Proof. This statement follows immediately from Lemma 3.1 and Theorem 2.3. ◻

Proof. Recall that we write

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D707}}(A)=\bigg\{x\in \mathbb{R}^{d}\,\bigg|\,\lim _{r{\searrow}0}\frac{\log \unicode[STIX]{x1D707}(B(x,r))}{\log r}\in A\bigg\}.\end{eqnarray}$$

We must now prove that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D707}}(A)$ is $\unicode[STIX]{x1D6F1}_{3}^{0}$ -hard and that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D707}}(A)\in \unicode[STIX]{x1D6F1}_{3}^{0}$ . We first note that it follows from Theorem 3.2 that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D707}}(A)$ is $\unicode[STIX]{x1D6F1}_{3}^{0}$ -hard. Also, since $A\in \unicode[STIX]{x1D6F1}_{2}^{0}$ , we conclude from Theorem 2.1 that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D707}}(A)\in \unicode[STIX]{x1D6F1}_{2+1}^{0}=\unicode[STIX]{x1D6F1}_{3}^{0}$ .◻

Proof. This follows from Theorem 3.4 since $\{\unicode[STIX]{x1D6FC}\}\in \unicode[STIX]{x1D6F1}_{2}^{0}$ .◻

Proof. We first prove the following two claims.

Claim 1. For $N\subseteq \mathbb{N}^{\mathbb{N}}$ , we have

$$\begin{eqnarray}(\mathbb{N}^{\mathbb{N}};N)\underset{\text{w}}{{\leqslant}}(\mathbb{R}^{d};\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(N)),\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(\mathbb{N}^{\mathbb{N}}\setminus N))).\end{eqnarray}$$

Proof of Claim 1. We must show that there is a continuous function $f:\mathbb{N}^{\mathbb{N}}\rightarrow \mathbb{R}^{d}$ such that $f(N)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(N))$ and $f(\mathbb{N}^{\mathbb{N}}\setminus N)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(N))$ . Since $\unicode[STIX]{x1D706}$ has a continuous selector, there is a continuous map ${\mathcal{S}}:I_{\unicode[STIX]{x1D706}}\rightarrow \mathbb{R}^{d}$ such that ${\mathcal{S}}(\unicode[STIX]{x1D6FC})\in \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})$ for all $\unicode[STIX]{x1D6FC}\in I_{\unicode[STIX]{x1D706}}$ . Now define $f:\mathbb{N}^{\mathbb{N}}\rightarrow \mathbb{R}^{d}$ by $f={\mathcal{S}}\circ h\circ \unicode[STIX]{x1D6F7}$ .

Since ${\mathcal{S}}$ , $h$ and $\unicode[STIX]{x1D6F7}$ are continues, we conclude that $f$ is continuous.

Next, we show that $f(N)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(N))$ . Indeed, if $\boldsymbol{\unicode[STIX]{x1D708}}\in N$ , then $f(\boldsymbol{\unicode[STIX]{x1D708}})={\mathcal{S}}(h(\unicode[STIX]{x1D6F7}(\boldsymbol{\unicode[STIX]{x1D708}})))\in \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(\boldsymbol{\unicode[STIX]{x1D708}}))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(N))$ , whence $f(N)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(N))$ .

Finally, we show that $f(\mathbb{N}^{\mathbb{N}}\setminus N)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(N))$ . If $\boldsymbol{\unicode[STIX]{x1D708}}\in \mathbb{N}^{\mathbb{N}}\setminus N$ , then $f(\boldsymbol{\unicode[STIX]{x1D708}})={\mathcal{S}}(h(\unicode[STIX]{x1D6F7}(\boldsymbol{\unicode[STIX]{x1D708}})))\in \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(\boldsymbol{\unicode[STIX]{x1D708}}))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(\mathbb{N}^{\mathbb{N}}\setminus N))$ , whence $f(\mathbb{N}^{\mathbb{N}}\setminus N)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(h\unicode[STIX]{x1D6F7}(N))$ . This completes the proof of Claim 1.

Claim 2. For $A\subseteq \mathbb{R}$ , we have

$$\begin{eqnarray}(\mathbb{N}^{\mathbb{N}};(h\unicode[STIX]{x1D6F7})^{-1}(A\cap U))\underset{\text{w}}{{\leqslant}}(\mathbb{R}^{d};\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A),\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\mathbb{R}\setminus A)).\end{eqnarray}$$

Proof of Claim 2. Applying the statement in Part (1) to the set $N=(h\unicode[STIX]{x1D6F7})^{-1}(A\cap U)$ noticing that in this case $h\unicode[STIX]{x1D6F7}(N)=A\cap U$ and $h\unicode[STIX]{x1D6F7}(\mathbb{N}^{\mathbb{N}}\setminus N)=(\mathbb{R}^{d}\setminus A)\cap U$ , we conclude that

(6.2) $$\begin{eqnarray}(\mathbb{N}^{\mathbb{N}};(h\unicode[STIX]{x1D6F7})^{-1}(A\cap U))\underset{\text{w}}{{\leqslant}}(\mathbb{R}^{d};\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A\cap U),\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}((\mathbb{R}\setminus A)\cap U)).\end{eqnarray}$$

However, since $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A\cap U)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ and $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}((\mathbb{R}\setminus A)\cap U)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\mathbb{R}\setminus A)$ and $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)\cap \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\mathbb{R}\setminus A)=\varnothing$ , it follows from (6.2) that

$$\begin{eqnarray}(\mathbb{N}^{\mathbb{N}};(h\unicode[STIX]{x1D6F7})^{-1}(A\cap U))\underset{\text{w}}{{\leqslant}}(\mathbb{R}^{d};\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A),\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\mathbb{R}\setminus A)).\end{eqnarray}$$

This completes the proof of Claim 2.

It follows immediately from Claim 2 that if $A\subseteq \mathbb{R}$ , then we have

(6.3) $$\begin{eqnarray}(\mathbb{N}^{\mathbb{N}};(h\unicode[STIX]{x1D6F7})^{-1}(A\cap U))\underset{\text{w}}{{\leqslant}}(\mathbb{R}^{d};\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)).\end{eqnarray}$$

The statements in parts (1) and (2) follow immediately from (6.3). ◻

Proof. For an ordinal $\unicode[STIX]{x1D6FE}$ with $\unicode[STIX]{x1D6FE}\geqslant 3$ , let $\unicode[STIX]{x1D6E4}$ denote $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}$ or let $\unicode[STIX]{x1D6E4}$ denote $\unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6FE}}^{0}$ . Let $A$ be a subset of $I_{\unicode[STIX]{x1D706}}$ and assume that $A$ is $\unicode[STIX]{x1D6E4}(\mathbb{R})$ -hard. We now prove the following three claims.

Claim 1. $\mathbb{R}\setminus A\not \in \unicode[STIX]{x1D6E4}(\mathbb{R})$ .

Proof of Claim 1. This is clear since $A$ is $\unicode[STIX]{x1D6E4}(\mathbb{R})$ -hard. This completes the proof of Claim 1.

Claim 2. $\mathbb{R}\setminus (A\cap U)\not \in \unicode[STIX]{x1D6E4}(\mathbb{R})$ .

Proof of Claim 2. Assume, in order to reach a contradiction, that $\mathbb{R}\setminus (A\cap U)\in \unicode[STIX]{x1D6E4}(\mathbb{R})$ . We have

(6.4) $$\begin{eqnarray}\mathbb{R}\setminus A=(\mathbb{R}\setminus (A\cap U))\,\cup \,(\mathbb{R}\setminus (A\setminus U)),\end{eqnarray}$$

where, by assumption,

(6.5) $$\begin{eqnarray}\mathbb{R}\setminus (A\cap U)\in \unicode[STIX]{x1D6E4}(\mathbb{R}).\end{eqnarray}$$

Also, since $A\setminus U$ is countable (because $A\setminus U\subseteq I_{\unicode[STIX]{x1D706}}\setminus U$ and $I_{\unicode[STIX]{x1D706}}\setminus U$ is countable), we conclude that

(6.6) $$\begin{eqnarray}\mathbb{R}\setminus (A\setminus U)\in {\mathcal{G}}_{\unicode[STIX]{x1D6FF}}(\mathbb{R})\subseteq \unicode[STIX]{x1D6E4}(\mathbb{R}).\end{eqnarray}$$

Combining (6.4)–(6.6), we conclude that $\mathbb{R}\setminus A\in \unicode[STIX]{x1D6E4}(\mathbb{R})$ contradicting Claim 1. This completes the proof of Claim 2.

Proof of Claim 3. For brevity write $M=\mathbb{R}\setminus (A\cap U)\in \unicode[STIX]{x1D6E4}(\mathbb{R})$ and note that $M\cap U=U\setminus (A\cap U)$ .

Assume now, in order to reach a contradiction, that $M\cap U=U\setminus (A\cap U)\in \unicode[STIX]{x1D6E4}(U)$ . We can therefore find $E\in \unicode[STIX]{x1D6E4}(\mathbb{R})$ such that $M\cap U=E\cap U$ , whence

(6.7) $$\begin{eqnarray}\displaystyle \mathbb{R}\setminus (A\cap U) & = & \displaystyle M\nonumber\\ \displaystyle & = & \displaystyle (M\cap U)\cup (M\setminus U)\nonumber\\ \displaystyle & = & \displaystyle (E\cap U)\,\cup (M\setminus U)\nonumber\\ \displaystyle & = & \displaystyle (E\cap U)\,\cup \,(\mathbb{R}\setminus U)\nonumber\\ \displaystyle & = & \displaystyle E\cup (\mathbb{R}\setminus U).\end{eqnarray}$$

Since $U$ is a homeomorphic image of $[0,1]\cap \mathbb{I}$ , we deduce that $U\in {\mathcal{G}}_{\unicode[STIX]{x1D6FF}}(\mathbb{R})$ , whence $\mathbb{R}\setminus U\in \unicode[STIX]{x1D6E4}(\mathbb{R})$ . Also, by assumption, $E\in \unicode[STIX]{x1D6E4}(\mathbb{R})$ . Since $\mathbb{R}\setminus U,E\in \unicode[STIX]{x1D6E4}(\mathbb{R})$ , we deduce that from (6.7) that $\mathbb{R}\setminus (A\cap U)=E\cup (\mathbb{R}\setminus U)\in \unicode[STIX]{x1D6E4}(\mathbb{R})$ contradicting Claim 2. This completes the proof of Claim 3.

Using the fact that $h\unicode[STIX]{x1D6F7}$ is a homeomorphism between $\mathbb{N}^{\mathbb{N}}$ and $U$ , we conclude from Claim 3 that

(6.8) $$\begin{eqnarray}\mathbb{N}^{\mathbb{N}}\setminus (h\unicode[STIX]{x1D6F7})^{-1}(A\cap U)=(h\unicode[STIX]{x1D6F7})^{-1}(U\setminus (A\cap U))\not \in \unicode[STIX]{x1D6E4}(\mathbb{N}^{\mathbb{N}}).\end{eqnarray}$$

Finally, since $\mathbb{N}^{\mathbb{N}}$ is a zero-dimensional Polish space, it follows from (6.8) and [Reference Kechris5, Theorem 22.10] that $(h\unicode[STIX]{x1D6F7})^{-1}(A\cap U)$ is $\unicode[STIX]{x1D6E4}(\mathbb{N}^{\mathbb{N}})$ -hard.◻

Proof. The statement in Theorem 2.2 follows immediately by combining Propositions 6.1 and 6.2. ◻

Proof. We must show that there is a continuous map $f:\{0,1\}^{\mathbb{N}}\rightarrow \mathbb{R}$ such that $f(\unicode[STIX]{x1D70C}^{-1}(H))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ and $f(\{0,1\}^{\mathbb{N}}\setminus \unicode[STIX]{x1D70C}^{-1}(H))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\mathbb{R}\setminus A)$ . By assumption there is a continuous map $\unicode[STIX]{x1D711}:\{0,1\}^{\mathbb{N}}\rightarrow I_{\unicode[STIX]{x1D706}}$ such that $\unicode[STIX]{x1D711}(H)\subseteq A$ and $h\unicode[STIX]{x1D711}(\{0,1\}^{\mathbb{N}}\setminus H)\subseteq I_{\unicode[STIX]{x1D706}}\setminus A$ . Also, since $\unicode[STIX]{x1D706}$ has a continuous selector, there is a continuous map ${\mathcal{S}}:I_{\unicode[STIX]{x1D706}}\rightarrow \mathbb{R}^{d}$ such that ${\mathcal{S}}(\unicode[STIX]{x1D6FC})\in \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D6FC})$ for all $\unicode[STIX]{x1D6FC}\in I_{\unicode[STIX]{x1D706}}$ . Now define $f:\{0,1\}^{\mathbb{N}}\rightarrow \mathbb{R}$ by $f={\mathcal{S}}\circ \unicode[STIX]{x1D711}\circ \unicode[STIX]{x1D70C}$ .

Since ${\mathcal{S}}$ , $\unicode[STIX]{x1D711}$ and $\unicode[STIX]{x1D70C}$ are continuous, we conclude that $f$ is continuous.

Next, we show that $f(\unicode[STIX]{x1D70C}^{-1}(H))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ . Indeed, if $\mathbf{u}\in \unicode[STIX]{x1D70C}^{-1}(H)$ , then $f(\mathbf{u})={\mathcal{S}}(\unicode[STIX]{x1D711}(\unicode[STIX]{x1D70C}(\mathbf{u})))\in \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D711}(\unicode[STIX]{x1D70C}(\mathbf{u})))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D711}(H))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ , whence $f(\unicode[STIX]{x1D70C}^{-1}(H))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ .

Finally, we show that $f(\{0,1\}^{\mathbb{N}}\setminus \unicode[STIX]{x1D70C}^{-1}(H))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\mathbb{R}\setminus A)$ . Namely, if $\mathbf{u}\in \{0,1\}^{\mathbb{N}}\setminus \unicode[STIX]{x1D70C}^{-1}(H)=\unicode[STIX]{x1D70C}^{-1}(\{0,1\}^{\mathbb{N}}\setminus H)$ , then $f(\mathbf{u})={\mathcal{S}}(\unicode[STIX]{x1D711}(\unicode[STIX]{x1D70C}(\mathbf{u})))\in \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D711}(\unicode[STIX]{x1D70C}(\mathbf{u})))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D711}(\{0,1\}^{\mathbb{N}}\setminus H))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(I_{\unicode[STIX]{x1D706}}\setminus A)\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\mathbb{R}\setminus A)$ , whence $f(\{0,1\}^{\mathbb{N}}\setminus \unicode[STIX]{x1D70C}^{-1}(H))\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(\mathbb{R}\setminus A)$ .◻

Proof. (1) We divide the proof into the following two parts.

Part 1. Assume that $3\leqslant \unicode[STIX]{x1D6FE}<\unicode[STIX]{x1D714}$ . For convenience write $n=\unicode[STIX]{x1D6FE}\in \mathbb{N}$ . Also, let $\unicode[STIX]{x1D70C}:\{0,1\}^{\mathbb{N}}\rightarrow \{0,1\}^{\mathbb{N}}$ be the map in (6.9), and let $H_{1},H_{2},\ldots$ be the sets in (6.10). We must now prove that $\unicode[STIX]{x1D6E5}_{n+1}(A)$ is $\unicode[STIX]{x1D6F1}_{n+1}^{0}$ -hard and that $\unicode[STIX]{x1D6E5}_{n+1}(A)$ in $\unicode[STIX]{x1D6F1}_{n+1}^{0}$ .

We first prove that $\unicode[STIX]{x1D6E5}_{n+1}(A)$ is $\unicode[STIX]{x1D6F1}_{n+1}^{0}$ -hard. Indeed, since $A$ is $\unicode[STIX]{x1D6F1}_{n}^{0}(I_{\unicode[STIX]{x1D706}})$ -hard and $H_{n}\in \unicode[STIX]{x1D6F1}_{n}^{0}$ , we conclude that

$$\begin{eqnarray}(\{0,1\}^{\mathbb{N}};H_{n})\underset{\text{w}}{{\leqslant}}(I_{\unicode[STIX]{x1D706}};A),\end{eqnarray}$$

and it therefore follows from Proposition 6.4 that

(6.11) $$\begin{eqnarray}(\{0,1\}^{\mathbb{N}};H_{n+1})=(\{0,1\}^{\mathbb{N}};\unicode[STIX]{x1D70C}^{-1}(H_{n}))\underset{\text{w}}{{\leqslant}}(\mathbb{R};\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)).\end{eqnarray}$$

Finally, since $H_{n+1}$ is $\unicode[STIX]{x1D6F1}_{n+1}^{0}$ -hard it now follows from (6.11) that $\unicode[STIX]{x1D6E5}_{n+1}(A)$ is $\unicode[STIX]{x1D6F1}_{n+1}^{0}$ -hard.

Next, we prove that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ in $\unicode[STIX]{x1D6F1}_{n+1}^{0}$ . Since $A\in \unicode[STIX]{x1D6F1}_{n}^{0}(I_{\unicode[STIX]{x1D706}})$ , we conclude that there is a set $E\in \unicode[STIX]{x1D6F1}_{n}^{0}(\mathbb{R})$ such that $A=E\cap I_{\unicode[STIX]{x1D706}}$ . Also, since $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(M)=\varnothing$ for all $M\subseteq \mathbb{R}\setminus I_{\unicode[STIX]{x1D706}}$ , we conclude that

(6.12) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E)=\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E\cap I_{\unicode[STIX]{x1D706}}).\end{eqnarray}$$

Furthermore, because $E\in \unicode[STIX]{x1D6F1}_{n}^{0}(\mathbb{R})$ , it follows from Theorem 2.1 that

(6.13) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E)\in \unicode[STIX]{x1D6F1}_{n+1}^{0}.\end{eqnarray}$$

Finally, combining (6.12) and (6.13) and using the fact that $A=E\cap I_{\unicode[STIX]{x1D706}}$ , we conclude that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)=\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E\cap I_{\unicode[STIX]{x1D706}})=\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E)\in \unicode[STIX]{x1D6F1}_{n+1}^{0}$ .

Part 2: Assume that $\unicode[STIX]{x1D714}\leqslant \unicode[STIX]{x1D6FE}.$ We must now prove that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ is $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}$ -hard and that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ in $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}$ .

We first prove that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ is $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}$ -hard. Indeed, since $A$ is $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}(I_{\unicode[STIX]{x1D706}})$ -hard, we conclude that $A$ is $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}(\mathbb{R})$ -hard, and it therefore follows from Theorem 2.2 that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ in $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}$ -hard.

Next, we prove that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)$ in $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}$ . Since $A\in \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}(I_{\unicode[STIX]{x1D706}})$ , we conclude that there is $E\in \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}(\mathbb{R})$ such that $A=E\cap I_{\unicode[STIX]{x1D706}}$ . Using an argument similar to the proof of (6.12) we conclude that

(6.14) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E)=\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E\cap I_{\unicode[STIX]{x1D706}}).\end{eqnarray}$$

Furthermore, because $E\in \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}(\mathbb{R})$ , it follows from Theorem 2.1 that

(6.15) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E)\in \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}.\end{eqnarray}$$

Finally, combining (6.14) and (6.15) and using the fact that $A=E\cap I_{\unicode[STIX]{x1D706}}$ , we conclude that $\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(A)=\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E\cap I_{\unicode[STIX]{x1D706}})=\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D706}}(E)\in \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FE}}^{0}$ .

(2) The proof of this statement is similar to the proof of the statement in Part (1) and is therefore omitted. ◻

Proof. Let $\mathbf{t}=t_{1}t_{2}\ldots \in (a_{\min },a_{\max })^{\mathbb{N}}$ . We define a sequence $(\mathbf{i}_{n}(\mathbf{t}))_{n}$ of strings $\mathbf{i}_{n}(\mathbf{t})\in I^{\ast }$ inductively such that for all $n$ , we have:

1. (1) $a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}(\mathbf{t})|k}$ is a monotone function of $k$ for $|\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})|\leqslant k\leqslant |\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}(\mathbf{t})|$ ;

2. (2) $|a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}(\mathbf{t})}-t_{n}|\leqslant c/n$ .

Below we construct the sequence $(\mathbf{i}_{n}(\mathbf{t}))_{n}$ inductively.

The start of the induction. Choose any $\mathbf{k}\in I^{\ast }$ and put $\mathbf{i}_{1}(\mathbf{t})=\mathbf{k}$ . Note that conditions (1) and (2) are satisfied. This completes the start of the induction.

The inductive step. Assume that the strings $\mathbf{i}_{1}(\mathbf{t}),\ldots ,\mathbf{i}_{n}(\mathbf{t})\in I^{\ast }$ have been chosen such that conditions (1) and (2) are satisfied. We will now construct the string $\mathbf{i}_{n+1}(\mathbf{t})$ .

If $a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})}<t_{n}$ , then the fact that $t_{n}\in (a_{\min },a_{\max })$ combined with the fact that the sequence $a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})i_{\max }^{k}}$ is increasing (by condition (i)) with $\lim _{k}a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})i_{\max }^{k}}=a_{\max }$ (by condition (iii)), imply that we can choose a unique positive $m_{n+1}(\mathbf{t})$ such that the following is satisfied: if we put $\mathbf{i}_{n+1}(\mathbf{t})=i_{\max }^{m_{n+1}(\mathbf{t})}$ and write $\mathbf{i}_{n+1}^{-}(\mathbf{t})=i_{\max }^{m_{n+1}(\mathbf{t})-1}$ , then

(7.7) $$\begin{eqnarray}a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}^{-}(\mathbf{t})}<t_{n}\leqslant a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}(\mathbf{t})}.\end{eqnarray}$$

If $t_{n}\leqslant a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})}$ , then the fact that $t_{n}\in (a_{\min },a_{\max })$ combined with the fact that the sequence $a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})i_{\min }^{k}}$ is decreasing (by condition (i)) with $\lim _{k}a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})i_{\min }^{k}}=a_{\min }$ (by condition (iii)), imply that we can choose a unique positive $m_{n+1}(\mathbf{t})$ such that the following is satisfied: if we put $\mathbf{i}_{n+1}(\mathbf{t})=i_{\min }^{m_{n+1}(\mathbf{t})}$ and write $\mathbf{i}_{n+1}^{-}(\mathbf{t})=i_{\min }^{m_{n+1}(\mathbf{t})-1}$ , then

(7.8) $$\begin{eqnarray}a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}(\mathbf{t})}<t_{n}\leqslant a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}^{-}(\mathbf{t})}.\end{eqnarray}$$

Next, we show that conditions (1) and (2) are satisfied. Indeed, it follows from the construction of $\mathbf{i}_{n+1}(\mathbf{t})$ and condition (i) that condition (1) is satisfied. It also follows from (7.7) and (7.8) and condition (ii) that $|a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}(\mathbf{t})}-t_{n}|\leqslant |a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}(\mathbf{t})}-a_{\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}^{-}(\mathbf{t})}|\leqslant c/|\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}^{-}(\mathbf{t})|\leqslant c/n$ (because $|\mathbf{i}_{1}(\mathbf{t})\ldots \mathbf{i}_{n}(\mathbf{t})\mathbf{i}_{n+1}^{-}(\mathbf{t})|\geqslant |\mathbf{i}_{1}(\mathbf{t})|+\cdots +|\mathbf{i}_{n}(\mathbf{t})|\geqslant n$ ), and condition (2) is therefore satisfied. This completes the inductive step.

We now define the function $F$ by

$$\begin{eqnarray}F(\mathbf{t})=\mathbf{i}_{1}(\mathbf{t})\,\mathbf{i}_{2}(\mathbf{t})\ldots .\end{eqnarray}$$

Below we show that $F$ is continuous and that $F$ satisfies (7.5) and (7.6).

We first prove that $F$ is continuous. Indeed, since the family $(a_{\mathbf{i}})_{\mathbf{i}\in I^{k}}$ is finite for all $k$ , for each positive integer $n$ there is an integer $N_{n}$ such that if $\mathbf{t}=t_{1}t_{2}\ldots ,\mathbf{u}=u_{1}u_{2}\ldots \in (\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max })^{\mathbb{N}}$ with $t_{i}=u_{i}$ for all $i\leqslant N_{n}$ , then $\mathbf{i}_{i}(\mathbf{t})=\mathbf{i}_{i}(\mathbf{u})$ for all $i\leqslant n$ . This clearly implies that $F$ is continuous.

Next, we note that it follows from the construction of $F$ that $F(tt\ldots )\rightarrow \mathbf{k}\,i_{\min }\,i_{\min }\ldots$ as $t{\searrow}a_{\min }$ and $F(tt\ldots )\rightarrow \mathbf{k}i_{\max }i_{\max }\ldots$ as $t{\nearrow}a_{\max }$ with $t\in (a_{\min },a_{\max })$ .

Finally, we note that it follows from conditions (1) and (2) that $\liminf _{n}a_{F(\mathbf{t})|n}=\liminf _{n}t_{n}$ and $\limsup _{n}a_{F(\mathbf{t})|n}=\limsup _{n}t_{n}$ for all $\mathbf{t}=t_{1}t_{2}\ldots \in (a_{\min },a_{\max })^{\mathbb{N}}$ .◻

Proof. For $\mathbf{i}\in \unicode[STIX]{x1D6F4}^{\mathbb{N}}$ , let

$$\begin{eqnarray}a_{\mathbf{i}}=\frac{\log p_{\mathbf{i}}}{\log r_{\mathbf{i}}},\end{eqnarray}$$

and choose $i_{\min },i_{\max }\in \{1,\ldots ,N\}$ such that

$$\begin{eqnarray}a_{i_{\min }}=\frac{\log p_{i_{\min }}}{\log r_{i_{\min }}}=\unicode[STIX]{x1D6FC}_{\min }\end{eqnarray}$$

and

$$\begin{eqnarray}a_{i_{\max }}=\frac{\log p_{i_{\max }}}{\log r_{i_{\min }}}=\unicode[STIX]{x1D6FC}_{\max }.\end{eqnarray}$$

It is easily seen that the family $(a_{\mathbf{i}})_{\mathbf{i}\in \unicode[STIX]{x1D6F4}^{\mathbb{N}}}$ satisfies conditions (i)–(iii) in Proposition 7.2, and we therefore conclude from Proposition 7.2 that there is a continuous function $F:(\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max })^{\mathbb{N}}\rightarrow \unicode[STIX]{x1D6F4}^{\mathbb{N}}$ and a string $\mathbf{k}\in \unicode[STIX]{x1D6F4}^{\ast }$ such that $F(tt\ldots )\rightarrow \mathbf{k}\,i_{\min }\,i_{\min }\ldots$ as $t{\searrow}a_{\min }$ and $F(tt\ldots )\rightarrow \mathbf{k}i_{\max }i_{\max }\ldots$ as $t{\nearrow}a_{\max }$ with $t\in (a_{\min },a_{\max })$ , and $\liminf _{n}a_{F(\mathbf{t})|n}=\liminf _{n}t_{n}$ and $\limsup _{n}a_{F(\mathbf{t})|n}=\limsup _{n}t_{n}$ for all $\mathbf{t}=t_{1}t_{2}\ldots \in (\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max })^{\mathbb{N}}$ . Now define $S:[\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max }]\rightarrow \unicode[STIX]{x1D6F4}^{\mathbb{N}}$ by

$$\begin{eqnarray}S(\unicode[STIX]{x1D6FC})=\left\{\begin{array}{@{}ll@{}}\mathbf{k}i_{\min }i_{\min }\ldots \quad & \text{for }\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FC}_{\min },\\ F(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FC}\ldots )\quad & \text{for }\unicode[STIX]{x1D6FC}\in (\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max }),\\ \mathbf{k}i_{\max }i_{\max }\ldots \quad & \text{for }\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FC}_{\max }.\end{array}\right.\end{eqnarray}$$

It is clear that $S$ is continuous. Also, if $\unicode[STIX]{x1D6FC}\in (\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max })$ , then we conclude that

$$\begin{eqnarray}\liminf _{n}a_{S(\unicode[STIX]{x1D6FC})|n}=\liminf _{n}a_{F(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FC}\ldots )|n}=\liminf _{n}\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FC}\end{eqnarray}$$

and

$$\begin{eqnarray}\limsup _{n}a_{S(\unicode[STIX]{x1D6FC})|n}=\limsup _{n}a_{F(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FC}\ldots )|n}=\limsup _{n}\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D6FC},\end{eqnarray}$$

whence $\log p_{S(\unicode[STIX]{x1D6FC})|n}/\text{log}\,r_{S(\unicode[STIX]{x1D6FC})|n}=a_{S(\unicode[STIX]{x1D6FC})|n}\rightarrow \unicode[STIX]{x1D6FC}.$ Finally, if $\unicode[STIX]{x1D6FC}\in \{\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max }\}$ , then clearly $\log p_{S(\unicode[STIX]{x1D6FC})|n}/\text{log}\,r_{S(\unicode[STIX]{x1D6FC})|n}\rightarrow \unicode[STIX]{x1D6FC}$ .◻

Proof. It follows from [Reference Mattila8] that $\unicode[STIX]{x1D707}$ is aspherical.

Next we prove that $\unicode[STIX]{x1D707}$ has a continuous selector. Indeed, it follows from Proposition 7.3 that we can find a continuous function $S:[\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max }]\rightarrow \unicode[STIX]{x1D6F4}^{\mathbb{N}}$ such that

$$\begin{eqnarray}S(\unicode[STIX]{x1D6FC})\in \bigg\{\mathbf{i}\in \unicode[STIX]{x1D6F4}^{\mathbb{N}}\,\bigg|\,\lim _{n}\frac{\log p_{\mathbf{i}\mid n}}{\log r_{\mathbf{i}\mid n}}=\unicode[STIX]{x1D6FC}\bigg\}\end{eqnarray}$$

for all $\unicode[STIX]{x1D6FC}\in [\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max }]$ . Let $\unicode[STIX]{x1D70B}:\unicode[STIX]{x1D6F4}^{\mathbb{N}}\rightarrow \mathbb{R}^{d}$ be the function in (7.4) and define ${\mathcal{S}}:[\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max }]\rightarrow \mathbb{R}^{d}$ by ${\mathcal{S}}=\unicode[STIX]{x1D70B}\circ S$ . The ${\mathcal{S}}$ is continuous and it follows from Proposition 7.2 that

$$\begin{eqnarray}\displaystyle {\mathcal{S}}(\unicode[STIX]{x1D6FC}) & = & \displaystyle \unicode[STIX]{x1D70B}(S(\unicode[STIX]{x1D6FC}))\in \unicode[STIX]{x1D70B}\bigg\{\mathbf{i}\in \unicode[STIX]{x1D6F4}^{\mathbb{N}}\,\bigg|\,\lim _{n}\frac{\log p_{\mathbf{i}\mid n}}{\log r_{\mathbf{i}\mid n}}=\unicode[STIX]{x1D6FC}\bigg\}\nonumber\\ \displaystyle & = & \displaystyle \bigg\{x\in \mathbb{R}^{d}\,\bigg|\,\lim _{r{\searrow}0}\frac{\log \unicode[STIX]{x1D707}(B(x,r))}{\log r}=\unicode[STIX]{x1D6FC}\bigg\}=\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D707}}(\unicode[STIX]{x1D6FC})\nonumber\end{eqnarray}$$

for all $\unicode[STIX]{x1D6FC}\in [\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max }]$ .

Finally, since $I_{\unicode[STIX]{x1D707}}=[\unicode[STIX]{x1D6FC}_{\min },\unicode[STIX]{x1D6FC}_{\max }]$ , we conclude that $\unicode[STIX]{x1D707}$ is admissible.◻

Proof. We must show that there is a continuous function $f:\mathbb{N}^{\mathbb{N}}\rightarrow \{0,1\}^{\mathbb{N}}$ such that $f(C)\subseteq \unicode[STIX]{x1D6EF}_{0}$ and $f(\mathbb{N}^{\mathbb{N}}\setminus C)\subseteq \unicode[STIX]{x1D6EF}$ . For $\mathbf{i}\in \{0,1\}^{\mathbb{N}}$ , let $a_{\mathbf{i}}=\unicode[STIX]{x1D70B}_{1}(\mathbf{i})$ . It is easily seen that the family $(a_{\mathbf{i}})_{\mathbf{i}\in \unicode[STIX]{x1D6F4}^{\mathbb{N}}}$ satisfies conditions (i)–(iii) in Proposition 7.2 with $a_{\min }=0$ and $a_{\max }=1$ , and we therefore conclude from Proposition 7.2 that there is a continuous function $F:(0,1)^{\mathbb{N}}\rightarrow \{0,1\}^{\mathbb{N}}$ such that $\liminf _{n}a_{F(\mathbf{t})|n}=\liminf _{n}t_{n}$ and $\limsup _{n}a_{F(\mathbf{t})|n}=\limsup _{n}t_{n}$ for all $\mathbf{t}=t_{1}t_{2}\ldots \in (0,1)^{\mathbb{N}}$ . Next, define the map $T:\mathbb{N}^{\mathbb{N}}\rightarrow (0,1)^{\mathbb{N}}$ by $T(\boldsymbol{\unicode[STIX]{x1D708}})=t_{1}(\boldsymbol{\unicode[STIX]{x1D708}})t_{2}(\boldsymbol{\unicode[STIX]{x1D708}})\ldots$ for $\boldsymbol{\unicode[STIX]{x1D708}}=\unicode[STIX]{x1D708}_{1}\unicode[STIX]{x1D708}_{2}\ldots \in \mathbb{N}^{\mathbb{N}}$ where

$$\begin{eqnarray}t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})=\left\{\begin{array}{@{}ll@{}}{\displaystyle \frac{1}{2n}}\quad & \text{for }n\text{ odd},\\ {\displaystyle \frac{1}{2\unicode[STIX]{x1D708}_{n}}}\quad & \text{for }n\text{ even},\end{array}\right.\end{eqnarray}$$

(the factor $2$ in the denominator in the definition of $t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})$ is included to ensure that $t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})\in (0,1)$ ). Finally, we define $f:\mathbb{N}^{\mathbb{N}}\rightarrow \{0,1\}^{\mathbb{N}}$ by $f=F\circ T$ .

Since $F$ and $T$ are continuous, we conclude that $f$ is continuous.

Next, we show that $f(C)\subseteq \unicode[STIX]{x1D6EF}_{0}$ . Indeed, if $\boldsymbol{\unicode[STIX]{x1D708}}\in C$ , then $t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})\rightarrow 0$ , and so $\liminf _{n}a_{f(\boldsymbol{\unicode[STIX]{x1D708}})|n}=\liminf _{n}a_{F(T(\boldsymbol{\unicode[STIX]{x1D708}}))|n}=\liminf _{n}t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})=0$ and $\limsup _{n}a_{f(\boldsymbol{\unicode[STIX]{x1D708}})|n}=\limsup _{n}a_{F(T(\boldsymbol{\unicode[STIX]{x1D708}}))|n}=\limsup _{n}t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})=0$ , whence $\unicode[STIX]{x1D70B}_{1}(f(\boldsymbol{\unicode[STIX]{x1D708}})|n)=a_{f(\boldsymbol{\unicode[STIX]{x1D708}})|n}\rightarrow 0$ , i.e. $f(\boldsymbol{\unicode[STIX]{x1D708}})\in \unicode[STIX]{x1D6EF}_{0}$ .

Finally, we show that $f(\mathbb{N}^{\mathbb{N}}\setminus C)\subseteq \unicode[STIX]{x1D6EF}$ . Namely, if $\boldsymbol{\unicode[STIX]{x1D708}}\in \mathbb{N}^{\mathbb{N}}\setminus C$ , then $\liminf _{n}t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})=0$ and $\limsup _{n}t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})>0$ , and so $\liminf _{n}a_{f(\boldsymbol{\unicode[STIX]{x1D708}})|n}=\liminf _{n}a_{F(T(\boldsymbol{\unicode[STIX]{x1D708}}))|n}=\liminf _{n}t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})=0$ and $\limsup _{n}a_{f(\boldsymbol{\unicode[STIX]{x1D708}})|n}=\limsup _{n}a_{F(T(\boldsymbol{\unicode[STIX]{x1D708}}))|n}=\limsup _{n}t_{n}(\boldsymbol{\unicode[STIX]{x1D708}})>0$ . It follows immediately from this that $f(\boldsymbol{\unicode[STIX]{x1D708}})\in \unicode[STIX]{x1D6EF}$ .◻

Proof. For each positive integer $n$ , write

$$\begin{eqnarray}\unicode[STIX]{x1D708}(\mathbf{u};n)=|\{m\,|\,\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(m)\leqslant n,u_{m}=1\}|.\end{eqnarray}$$

Next, note that it follows from the definitions of $\unicode[STIX]{x1D708}(\mathbf{u};n)$ and $\mathbf{j}=f_{i,\mathbf{i},j}(\mathbf{u})$ that

(7.9) $$\begin{eqnarray}\frac{\log p_{\mathbf{j}|n}}{\log r_{\mathbf{j}|n}}=\frac{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};n)\log p_{k}+(\unicode[STIX]{x1D708}(\mathbf{u};n)/n)(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};n)\log r_{k}+(\unicode[STIX]{x1D708}(\mathbf{u};n)/n)(\log r_{j}-\log r_{i})}.\end{eqnarray}$$

We can now prove statements (1)–(5).

(1) This is clear from the construction.

(2) Assume now that $\unicode[STIX]{x1D70B}_{1}(\mathbf{u};n)\rightarrow 0$ . For brevity write $N(\mathbf{u};n)=|\{m\,|\,\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(m)\leqslant n\}|$ , and note that $N(\mathbf{u};n)\leqslant n$ , whence

$$\begin{eqnarray}\frac{\unicode[STIX]{x1D708}(\mathbf{u};n)}{n}\leqslant \frac{\unicode[STIX]{x1D708}(\mathbf{u};n)}{N(\mathbf{u};n)}=\unicode[STIX]{x1D70B}_{1}(\mathbf{u};\,N(\mathbf{u};n))\rightarrow 0\end{eqnarray}$$

(because $\unicode[STIX]{x1D70B}_{1}(\mathbf{u};n)\rightarrow 0$ ). Using (7.9), it follows from this and the fact that $\boldsymbol{\unicode[STIX]{x1D70B}}(\mathbf{i};n)\rightarrow \mathbf{q}$ (i.e. $\unicode[STIX]{x1D70B}_{k}(\mathbf{i};n)\rightarrow q_{k}$ for all $k$ ), that

(7.10) $$\begin{eqnarray}\displaystyle \frac{\log p_{\mathbf{j}|n}}{\log r_{\mathbf{j}|n}} & = & \displaystyle \frac{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};n)\log p_{k}+(\unicode[STIX]{x1D708}(\mathbf{u};n)/n)(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};n)\log r_{k}+(\unicode[STIX]{x1D708}(\mathbf{u};n)/n)(\log r_{j}-\log r_{i})}\nonumber\\ \displaystyle & \rightarrow & \displaystyle \frac{\mathop{\sum }_{k}q_{k}\log p_{k}}{\mathop{\sum }_{k}q_{k}\log r_{k}}.\end{eqnarray}$$

Finally, it follows from condition (i) and assumption (7.10) that $\log p_{\mathbf{j}|n}/\text{log}\,r_{\mathbf{j}|n}\rightarrow \unicode[STIX]{x1D6FC}$ .

(3) Assume now that there is a strictly increasing sequence $(m_{n})_{n}$ of positive integers such that $\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m_{n})\rightarrow v$ . For brevity write $\unicode[STIX]{x1D70E}(m)=\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(m)$ . Noticing that $\unicode[STIX]{x1D708}(\mathbf{u};\unicode[STIX]{x1D70E}(m))=m\,\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m)$ for all $m$ , we conclude from (7.9) that

(7.11) $$\begin{eqnarray}\displaystyle \frac{\log p_{\mathbf{j}|\unicode[STIX]{x1D70E}(m)}}{\log r_{\mathbf{j}|\unicode[STIX]{x1D70E}(m)}} & = & \displaystyle \frac{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\log p_{k}+(\unicode[STIX]{x1D708}(\mathbf{u};\unicode[STIX]{x1D70E}(m))/\unicode[STIX]{x1D70E}(m))(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\log r_{k}+(\unicode[STIX]{x1D708}(\mathbf{u};\unicode[STIX]{x1D70E}(m))/\unicode[STIX]{x1D70E}(m))(\log r_{j}-\log r_{i})}\nonumber\\ \displaystyle & = & \displaystyle \frac{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\log p_{k}+(m/\unicode[STIX]{x1D70E}(m))\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m)(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\log r_{k}+(m/\unicode[STIX]{x1D70E}(m))\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m)(\log r_{j}-\log r_{i})}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Next, since clearly $m/\unicode[STIX]{x1D70E}(m)=\unicode[STIX]{x1D70B}_{i}(\mathbf{i};\unicode[STIX]{x1D70E}(m))$ , it follows from (7.11) that

(7.12) $$\begin{eqnarray}\displaystyle \frac{\log p_{\mathbf{j}|\unicode[STIX]{x1D70E}(m)}}{\log r_{\mathbf{j}|\unicode[STIX]{x1D70E}(m)}} & = & \displaystyle \frac{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\log p_{k}+(m/\unicode[STIX]{x1D70E}(m))\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m)\,(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\log r_{k}+(m/\unicode[STIX]{x1D70E}(m))\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m)\,(\log r_{j}-\log r_{i})}\nonumber\\ \displaystyle & = & \displaystyle \frac{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\log p_{k}+\unicode[STIX]{x1D70B}_{i}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m)(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\log r_{k}+\unicode[STIX]{x1D70B}_{i}(\mathbf{i};\unicode[STIX]{x1D70E}(m))\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m)(\log r_{j}-\log r_{i})}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Finally, since $\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m_{n}))\rightarrow q_{k}$ for all $k$ (because $\boldsymbol{\unicode[STIX]{x1D70B}}(\mathbf{i};n)\rightarrow \mathbf{q}$ by assumption) and $\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m_{n})\rightarrow v$ , we deduce from (7.12) that

$$\begin{eqnarray}\displaystyle \frac{\log p_{\mathbf{j}|\unicode[STIX]{x1D70E}(m_{n})}}{\log r_{\mathbf{j}|\unicode[STIX]{x1D70E}(m_{n})}} & = & \displaystyle \frac{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m_{n}))\log p_{k}+\unicode[STIX]{x1D70B}_{i}(\mathbf{i};\unicode[STIX]{x1D70E}(m_{n}))\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m_{n})(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}\unicode[STIX]{x1D70B}_{k}(\mathbf{i};\unicode[STIX]{x1D70E}(m_{n}))\log r_{k}+\unicode[STIX]{x1D70B}_{i}(\mathbf{i};\unicode[STIX]{x1D70E}(m_{n}))\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m_{n})(\log r_{j}-\log r_{i})}\nonumber\\ \displaystyle & \rightarrow & \displaystyle \frac{\mathop{\sum }_{k}q_{k}\log p_{k}+q_{i}v(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}q_{k}\log r_{k}+q_{i}v(\log r_{j}-\log r_{i})}.\nonumber\end{eqnarray}$$

(4) This statement follows from part (2).

(5) Since $\mathbf{u}\in \unicode[STIX]{x1D6EF}$ , we can find two strictly increasing sequences $(m_{n})_{n}$ and $(l_{n})_{n}$ of positive integers and a real number $v$ with $v>0$ such that $\unicode[STIX]{x1D70B}_{1}(\mathbf{u};m_{n})\rightarrow v$ and $\unicode[STIX]{x1D70B}_{1}(\mathbf{u};l_{n})\rightarrow 0$ . We conclude from this and (2) that

$$\begin{eqnarray}\frac{\log p_{\mathbf{j}|\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(m_{n})}}{\log r_{\mathbf{j}|\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(m_{n})}}\rightarrow \frac{\mathop{\sum }_{k}q_{k}\log p_{k}+q_{i}v(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}q_{k}\log r_{k}+q_{i}v(\log r_{j}-\log r_{i})}\end{eqnarray}$$

and that

$$\begin{eqnarray}\frac{\log p_{\mathbf{j}|\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(l_{n})}}{\log r_{\mathbf{j}|\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(l_{n})}}\rightarrow \frac{\mathop{\sum }_{k}q_{k}\log p_{k}+q_{i}0(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}q_{k}\log r_{k}+q_{i}0(\log r_{j}-\log r_{i})}=\frac{\mathop{\sum }_{k}q_{k}\log p_{k}}{\mathop{\sum }_{k}q_{k}\log r_{k}}=\unicode[STIX]{x1D6FC}.\end{eqnarray}$$

Finally, since $q_{i}v>0$ , it follows from condition (ii) that

$$\begin{eqnarray}\frac{\mathop{\sum }_{k}q_{k}\log p_{k}+q_{i}v(\log p_{j}-\log p_{i})}{\mathop{\sum }_{k}q_{k}\log r_{k}+q_{i}v(\log r_{j}-\log r_{i})}\not =\unicode[STIX]{x1D6FC},\end{eqnarray}$$

and we therefore deduce that

$$\begin{eqnarray}\lim _{n}\frac{\log p_{\mathbf{j}|\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(k_{n})}}{\log r_{\mathbf{j}|\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(k_{n})}}\not =\lim _{n}\frac{\log p_{\mathbf{j}|\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(l_{n})}}{\log r_{\mathbf{j}|\unicode[STIX]{x1D70E}_{i,\mathbf{i}}(l_{n})}}.\end{eqnarray}$$

In particular, this shows that the sequence $(\log p_{\mathbf{j}|n}/\text{log}\,r_{\mathbf{j}|n})_{n}$ diverges.◻

Proof. We must show that there is a continuous function $f:\{0,1\}^{\mathbb{N}}\rightarrow \mathbb{R}^{d}$ such that $f(\unicode[STIX]{x1D6EF}_{0})\subseteq \unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D707}}(A)$ and $f(\unicode[STIX]{x1D6EF})\subseteq D_{\unicode[STIX]{x1D707}}$ . Let $\unicode[STIX]{x1D70B}:\unicode[STIX]{x1D6F4}^{\mathbb{N}}\rightarrow \mathbb{R}^{d}$ be the function in (7.4) and define $f:\{0,1\}^{\mathbb{N}}\rightarrow \mathbb{R}^{d}$ by $f=\unicode[STIX]{x1D70B}\circ f_{i,\mathbf{i},j}$ . Then $f$ is continuous and it follows from Propositions 7.1 and 7.5 that

$$\begin{eqnarray}\displaystyle f(\unicode[STIX]{x1D6EF}_{0}) & = & \displaystyle \unicode[STIX]{x1D70B}(f_{i,\mathbf{i},j}(\unicode[STIX]{x1D6EF}_{0}))\subseteq \unicode[STIX]{x1D70B}\bigg\{\mathbf{i}\in \unicode[STIX]{x1D6F4}^{\mathbb{N}}\,\bigg|\,\frac{\log p_{\mathbf{i}\mid n}}{\log r_{\mathbf{i}\mid n}}\rightarrow \unicode[STIX]{x1D6FC}\bigg\}\nonumber\\ \displaystyle & = & \displaystyle \bigg\{x\in \mathbb{R}^{d}\,\bigg|\,\lim _{r{\searrow}0}\frac{\log \unicode[STIX]{x1D707}(B(x,r))}{\log r}=\unicode[STIX]{x1D6FC}\bigg\}\nonumber\\ \displaystyle & \subseteq & \displaystyle \bigg\{x\in \mathbb{R}^{d}\,\bigg|\,\lim _{r{\searrow}0}\frac{\log \unicode[STIX]{x1D707}(B(x,r))}{\log r}\in A\bigg\}=\unicode[STIX]{x1D6E5}_{\unicode[STIX]{x1D707}}(A)\nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle f(\unicode[STIX]{x1D6EF}) & = & \displaystyle \unicode[STIX]{x1D70B}(f_{i,\mathbf{i},j}(\unicode[STIX]{x1D6EF}))\subseteq \unicode[STIX]{x1D70B}\bigg\{\mathbf{i}\in \unicode[STIX]{x1D6F4}^{\mathbb{N}}\,\bigg|\,\frac{\log p_{\mathbf{i}\mid n}}{\log r_{\mathbf{i}\mid n}}\text{diverges as }n\rightarrow \infty \bigg\}\nonumber\\ \displaystyle & = & \displaystyle \bigg\{x\in \mathbb{R}^{d}\,\bigg|\,\frac{\log \unicode[STIX]{x1D707}(B(x,r))}{\log r}\text{ diverges as }r{\searrow}0\bigg\}=D_{\unicode[STIX]{x1D707}}.\nonumber\end{eqnarray}$$

This completes the proof. ◻

Proof. (1)  $\Rightarrow$  (2) This implication is trivial.

(2)  $\Rightarrow$  (3) For brevity write $x=\sum _{k}q_{k}\log p_{k}$ and $y=\sum _{k}q_{k}\log r_{k}$ . We first prove the following claim.

Claim 1. For all $i,j\in \{1,\ldots ,N\}$ , we have $p_{i}/r_{i}^{\unicode[STIX]{x1D6FC}}=p_{j}/r_{j}^{\unicode[STIX]{x1D6FC}}$ .

Proof of Claim 1. Fix $i,j\in \{1,\ldots ,N\}$ . It follows from statement (2) that there is a real number $a$ with $a>0$ such that we have

$$\begin{eqnarray}\frac{\mathop{\sum }_{k}q_{k}\log p_{k}+a(\log p_{i}-\log p_{j})}{\mathop{\sum }_{k}q_{k}\log r_{k}+a(\log r_{i}-\log r_{j})}=\unicode[STIX]{x1D6FC},\end{eqnarray}$$

i.e.

$$\begin{eqnarray}\frac{x+a(\log p_{i}-\log p_{j})}{y+a(\log r_{i}-\log r_{j})}=\frac{x}{y}.\end{eqnarray}$$

Rearranging this equality and using the fact that $a\not =0$ gives $y\log (p_{i}/p_{j})=x\log (r_{i}/r_{j})$ , and so $\log (p_{i}/p_{j})=(x/y)\log (r_{i}/r_{j})=\unicode[STIX]{x1D6FC}\log (r_{i}/r_{j})=\log (r_{i}^{\unicode[STIX]{x1D6FC}}/r_{j}^{\unicode[STIX]{x1D6FC}})$ . We conclude from this that $p_{i}/p_{j}=r_{i}^{\unicode[STIX]{x1D6FC}}/r_{j}^{\unicode[STIX]{x1D6FC}}$ , i.e. $p_{i}/r_{i}^{\unicode[STIX]{x1D6FC}}=p_{j}/r_{j}^{\unicode[STIX]{x1D6FC}}$ . This completes the proof of Claim 1.

It follows from Claim 1, that there is a constant $c>0$ such that $p_{1}/r_{1}^{\unicode[STIX]{x1D6FC}}=\cdots =p_{N}/r_{N}^{\unicode[STIX]{x1D6FC}}=c$ , i.e.

(7.13) $$\begin{eqnarray}p_{i}=cr_{i}^{\unicode[STIX]{x1D6FC}}\quad \text{for all }i=1,\ldots ,N.\end{eqnarray}$$

Next, we prove that

(7.14) $$\begin{eqnarray}c=1.\end{eqnarray}$$

Indeed, using the fact that $p_{i}=cr_{i}^{\unicode[STIX]{x1D6FC}}$ for all $i=1,\ldots ,N$ , we conclude that

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\frac{\mathop{\sum }_{i}q_{i}\log p_{i}}{\mathop{\sum }_{i}q_{i}\log r_{i}}=\frac{\mathop{\sum }_{i}q_{i}\log (cr_{i}^{\unicode[STIX]{x1D6FC}})}{\mathop{\sum }_{i}q_{i}\log r_{i}}=\frac{\log c}{\mathop{\sum }_{i}q_{i}\log r_{i}}+\unicode[STIX]{x1D6FC},\end{eqnarray}$$

and so $c=1$ .

Finally, it follows from (7.13) and (7.14) that $p_{i}=cr_{i}^{\unicode[STIX]{x1D6FC}}=r_{i}^{\unicode[STIX]{x1D6FC}}$ for all $i=1,\ldots ,N$ .

(3)  $\Rightarrow$  (1) This implication follows from a simple and direct calculation.◻

Proof. Let $\mathbb{P}^{N}$ denote the simplex of $N$ dimensional probability vectors in  $\mathbb{R}^{N}$ .

Choose $i_{\min },i_{\max }\in \{1,\ldots ,N\}$ such that $\log p_{i_{\min }}/\text{log}\,r_{i_{\min }}=\unicode[STIX]{x1D6FC}_{\min }$ and $\log p_{i_{\max }}/\text{log}\,r_{i_{\max }}=\unicode[STIX]{x1D6FC}_{\max }$ , and put $\mathbf{q}_{\min }=(\unicode[STIX]{x1D6FF}_{i,i_{\min }})_{i=1,\ldots ,N}\in \mathbb{P}^{N}$ and $\mathbf{q}_{\max }=(\unicode[STIX]{x1D6FF}_{i,i_{\max }})_{i=1,\ldots ,N}\in \mathbb{P}^{N}$ (where $\unicode[STIX]{x1D6FF}_{i,j}$ denotes the Kronecker delta). Next, let $\mathbf{u}=(1/N,\ldots ,1/N)$ denote the uniform $N$ dimensional probability vector in $\mathbb{R}^{N}$ and let $f_{\min },f_{\max }:[0,1]\rightarrow [0,1]$ be continuous functions with $f_{\min }(0)=1$ , $f_{\min }(1)=0$ , $f_{\max }(0)=0$ , $f_{\max }(1)=1$ and $t(1-t)+f_{\min }(t)+f_{\max }(t)=1$ for all $t\in [0,1]$ (for example, we may put $f_{\min }(t)=(1-t)^{2}$ and $f_{\max }(t)=t$ for $t\in [0,1]$ ). Finally, for $t\in [0,1]$ , define the probability vector $\mathbf{q}_{t}$ by $\mathbf{q}_{t}=t(1-t)\mathbf{u}+f_{\min }(t)\mathbf{q}_{\min }+f_{\max }(t)\mathbf{q}_{\max }$ .