2. Preliminaries

2.1. Outline of paper

We will prove Theorem 1·1 and Corollary 1·2 for the particular mapping

\begin{equation*}f(z)\,:\!=\,f_{-1}(z)=e^z-1.\end{equation*}

The proof for all other Fatou parameters will then follow from the conjugacy [Reference Rempe19, theorem 1·2]. Theorem 1·3 and Corollary 1·4 will essentially follow from the case $a=-1$ as well.

Instead of working directly in the Julia set $J(f)\subset \mathbb C$ , we will work in a topologically equivalent subset of $\mathbb R^2$ known as a brush. In this section we will give the definition of a brush and its connection to exponential Julia sets. We will then define a specific brush which is homeomorphic to J(f), together with a mapping which models f.

In Section 3 we will prove three easy lemmas for the model mapping. In Section 4 we will show that certain endpoint sets in brushes cannot be written as countable unions of nowhere dense C-sets, given the existence of certain Jordan curves. In Section 5 we will prove Theorem 1·1 by showing that such curves exist with respect to our model of $\dot E(f)$ . Theorem 1·1 is strengthened by Remark 6·2 in Section 6. Finally, the proofs of Theorem 1·3 and Corollary 1·4 are given in Section 7.

2.2. Brushes and Cantor bouquets

Let $\mathbb P=\mathbb R\setminus \mathbb Q$ . A brush is a closed subset of $\mathbb R^2$ of the form

\begin{equation*}B=\bigcup_{y\in Y} [t_y,\infty)\times \{y\},\end{equation*}

where $Y\subset \mathbb P$ and $t_y\in \mathbb R$ . The endpoints of B are the points $\langle t_y,y\rangle$ .

When $a\in ({-}\infty,-1]$ , the Julia set $J(f_a)$ is a Cantor bouquet [Reference Aarts and Oversteegen1, Reference Alhabib and Rempe-Gillen2] which is homeomorphic to a brush with a dense set of endpoints. For all other Fatou parameters, $J(f_a)$ is a pinched Cantor bouquet that is homeomorphic to a brush modulo a closed equivalence relation on its set of endpoints; see [Reference Rempe19, corollary 9·3] and [Reference Alhamed, Rempe and Sixsmith3]. The relation establishes a one-to-one correspondence between endpoints of J(f) and equivalence classes of endpoints ofB [Reference Lipham13, section 3·2].

2.3. Brush model of J(f)

We will now define a mapping $\mathcal F$ and a brush $J(\mathcal F)$ that model f and $J(\mathcal F)$ . These objects were studied extensively in [Reference Alhabib and Rempe-Gillen2, Reference Rempe19].

Let $\mathbb Z ^\omega$ denote the space of integer sequences $\underline{s}=s_0s_1s_2\cdots$ in the product (or lexicographic order) topology. Define $\mathcal F\,:\,[0,\infty)\times \mathbb Z ^\omega\to \mathbb R\times \mathbb Z ^\omega$ by

\begin{equation*}\langle t,\underline{s}\rangle\longmapsto \langle F(t)-2\pi|s_0|,\sigma(\underline{s})\rangle,\end{equation*}

where $F(t)=e^t-1$ and $\sigma$ is the shift map on $\mathbb Z ^\omega$ ; i.e.

\begin{equation*}\sigma(s_0s_1s_2\cdots)=s_1s_2s_3\cdots.\end{equation*}

For each $x=\langle t,\underline{s}\rangle\in [0,\infty)\times \mathbb Z ^\omega$ put $T(x)=t$ and $\underline s(x)=\underline{s}$ . The integer sequence $\underline{s}(x)$ is called the external address of x.

Define

\begin{equation*}J(\mathcal F)=\left\{x\in [0,\infty)\times \mathbb Z ^\omega\,:\,T\!\left(\mathcal F^n(x)\right)\geq 0\text{ for all }n\geq 0\right\}.\end{equation*}

Continuity of $\mathcal F$ implies that $J(\mathcal F)$ is closed in $[0,\infty)\times \mathbb Z ^\omega$ . Now let

\begin{equation*}\mathbb S=\left\{\underline s\in \mathbb Z^\omega\,:\,\text{ there exists } t\geq 0\text{ such that }\langle t,\underline{s}\rangle\in J(\mathcal F)\right\},\end{equation*}

and for each $\underline s\in \mathbb S$ put $t_{\underline{s}}=\min\{t\geq 0\,:\,\langle t,\underline{s}\rangle\in J(\mathcal F)\}.$ Then

\begin{equation*}J(\mathcal F)=\bigcup_{\underline s\in \mathbb S} [t_{\underline s},\infty)\times \{\underline s\}.\end{equation*}

Finally, $\mathbb Z^\omega$ can be identified with $\mathbb P$ using an order isomorphism between $\mathbb Z ^\omega$ in the lexicographic ordering and $\mathbb P$ in the real ordering [Reference Alhabib and Rempe-Gillen2, observation 3·2]. Under this identification, $J(\mathcal F)\subset \mathbb R\times \mathbb P$ . Hence $J(\mathcal F)$ is a brush. See Figure 1.

Fig. 1.

Three endpoints and curves in J(f) (left) and in $J(\mathcal F)$ (right).

The restrictions $\mathcal F\restriction J(\mathcal F)$ and $f\restriction J(f)$ are topologically conjugate [Reference Rempe19, section 9]. Hence $J(\mathcal F)$ is homeomorphic to J(f), and the action of $\mathcal F$ on $J(\mathcal F)$ captures the essential dynamics of f on J(f); see also [Reference Alhabib and Rempe-Gillen2, p.74].

3. Lemmas for $\mathcal F$

In orderto prove Theorem 1·1 we need a few basic lemmas regarding the model above. Define $F^{-1}(t)=\ln(t+1)$ for $t\geq 0$ , so that $F^{-1}$ is the inverse of F. For each $n\geq 1$ , the n-fold composition of $F^{-1}$ is denoted $F^{-n}$ .

Lemma 3·1 (Iterating between double squares)Let $x\in J(\mathcal F)$ . If $k\geq 5$ and

\begin{equation*}T\!\left(\mathcal F^{2k^2}(x)\right)\geq F^{k^2}(1),\end{equation*}

then $T(\mathcal F^{n}(x))\geq F^{k}(1)$ for all $n\in [2(k-1)^2,2k^2]$ .

Proof. Suppose $k\geq 5$ and $T(\mathcal F^{2k^2}(x))\geq F^{k^2}(1)$ . Let $n\in [2(k-1)^2,2k^2]$ . Then there exists $i\leq 4k-2$ such that $n=2k^2-i$ . We have

\begin{equation*}F^i\!\left(T\!\left(\mathcal F^{n}(x)\right)\right)=F^i\!\left(T\!\left(\mathcal F^{2k^2-i}(x)\right)\right)\geq T\!\left(\mathcal F ^i \!\left(\mathcal F^{2k^2-i}(x)\right)\right)=T\!\left(\mathcal F ^{2k^2}(x)\right)\geq F^{k^2}(1).\end{equation*}

Applying $F^{-i}$ to each side of the inequality shows that

\begin{equation*}T\!\left(\mathcal F^{n}(x)\right)\geq F^{k^2-i}(1)> F^{k}(1),\end{equation*}

where we used the fact $k^2-i>k$ for all $k\geq 5$ and $i\leq 4k-2$ .

Proof. Suppose $\underline s(y)=\underline s(x)$ and $T(y)> T(x)$ . Let $\varepsilon=T(y)-T(x)$ and $\delta=T(\mathcal F^n(y))-T(\mathcal F^n(x))$ . Note that $\delta>0$ . So $T(y)\leq T(x)+F^{-n}(\delta)$ by [Reference Alhabib and Rempe-Gillen2, observation 3·9]. Therefore $\delta \geq F^n(\varepsilon)$ . We have

\begin{equation*}T\!\left(\mathcal F^n(y)\right)\geq T\!\left(\mathcal F^n(y)\right)-T\!\left(\mathcal F^n(x)\right)=\delta \geq F^n\!\left(\varepsilon\right) =F^n\!\left(T(y)-T(x)\right)\end{equation*}

as desired.

Lemma 3·1 implies that $T(\mathcal F^n(x))\to\infty$ if the double square iterates $T(\mathcal F^{2k^2}(x))$ increase at a sufficient rate. Lemma 3·2 will be used to show that $T(\mathcal F^{2k^2}(x))$ increases at that rate for certain points $x\in J(\mathcal F)$ .

Proof. The proof is by induction on n. If $n=1$ then we have $F^{-n}(1)=F^{-1}(1)=\ln\!(2)<3=3/n$ . Now suppose the inequality holds for a given n. Then

\begin{equation*} F^{-(n+1)}(1)=F^{-1}(F^{-n}(1))<F^{-1}(3/n)=\ln\left(\frac{3}{n}+1\right)<\frac{3}{n+1}\end{equation*}

by calculus.

Although the series $\sum_{k=1}^\infty F^{-k}(1)$ diverges, Lemma 3·3 implies that $\sum_{k=1}^\infty F^{-k^2}(1)$ converges (to something less than $\pi ^2/2$ ).

4. Jordan curve lemma

The following topological lemma will also be required to prove Theorem 1·1. Recall that a C-set in a space X is an intersection of clopen subsets of X.

Lemma 4·1. Let

\begin{equation*}B=\bigcup_{y\in Y} [t_y,\infty)\times \{y\}\end{equation*}

be a brush, and let $E=\{\langle t_y,y\rangle\,:\,y\in Y\}$ denote the set of endpoints of B. Suppose $\tilde E\subset E$ and $\tilde E\cup \{\infty\}$ is connected. If there exists an open set $U\subset \mathbb R ^2$ such that:

1. (i) $U\cap \tilde E\neq\varnothing$ ;

2. (ii) $\partial U$ is a simple closed curve; and

3. (iii) $\partial U\cap E\subset \tilde E$ ,

then $\tilde E$ cannot be written as a countable union of nowhere dense C-sets.

Proof. Suppose that $\tilde E$ , U, and $\beta\,:\!=\,\partial U$ satisfy all of the hypotheses. Let

\begin{equation*}A\,:\!=\,\{y\in Y\,:\, \beta\text{ contains an interval of }[t_y,\infty)\times \{y\}\}.\end{equation*}

Since every collection of pairwise disjoint arcs of a simple closed curve is countable,A is countable. So $\tilde E\cap (\mathbb R\times A)$ is countable. Observe also that since B is closed in $ \mathbb R\times Y$ and $Y\subset \mathbb P$ , the space E has a neighbourhood basis of C-sets of the form $E\cap ({-}\infty,t]\times W$ , where $t\in \mathbb R$ and W is clopen in Y. By [Reference Dijkstra and Lipham4, theorem 4·7] and the assumption that $\tilde E\cup \{\infty\}$ is connected, we see that $\tilde E\setminus (\mathbb R\times A)\cup \{\infty\}$ is connected.

Now aiming for a contradiction, suppose that $\tilde E=\bigcup \{C_n\,:\,n<\omega\}$ where each $C_n$ is a nowhere dense C-set in $\tilde E$ . Let V be the unbounded component of $\mathbb R^2 \setminus \beta$ . Since $\tilde E\setminus (\mathbb R\times A)$ $\cup \{\infty\}$ is connected,

\begin{equation*}\tau\,:\!=\,\overline{U\cap \tilde E}\cap \overline{V\cap \tilde E}\cap \tilde E\setminus (\mathbb R\times A)\end{equation*}

is non-empty. Note also that $\tau$ is a relatively closed subset of

\begin{equation*}\beta\cap \tilde E\setminus (\mathbb R\times A)=\beta\cap E\setminus (\mathbb R\times A),\end{equation*}

which is a $G_\delta$ -subset of E. And E is a $G_\delta$ -subset of $\mathbb R \times \mathbb P$ because B is closed and $B\setminus E$ is the union of countably many closed sets $B+\langle {1}/{n},0\rangle$ . Therefore $\tau$ is completely metrisable. By Baire’s theorem there is an open rectangle $(x_1,x_2)\times (y_1,y_2)$ and $n<\omega$ such that $\varnothing\neq \tau\cap ((x_1,x_2)\times (y_1,y_2))\subset C_n.$ Let

\begin{equation*}\langle t_y,y\rangle\in \tau\cap ((x_1,x_2)\times (y_1,y_2)).\end{equation*}

See Figure 2.

Fig. 2.

Construction of clopen set in Lemma 4·1.

Since $y\notin A$ , there exist $x_3\in (t_y,x_2)$ and $y_3,y_4\in (y_1,y_2)\cap \mathbb Q$ such that $y_3<y<y_4$ and $\{x_3\}\times [y_3,y_4]\subset \mathbb R ^2\setminus \beta$ . Without loss of generality, assume $\{x_3\}\times [y_3,y_4]\subset V$ . Note that $U\cap\tilde E\cap ((x_1,x_3)\times (y_3,y_4))\neq\varnothing$ because

\begin{equation*}\langle t_y,y\rangle\in \overline{U\cap \tilde E}\cap ((x_1,x_3)\times (y_3,y_4)). \end{equation*}

Further, $C_n\cup (\tilde E\cap (\mathbb R\times A))$ is nowhere dense in $\tilde E$ , so there exists

\begin{equation*}z\in U\cap\tilde E\cap ((x_1,x_3)\times (y_3,y_4))\setminus (C_n\cup (\mathbb R\times A)).\end{equation*}

Since z is an endpoint and B is closed in $\mathbb R\times Y$ , we can find $y_5,y_6\in (y_3,y_4)\cap \mathbb Q$ and $x_4>x_1$ such that $y_5<y_6$ , $z\in(x_4,\infty)\times (y_5,y_6)$ , and

\begin{equation*}(\{x_4\}\times [y_5,y_6])\cup ([x_4,\infty)\times \{y_5,y_6\})\cap B=\varnothing.\end{equation*}

Further, since $z\notin C_n$ there is a relatively clopen subset O of $\tilde E\setminus (\mathbb R\times A)$ such that $z\in O$ and $C_n\cap O=\varnothing$ . Then

\begin{equation*}U\cap O \cap ([x_4,x_3]\times [y_5,y_6])\end{equation*}

is a non-empty bounded clopen subset of $\tilde E\setminus (\mathbb R\times A)$ . This contradicts the previously established fact that $\tilde E\setminus (\mathbb R\times A)\cup \{\infty\}$ is connected.

5. Proof of Theorem 1·1

As indicated in Section 2.2, we will identify $\mathbb Z ^\omega$ with $\mathbb P$ , so that

\begin{equation*}J(\mathcal F)\subset \mathbb R\times \mathbb P\subset \mathbb R ^2.\end{equation*}

Let $E(\mathcal F)=\{\langle t_{\underline{s}},\underline{s}\rangle\,:\,s\in \mathbb S\}$ denote the set of endpoints of $J(\mathcal F)$ . Define

\begin{align*}I(\mathcal F)&=\{x\in J(\mathcal F)\,:\,T(\mathcal F^n(x))\longrightarrow\infty\} \text{ and}\\\tilde{E}(\mathcal F)&= I(\mathcal F)\cap E(\mathcal F).\end{align*}

The conjugacy [Reference Rempe19, theorem 9·1] shows that $\tilde{E}(\mathcal F) \simeq \dot E(f)$ , and $\tilde{E}(\mathcal F)\cup \{\infty\}$ is connected by [Reference Alhabib and Rempe-Gillen2, theorem 3·4]. Thus to reach the conclusion that $\dot E(f)$ cannot be written as a countable union of nowhere dense C-sets, by Lemma 4·1 we only need to find a simple closed curve $\beta\subset \mathbb R ^2$ such that:

1. (i) if U is the bounded component of $\mathbb R ^2\setminus \beta$ then $U\cap \tilde E(\mathcal F)\neq\varnothing$ , and

2. (ii) $\beta\cap J(\mathcal F)\subset I(\mathcal F)$ (in particular, $\beta\cap E(\mathcal F)\subset \tilde E(\mathcal F)$ ).

The $\beta$ that we construct will essentially be the boundary of a union of rectangular regions in $\mathbb R ^2$ . We recursively define the collections of rectangles (or “boxes”) as follows. Choose n sufficiently large so that $({-}n,n)^2\cap \tilde{E}(\mathcal F)\neq\varnothing$ . For the sake of simplicity, let us assume $n=1$ . Let $\mathscr{B}_0=\{[{-}1,1]^2\}$ .

Proof. Suppose $\mathscr{B}_{k-1}$ has already been defined. Let

\begin{equation*}K=\bigcup_{B'\in \mathscr{B}_{k-1}} J(\mathcal F)\cap \{b'\}\times [c',d'].\end{equation*}

Since $J(\mathcal F)$ is closed in $\mathbb R^2$ [Reference Alhabib and Rempe-Gillen2, theorem 3·3] and $\mathscr{B}_{k-1}$ is finite, K is compact. Given a segment of integers $\langle s_0,s_1,...,s_{2k^2-1}\rangle$ of length $2k^2$ , observe that the set $\{\underline s\in \mathbb Z^\omega\,:\,\underline s\restriction 2k^2=\langle s_0,s_1,...,s_{2k^2-1}\rangle\}$ is clopen and convex in the lexicographic ordering on $\mathbb Z^\omega$ , and thus corresponds to the intersection of $\mathbb P$ with an interval whose endpoints are in $\mathbb Q$ . By compactness of K, the projection

\begin{equation*}\pi_1[K]=\{y\in \mathbb R \,:\,[0,\infty)\times \{y\}\cap K\neq\varnothing\}\end{equation*}

can be covered by finitely many of those intervals. Let $\mathcal I$ be a finite cover of $\pi_1[K]$ consisting of such intervals. Let

\begin{equation*}-1=q_0<q_1<q_2< \cdots <q_n=1\end{equation*}

be the increasing enumeration of all c’s and d’s used in $\mathscr{B}_{k-1}$ , together with the endpoints of all intervals in $\mathcal I$ . If necessary, we can insert a few more rationals to obtain $q_{i+1}-q_i\leq F^{-k^2}(1)$ for each $i< n$ . Since $\pi_1[K]$ is a compact set missing $\mathbb Q$ , we can also guarantee that for every $i<n-1$ , $[q_i,q_{i+1}]\cap \pi_1[K]=\varnothing$ or $[q_{i+1},q_{i+2}]\cap \pi_1[K]=\varnothing$ .

Note that b ^′ is the same for each box $B'\in \mathscr{B}_{k-1}$ (in fact, $b'=1+\sum_{i=1}^{k-1}F^{-i^2}(1)$ ). If $\{b'\}\times [q_i,q_{i+1}]\cap J(\mathcal F)\neq\varnothing$ then add the box $[b',b'+F^{-k^2}(1)]\times [q_i,q_{i+1}]$ to the collection $\mathscr{B}_k$ . It can be easily shown that defining $\mathscr{B}_k$ in this manner will satisfy conditions (i) through (viii).

Let $\mathscr{B}=\bigcup \{\mathscr{B}_k\,:\,k<\omega\}$ . For each $B=[a,b]\times [c,d]\in \mathscr{B}$ let

\begin{equation*}h_B\,:\,\{a\}\times [c,d]\longrightarrow ([a,b]\times \{c,d\})\cup (\{b\}\times [c,d])\end{equation*}

be a homeomorphism with fixed points $\langle a,c\rangle$ and $\langle a,d\rangle$ .

Claim 5·2. There is a continuous mapping

\begin{equation*}g\,:\,[0,1] \longrightarrow [1,\infty)\times [{-}1,1]\end{equation*}

such that $g(0)=\langle 1,-1\rangle$ , $g(1)=\langle 1,1\rangle$ , and $g[0,1]\cap J(\mathcal F)\subset I(\mathcal F)$ .

Proof. The mapping g will be the pointwise limit of a uniformly Cauchy sequence of continuous functions. To begin, define $g_0(t)=\langle 1,2t-1\rangle$ for all $t\in [0,1]$ . Now suppose $k\geq 1$ is given and $g_{k-1}$ has been defined. Let $t\in [0,1]$ . We set

\begin{equation*}g_k(t)=g_{k-1}(t)\end{equation*}

if $g_{k-1}(t)$ is not in any member of $\mathscr{B}_k$ . Otherwise, by item (vi) there is a unique box $B=[a,b]\times[c,d]\in \mathscr{B}_k$ which contains $g_{k-1}(t)$ . Inductively $g_{k-1}(t)$ is contained in some element of $\mathscr{B}_0\cup\ldots\cup \mathscr{B}_{k-1}$ , so by (iv) we have $g_{k-1}(t)\in \{a\}\times [c,d]$ . Put

\begin{equation*}g_k(t)=h_{B}(g_{k-1}(t)).\end{equation*}

The mapping $g_k$ defined in this manner is easily seen to be continuous.

Note that the image of $g_k$ is contained in $[1,\infty)\times [{-}1,1]$ , $g_k(0)=\langle 1,-1\rangle$ , and $g_k(1)=\langle 1,1\rangle$ . To see that the sequence $(g_k)$ is uniformly Cauchy, fix $\varepsilon>0$ . Let N be such that $\sum_{k=N}^\infty 1/k^2<\varepsilon/5$ . For any $t\in [0,1]$ , if $g_k(t)\neq g_{k-1}(t)$ then $g_k(t)$ and $g_{k-1}(t)$ belong to the same element of $\mathscr{B}_k$ . Hence, by items (ii) and (iii), the distance between $g_k(t)$ and $g_{k-1}(t)$ is at most $\sqrt{2}F^{-k^2}(1)$ . Combined with Lemma 3·3 we have

\begin{equation*} |g_k(t)-g_{k-1}(t)|\leq \sqrt{2}F^{-k^2}(1)<\sqrt{2}\frac{3}{k^2}<\frac{5}{k^2}\end{equation*}

for every $k<\omega$ and $t\in [0,1]$ . Thus for any $i> j\geq N$ and $t\in [0,1]$ ,

\begin{equation*} |g_i(t)-g_{j}(t)|\leq \sum_{k=j}^{i-1} |g_{k+1}(t)-g_{k}(t)| <\sum_{k=N}^\infty \frac{5}{k^2}<\varepsilon.\end{equation*}

This proves that $(g_k)$ is uniformly Cauchy. Therefore $(g_k)$ converges uniformly to a continuous function $g\,:\,[0,1]\to[1,\infty)\times [{-}1,1]$ which satisfies $g(0)=\langle 1,-1\rangle$ and $g(1)=\langle 1,1\rangle$ .

It remains to show that $g[0,1]\cap J(\mathcal F)\subset I(\mathcal F)$ . Let $x=g(t)\in g[0,1]\cap J(\mathcal F)$ . ByLemma 3·1 and the fact $F^k(1)\to\infty$ , to prove $x\in I(\mathcal F)$ it suffices to show $T(\mathcal F^{2j^2}(x))\geq F^{j^2}(1)$ for all $j\geq 1$ . Fix $j\geq 1$ . To prove $T(\mathcal F^{2j^2}(x))\geq F^{j^2}(1)$ , by continuity of $T\circ \mathcal F^{2j^2}$ we only need to show that every neighbourhood of x contains a point w such that $T(\mathcal F^{2j^2}(w))\geq F^{j^2}(1)$ .

Let W be a neighbourhood of x. Observe that since c and d are rational in item (i), the top and bottom edges of all boxes miss $J(\mathcal F)$ . So if $g_k(t)\in J(\mathcal F)$ , then $g_k(t)$ belongs to the right edge of an element of $\mathscr{B}_{k}$ . Then by the construction of $g_{k+1}$ and items (iv) and (viii) we get $g_{k+1}(t)\neq g_k(t)$ . Since $g_k(t)\to x$ , it follows that the sequence $g_0(t),g_1(t),\ldots$ is not eventually constant. So for every k there exists $B\in \mathscr{B}_k$ such that $g_k(t)\in B$ . By (ii) and (iii), the diameter of each box in $\mathscr{B}_k$ is at most $\sqrt{2} F^{-k^2}(1)$ , which goes to 0 as $k\to\infty$ (e.g. by Lemma 3·3). Since $g_k(t)\to x$ and x belongs to the interior of W, there exists $k>j$ such that W contains a box $B(k)\in \mathscr{B}_k$ . By (v) there exists $w\in B(k)\cap W\cap J(\mathcal F)$ .

By (iv), for each $i<k$ there exists a unique $B(i)\in \mathscr{B}_i$ such that $\pi_1[B(k)]\subset \pi_1[B(i)]$ . By (iv) and (v) there exists $y\in B(j)\cap B(j-1)\cap J(\mathcal F)$ . Then y is on the left edge of B(j). Let z be the point on the right edge of B(j) such that $\underline s(z)=\underline s(y)$ (see Figure 3). Then $z\in J(\mathcal F)$ , and by (i) we have $T(z)-T(y)= F^{-j^2}(1)$ . Thus, by Lemma 3·2

\begin{equation*}T\!\left(\mathcal F^{2j^2}(z)\right)\geq F^{2j^2}\!\left(T(z)-T(y)\right)=F^{2j^2}\!\left(F^{-j^2}(1)\right)=F^{j^2}(1).\end{equation*}

Note also that $\underline s(z)\restriction 2j^2= \underline s(w)\restriction 2j^2$ because z and w are contained in the same horizontal strip where the first $2j^2$ coordinates agree; see items (vii) and (iv). The preceding equations together with $T(z)\leq T(w)$ imply that

\begin{equation*}T\!\left(\mathcal F^{2j^2}(w)\right)\geq F^{j^2}(1),\end{equation*}

as desired.

Fig. 3.

Illustration of ${\color{blue}g_0},{\color{cyan}g_1} \textrm{ and}\; \color{red}g_2$ .

Fig. 4.

Proof of Claim 5·2.

By [Reference Nadler18, corollary 8·17] and [Reference Nadler18, theorem 8·23], there is an arc $\alpha\,:\,[0,1]\hookrightarrow g[0,1]$ such that $\alpha(0)=g(0)=\langle 1,-1\rangle$ and $\alpha(1)=g(1)=\langle 1,1\rangle$ . Let $\beta=(\{-1\}\times [{-}1,1])\cup ([{-}1,1]\times\{-1,1\}) \cup \alpha[0,1].$ Then $\beta$ is a simple closed curve, and by Claim 5·2 we have $\beta\cap E(\mathcal F)\subset \tilde E(\mathcal F).$ If U is the bounded component of $\mathbb R ^2\setminus\beta$ then $({-}1,1)^2\subset U$ , hence $U\cap \tilde{E}(\mathcal F)\neq\varnothing$ . Now we can apply Lemma 4·1 to see that $\tilde{E}(\mathcal F)$ cannot be written as a countable union of nowhere dense C-sets. This concludes the proof of Theorem 1·1.

6. Remarks on Theorem 1·1

Remark 6·2. The statement of Theorem 1·1 can be strengthened to: No neighbourhood in $\dot E(f)$ can be covered by countably many nowhere dense C-sets of $\dot E(f)$ . To see this, let $x_0\in\mathbb R ^2$ and $\varepsilon>0$ . Since $J(\mathcal F)$ is closed, there is a box $[a,b]\times [c,d]\subset B(x_0,\varepsilon/2)$ containing $x_0$ such that

\begin{equation*}(\{a\}\times [c,d]\cup[a,b]\times \{c,d\})\cap J(\mathcal F)=\varnothing.\end{equation*}

Begin the construction of the $\mathscr{B}_k$ ’s with $\mathscr{B}_0=\{[a,b]\times [c,d]\}$ , and choose l large enough so that

\begin{equation*}\sum_{k=1}^\infty F^{-(l+k)^2}(1)<\frac{\varepsilon}{2}.\end{equation*}

Construct $\mathscr{B}_k$ with $l+k$ replacing each k in items (ii), (iii) and (vii). By the arguments in Claim 5·2, if $x\in J(\mathcal F)$ lies in the limit of $(g_k)$ (constructed using the new $\mathscr{B}_k$ ’s), then $T(\mathcal F^{2n^2}(x))\geq F^{n^2}(1)$ for all $n>l$ and consequently $x\in I(\mathcal F)$ . As in Claim 5·2 we can construct an arc $\alpha$ such that $\beta\,:\!=\,(\{a\}\times [c,d]\cup[a,b]\times \{c,d\})\cup \alpha$ is a simple closed curve in $B(x_0,\varepsilon)$ , the bounded component of $\mathbb R ^2\setminus \beta$ contains $x_0$ , and $\beta\cap J(\mathcal F)\subset I(\mathcal F)$ . Lemma 4·1 now shows that $B(x_0,\varepsilon)$ cannot be covered by countably many nowhere dense C-sets of $\tilde E(\mathcal F)$ .

7. The space $F(f_a)\cup I(f_a)$ when $a\in ({-}\infty,-1]$

We are now ready to prove Theorem 1·3 and Corollary 1·4. Fix $a\in ({-}\infty,-1]$ .

Proof of Theorem 1·3. Let $z\in \mathbb C$ . By [Reference Rempe19, section 9] and [Reference Alhabib and Rempe-Gillen2, theorem 2·8], there is a homeomorphism $\varphi\,:\,\mathbb R ^2\to \mathbb C$ such that $\varphi[J(\mathcal F)]=J(f_a)$ and $\varphi[I(\mathcal F)]=I(f_a)$ . Let $x=\varphi^{-1}(z)$ . We have shown that there is a Jordan curve $\beta\subset(\mathbb R ^2\setminus J(\mathcal F))\cup I(\mathcal F)$ around x; see Remark 6·2, or simply begin the construction in Section 6 with $\mathscr{B}_0=\{[{-}n,n]^2\}$ where $n\in \mathbb N$ is such that $x\in ({-}n,n)^2$ . Then $\varphi[\beta]\subset F(f_a)\cup I(f_a)$ is a Jordan curve which separates z from $\infty$ .

In the proof below we will make use of two well-known facts: (a) $F(f_a)$ is path-connected and (b) each component of $J(f_a)$ is contained in $I(f_a)$ , with the possible exception of its endpoint.

Proof of Corollary 1·4. Let $z_0,z_1\in F(f_a)\cup I(f_a)$ . We will find a path in $F(f_a)\cup I(f_a)$ from $z_0$ to $z_1$ . There are three cases to consider.

The case $z_0,z_1\in F(f_a)$ is trivial since $F(f_a)$ is path-connected.

The next case is that $z_0\in F(f_a)$ and $z_1\in I(f_a)$ . Let $\beta$ be a simple closed curve around the point $z_1$ , such that $\beta\cap J(f_a)\subset I(f_a)$ (apply Theorem 1·3). Let $\gamma$ be the component of $J(f_a)$ containing $z_1$ . There exists $z_2\in \beta\setminus J(f_a)$ and $z_3\in \gamma\cap \beta$ . There are paths $\alpha_1\subset F(f_a)$ from $z_0$ to $z_2$ , $\alpha_2\subset \beta$ from $z_2$ to $z_3$ , and $\alpha_3\subset \gamma$ from $z_3$ to $z_1$ . It is clear that $\alpha_3$ can be constructed to avoid the endpoint of $\gamma$ , so that $\alpha_3\subset I(f_a)$ . Then $\alpha_1\cup \alpha_2\cup \alpha_3\subset F(f_a)\cup I(f_a)$ contains a path from $z_0$ to $z_1$ .

The third and final case $z_0,z_1\in I(f_a)$ can be handled by connecting each point $z_0$ and $z_1$ to a third point of $F(f_a)$ , as was done the second case.