3·1. Proof of Theorem 9·9

We begin by proving our result on dot product trees. As discussed in the introduction, dot products are much simpler than distances because thickness is preserved under affine transformations. As a consequence, Theorem 1·9 is the simplest of our results.

Theorem 1·9 is an immediate consequence of Theorem 2·4 and the following lemma.

Proof. For any $x=(x^1, x^2)\in \mathbb{R}^2$ , we have $t\in \Pi_x(K_1\times K_2)$ if and only if $(t-x_1K_1)\cap (x_2K_2)\neq\phi$ . Since $\tau(t-x_1K_1)\cdot \tau(x_2K_2)=\tau(K_1)\cdot \tau(K_2)\geq 1$ , by the Newhouse gap lemma, this intersection will be non-empty for any x and t such that the sets $(t-x_1K_1)$ and $x_2K_2$ are linked. Denote the convex hull of $x_jK_j$ by $[a_j,a_j+\ell|x_j|]$ , and without loss of generality assume $\ell_1|x_1|\geq \ell_2|x_2|$ . The sets $t-x_1K$ and $x_2K$ are linked whenever

(*) \begin{align}a_1+a_2+\ell_1 |x_1|<t<a_1+a_2+\ell_1 |x_1|+\ell_2 |x_2|.\end{align}

The set of t satisfying ( $*$ ) is an interval of length $\ell_2 |x_2|$ , and so (i) follows immediately. To prove (ii), assume $(x_1,x_2)\in Q$ and therefore $|x_j-x_j^0|<\delta$ for each j. The value t satisfies ( $*$ ) for all such $x_1,x_2$ provided

\begin{align*}a_1+a_2+\ell_1 |x_1^0|+\ell_1\delta<t<a_1+a_2+\ell_1 |x_1^0|+\ell_2 |x_2^0|-\ell_1\delta-\ell_2\delta.\end{align*}

This inequality determines an interval of length $\ell_2|x_2^0|-(\ell_2+2\ell_1)\delta$ .

Note that when we apply Theorem 2·4, we can start with any skeleton $x^1,\dots,x^{k+1}\in K\times K$ such that none of the points $x^i$ are on the axes.

3·2. Proof of Theorem 1·7

As in the previous section, the proof will rely on the mechanism established in Theorem 2·4 coupled with a pin wiggling lemma. The difference is that the lemma of this section will not follow directly from the linear theory and some preliminary set up is required.

First, observe that given a pin $x\in \mathbb{R}^2$ and distance $t\in \mathbb{R}$ , we have $t\in \Delta_x(K\times K)$ whenever $y_2=g_{x,t}(y_1)$ for some $y = (y_1, y_2) \in K\times K$ , where

\begin{align*}g_{x,t}(z)=x_2+\sqrt{t^2-(z-x_1)^2}.\end{align*}

We would like to apply the Newhouse gap lemma to K and $g_{x,t}(K)$ to prove a pin wiggling lemma for distance sets (Lemma 3·5 below), then conclude that Theorem 1·7 follows (by Theorem 2·4). However, there is no thickness assumption on K which would guarantee $\tau(g_{x,t}(K))\geq 1$ , so we cannot apply Newhouse directly. However, if I is a sufficiently small interval about a non-singular point of $g_{x,t}$ , then the thickness of $g_{x,t}(K\cap I)$ is not too much smaller than that of K. This can be proved using a generalization of thickness which was introduced in [ Reference Simon and Taylor23 ], which we describe here.

Definition 3·2 ( $\epsilon$ -thickness). Let $K\subset\mathbb{R}$ be a Cantor set, let u be a right endpoint of a bounded gap, G, and let $\epsilon>0$ . Let (a, b) be the closest gap to G with the property that $a>u$ and $(b-a)>(1-\epsilon)|G|$ . The $\epsilon$ -bridge of u, denoted $B_\epsilon(u)$ , is the interval (u, a). We make analogous definitions for left endpoints. The $\epsilon$ -thickness of K at u is the quantity

\begin{align*}\tau_\epsilon(K,u)\,{:\!=}\, \frac{|B_\epsilon(u)|}{|G|}.\end{align*}

Finally, the $\epsilon$ -thickness of the Cantor set K is the quantity

\begin{align*}\tau_\epsilon(K)\,{:\!=}\,\inf_u \tau_\epsilon(K,u),\end{align*}

where the infimum is taken over all gap endpoints u.

We record some easily verifiable properties of $\epsilon$ -thickness in the following proposition.

With these definitions in place, we can prove that the image of a thick Cantor set must at least contain a thick Cantor set. More precisely, we have the following lemma, which is essentially [ Reference Simon and Taylor23 , lemma 3·8]. We include a proof here for completeness.

Lemma 3·4 (Thickness of the image is nearly preserved). Let $K\subset\mathbb{R}$ be a Cantor set, let u be a right endpoint of some gap of K, and let g be a function which is continuously differentiable on a neighbourhood of u and satisfies $g'(u)\neq 0$ . For every $\epsilon>0$ , there exists $\delta>0$ such that

\begin{align*}\tau(g(K\cap [u,u+\delta]))>\tau_\epsilon(K)(1-\epsilon).\end{align*}

Proof. Fix $\epsilon>0$ . By continuity of g $^{\prime}$ , we may choose $\delta$ such that for all $x_1,x_2\in[u,u+\delta]$ we have

\begin{align*}\left|\frac{g'(x_1)}{g'(x_2)}-1\right|<\epsilon.\end{align*}

Note that our choice of $\delta$ guarantees that g is monotone on the interval $[u,u+\delta]$ , so for any subinterval I the mean value theorem guarantees the existence of some $x_I\in I$ such that $|g(I)|=|I|\cdot |g'(x_I)|$ . Let v be the endpoint of some gap G in $K\cap [u,u+\delta]$ . We first observe $g(B_\epsilon(v))\subset B_{\epsilon^2}(g(v))$ . To prove this, note that any gap in $g(K\cap [u,u+\delta])$ is the image of a gap in $K\cap [u,u+\delta]$ . Therefore, it suffices to prove that any gap $H\subset B_\epsilon(v)$ satisfies $|g(H)|<(1-\epsilon^2)|g(G)|$ . To this end, observe that if H is a gap in $B_\epsilon(v)$ , then $|H| \le (1-\epsilon)|G|$ , and we have

\begin{align*}|g(H)|&=|H|\cdot |g'(x_H)| \\&\le(1-\epsilon)|G|\cdot\frac{|g'(x_H)|}{|g'(x_G)|}\cdot |g'(x_G)| \\&<(1-\epsilon)|G|\cdot (1+\epsilon)\cdot |g'(x_G)| \\&=(1-\epsilon^2)|g(G)|.\end{align*}

It follows by definition then that the thickness of the image at the point g(v) satisfies

\begin{align*}\tau_{\epsilon^2}(g(K\cap[u,u+\delta]),g(v)) &\geq \frac{|g(B_\epsilon(v))|}{|g(G)|}\\&=\frac{|B_\epsilon(v)|}{|G|}\cdot \left|\frac{g'(x_{B_\epsilon(v)} )}{g'(x_G)}\right| \\&> \tau_\epsilon(K\cap[u,u+\delta],v)\cdot(1-\epsilon).\end{align*}

Taking the infimum over v, we have

\begin{align*}\tau(g(K\cap[u,u+\delta]))\geq \tau_{\epsilon^2}(g(K\cap[u,u+\delta])) > \tau_\epsilon(K)(1-\epsilon).\end{align*}

This completes the proof of the lemma.

We are now prepared to state and prove the key lemma for distances.

Lemma 3·5 (Pin wiggling for distances). Let $K_1,K_2$ be Cantor sets satisfying $\tau(K_1)\cdot \tau(K_2) > 1$ . For any $x^0\in \mathbb{R}^2$ , there exists an open S about $x^0$ so that

\begin{align*}\bigcap_{x\in S} \Delta_{x}(K_1\times K_2)\end{align*}

has non-empty interior.

Proof. For $(x,t)\in\mathbb{R}^2\times (0,\infty)$ , define

(3·1) \begin{equation}g_{x,t}(z)=x_2+\sqrt{t^2-(z-x_1)^2},\end{equation}

and observe that $t\in\Delta_{x}(K_1\times K_2)$ provided $K_2\cap g_{x,t}(K_1)\neq\phi$ . Let $u_j$ be a right endpoint of a bounded gap of $K_j$ , and without loss of generality assume $u_j>x_j^0$ , where $x^0 = \left(x^0_1, x^0_2\right)$ . We choose small subsets $\widetilde{K_j}\subset K_j$ with left endpoint $u_j$ , and focus on the box $\widetilde{K_1}\times \widetilde{K_2}$ (Figure 2). Let $t_0=|x^0-(u_1,u_2)|$ . Let $\widetilde{K_j}=K_j\cap[u_j,u_j+\delta_j]$ for some small $\delta_1,\delta_2$ . In particular, we choose $\delta_j>0$ small enough that $\tau(\widetilde{K_2})\cdot \tau(g(\widetilde{K_1}))>1$ (this is possible by Lemma 3·4), and so that $\widetilde{K_1}$ is in the domain of $g_{x,t}$ whenever (x, t) is sufficiently close to $(x^0,t_0)$ . We will also assume $u_j+\delta_j\in K_j$ .

Fig. 2. The box containing $\widetilde{K_1}\times \widetilde{K_2}$ .

By the Newhouse gap lemma (Proposition 2·3), we will have $\widetilde{K_2}\cap g_{x,t}(\widetilde{K_1})\neq\phi$ whenever the parameters (x, t) are such that $\widetilde{K_2}$ and $g_{x,t}(\widetilde{K_1})$ are linked. To this end, consider the set

\begin{align*}U=\{(x,t)\in\mathbb{R}^2\times\mathbb{R}\,{:}\,g_{x,t}(u_1+\delta_1)<u_2<g_{x,t}(u_1)<u_2+\delta_2\}.\end{align*}

By construction, for any $(x,t)\in U$ we have $K_2$ and $g_{x,t}(K_1)$ linked (Figure 3) and hence $t\in\Delta_x(K_1\times K_2)$ . We claim that U is an open set containing a point of the form $(x^0,t)$ for some t. Lemma 3·5 follows from this claim, as we can then take open neighbourhoods S, T of $x^0,t$ respectively such that

\begin{align*}T\subset \bigcap_{x\in S} \Delta_{x}(K_1\times K_2).\end{align*}

Fig. 3. Graphs of g.

To prove the claim, we first observe that U is open, since for fixed z the quantity $g_{x,t}(z)$ is a continuous function of (x, t). To finish the proof, we must find a t such that $(x^0,t)\in U$ . By construction, we have $g_{x^0,t_0}(u_1)=u_2$ . Since the quantity $g_{x,t}(z)$ is strictly increasing in t, for any $t>t_0$ we will have $g_{x^0,t}(u_1)>u_2$ . On the other hand, by continuity in t we will also have $g_{x^0,t}(u_1+\delta_1)<u_2$ and $g_{x^0,t}(u_1)<u_2+\delta_2$ whenever t is sufficiently close to $t_0$ . Therefore, we can choose t with the property that $(x^0,t_0)\in U$ .

Theorem 1·7 follows from Lemma 3·5 and Theorem 2·4. Note that when we apply Theorem 2·4, we can start with any skeleton $x^1,\dots,x^{k+1}\in K_1\times K_2$ provided no two points $x^i,x^j$ share a coordinate.

3·3. Proof of Theorem 1·14

As in the previous two sections, Theorem 1·14 on $\phi$ distance trees is an immediate consequence of Theorem 2·4 and the following lemma.

Lemma 3·6 (Pin wiggling for $\phi$ distance trees). Let $K_1,K_2$ be Cantor sets satisfying $\tau(K_1)\cdot \tau(K_2) > 1$ Suppose $\phi\,{:}\,\mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$ satisfies the derivative condition of Definition 1·12 on $A\times B$ for open $A,B\subset \mathbb{R}^2$ , each of which intersects $K_1\times K_2$ . For any $x^0 \in A$ , there exists an open S about $x^0$ so that

\begin{align*}\bigcap_{x\in S} \Delta_{\phi, x}(K_1\times K_2)\end{align*}

has non-empty interior, where $\Delta_{\phi, x}(K_1\times K_2) = \{ \phi(x,y)\,{:}\, y \in K_1\times K_2\}.$

The strategy for establishing Lemma 3·6 is as follows: given a pin $x\in \mathbb{R}^2$ and distance $t\in \mathbb{R}$ , we note that $t\in \Delta_{\phi,x}(K_1\times K_2)$ whenever $\phi(x,y) = t$ for some $y=(y_1,y_2)\in K_1\times K_2$ . We then use the implicit function theorem to solve for $y_2$ in terms of $(x, y_1,t)$ and call the resulting function g. Observing that g behaves like the function $g_{x,t}$ introduced in the proof of Lemma 3·5 from the previous section, the lemma then follows from the exact proof used for Euclidean distances. The only real effort of the proof then is setting up the implicit function theorem.

Proof. Let $\varphi_x(y) = \phi(x,y)$ as in Definition 1·12 so that for all $(x,y) \in A\times B$ ,

(3·2) \begin{equation}\frac{\partial (\varphi_x)}{\partial y_i} (y)\neq 0 \text{ for } i =1,2.\end{equation}

Define $F\,{:}\,\mathbb{R}^2\times \mathbb{R}^2 \times \mathbb{R} \rightarrow \mathbb{R}$ by

(3·3) \begin{equation}F(x, y,t) = \phi(x,y) -t.\end{equation}

Then F is continuously differentiable on $A\times B\times \mathbb{R}$ . Moreover, it follows by (3·2) that the partial derivative of F in $y_2$ is non-vanishing on $A\times B\times \mathbb{R}$ . Choose $x^0 \in A$ and $u=(u_1,u_2) \in B\cap (K_1 \times K_2)$ . Set

\begin{align*}t_0 = \phi\!\left( x^0, u\right)\end{align*}

so that

\begin{align*}F\!\left(x^0,u, t_0\right) = 0.\end{align*}

By the implicit function theorem, there exists a real-valued function g and a $\delta$ -ball about the point $(x^0, u_1, t_0)$ , denoted by $B_\delta= B_{\delta}(x^0, u_1, t_0)$ , so that:

1. (i) g is continuously differentiable on $B_{\delta}$ ;

2. (ii) $g(x^0,u_1, t_0) = u_2$ ;

3. (iii) if $(x, y_1, t) \in B_{\delta}$ , then $(x, y) \in A\times B$ for all $y= (y_1, g(x,y_1, t) )$ ;

4. (iv) $\phi(x, y) = t$ for all $y= (y_1, g(x,y_1, t) ) $ when $(x, y_1, t) \in B_{\delta}$ .

Moreover, if $g_{x,t}(y_1) = g(x,y_1, t)$ , then

\begin{align*}g_{x,t}^{\prime} (y_1) = -\left( \frac{\partial (\varphi_x)}{\partial y_1}(y_1, g(x, y_1, t))\right) / \left( \frac{\partial (\varphi_x)}{\partial y_2}(y_1, g(x, y_1, t))\right) \neq 0\end{align*}

for $(x, y_1, t) \in B_\delta= B_{\delta}(x^0,u_1, t_0)$ . In other words, there is a neighbourhood of $(x^0, t_0) $ so that the derivative of $g_{x,t}$ is non-vanishing in a neighbourhood of $u_1$ .

Replacing the function in (3·1) of Lemma 3·5 with this new $g_{x,t}$ , the proof proceeds as in the proof of Lemma 3·5.

3·4. Distance trees in the middle thirds cantor set

In this section, we prove Theorem 1·10. The new twist in this context is that the middle thirds Cantor set has thickness equal to 1, not strictly greater than 1. Our method uses Lemma 3·4 which controls how much smaller the thickness of $g(\widetilde{K})$ can be compared to K, where $\widetilde{K}$ is a choosen subset of K. If $\tau(K)=1$ , then we cannot assume $\tau(g(\widetilde{K}))\geq 1$ and therefore cannot apply the Newhouse gap lemma directly. However, the self similarity of the middle thirds Cantor set provides a tool that allows one to adapt the proof of the Newhouse gap lemma in this special setting. This observation was used in [ Reference Simon and Taylor23 ] to prove that the pinned distance set $\Delta_x(C_{1/3}\times C_{1/3})$ has non-empty interior. We use this idea to prove a pin wiggling lemma for the middle thirds Cantor set.

Before proceeding, we introduce some terminology. Let $C_{1/3}$ denote the standard middle thirds Cantor set in the interval [0, 1]. The standard construction of this set is given by defining $C_{1/3}$ as the intersection of a family of sets $\{C_n\}$ , where each $C_n$ is the union of $2^n$ closed intervals of length $1/3^n$ . Define a section of $C_{1/3}$ to be the intersection of $C_{1/3}$ with any one of the intervals making up any of the sets $C_n$ .

Proof. We begin with a simple observation. Let $I_1,I_2$ denote the convex hulls of $K_1, K_2$ , respectively. By the mean value theorem, for any interval J there exists $x_J\in J$ such that $|g(J)|=|J|\cdot|g'(x_J)|$ . In particular, if U and V are bounded gaps of $K_1$ and $|U|>|V|$ , we have $|U|\geq 3|V|$ and therefore

\begin{align*}\\[-20pt]|g(U)|&=|U|\cdot |g'(x_U)| \\[3pt] &> 3|V|\cdot 1 \\[3pt] &> |g'(x_V)|\cdot |V| \\[3pt] &= |g(V)|.\end{align*}

It follows that the bridges of the images are the images of the bridges. More precisely, the bridge of gap g(U) in $g(K_1)$ is g(B), where B is the bridge next to the gap U.

Now, to prove the lemma, we proceed by contradiction and assume $K_2\cap g(K_1)=\phi$ . The strategy of the proof is a variant of the strategy used to prove the original Newhouse gap lemma. We construct sequences of gaps $U_n, V_n$ satisfying the following conditions:

• (i) $U_n$ is a bounded gap of $K_2$ and $g(V_n)$ is a bounded gap of $g(K_1)$ ;

• (ii) $U_n$ and $g(V_n)$ are linked for every n;

• (iii) for every n, either

  1. (a) $U_{n+1}=U_n$ and $|g(V_{n+1})|<|g(V_n)|$ , or

  2. (b) $V_{n+1}=V_n$ and $|U_{n+1}|<|U_n|$ .

Thus, at each stage we are replacing one of the gaps $U_n,g(V_n)$ with a strictly smaller one and leaving the other unchanged. In particular, this means that one of the two gap sequences must have a subsequence which is strictly decreasing in length. Since gaps of different sizes are automatically disjoint and the total length of the gaps is bounded, we must have either $|U_n|\to 0$ or $|g(V_n)|\to 0$ as $n\to\infty$ . However, because $U_n$ and $g(V_n)$ are linked, the closure of $U_n$ contains points of both $K_2$ and $g(K_1)$ , and similarly for $g(V_n)$ . This implies that the distance between $K_2$ and $g(K_1)$ is zero, which contradicts the assumption $K_2\cap g(K_1)= \phi$ .

We construct our sequences $\{U_n\}$ and $\{V_n\}$ recursively. First, we construct $U_1$ and $V_1$ . Recall that $I_i$ denotes the convex hull of $K_i$ for $i=1,2$ . Since $I_2$ and $g(I_1)$ are linked by assumption, there is a bounded gap $U_1$ of $K_2$ which is a subset of $I_2\cap g(I_1)$ . Let u be an endpoint of $U_1$ . In particular we have $u\in K_2$ , and therefore, by our assumption that $K_2\cap g(K_1)=\phi$ , it follows that u is in some bounded gap $g(V_1)$ of $g(K_1)$ .

Next, suppose we have constructed $U_n$ and $V_n$ satisfying the properties in the first two bullet points above. We construct $U_{n+1}$ and $V_{n+1}$ satisfying the third bullet point. By construction, $U_n$ and $g(V_n)$ are linked. Let $B_n^U$ be the bridge corresponding to $U_n$ on the same side as $g(V_n)$ (see Figure 4), and let $g\left(B_n^V\right)$ be the bridge of $g(V_n)$ on the same side as $U_n$ . We claim that one of the following two inequalities must hold:

\begin{align*}|g(B_n^V)|&>|U_n| \\[4pt] |B_n^U|&>|g(V_n)|.\end{align*}

This follows because we are working with the middle thirds Cantor set, so bounded gaps and corresponding bridges have the same length, i.e., we have $|B_n^U|=|U_n|$ and $|B_n^V|=|V_n|.$ The middle thirds Cantor set also has the property that if U and V are gaps with $|U|>|V|$ , we automatically have $|U|\geq 3|V|$ . Finally, by our assumptions on g, we have $|J|< |g(J)|< 3|J|$ for every interval $J\subset I_1$ . The claim then follows, as $|U_n|>|V_n|$ implies the second inequality and $|U_n|\leq |V_n|$ implies the first.

Fig. 4. Construction of gap sequence (gaps are red, bridges are blue).

Assume the second inequality holds (an analogous argument will apply when the first holds). This means one endpoint of $g(V_n)$ is in $U_n$ and the other is in $B_n^U$ (see Figure 4 again). Recall that we are proving the lemma by contradiction, assuming $K_2\cap g(K_1)=\phi$ . This assumption means the endpoint of $g(V_n)$ which is in $B_n^U$ must be contained in some gap $U_{n+1}$ of $K_2$ , and by definition of a bridge we must have $|U_{n+1}|<|U_n|$ . We then take $V_{n+1}=V_n$ . This completes the construction.

Theorem 3·8. Given a point $x^0\in \mathbb{R}^2$ , let $W_{x^0}$ denote the open wedge

\begin{align*}W_{x^0}=\left\{x\in\mathbb{R}^2\,{:}\,\left(x_2-x_2^0\right)<\left(x_1-x_1^0\right)<3\left(x_2-x_2^0\right)\right\}.\end{align*}

Let $K_1,K_2$ be sections of the middle thirds Cantor set, and suppose $K_1\times K_2\subset W_{x^0}$ (Figure 5 ). Then, there exists an open neighbourhood S of $x^0$ such that

\begin{align*}\bigcap_{x\in S}\Delta_x(K_1\times K_2)\end{align*}

has non-empty interior.

Fig. 5. The wedge of acceptable boxes.

Proof. For $x\in\mathbb{R}^2$ and $t\in\mathbb{R}$ , consider the function

\begin{align*}g_{x,t}(z)=x_2+\sqrt{t^2-(z-x_1)^2}.\end{align*}

We have

\begin{align*}g_{x,t}^{\prime}(z)=-\frac{z-x_1}{g_{x,t}(z)-x_2},\end{align*}

so $1<|g_{x,t}^{\prime}(z)|<3$ whenever $(z,g_{x,t}(z))\in W_x$ . For $j=1,2$ , let $u_j=\min K_j$ , so that $u=(u_1,u_2)$ denotes the lower left corner of $K_1\times K_2$ , and let $t_0=|x^0-u|$ (thus, we have $g_{x^0,t_0}(u_1)=u_2$ ). For every $\delta>0$ , define $\widetilde{K_j}=K_j\cap [u_1,u_1+\delta]$ and consider the set

\begin{align*}U_\delta=\{(x,t) \,{:}\,\widetilde{K_2}\text{ and }g_{x,t}(\widetilde{K_1}) \text{ are linked, and }(z,g_{x,t}(z))\in W_x \text{ for all }z\in[u_1,u_1+\delta]\}.\end{align*}

By Lemma 3·7, if $(x,t)\in U_\delta$ then $t\in \Delta_x(K_1\times K_2)$ . Clearly $U_\delta$ is open, so as in the previous proofs it suffices to show that $U_\delta$ contains a point of the form $(x^0,t)$ . We first observe that if $\delta$ is sufficiently small, the sets $\widetilde{K_2}$ and $g_{x,t}(\widetilde{K_1})$ are linked for all $t\in(t_0,t_0+\delta)$ . This is because for any $t>t_0$ we have $g_{x^0,t}(u_1)>u_2$ , and for any $t<u_1+\delta$ we will also have $g_{x^0,t}(u_1)\gt u_2+\delta$ and $g_{x^0,t}(u_1+\delta)\gt u_2$ . Finally, since $u\in W_{x^0}$ by assumption, if $\delta'$ is sufficiently small we will have $(z,g_{x^0,t}(z))\in W_{x^0}$ for all $z\in [u_1,u_1+\delta]$ and all $t\in (t_0,t_0+\delta')$ .

Proof of Theorem 1·10. By Theorem 2·4, it suffices to find a skeleton $x^1,...,x^{k+1}\in C_{1/3}\times C_{1/3}$ such that whenever $i\gt j$ we have $x^j\in W_{x^i}$ . Wedge membership is transitive in the sense that $x^3\in W_{x^2}$ and $x^2\in W_{x^1}$ implies $x^3\in W_{x^1}$ , so it suffices to construct our skeleton so that $x^{i+1}\in W_{x^i}$ for every i. We construct such a sequence recursively as follows. Let $x^1$ be the origin. One can check that $W_{x^1}$ contains the box $[8/9,1]\times [6/9,7/9]$ (refer again to Figure 5). Let $x^2=(8/9,6/9)$ . Since $x^2$ is the lower left corner of a similar copy of $C_{1/3}\times C_{1/3}$ , one can run the same argument by symmetry and take $x^3$ to be the lower left corner of a box contained in $W_{x^2}$ . This process can be repeated as many times as needed.