{(n)}$ contains a neighborhood of 0. Then

  $$ \begin{align*} {\mathrm{diam}}(T^{m-1}B_1^{(m)})& \geq {\mathrm{diam}}\bigg(\bigg(0,\frac{1}{N_2}\bigg]\bigg)\cdot \frac{\beta}{4} =\frac{1}{N_2}\cdot \frac{\beta}{4}, \end{align*} $$
  but this contradicts that level m is of Type II.

• • Case 2: Suppose that $T^{n-1}B_2^{(n)}$ contains a neighborhood of 1. Then

  $$ \begin{align*} {\mathrm{diam}}(T^{m-1}B_1^{(m)})& \geq {\mathrm{diam}}\bigg(\bigg[\frac{1}{N_2},1\bigg)\bigg)\geq {\mathrm{diam}}\bigg(\bigg[\frac{1}{2},1\bigg)\bigg)\cdot \frac{\beta}{4} \geq \frac{1}{N_2}\cdot \frac{\beta}{4}, \end{align*} $$
  but this contradicts that level m is of Type II.

Therefore, we have proved the lemma by contradiction.

Then by Lemma 2 for System II, there is at most one Type II level between level $J^{(M)}$ and level M, so

$$ \begin{align*} \rho_1^{(M)}\geq \beta\cdot {\mathrm{diam}} (T^{J^{(M)}}B_1^{(J^{(M)}+1)})\overset{(13), (18)}{\geq} \beta\cdot \rho. \end{align*} $$

Since Statement 3 is assumed for all $0<n<M$ , by the expanding property of T (Lemma 3), note that for all $0<n<M$ ,

$$ \begin{align*} {\mathrm{diam}}(T^{n+\unicode{x3bb}-1}B_1^{(n+\unicode{x3bb})})\geq \beta \cdot \tfrac{1}{4}\cdot R\cdot \psi^{\unicode{x3bb}-1}\cdot {\mathrm{diam}}(T^n B_1^{(n)})\geq \beta \cdot \tfrac{1}{4}\cdot R\cdot \psi^{\unicode{x3bb}-1}\cdot \rho\geq 1. \end{align*} $$

meaning that $T^{\lceil \unicode{x3bb} \rceil } B_1^{(n)}=[0,1]$ by its diameter, so within $\unicode{x3bb} $ levels any Type III level $(n)$ , there must be another Type III level, and proving Statement 3 for all n.

From (9) and (16), again we obtain the most crucial result: Proposition 5. We will finish our proof of Theorem 4 in the concluding subsection.

3.4 Completing the proofs

We are now ready to prove Theorem 4.

Let $\{w\}=\bigcap _{k,n} B_{k}^{(n)}$ . Then by Proposition 5, there exists some $\delta>0$ such that

$$ \begin{align*} {\mathrm{dist}}\ (T^n w,f_n(c^{(n)}))\geq \delta \quad \text{ for all } n\in \mathbb{N}. \end{align*} $$

On the other hand, since the sequence $(f_n)_n$ is equicontinuous, there exists some $\delta '>0$ so that as long as $|x-y|<\delta '$ , $|f_n(x)-f_n(y)|< {\delta }/{2}$ .

Since the sequence $(c^{(n)})_n$ converges to w, we may find N large enough such that for all $n\geq N$ , we have ${\mathrm {dist}}\ (c^{(n)},w)<\delta '$ . Hence, for all $n\geq N$ ,

$$ \begin{align*} {\mathrm{dist}}\ (T^n w,f_n(w))\geq {\mathrm{dist}}\ (T^n w,f_n(c^{(n)}))-{\mathrm{dist}}\ (f_n(c^{(n)}),f_n(w)) \geq \frac{\delta}{2}. \end{align*} $$