2·1. Lipschitz decomposable relations

We recall from Section 1·3 that the limit set K comes equipped with a Lipschitz coding map $\pi\colon\Delta\to K$ . It would be convenient if the map $\pi$ would preserve Assouad dimension: however, Lipschitz maps may both decrease and increase Assouad dimension. On the other hand, bi-Lipschitz maps do preserve Assouad dimension. However, the coding space $\Delta$ is always totally disconnected, whereas K could be a connected set, so in such a situation the coding map cannot be bi-Lipschitz.

The following notion is a generalisation of a bi-Lipschitz function between metric spaces. First, given a relation $\mathcal{R}\subset X\times Y$ , its domain is the set

\begin{equation*} \mathrm{dom}\mathcal{R}=\bigl\{x\in X:(\{x\}\times Y)\cap\mathcal{R}\neq\varnothing\bigr\}.\end{equation*}

The image of a given set $A \subset X$ under the relation $\mathcal{R}$ is the set

\begin{equation*} \mathcal{R}(A)=\bigl\{y\in Y:A\times \{y\}\cap\mathcal{R}\neq\varnothing \bigr\}.\end{equation*}

We also define the inverse relation $\mathcal{R}^{-1}=\{(y,x):(x,y)\in\mathcal{R}\}\subset Y\times X$ .

Definition 2·1. Let $(X,d_1)$ and $(Y,d_2)$ be non-empty metric spaces and let $\mathcal{R}\subset X\times Y$ be a relation. We say that $\mathcal{R}$ is Lipschitz decomposable if $\mathrm{dom}\mathcal{R}=X$ and there are constants $M\in{\mathbb{N}}$ and $c{\gt} 0$ so that for all $x\in X$ and $r{\gt}0$ , there are $y_1,\ldots,y_M\in Y$ so that

\begin{equation*} \mathcal{R}(B(x,r))\subset\bigcup_{i=1}^M B(y_i, cr). \end{equation*}

We say that $\mathcal{R}$ is bi-Lipschitz decomposable if $\mathcal{R}$ and $\mathcal{R}^{-1}$ are both Lipschitz decomposable.

If we wish to indicate the dependence on the numbers M and c, we will say that a relation is (M,c)-Lipschitz decomposable. The following fact is easy to verify.

In general, pairs of metric spaces which support bi-Lipschitz decomposable relations will have the same dimensions, for most notions of dimensions defined using covers. Bi-Lipschitz decomposable maps also preserve (up to a multiplication by uniformly bounded constants) quantities defined in terms of covers, such as Hausdorff measure. For completeness, we give the elementary proof for Assouad dimension and leave consideration of other dimensions to the motivated reader.

Let $\varepsilon{\gt}0$ , $x\in X$ , and $0{\lt}r\leq R{\lt}c^{-1}$ be arbitrary. By the assumption, get $y_1,\ldots,y_M\in Y$ so that

(2·1) \begin{equation} \mathcal{R}(B(x,R))=\{y\in Y:(B(x,R)\times \{y\})\cap\mathcal{R}\neq\varnothing\}\subset\bigcup_{i=1}^M B(y_i, cR). \end{equation}

For each $i=1,\ldots,M$ , by definition of the Assouad dimension, get a family of balls $\{B(y_{i,\;j},c^{-1}r)\}_{i=1}^{N}$ covering $B(y_i,cR)$ where

(2·2) \begin{equation} N\lesssim_\varepsilon\left(\frac{cR}{c^{-1}r}\right)^{\mathrm{dim_A} Y+\varepsilon}\lesssim \left(\frac{R}{r}\right)^{\mathrm{dim_A} Y+\varepsilon}. \end{equation}

Then for each $B(y_{i,\;j},c^{-1} r)$ , since $\mathcal{R}^{-1}$ is also Lipschitz decomposable, there are balls $\{B(x_{i,\;j,k},r)\}_{k=1}^M$ so that

\begin{equation*} \{x\in X:(\{x\}\times B(y_{i,\;j},c^{-1}r))\cap\mathcal{R}\neq\varnothing\}\subset\bigcup_{k=1}^M B(x_{i,\;j,k}, r). \end{equation*}

But (2·1) and (2·2) together imply that the family of balls

\begin{equation*} \{B(x_{i,\;j,k},r):i=1,\ldots,M;\, j=1,\ldots, N;\, k=1,\ldots,M\} \end{equation*}

is a cover for B(x,R) with cardinality $M^2 N\lesssim (R/r)^{\mathrm{dim_A} Y+\varepsilon}$ . Since $\varepsilon{\gt}0$ was arbitrary, it follows that $\mathrm{dim_A} X\leq\mathrm{dim_A} Y$ .

Of course, the identical argument applied to the inverse $\mathcal{R}^{-1}$ implies that $\mathrm{dim_A} Y\leq \mathrm{dim_A} X$ , as required.

It is now straightforward to obtain our result on the Assouad dimension.

Proof. First, suppose the IFS satisfies the bounded neighbourhood condition. Since $\pi\colon\Delta\to K$ is Lipschitz, it is Lipschitz decomposable. Next, let M be as in the definition of the bounded neighbourhood condition. Suppose $y\in K$ and $0{\lt}r{\lt}1$ are arbitrary: then

\begin{equation*} \#\{Q\in\mathcal{T}(r):\pi(Q)\cap B(y,r)\neq\varnothing\}\leq M. \end{equation*}

But $\mathrm{diam} Q\leq r$ , so in fact $\mathcal{R}^{-1}$ is (M,1)-Lipschitz decomposable.

Conversely, suppose $\pi$ is bi-Lipschitz decomposable and let $r{\gt}0$ and $y\in K$ be arbitrary. Let M and c be such that $\pi^{-1}$ is Lipschitz decomposable. Since $\mathbb{R}^d$ is doubling, there exists a constant $N=N(c)$ such that we may cover any ball B(y,r) by N balls of radius $c^{-1} r$ . Now fix some ball B(y,r) and cover it with balls $B(y_i, c^{-1} r)$ for $i=1,\ldots, N$ . For each i, since $\pi^{-1}$ is Lipschitz decomposable, there exist points $\{x_{i,1},\ldots,x_{i,M}\}\subset\Delta$ such that

\begin{equation*} \pi^{-1}(B(y_i,c^{-1}r))\subset\bigcup_{j=1}^M B(x_{i,\;j}, r). \end{equation*}

For each $x_{i,\;j}$ , let $Q_{i,\;j}\in\mathcal{T}(r)$ be the unique cylinder containing $x_{i,\;j}$ . Moreover, since $\mathrm{diam} \widehat{Q_{i,\;j}} {\gt} r$ , if $x\notin Q_{i,\;j}$ , then $d(x,x_{i,\;j}){\gt}r$ . Therefore, $B(x_{i,\;j},r)=Q_{i,\;j}$ and

\begin{equation*} K \cap B(y,r)\subset K \cap \bigcup_{i=1}^N B(y_i,c^{-1}r))\subset \bigcup_{i=1}^N\bigcup_{j=1}^M\pi(Q_{i,\;j}). \end{equation*}

Since $r{\gt}0$ and $y\in K$ were arbitrary, it follows that the IFS satisfies the bounded neighbourhood condition with constant $N\cdot M$ .

In particular, if either of these conditions hold, since bi-Lipschitz decomposable maps preserve Assouad dimension by Lemma 2·3, it follows that $\mathrm{dim_A} K=\mathrm{dim_A}\Delta$ .

2.2 Consequences of the bounded neighbourhood condition

In this section, we discuss some of the implications of the bounded neighbourhood condition. First, we introduce a local non-degeneracy condition similar to (1·1).

Definition 2·6. We say that the IFS is locally contracting if

\begin{equation*} \lim_{m\to\infty}\sup_{n\in{\mathbb{N}}}\max\{r_{n,\;j_0}\cdots r_{n+m-1,\;j_{m-1}}:\,j_i\in\mathcal{J}_{n+i}\text{ for each }i=0,\ldots,m-1\}= 0. \end{equation*}

Recall also the definition of bounded branching from Definition 1·3. We then have the following result. This proves the first part of Theorem 1·2.

Proof. Throughout, let $M\in{\mathbb{N}}$ be the constant with respect to which the IFS satisfies the bounded neighbourhood condition. We first show that the IFS has bounded branching. Let $r{\gt}0$ and $Q\in\mathcal{T}(r)$ be arbitrary and denote the set of geometric siblings by

\begin{equation*} \mathcal{S} = \{Q'\in\mathcal{T}(r):Q'\subset\widehat{Q}\}. \end{equation*}

Next, set $r_0 = \rho(\widehat{Q}){\gt}r$ , and let $r_0{\gt}r'{\gt}\max\{r_0/2, r\}$ be arbitrary. Since $\mathrm{diam} \pi(\widehat{Q}) \leq r_0\mathrm{diam} X$ , there is a constant $N\in{\mathbb{N}}$ depending only on $\mathrm{diam} X$ so that $\pi(Q)$ can be covered by N balls of radius r’, say $\{B(x_i, r')\}_{i=1}^N$ . Now, if $Q'\in\mathcal{S}$ , then $Q'\in\mathcal{N}(x_i,r')$ for some i since $r{\lt}r'{\lt}\rho(\widehat{Q})$ . Therefore $\#\mathcal{S}\leq N M$ , so the IFS has bounded branching.

Next, suppose for contradiction that the IFS is not locally contracting. Then there is a $\delta{\gt}0$ so that for all $m\in{\mathbb{N}}$ , there is an $n\in{\mathbb{N}}$ and $(i_0,\ldots,i_{m-1})\in\mathcal{J}_n\times\cdots\times\mathcal{J}_{n+m-1}$ such that

\begin{equation*} r_{n,i_0}\cdots r_{n+m-1,i_{m-1}} \geq \delta. \end{equation*}

Fix $m\in{\mathbb{N}}$ and the corresponding $n\in{\mathbb{N}}$ and $(i_0,\ldots,i_{m-1})$ as above; we may moreover assume that $r_{n,i_0}\cdots r_{n+m-1,i_{m-1}}$ is maximal. Also, choose some arbitrary initial segment $(\;j_1,\ldots,\;j_{n-1})$ , let $Q_0 = [\;j_1,\ldots,\;j_{n-1}]$ , and let $Q = [\;j_1,\ldots,\;j_{n-1},i_0,\ldots,i_{m-1}]$ .

Write $r=\rho(Q)$ . Since $\#\mathcal{J}_k\geq 2$ for all $k\in{\mathbb{N}}$ , for each $\ell=0,\ldots,m-1$ , there is a cylinder

\begin{equation*} Q'_\ell = [\;j_1,\ldots,\;j_{n-1},i_0,\ldots,i_\ell'], \end{equation*}

where $i_\ell' \neq i_\ell$ . Moreover, by maximality of the choice of $[i_0,\ldots,i_{m-1}]$ , each $Q'_\ell$ is either itself an element of $\mathcal{T}(r)$ or has some some offspring $Q_\ell\in\mathcal{T}(r)$ . In either case, denote this element by $Q_\ell$ . Note that all of the $Q_\ell$ are distinct. But now since $\delta{\gt}0$ is fixed and $\rho(Q)/\rho(Q_0)\geq \delta$ , there is a fixed constant $N_\delta$ (independent m) so that the cylinder $Q_0$ can be covered by $N_\delta$ balls B(x,r) for $x\in K$ . By the pigeonhole principle, one of these balls must intersect $m/N_\delta$ cylinders $Q_\ell$ . Since $m\in{\mathbb{N}}$ was arbitrary, this contradicts the bounded neighbourhood condition.

2·3. Open set condition and bounds on branching and contraction

In this section, we discuss the difference between the bounded neighbourhood condition and the open set condition. Recall the definitions of the open set condition Definition 1·2 and bounded branching Definition 1·3 from the introduction.

We first give the quick proof relating the bounded branching condition with uniform bounds on $\#\mathcal{J}_n$ .

The following proposition summarises the relationship between these two conditions as well as the bounded neighbourhood condition. In particular, along with the proof of Lemma 2·7, this completes the proof of Theorem B.

Proof. Assume that the IFS satisfies the open set condition. For notational simplicity, given a cylinder $Q=[\;j_1,\ldots,\;j_n]$ , we write

\begin{equation*} U(Q) = S_{1,\;j_1}\circ\cdots\circ S_{n,\;j_n}(X^\circ). \end{equation*}

Let $x\in K$ and $r{\gt}0$ be arbitrary: we need to control the size of $\mathcal{N}(x,r)$ . We do this under the three assumptions in order.

First, suppose 2·1 holds. Let $N= \sup_{r{\gt}0}\beta(r)$ be the constant from the bounded branching assumption, and moreover observe that it follows from the cone condition that

\begin{equation*} c\,:\!= \inf_{x\in X}\inf_{r\in(0,\mathrm{diam} X)}\frac{\mathcal{L}^d\bigl(B(x,r)\cap X^\circ\bigr)}{\mathcal{L}^d\bigl(B(x,r)\bigr)}{\gt}0. \end{equation*}

Write $\mathcal{E}=\{\widehat{Q}:Q\in\mathcal{N}(x,r)\}$ . Now, let $Q\in\mathcal{N}(x,r)$ be arbitrary and fix some $y_Q\in \pi(Q)\cap B(x,r)$ . Then since $y_Q\in \pi(\widehat{Q})$ and $\rho(\widehat{Q}) {\gt} r$ , writing $y_Q = S_{1,\;j_1}\circ\cdots\circ S_{n-1, j_{n-1}}(z_Q)$ for some $z_Q\in X$ and applying the cone condition to the ball $B(z_Q, r / \rho(\widehat{Q}))$ ,

\begin{equation*} \mathcal{L}^d\bigl(B(y_Q,r)\cap U(\widehat{Q})\bigr)\geq c\cdot \mathcal{L}^d\bigl(B(x,r)\bigr). \end{equation*}

Moreover, the sets $U(\widehat{Q})\cap B(y_Q,r)\subset B(x, 2r) $ are disjoint for distinct choices of $\widehat{Q}$ by the open set condition, so

\begin{align*} c\cdot\#\mathcal{E}\cdot \mathcal{L}^d\bigl(B(x,r)\bigr) &\leq \sum_{\widehat{Q}\in \mathcal{E}}\mathcal{L}^d\bigl(B(y_Q,r)\cap U(\widehat{Q})\bigr)\\ &\leq \mathcal{L}^d\bigl(B(x,2r)\bigr)\\ &= 2^d\mathcal{L}^d\bigl(B(x,r)\bigr). \end{align*}

In other words, $\# \mathcal{E}\leq c^{-1} 2^d$ so by the bounded branching condition, $\#\mathcal{N}(x,r) \leq c^{-1}2^d N$ .

Next, suppose (b) holds. Let $N=\sup_{r{\gt}0}\beta(r)$ be as before and again set $\mathcal{E}=\{\widehat{Q}:Q\in\mathcal{N}(x,r)\}$ . Since the IFS is homogeneous, $\rho$ has constant value $r_0$ on $\mathcal{E}$ ; in particular, if $\widehat{Q}\in \mathcal{E}$ , then $\mathcal{L}^d\bigl(U(\widehat{Q})\bigr) = r_0^d \mathcal{L}^d\bigl(X^\circ\bigr)$ . Moreover, the sets $U(\widehat{Q})$ are disjoint and, since $\pi(\widehat{Q})\cap B(x,r_0)\neq\varnothing$ , it follows that $U(\widehat{Q})\subset B(x,2r_0\cdot\mathrm{diam} X)$ . Therefore

\begin{align*} \#\mathcal{E}\cdot r_0^d\mathcal{L}^d\bigl(X^\circ\bigr) &= \sum_{\widehat{Q}\in \mathcal{E}} \mathcal{L}^d \bigl(U(\widehat{Q})\bigr)\\ &\leq \mathcal{L}^d \bigl(B(x,2r_0\cdot \mathrm{diam} X)\bigr)\\ & = (2\cdot \mathrm{diam} X)^d r_0^d\mathcal{L}^d(B(0,1)). \end{align*}

Thus $\#\mathcal{N}(x,r)\leq N\cdot (2\cdot\mathrm{diam} X)^d ({\mathcal{L}^d(B(0,1))}/{\mathcal{L}^d(X^\circ)})$ .

Finally, suppose (c) holds. Now, if $\widehat{Q}\in\mathcal{E}$ , since $r r_{\min}\leq \rho(Q)\leq r$ , $\pi(Q)\subset B(x,2r\cdot \mathrm{diam} X)$ , and the U(Q) are disjoint for distinct Q, then

\begin{align*} \#\mathcal{N}(x,r)\cdot (r r_{\min})^d \mathcal{L}^d\bigl(X^\circ\bigr)^d &\leq \sum_{Q\in\mathcal{N}(x,r)}\mathcal{L}^d \bigl(U(Q)\bigr)\\ &\leq \mathcal{L}^d\bigl(B(x,2r\cdot\mathrm{diam} X)\bigr)\\ & = (2\cdot\mathrm{diam} X)^d r^d\mathcal{L}^d(B(0,1)). \end{align*}

We conclude that $\#\mathcal{N}(x,r)\leq r_{\min}^{-d} (2\mathrm{diam} X)^d ({\mathcal{L}^d(B(0,1))}/{\mathcal{L}^d(X^\circ)})$ .

Since $x\in K$ and $r{\gt}0$ were arbitrary, in any of the cases, it follows that the bounded neighbourhood condition holds.

3·1. A disc-packing formulation of Assouad dimension

In this section, we observe that in the definition of the Assouad dimension one may replace the exponent associated to localised coverings of balls of the same size by an exponent coming from localised packings of balls which may have different sizes. This is essentially the same as the disc-packing formulation for box dimension (see, for example, [ Reference Bishop and Peres3 , Section 2·6]) and our result also proceeds by a similar proof. This is useful since the natural covers appearing from the symbolic representation of K consist of cylinders which may have very non-uniform diameters when indexed by length.

First, for a metric space X, $x\in X$ , and $R\in(0,1)$ , denote the family of all localised centred packings by

\begin{equation*} \operatorname{pack}(X,x,R)=\left\{\{B(x_i,r_i)\}_i:\begin{matrix}0{\lt}r_i\leq R,x_i\in X,B(x_i,r_i)\subset B(x,R),\\B(x_i,r_i)\cap B(x_j,r_j)=\varnothing\text{ for all }i\neq j\end{matrix}\right\}.\end{equation*}

Here the collections may be finite or countably infinite. We now have the following result.

Proposition 3·1. Let X be a bounded metric space. Then

\begin{align*} \mathrm{dim_A} X=\inf\Bigl\{\alpha:\forall 0{\lt}R{\lt}1\,\forall x\in X\,&\forall\{B(x_i,r_i)\}_i\in\operatorname{pack}(X,x,R)\nonumber \\ &\sum_i r_i^\alpha\lesssim_\alpha R^\alpha\Bigr\}. \end{align*}

We now show the lower bound by pigeonholing. First, for a centred packing $\mathcal{B}=\{B(x_i,r_i)\}_i\in\operatorname{pack}(X,x,R)$ , we write

\begin{equation*} \mathcal{B}_n = \bigl\{B(x_i,r_i)\in\mathcal{B}:2^{-n}\leq r_i/R {\lt} 2^{-n+1}\bigr\}. \end{equation*}

Now let $s{\lt}t$ and $r_0{\gt}0$ be arbitrary. Then get some $x\in X$ , $0{\lt}R{\lt}1$ , and a centred packing $\mathcal{B}$ as above such that $r_i\leq r_0$ for all i and moreover

\begin{equation*} 1\lesssim_s \sum_n \#\mathcal{B}_n 2^{-ns}. \end{equation*}

By the pigeonhole principle with respect to the sequence of weights $(n^{-2})_{n=1}^\infty$ , since the weights are summable, there is a $\gamma{\gt}0$ depending only on s so that there is some $m\in{\mathbb{N}}$ with $\gamma\leq \#\mathcal{B}_m m^2 2^{-ms}$ . Taking $r_0$ sufficiently small, $\#\mathcal{B}_m=0$ for all small m, so that m must be arbitrarily large. Since for every $\varepsilon{\gt}0$ there is a constant $C_\varepsilon$ so that $m^2 \leq C_\varepsilon 2^{m\varepsilon}$ for all $m\in{\mathbb{N}}$ , we conclude that $\mathrm{dim_A}\ K\geq s$ . Since $s{\lt}t$ was arbitrary, the proof is complete.

3·2. Subadditivity and submaximality of covering numbers

In this section, we study a weaker variant of subadditivity, which we call submaximality. Throughout this section, we fix a set $A = G\cap(0,\infty)$ where G is a non-trivial additive subgroup of ${\mathbb{R}}$ . For concreteness, one might simply fix $A=(0,\infty)$ or $A=\{\kappa n:n\in{\mathbb{N}}\}$ for some $\kappa{\gt}0$ .

(i) for all $y,z\in A$ ,

\begin{equation*} g(y+z)\leq \max\{g(y),g(z)\}; \end{equation*}

(ii) for all $\varepsilon{\gt}0$ and $a\in A$ , there is an $N=N(\varepsilon, a)$ so that for all $y\geq N$ and $t\leq a$ ,

\begin{equation*} g(y+t)\leq g(y)+\varepsilon. \end{equation*}

To motivate this definition, we observe the following version of Fekete’s subadditive lemma.

Lemma 3·3. Let $g\colon A\to {\mathbb{R}}\cup\{-\infty\}$ be submaximal. Then

\begin{equation*} \limsup_{y\to\infty}g(y)=\inf_{y\in A}g(y). \end{equation*}

Proof. Let $\varepsilon{\gt}0$ and $a\in A$ be arbitrary, and let $N=N(\varepsilon, a)$ be chosen to satisfy the conclusion of Definition 3·2 (ii). Let $y\geq N$ be arbitrary and write $y=\ell a+t$ for $\ell\in{\mathbb{N}}$ and $0\leq t {\lt} a$ . Then applying 3·1 followed by 3·1 $\ell-1$ times,

\begin{equation*} g(y) = g(\ell a+t)\leq g(\ell a)+\varepsilon\leq g(a)+\varepsilon. \end{equation*}

Thus

\begin{equation*} \limsup_{y\to\infty}g(y)\leq g(a)+\varepsilon. \end{equation*}

But $a\in A$ and $\varepsilon{\gt}0$ were arbitrary, so the desired result follows.

Note that assumption Definition 3·2 (i) is not sufficient by itself to guarantee the existence of the limit $\lim_{y\to\infty}g(y)$ . For example, the function $g\colon{\mathbb{N}}\to{\mathbb{R}}$ defined by

\begin{equation*} g(n)=\begin{cases} 1 &: n\text{ odd},\\ 0 &: n\text{ even}, \end{cases} \end{equation*}

satisfies $g(y+z)\leq \max\{g(y),g(z)\}$ for all $y,z \in {\mathbb{N}}$ but has no limit.

Similarly to before, we say that a function $f\colon A\times A\to{\mathbb{R}}\cup\{-\infty\}$ is submaximal if it satisfies the following two assumptions:

(i) for all $x,y,z\in A$ ,

\begin{equation*} f(x,y+z)\leq \max\{\;f(x,y),f(x+y,z)\}; \end{equation*}

(ii) for all $\varepsilon{\gt}0$ and $a\in A$ , there is an $N=N(\varepsilon, a)$ so that for all $x,y,t\in A$ with $y\geq N$ , $t\leq a$ , and $x\leq x'\leq x+t$ ,

\begin{equation*} f(x, y+t)\leq f(x',y)+\varepsilon. \end{equation*}

It is this form that will be essential for us in applications. We note that the above assumptions are satisfied by the function $f(y,x)/x$ , where $f \colon {\mathbb{N}} \cup \{0\} \times {\mathbb{N}} \to {\mathbb{R}}\cup\{-\infty\}$ satisfies the generalised subadditive condition introduced in [ Reference Käenmäki10 , Section 2]. In Lemma 3·2, we generalise this observation. But first we acquire the following two-parameter generalisation of Lemma 3·1.

Proposition 3·5. Suppose $f\colon A\times A\to{\mathbb{R}}\cup\{-\infty\}$ is submaximal. Then

\begin{align*} \beta\,:\!={}&\limsup_{y\to\infty}\limsup_{x\to\infty}f(x,y)\\ ={}&\lim_{y\to\infty}\limsup_{x\to\infty}f(x,y)\\ ={}&\lim_{y\to\infty}\sup_{x\in A}f(x,y)\\ ={}&\inf_{y\in A}\sup_{x\in A}f(x,y). \end{align*}

Moreover, if $B\subset A$ is of the form $B=\{\kappa n:n\in{\mathbb{N}}\}$ for some $\kappa{\gt}0$ , then

\begin{equation*} \beta=\lim_{\substack{y\to\infty\\y\in B}}\sup_{x\in B}f(x,y). \end{equation*}

Proof. We assume that $\beta{\gt}-\infty$ : the proof for $\beta=-\infty$ is similar (and substantially easier). Write $g(y)=\limsup_{x\to\infty}f(x,y)$ . We first show that g is submaximal, so that $\beta=\lim_{y\to\infty}g(y)$ . First,

\begin{align*} g(y_1+y_2)&=\limsup_{x\to\infty}f(x,y_1+y_2)\\ &\leq \limsup_{x\to\infty}\max\{\;f(x,y_1),f(x+y_1,y_2)\}\\ &\leq \max\{g(y_1),g(y_2)\}. \end{align*}

Moreover, if $\varepsilon{\gt}0$ and $a\in A$ are arbitrary, taking N to satisfy the conclusions of (ii) and $y,t\in A$ with $t\leq a$ and $y\geq N$ ,

\begin{equation*} g(y+t) = \limsup_{x\to\infty}f(x,y+t)\leq \limsup_{x\to\infty}f(x,y)+\varepsilon = g(y)+\varepsilon. \end{equation*}

The same argument with respect to the function $y\mapsto \sup_{x\in A}f(x,y)$ gives that

\begin{equation*} \limsup_{y\to\infty}\sup_{x\in A}f(x,y)=\inf_{y\in A}\sup_{x\in A}f(x,y). \end{equation*}

To complete the proof of the various equalities involving $\beta$ , it remains to show that

(3·1) \begin{equation} \inf_{y\in A}\sup_{x\in A}f(x,y)\leq \beta. \end{equation}

Let $\varepsilon{\gt}0$ be arbitrary. By definition of $\beta$ , there are $y_0$ and K so that for all $x\geq K$ , $f(x,y_0)\leq\beta+\varepsilon$ . Let $N_1=N(\varepsilon, y_0)$ be chosen to satisfy the conclusions of (ii) and let $y\in A$ with $y\geq N_1$ be arbitrary. Write $y=\ell y_0+t$ for some $\ell\in{\mathbb{N}}$ and $0\leq t {\lt} y_0$ . Then by (i), for all $x\geq K$ and $y\geq N_1$ ,

(3·2) \begin{equation} \begin{aligned} f(x,y) &= f(x,\ell y_0+t) \leq f(x,\ell y_0)+\varepsilon\\ &\leq \max_{i=0,\ldots,\ell-1}f(x+i y_0, y_0)+\varepsilon\\ &\leq \beta+2\varepsilon. \end{aligned} \end{equation}

Now let $N_2=N(\varepsilon, K)$ be chosen to satisfy the conclusions of (ii). Let $x\in(0,K)\cap A$ and $y\geq \max\{N_1 + K, N_2\}$ . Then since $y\geq N_2$ and (3·2) (which applies since $y+x-K\geq N_1$ ), for all $x\in A$ ,

\begin{equation*} f(x,y) \leq f(K, y+x-K) + \varepsilon \leq \beta+3\varepsilon. \end{equation*}

Since $\varepsilon{\gt}0$ was arbitrary, this proves (3·1).

Finally, suppose $B\subset A$ is of the form $B=\{\kappa n:n\in{\mathbb{N}}\}$ for some $\kappa{\gt}0$ . First, note that since $B\subset A$ ,

\begin{equation*} \beta\geq \lim_{\substack{y\to\infty\\y\in B}}\sup_{x\in B}f(x,y) \end{equation*}

and moreover the limit exists as proven above. Conversely, let $(x,y)\in A\times A$ be arbitrary with $y\geq 2\kappa$ and get $(x_0,y_0)\in B\times B$ such that $x\leq x_0{\lt}x+\kappa$ and $x+y-\kappa{\lt}x_0+y_0\leq x+y$ . Let $\varepsilon{\gt}0$ be arbitrary and let $N=N(\varepsilon,\kappa)$ satisfy the conclusion of (ii). Then for $y\geq N$ , applying (ii) twice,

\begin{equation*} f(x_0,y_0) \geq f(x,y)-2\varepsilon. \end{equation*}

Since $\varepsilon{\gt}0$ and $(x,y)\in A\times A$ were arbitrary, it follows that

\begin{equation*} \limsup_{\substack{y_0\to\infty\\y_0\in B}}\sup_{x_0\in B} f(x_0,y_0)\geq \limsup_{y\to\infty}\sup_{x\in A} f(x,y)=\beta \end{equation*}

as required.

Finally, we show that the hypotheses of Lemma 3·2 are satisfied by functions satisfying a two-parameter version of subadditivity.

Lemma 3·6. Suppose $f\colon A\times A\to\{-\infty\}\cup{\mathbb{R}}$ is any function such that for all $x,y,z\in A$ ,

(3·3) \begin{equation} f(x,y+z)\leq \frac{ y \;f(x,y) + z \;f(x+y,z)}{y+z}. \end{equation}

Then:

1. (i) for all $x,y,z\in A$ ,

   \begin{equation*} f(x,y+z)\leq \max\{\;f(x,y), f(x+y,z)\}; \end{equation*}

2. (ii) suppose moreover that f is bounded from above by some constant $C{\gt}0$ . Then for all $\varepsilon{\gt}0$ and $x,y,t\in A$ with $t\leq \varepsilon C^{-1} y$ and $x\leq x'\leq x+t$ ,

   \begin{equation*} f(x, y+t)\leq f(x',y)+\varepsilon. \end{equation*}

In particular, f is submaximal.

Proof. Of course, (i) is immediate. To see (ii), let $C\in{\mathbb{R}}$ be such that $f(x,y)\leq C$ for all $x,y\in A$ . Let $\varepsilon{\gt}0$ . Then for all $x,y,t\in A$ with $t\leq \varepsilon C^{-1} y$ and $x\leq x'\leq x+t$ , applying 3·3 twice,

\begin{align*} f(x,y+t)&\leq \frac{ (x'-x) f(x,x'-x) + (y+t+x-x') f(x',y+t+x-x')}{y+t}\\ &\leq \frac{ (x'-x) f(x,x'-x) + yf(x',y)+(t+x-x')f(x'+y,t+x-x')}{y+t}\\ &\leq \frac{y}{y+t} f(x',y)+\frac{t C}{y+t}\\ &\leq f(x',y) + \varepsilon \end{align*}

as claimed.

To motivate why the hypotheses of Lemma 3·5 and Lemma 3·6 are related to dimensions, we present the observation that the Assouad dimension can be reformulated in a way reminiscent of a notion of dimension studied by Larman [ Reference Larman13 ]. Let X be a bounded doubling metric space and for $\delta\in(0,1)$ and $r\in(0,1)$ , write

\begin{equation*} \psi(r,\delta) = \sup_{x\in X}N_{r\delta}\bigl(B(x,r)\cap K\bigr)\end{equation*}

and then set

\begin{equation*} \Psi(r,\delta) = \frac{\log \psi(r,\delta)}{\log (1/\delta)}.\end{equation*}

One can think of $\Psi(r,\delta)$ is the best guess for the Assouad dimension of X at scales $0{\lt}r\delta{\lt}\delta{\lt}1$ . This heuristic is made precise in the following result.

Corollary 3·6. Let X be a bounded doubling metric space. Then

(3·4) \begin{equation} \mathrm{dim_A} X=\limsup_{\delta\to 0}\limsup_{r\to 0}\Psi(r,\delta)=\lim_{\delta\to 0}\sup_{r\in(0,1)}\Psi(r,\delta). \end{equation}

Proof. Since X is doubling, there is an $M\geq 0$ so that $\Psi(r,\delta)\in[0,M]$ . Moreover, given $r,\delta_1,\delta_2\in(0,1)$ , by covering balls $B(x,r\delta_1)$ by balls of radius $r\delta_1\delta_2$ ,

\begin{equation*} \psi(r,\delta_1\delta_2)\leq\psi(r,\delta_1)\psi(r\delta_1,\delta_2) \end{equation*}

and therefore

\begin{align*} \Psi(r,\delta_1\delta_2) &= \frac{\log\psi(r,\delta_1\delta_2)}{\log(1/\delta_1\delta_2)}\\ &\leq\frac{\log\psi(r,\delta_1)+\log\psi(r\delta_1,\delta_2)}{\log(1/\delta_1\delta_2)}\\ &=\frac{\log(1/\delta_1)\Psi(r,\delta_1)+\log(1/\delta_2)\Psi(r\delta_1,\delta_2)}{\log(1/\delta_1)+\log(1/\delta_2)}. \end{align*}

Thus with the change of coordinate $g(x,y)=(e^{-x},e^{-y})$ , the second equality in (3·4) follows by applying Lemma 3·2 and Proposition 3·5 to the function $\Psi\circ g$ .

To see the first equality in (3·4), it is a direct consequence of the definition of the Assouad dimension that

\begin{equation*} \limsup_{\delta\to 0}\limsup_{r\to 0}\Psi(r,\delta)\leq\mathrm{dim_A} K \end{equation*}

and that there are sequences $(\delta_n)_{n=1}^\infty$ and $(r_n)_{n=1}^\infty$ with $\lim_{n\to\infty}\delta_n=0$ such that

\begin{equation*} \lim_{\delta\to 0}\sup_{r\in(0,1)}\Psi(r,\delta)\geq\limsup_{n\to\infty}\Psi(r_n,\delta_n)\geq\mathrm{dim_A} K, \end{equation*}

as required.

4·1. Regularity of the numbers $\theta(n,m)$

Recall the definition of the pressure functions $\phi_{n,m}$ for $n,m\in{\mathbb{N}}$ in Section 1·2. In particular, these functions are convex, differentiable, strictly decreasing, and have unique zero $\theta(n,m)$ . We begin with a upper bound on the derivative of the pressure function at its unique zero.

Lemma 4·1. Let $(\Phi_n)_{n=1}^\infty$ be a locally contracting non-autonomous IFS. Then there is an $M\in{\mathbb{N}}$ and a constant $c{\gt}0$ so that for all $m\geq M$ ,

\begin{equation*} \phi_{n,m}'(\theta(n,m)) \leq -c. \end{equation*}

Proof. Since the IFS is locally contracting, there is an $M\in{\mathbb{N}}$ and a $\delta\in(0,1)$ so that for all $n\in{\mathbb{N}}$ , and $(\;j_0,\ldots,\;j_{M-1})\in\mathcal{J}_n\times\cdots\times\mathcal{J}_{n+M-1}$ , we have

\begin{equation*} r_{n, j_0}\cdots r_{n+M-1, j_{M-1}} \leq \delta. \end{equation*}

Then for $m\geq M$ , write $m = \ell M + j$ for $\ell\in{\mathbb{N}}$ and we compute

\begin{align*} \phi_{n,m}'(\theta(n,m)) &\leq \frac{1}{m} \sum_{j_0\in\mathcal{J}_n}\cdots \sum_{j_{m-1}\in\mathcal{J}_{n+m-1}}(r_{n, j_0}\cdots r_{n+m-1, j_{m-1}})^{\theta(n,m)}\log \delta^{\ell}\\ &= \log(\delta)\cdot \frac{\ell}{\ell M + j}\\ &\leq - \frac{\log(1/\delta)}{2M} \end{align*}

as claimed.

Using this, we can now establish submaximality of the function $\theta(n,m)$ . Actually, we will prove a somewhat stronger version of the continuity hypothesis.

Lemma 4·2. Let $(\Phi_n)_{n=1}^\infty$ be a non-autonomous IFS. Then for all $n,m,k\in{\mathbb{N}}$ ,

(4·1) \begin{equation} \theta(n,m+k) \leq \max\{\theta(n,m),\theta(n+m, k)\}. \end{equation}

Suppose moreover that the IFS satisfies the bounded neighbourhood condition. Then there is a constant $C{\gt}0$ and an $M\in{\mathbb{N}}$ so that for all $n,m,k\in{\mathbb{N}}$ with $m\geq M$ and $n \leq n' \leq n+k$

\begin{equation*} \theta(n, m+k) - \theta(n', m) \leq C\frac{k}{m}. \end{equation*}

In particular, the function $\theta(n,m)$ is submaximal.

Proof. Let n,m,k be arbitrary. Writing $s=\max\{\theta(n,m),\theta(n+m, k)\}$ , observe that

\begin{align*} 1&=\sum_{j_0\in\mathcal{J}_{n}}\cdots\sum_{j_{m+k-1}\in\mathcal{J}_{n+m+k-1}}\left(\prod_{\ell=0}^{m-1} r_{n+\ell,\;j_\ell}\right)^{\theta(n,m)}\left(\prod_{\ell=m}^{m+k-1}r_{n+\ell,\;j_\ell}\right)^{\theta(n+m,\ell)}\\ &\geq\sum_{j_0\in\mathcal{J}_{n}}\cdots\sum_{j_{m+k-1}\in\mathcal{J}_{n+m+k-1}}\left(\prod_{\ell=0}^{m+k-1} r_{n+\ell,\;j_\ell}\right)^{s}\\ &= \exp\bigl((m+k)\cdot\phi_{n,m+k}(s)\bigr). \end{align*}

Since $\phi_{n,m+k}(\theta(n,m+k))=0$ and $\phi_{n,m+k}$ is strictly decreasing, it follows that $\theta(n,m+k)\leq s$ , which is (4·1).

We now proceed with the continuity bounds. First, since the bounded neighbourhood condition implies bounded branching by Lemma 2·7, there is an $N\in{\mathbb{N}}$ so that $\#\mathcal{J}_n\leq N$ for all $n\in{\mathbb{N}}$ . Now let $n,m\in{\mathbb{N}}$ be arbitrary. To begin, observe that

\begin{align*} \phi_{n,m+1}&(\theta(n,m))\\ &=\frac{1}{m+1}\log\sum_{j_0\in\mathcal{J}_{n}}\cdots\sum_{j_{m-1}\in\mathcal{J}_{n+m-1}}\sum_{i\in\mathcal{J}_{n+m}} r_{n,\;j_0}^{\theta(n,m)}\cdots r_{n+m-1,\; j_{m-1}}^{\theta(n,m)}\cdot r_{n+m,i}^{\theta(n,m)}\\ &= \frac{1}{m+1}\log \sum_{i\in\mathcal{J}_{n+m}} r_{n+m,i}^{\theta(n,m)}\\ &\leq \frac{\log N}{m+1}. \end{align*}

Similarly,

\begin{align*} \phi_{n,m+1}(\theta(n+1,m)) &= \frac{1}{m+1}\log \sum_{i\in\mathcal{J}_{n}} r_{n,i}^{\theta(n,m+1)}\leq \frac{\log N}{m+1}. \end{align*}

Next, since the bounded neighbourhood condition implies that the IFS is locally contracting by Lemma 2·7, applying Lemma 4·1, get a constant $c{\gt}0$ and $M\in{\mathbb{N}}$ so that for all $m\geq M$ , $\phi_{n,m+1}'(\theta(n,m+1)) \leq -c {\lt} 0$ . Since $\phi_{n,m+1}$ is convex, differentiable, and strictly decreasing, since $\phi_{n,m+1} \leq (\log N)/(m+1)$ ,

\begin{equation*} \min\{\theta(n,m), \theta(n+1,m)\} \geq \theta(n,m+1) - \frac{\log N}{c(m+1)}. \end{equation*}

Applying this 1-step bound k times for general $n,m,k\in{\mathbb{N}}$ with $m\geq M$ yields the desired result by taking $C = 2(\log N)/c$ .

4·2. Proof of the Assouad dimension formula

Finally, we establish our main formula for the Assouad dimension of the non-autonomous self-similar set K.

Theorem 4·3. Let $(\Phi_n)_{n=1}^\infty$ be a non-autonomous IFS satisfying the bounded neighbourhood condition. Denote the associated non-autonomous self-similar set by K. Then

(4·2) \begin{equation} \mathrm{dim_A} K = \lim_{m\to\infty}\sup_{n\in{\mathbb{N}}}\theta(n,m) = \lim_{m\to\infty}\limsup_{n\to\infty}\theta(n,m) = \inf_{m\in{\mathbb{N}}}\limsup_{n\to\infty}\theta(n,m). \end{equation}

Proof. Recall that the limit in (4·2) exists by Lemma 4·2 and Proposition 3·5, and moreover

\begin{equation*} s\,:\!= \lim_{m\to\infty}\sup_{n\in{\mathbb{N}}}\theta(n,m) = \lim_{m\to\infty}\limsup_{n\to\infty}\theta(n,m) = \inf_{m\in{\mathbb{N}}}\limsup_{n\to\infty}\theta(n,m). \end{equation*}

We verify the lower and upper bounds separately.

First, recall from Proposition 2·5 that $\mathrm{dim_A} K=\mathrm{dim_A}\Delta$ , where $\Delta$ denotes the infinite product space associated with K. Let $\varepsilon{\gt}0$ be fixed and let M be sufficiently large so that for all $m\geq M$ , there is an $n\in{\mathbb{N}}$ so that

\begin{equation*} |\theta(n,m)-s|\leq\varepsilon. \end{equation*}

Now fix a cylinder $[\;j_1,\ldots,\;j_n]\subset\Delta$ for some $(\;j_1,\ldots,\;j_n)\in\mathcal{J}_1\times\cdots\times\mathcal{J}_n$ and write $R=\mathrm{diam}([\;j_1,\ldots,\;j_n])=r_{1,\;j_1}\cdots r_{n,\;j_n}$ . Note that if $m\geq M$ , by definition of $\theta(n,m)$

\begin{equation*} \sum_{j_{n+1}\in\mathcal{J}_{n+1}}\cdots\sum_{j_{n+m}\in\mathcal{J}_{n+m}}\prod_{k=1}^{n+m} r_{k,\;j_k}^{\theta(n,m)}=R^{\theta(n,m)}. \end{equation*}

But the family of cylinders

\begin{equation*} \bigl\{[\;j_1,\ldots,\;j_{n+m}]:(\;j_{n+1},\ldots,\;j_{n+m})\in\mathcal{J}_{n+1}\times\cdots\times\mathcal{J}_{n+m}\bigr\} \end{equation*}

forms a packing of B(x,R). Moreover, since $m\geq M$ is arbitrary, since the IFS is locally contracting by Lemma 2·7, the width of each cylinder in this family relative to $[\;j_1,\ldots,\;j_n]$ converges uniformly to 0. Thus by Proposition 3·1, $\mathrm{dim_A} K\geq s-\varepsilon$ , so the lower bound follows since $\varepsilon{\gt}0$ was arbitrary.

Conversely, let us upper bound $\mathrm{dim_A} K$ . Let $\varepsilon{\gt}0$ be arbitrary and let M be as before, and let $m\geq M$ . Now let $0{\lt}r\leq R{\lt}1$ and fix a ball $B(x,R)\subset\Delta$ . By definition of the metric on $\Delta$ , $B(x,R)=[\;j_1,\ldots,\;j_p]$ where $r_{1,\;j_1}\cdots r_{p,\;j_p}\leq R$ . We inductively build a sequence of covers $(\mathcal{B}_k)_{k=1}^\infty$ for B(x,R) such that each $\mathcal{B}_k$ is composed only of cylinder sets and

(4·3) \begin{equation} \sum_{[i_1,\ldots,i_\ell]\in\mathcal{B}_k}(r_{1,i_1}\cdots r_{\ell,i_\ell})^{s+\varepsilon}\leq R^{s+\varepsilon}. \end{equation}

Begin with $\mathcal{B}_1=\{[\;j_1,\ldots,\;j_p]\}$ , which clearly satisfies the requirement. Now suppose we have constructed $\mathcal{B}_k$ for some $k\in{\mathbb{N}}$ . Let $[i_1,\ldots,i_\ell]\in\mathcal{B}_k$ be an arbitrary cylinder set. If there exists $(\;j_1,\ldots,\;j_m) \in \mathcal{J}_{\ell+1}\times\cdots\times\mathcal{J}_{\ell+m}$ such that $r_{1,i_1}\cdots r_{\ell,i_\ell} r_{\ell+1,\;j_1} \cdots r_{\ell+m,\;j_m} \leq r$ , do nothing. Otherwise, replace the cylinder $[i_1,\ldots,i_\ell]$ with the family of cylinders

\begin{equation*} \{[i_1,\ldots,i_\ell,\;j_1,\ldots,\;j_m]:(\;j_1,\ldots,\;j_m)\in\mathcal{J}_{\ell+1}\times\cdots\times\mathcal{J}_{\ell+m}\}. \end{equation*}

The choice of $m\geq M$ and the definition of $\theta(\ell,m)$ ensures that (4·3) holds. Repeat this process until there are no more cylinders in $\mathcal{B}_k$ that can be replaced. That this process terminates at a finite level k is guaranteed by (1·1).

Since the IFS has bounded branching by Lemma 2·7, there exists a constant $D = \sup_{a{\gt}0} \beta(a) {\lt} \infty$ such that

\begin{equation*} \#\{ Q' \in \mathcal{T}(a) : Q' \subset \widehat{Q} \} \le D \end{equation*}

for all $Q \in \mathcal{T}(a)$ .

Now, fix some $Q_0 = [i_1,\ldots,i_\ell] \in \mathcal{B}_k$ and choose $(\;j_1,\ldots,\;j_m) \in \mathcal{J}_{\ell+1}\times\cdots\times\mathcal{J}_{\ell+m}$ such that $\mathrm{diam}([i_1,\ldots,i_\ell,\;j_1,\ldots,\;j_m]) \leq r$ . Then for each $q=1,\ldots,m$ , write $Q_q = [i_1,\ldots,i_\ell,\;j_1,\ldots,\;j_q]$ and $a_q = \mathrm{diam}(Q_q)$ . By the bounded branching condition, there are at most D cylinders in $\mathcal{T}(a_1)$ which intersect $Q_0$ . Each of these cylinders can further be covered by at most D many cylinders of $\mathcal{T}(a_2)$ . Since $a_m\leq r$ , there are at most $D^m$ cylinders of $\mathcal{T}(r)$ that intersect $Q_0$ .

Finally, let $\mathcal{C}_k$ denote the set of all cylinders in $\mathcal{T}(r)$ which intersect some $Q_0\in\mathcal{B}_k$ . Thus replacing each cylinder $[i_1,\ldots,i_\ell] \in \mathcal{C}_k$ with a ball $B(x_{i_1,\ldots,i_\ell},r)$ for some $x_{i_1,\ldots,i_\ell}\in[i_1,\ldots,i_\ell]$ , by (4·3), the corresponding cover has

\begin{equation*} \#\mathcal{C}_k r^{s+\varepsilon} \le D^m \#\mathcal{B}_k r^{s+\varepsilon} \leq D^m\sum_{[i_1,\ldots,i_\ell]\in\mathcal{B}_k}(r_{1,i_1}\cdots r_{\ell,i_\ell})^{s+\varepsilon}\lesssim R^{s+\varepsilon}. \end{equation*}

Since $0{\lt}r\leq R{\lt}1$ and $x\in K$ were arbitrary, it follows that $\mathrm{dim_A} K\leq s+\varepsilon$ , as required.

4·3. Some examples exhibiting exceptional behaviour

In this section, we give two examples of non-autonomous IFSs in ${\mathbb{R}}$ with respect to the invariant set $X=[0,1]$ with various properties. In order to lower bound the Assouad dimension in these examples, we find it convenient to use the following lemma. Let $p_{\mathcal{H}}$ denote the one-sided Hausdorff metric on compact subsets of ${\mathbb{R}}$ :

\begin{equation*} p_{\mathcal{H}}(E_1;E_2) = \inf\{\delta{\gt}0:\mathrm{dist}(x, E_1) {\lt}\delta\text{ for all }x\in E_2\}.\end{equation*}

The following lemma is standard (see, for instance, [ Reference Mackay and Tyson15 , Proposition 6·1·5]).

Lemma 4·4. Let $K\subset{\mathbb{R}}$ be compact, and suppose E is a compact set for which there are numbers $\lambda_n \geq 1$ and $x_n\in{\mathbb{R}}$ so that

\begin{equation*} \lim_{n\to\infty} p_{\mathcal{H}}(E; \lambda_n(K - x_n)) = 0. \end{equation*}

Then $\mathrm{dim_A} K \geq \mathrm{dim_H} E$ .

We now begin with an example demonstrating that the bounded neighbourhood condition is necessary. First, for $r\in(0,1)$ , set

\begin{equation*} M(r)=\sup_{x\in K}\#\mathcal{N}(x,r).\end{equation*}

Then the bounded neighbourhood condition says that $\limsup_{r\to 0}M(r){\lt}\infty$ .

Let us first note some of the implications of the above properties. If $\varepsilon{\lt}1$ , then Theorem 4·3 together with (iv) and (v) shows that the bounded neighbourhood condition fails. By Theorem 2·9, this together with 4·1 implies that the IFS does not have bounded branching. Therefore, by Lemma 2·4 and (iii), such an IFS cannot be homogeneous.

Now, we start with the main observation underlying this example, which is to exploit inhomogeneity in a way which makes the numbers $\theta(n,m)$ very small, while still allowing large contraction ratios for flexibility in the construction. Given $r_1,r_2\in(0,1)$ , set let $s(r_1,r_2)$ denote the unique solution to the equation

\begin{equation*} r_1^{s(r_1,r_2)}+r_2^{s(r_1,r_2)} = 1. \end{equation*}

Then the following is easy to verify: for any $r_1\in(0,1)$ ,

(4·4) \begin{equation} \lim_{r_2\to 0}s(r_1,r_2) = 0. \end{equation}

We now begin the main construction. First, fix a number $\ell\in{\mathbb{N}}$ with $\ell\geq 2$ and a small number $\varepsilon{\gt}0$ : we define IFSs $\Phi^{\;j}_{\ell,\varepsilon}$ for $j=1,\ldots,\ell-1$ . First, let $r^1_1 = 1 - 1/\ell$ , and applying (4·4) let $0{\lt}r^1_2{\lt} 1-r^1_1$ be sufficiently small so that $s\big(r^1_1,r^1_2\big)\leq \varepsilon$ , and define $\Phi^1_{\ell,\varepsilon} = \big\{S^1_1,S^1_2\big\}$ by

\begin{equation*} S^1_1(x) = r^1_1 x\qquad\text{and}\qquad S^1_2(x) = r^1_2 x + r^1_1. \end{equation*}

Now suppose we have defined $\Phi^{\;j}_{\ell,\varepsilon}$ for some $j\in{\mathbb{N}}$ with $j{\lt}\ell-1$ . Let $r^{\;j+1}_1$ be chosen so that $r^1_1\cdots r^{\;j}_1\cdot r^{\;j+1}_1 = 1 - {j+1}/{\ell}$ , and let $r^{\;j+1}_2 {\lt} 1-r^{\;j+1}_1$ be chosen so that $s(r^{\;j+1}_1, r^{\;j+1}_2) \leq\varepsilon$ , and define $\Phi^{\;j+1}_{\ell,\varepsilon} = \big\{S^{\;j+1}_1,S^{\;j+1}_2\big\}$ by

\begin{equation*} S^{\;j+1}_1(x) = r^{\;j+1}_1 x\qquad\text{and}\qquad S^{\;j+1}_2(x) = r^{\;j+1}_2 x + r^{\;j+1}_1. \end{equation*}

Note for each $j=1,\ldots,\ell-1$ that

(4·5) \begin{equation} S^1_1\circ\cdots\circ S^{\;j-1}_1\circ S^{\;j}_2(0) = r^1_1\cdots r^{\;j-1}_1 r^{\;j}_1 = 1 - \frac{j}{\ell}. \end{equation}

Observe that for every $j=1,\ldots,\ell-1$ the IFS $\Phi^{\;j}_{\ell,\varepsilon} = \{S^{\;j}_1, S^{\;j}_2\}$ we have defined satisfies the open set condition with respect to the interval (0,1) since $r^{\;j}_1+r^{\;j}_2 {\lt}1$ and the similarity dimension satisfies $s(r^{\;j}_1, r^{\;j}_2) \leq\varepsilon$ .

Next, for each $\ell\in{\mathbb{N}}$ with $\ell\geq 2$ , define the tuple of IFSs

\begin{equation*} \Phi^\ell = \bigl(\Phi^1_{\ell, \varepsilon},\ldots,\Phi^{\ell-1}_{\ell, \varepsilon}\bigr). \end{equation*}

Our non-autonomous IFS will be a concatenation of the sequence of tuples of IFSs $(\Phi^{\ell_k})_{k=1}^\infty$ , for numbers $\ell_k$ with $\limsup_{k\to\infty}\ell_k = \infty$ . We will choose these numbers in blocks $(\ell_1,\ldots,\ell_{m_k})$ for $m_k\in{\mathbb{N}}$ . To begin, the IFS $\Phi^1_{2,\varepsilon}$ satisfies the bounded neighbourhood condition with some constant $N_1$ . Let $r_1$ be such that $f(r) \geq 2^2 \cdot N_1$ for all $0{\lt}r\leq r_1$ , and choose $m_1$ sufficiently large so that with

\begin{equation*} (\ell_1,\ldots,\ell_{m_1}) = (2,\ldots,2) \end{equation*}

every cylinder corresponding to a composition from $(\Phi^{\ell_1},\ldots,\Phi^{\ell_{m-1}})$ has contraction ratio at most $r_1$ .

Now suppose we have chosen $(\ell_1,\ldots,\ell_{m_k})$ for some $k\in{\mathbb{N}}$ . By Theorem 2·1, the non-autonomous IFS formed by repeating the maps in $(\Phi^{\ell_1},\ldots,\Phi^{\ell_{m_k}})$ satisfies the bounded neighbourhood condition with some constant $N_k$ . Since f(r) diverges to infinity, get $0{\lt}r_{k+1}\leq r_k$ so that $f(r)\geq N_k 2^{k+1}$ for all $0{\lt}r\leq r_{k+1}$ . Then let $n_k$ be sufficiently large so that every cylinder corresponding to a composition in $(\ell_1,\ldots,\ell_{m_k})^{n_k}$ has contraction ratio at most $r_{k+1}$ . Then let $m_{k+1} = n_k m_k + 1$ and define

\begin{equation*} (\ell_1,\ldots,\ell_{m_{k+1}}) = (\ell_1,\ldots,\ell_{m_k},\ldots,\ell_1,\ldots,\ell_{m_k}, k+2). \end{equation*}

Let $\Phi$ denote the infinite concatenation of $(\Phi^{\ell_k})_{k=1}^\infty$ have corresponding function M(r). For each $k\in{\mathbb{N}}$ , since the IFS corresponding to the tuple $\Phi^{k+2}$ has $2^{k+1}$ maps, it follows that $N_{k+1} \leq N_k 2^{k+1}$ . Moreover, by construction, for all $0{\lt}r_{k+1}{\lt}r \leq r_k$ , $M(r) \leq N_k$ ; and on the other hand, the $r_k$ were chosen precisely so that $f(r) \geq N_{k-1} 2^k \geq N_k$ for all $0{\lt}r\leq r_k$ . Therefore, $M(r) \leq f(r)$ for all $0{\lt}r\leq r_1$ , as required.

Moreover, by construction, this non-autonomous IFS satisfies the open set condition and $\#\mathcal{J}_n=2$ for all n. Let K denote the corresponding limit set.

Observe that $0\in K$ since 0 is a fixed point of the first map in each component IFS. Next, for $k\in{\mathbb{N}}$ , let g denote the map corresponding to any cylinder from $(\Phi^{\ell_1},\ldots,\Phi^{\ell_{m_k-1}})$ . Since $0\in K$ , by (4·5),

\begin{equation*} \left\{\frac{1}{k+1},\ldots,\frac{k}{k+1}\right\}\subset g^{-1}(K)\cap [0,1]. \end{equation*}

Since $k\in{\mathbb{N}}$ was arbitrary, by Lemma 4·4, $\mathrm{dim_A} K=1$ .

Finally, to verify the condition concerning the numbers $\theta(n,m)$ , by construction $\theta(n,1)\leq\varepsilon$ for all $n\in{\mathbb{N}}$ , so by Lemma 4·2, $\theta(n,m)\leq\varepsilon$ for all n,m.

Next, we give an example which is homogeneous and satisfies the open set condition and the cone condition for which the Assouad dimension and the symbolic formula for the Assouad dimension can be specified arbitrarily.

Observe that if $s{\lt}t$ , then Theorem 4·3 together with (iv) and (v) implies that the bounded neighbourhood condition fails. Alternatively, if $\limsup_{n\to\infty}\#\mathcal{J}_n=\infty$ , then Lemma 2·8 shows that the bounded branching does not hold which, by Lemma 2·7, implies that the bounded neighbourhood condition fails.

The idea in this construction is to successively approximate a Cantor set of dimension t at large levels n, using maps with ratios corresponding to a Cantor set of dimension s. Let $r_1$ be chosen so that $\log 2 / \log\big(1/r_1\big) = s$ and let $r_2$ be chosen so that $\log 2/ \log(1/r_2)=t$ . Also, let $\Psi = \{S_1,S_2\}$ denote the Cantor IFS where $S_1(x) = r_2 x$ and $S_2(x) = r_2 x + (1-r_2)$ , and let $E_n$ denote the set of left endpoints at level n:

\begin{equation*} E_n = \bigl\{S_{i_1}\circ\cdots\circ S_{i_n}(0):(i_1,\ldots,i_n)\in\{1,2\}^n\bigr\}. \end{equation*}

We then denote corresponding maps

\begin{equation*} \Phi_n = \{ x\mapsto r_1^n x + y: y\in E_n\}. \end{equation*}

Finally, let $(m_n)_{n=1}^\infty$ be a sequence of natural numbers with $\limsup_{n\to\infty}m_n = \infty$ , and $2^{m_n} \leq k_n$ for all $n\in{\mathbb{N}}$ . Such a choice is possible since, by assumption, $\limsup_{n\to\infty}k_n = \infty$ . We then define the non-autonomous IFS

\begin{equation*} \Phi = (\Phi_{m_n})_{n=1}^\infty. \end{equation*}

Since $\#\mathcal{J}_n = 2^{m_n} \leq k_n$ , the growth rate condition is satisfied, and the IFS is also clearly homogeneous since all of the contraction ratios in $\Phi_{m_n}$ are exactly $r_1^{m_n}$ . Moreover, observe that $\theta(n,1)=s$ by choice of $r_1$ for all $n\in{\mathbb{N}}$ , so in fact $\theta(n,m) = s$ for all $n,m\in{\mathbb{N}}$ .

Let K denote the corresponding limit set: it remains to show that $\mathrm{dim_A} K=t$ . First note that $0\in K$ , since $0\in E_n$ for all $n\in{\mathbb{N}}$ . Now let $M\in{\mathbb{N}}$ be arbitrary and get n such that $m_n \geq M$ . Let $f= f_1\circ\cdots\circ f_{n-1}$ for $f_j \in \Psi_j$ . Since $0\in K$ , $E_M\subset E_{m_n} \subset f^{-1}(K)\cap [0,1]$ . But $\lim_{M\to\infty} E_M = E$ in the Hausdorff metric, where E is the attractor of the IFS $\Psi$ . Therefore by Lemma 4·4, $\mathrm{dim_A} K \geq \mathrm{dim_H} E = t$ .

Now, we obtain the upper bound. Write $\rho_n = r_1^{m_1+\cdots+m_n}$ . We use a different covering strategy between different pairs of scales.

1. (1) We first observe between pairs of scales $\rho_n$ and $\rho_{n+k}$ that K has dimension approximately s. More precisely, let $f\in\Phi_1\circ\cdots\circ\Phi_n$ and consider the interval $I = f([0,1])$ . Then I can be covered by $2^{m_{n+1}+\cdots+m_{n+k}}$ intervals of side-length $\rho_{n+k}$ , so that

   \begin{equation*} \frac{\log N_{\rho_{n+k}}(K\cap I)}{\log(\rho_n/\rho_{n+k})} \leq \frac{(m_{n+1}+\cdots m_{n+k})\log 2}{(m_{n+1}+\cdots m_{n+k})\log(1/r_1)} = s. \end{equation*}
   But any ball $B(x,\rho_n)$ can be covered by at most 2 intervals $I=f([0,1])$ , so
   (4·6) \begin{equation} N_{\rho_{n+k}}(K\cap B(x, \rho_n)) \leq 2\left(\frac{\rho_n}{\rho_{n+k}}\right)^s. \end{equation}

2. (2) Next, let $\rho_n \geq R\geq r \geq \rho_{n+1}$ and consider some interval $I = f([0,1])$ where $f\in\Phi_1\circ\cdots\circ\Phi_n$ . Since E is a self-similar set satisfying the strong separation condition with dimension t, there is a constant $C_0{\gt}0$ so that

   (4·7) \begin{equation} N_{r/\rho_n}\bigl(E \cap B(x,R/\rho_n)\bigr) \leq C_0\left(\frac{R}{r}\right)^t. \end{equation}
   Observe moreover that $K\cap f([0,1])$ is contained in a $\rho_{n+1}$ -neighbourhood of $f(E_{m_n})\subset f(E)$ . But B(x,R) is contained in at most 2 intervals of the form $I = f([0,1])$ , so there is a constant $C{\gt}0$ so that
   \begin{equation} N_r(B(x,R)\cap K) \leq C\left(\frac{R}{r}\right)^t. \end{equation}

Let $0{\lt}r\leq R{\lt}1$ be arbitrary and let $x\in K$ . Let $n_1\in{\mathbb{N}}$ be maximal so that $\rho_{n_1}\geq R$ and let $n_2\in{\mathbb{N}}$ be maximal so that $\rho_{n_2}\geq r$ . If there is an n such that $\rho_n \geq R\geq r \geq\rho_{n+1}$ , we are simply done by (4·6). Otherwise, there is a minimal n and a maximal k so that $R \geq \rho_n \geq \rho_{n+k} \geq r$ . Applying (4·7) to the pairs $R\geq \rho_n$ and $\rho_{n+k}\geq r$ , and (4·6) between $\rho_n\geq \rho_{n+k}$ yields the bound

\begin{align*} N_r\bigl(B(x,R)\cap K\bigr) &\leq C \left(\frac{R}{\rho_n}\right)^t \cdot 2\left(\frac{\rho_n}{\rho_{n+k}}\right)^s \cdot C \left(\frac{\rho_{n+k}}{r}\right)^s\\ & \leq 2 C^2 \left(\frac{R}{r}\right)^t. \end{align*}

Therefore $\mathrm{dim_A} K\leq t$ , as claimed.