Proof. Let $T\in {\mathrm {Tan}}(X)$ . By definition, we may choose a sequence $(\mathtt {i}_n)_{n\in \mathbb {N}}$ of infinite words and a sequence $(r_n)_{n\in \mathbb {N}}$ of positive real numbers converging to $0$ such that

$$ \begin{align*} M_{\pi\mathtt{i}_n,r_n}(X)\cap B(0,1)\to T \end{align*} $$

in Hausdorff distance. Since X satisfies the WBNC, there exists $M>0$ such that

$$ \begin{align*} \#\Phi(\pi\mathtt{i}_n,r_n)\leqslant M, \end{align*} $$

for all $n \in \mathbb {N}$ . Hence, there is $K \in \{1,\ldots ,M\}$ such that $\#\Phi (\pi \mathtt {i}_n,r_n)=K$ for infinitely many n. In other words, there exists a sequence $(n_k)_{k \in \mathbb {N}}$ of natural numbers such that $\#\Phi (\pi \mathtt {i}_{n_k},r_{n_k})=K$ for all $k\in \mathbb {N}$ . Write

$$ \begin{align*} \Phi(\pi\mathtt{i}_{n_k},r_{n_k}) = \{\varphi_{\mathtt{j}_{n_k}^i}\}_{i=1}^K \end{align*} $$

for all $k \in \mathbb {N}$ . By passing to a subsequence if necessary, we see that for every $i\in \{1,\ldots ,K\}$ , there exists a set $T_i$ such that

$$ \begin{align*} (M_{\pi\mathtt{i}_{n_k},r_{n_k}} \circ \varphi_{\mathtt{j}_{n_k}^1},\ldots,M_{\pi\mathtt{i}_{n_k},r_{n_k}} \circ \varphi_{\mathtt{j}_{n_k}^K})(X^K) \cap B(0,1)^K \to T_1 \times \cdots \times T_K \end{align*} $$

in Hausdorff distance. Noting that

$$ \begin{align*} M_{\pi\mathtt{i}_{n_k},r_{n_k}}(X)\cap B(0,1)= \bigcup_{i=1}^KM_{\pi\mathtt{i}_{n_k},r_{n_k}}\circ \varphi_{\mathtt{j}_{n_k}^{i}}(X)\cap B(0,1) \end{align*} $$

for all $k \in \mathbb {N}$ , we see that (1) holds.

Since $X^K$ is compact, we may assume that

$$ \begin{align*} (\varphi_{\mathtt{j}_{n_k}^{1}}^{-1}(\pi\mathtt{i}_{n_k}),\ldots,\varphi_{\mathtt{j}_{n_k}^{K}}^{-1}(\pi\mathtt{i}_{n_k})) \to (y_1,\ldots,y_K)\in X^K, \end{align*} $$

and therefore for each $i \in \{1,\ldots ,K\}$ there exists a linear map $G_i$ such that

$$ \begin{align*} (\varphi_{\mathtt{j}_{n_k}^{1}}^{-1}\circ M_{\pi\mathtt{i}_{n_k},r_{n_k}}^{-1},\ldots,\varphi_{\mathtt{j}_{n_k}^{K}}^{-1}\circ M_{\pi\mathtt{i}_{n_k},r_{n_k}}^{-1}) \to (G_1+y_1,\ldots,G_K+y_K) \end{align*} $$

in the uniform convergence in $X^K$ . Clearly,

$$ \begin{align*} \varphi_{\mathtt{j}_{n_k}^{i}}^{-1}\circ M_{\pi\mathtt{i}_{n_k},r_{n_k}}^{-1}(M_{\pi\mathtt{i}_{n_k},r_{n_k}}\circ \varphi_{\mathtt{j}_{n_k}^{i}}(X)\cap B(0,1))\subset X, \end{align*} $$

so by taking the limit, we see that $G_i(T_i)+y_i\subset X$ which proves (2).

Finally, to prove (3), denote by $A_{\mathtt {j}_{n_k}^{i}}$ the linear part of $\varphi _{\mathtt {j}_{n_k}^{i}}$ . Then, by the definition of $\Phi (\pi \mathtt {i}_{n_k},r_{n_k})$ , we have that

$$ \begin{align*} \|r_{n_k}A_{\mathtt{j}_{n_k}^{i}}^{-1}\| = r_{n_k}\alpha_2(A_{\mathtt{j}_{n_k}^{i}})^{-1} \geqslant 1. \end{align*} $$

Since $A\mapsto \|A\|$ is continuous, we have $\|G_i\|=\lim _{k\to \infty }\|r_{n_k}A_{\mathtt {j}_{n_k}^{i}}^{-1}\|\geqslant 1>0$ and, in particular, $\mathrm {rank}(G_i)\geqslant 1$ for all $i \in \{1,\ldots ,K\}$ .

Let us next assume that X is dominated. Fix $i\in I$ and, for simplicity, denote $\mathtt {j}_{n_k}^{i}$ by $\mathtt {j}_k$ . First observe that the sequence $(|\mathtt {j}_k|)_{k \in \mathbb {N}}$ is unbounded, since if it were bounded by some number $L\in \mathbb {N}$ , we would have

$$ \begin{align*} r_{n_k} \geqslant \alpha_2(A_{\mathtt{j}_k})\geqslant \Big(\min_{j \in \{1,\ldots,N\}} \alpha_2(A_j)\Big)^L> 0 \end{align*} $$

for all $k\in \mathbb {N}$ , contradicting the fact that $\lim _{k \to \infty } r_{n_k} = 0$ . By domination, there exist ${C>0}$ and $0<\tau <1$ such that

$$ \begin{align*} |\!\det(r_{n_k}A_{\mathtt{j}_k}^{-1})|&=\frac{r_{n_k}^2}{\det(A_{\mathtt{j}_k})}=\frac{r_{n_k}^2}{\alpha_1(A_{\mathtt{j}_k})\alpha_2(A_{\mathtt{j}_k})}\\ &\leqslant\frac{\alpha_2(A_{\mathtt{j}_k})}{\alpha_1(A_{\mathtt{j}_k})\min_{j\in\{1,\ldots,N\}}\alpha_2(A_j)} \leqslant \frac{C}{\min_{j\in\{1,\ldots,N\}}\alpha_2(A_j)}\tau^{|\mathtt{j}_k|}. \end{align*} $$

Since $|\mathtt {j}_k|$ is unbounded, we see that $\mathrm {rank}(G_i)=1$ . Finally, since $r_{n_k}A_{\mathtt {j}_{n_k}^i}^{-1}$ converges to the linear map $G_i$ , which is a rank-one map, ${\mathrm {im}}(G_i)\in X_F$ .

Proof of Proposition 3.1

Let $T\in {\mathrm {Tan}}(X)$ and $\{T_i\}_{i\in I}$ be a tangent decomposition of T given by Lemma 3.2. Notice that, since $T=\bigcup _{i\in I}T_i$ , we have ${\mathrm {dim}_{\mathrm {H}}}(T)=\max _{i\in I}{\mathrm {dim}_{\mathrm {H}}}(T_i)$ . Let $i \in I$ be the index which achieves this maximum. By Lemma 3.2, we have $G_i(T_i)+y_i\subset X\cap ({\mathrm {im}}(G_i)+y_i)$ , and hence

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}(G_i(T_i)+y_i) \leqslant {\mathrm{dim}_{\mathrm{H}}}(X \cap ({\mathrm{im}}(G_i)+y_i)). \end{align*} $$

If $\mathrm {rank}(G_i)=2$ then ${\mathrm {dim}_{\mathrm {H}}}(G_i(T_i)+y_i)\leqslant {\mathrm {dim}_{\mathrm {H}}}(X)$ . On the other hand, if $\mathrm {rank}(G_i)=1$ then $x \mapsto G_ix+y_i$ is bi-Lipschitz equivalent to ${\mathrm {proj}}_{\ker (G_i)^\bot }$ , thus we have

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}(T_i) &\leqslant {\mathrm{dim}_{\mathrm{H}}}(\mathbb{R} \times {\mathrm{proj}}_{\ker(G_i)^\bot}(T_i)) \\ &= 1+{\mathrm{dim}_{\mathrm{H}}}({\mathrm{proj}}_{\ker(G_i)^\bot}(T_i)) = 1+{\mathrm{dim}_{\mathrm{H}}}(G_i(T_i)+x). \end{align*} $$

Therefore,

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}(T) = {\mathrm{dim}_{\mathrm{H}}}(T_i) \leqslant 1+{\mathrm{dim}_{\mathrm{H}}}(X \cap ({\mathrm{im}}(G_i)+y_i)) \end{align*} $$

and we have shown the first claim. By Lemma 2.1, the second claim follows immediately from the first claim.

If X is dominated, then $\mathrm {rank}(G_i)=1$ above by the last assertion of Lemma 3.2 and the last claim follows.

Proof. Let $(n_k)_{k \in \mathbb {N}}$ be a strictly increasing sequence of integers such that the limit exists and let $\varepsilon>0$ . By Lemma 2.2(1), $\overline {\vartheta }_1(\mathtt {i}_k)$ is well defined for every $k\in \mathbb {N}$ and, by the uniform convergence, we may choose $k_0\in \mathbb {N}$ large enough such that

$$ \begin{align*} \sphericalangle(\vartheta_1(\mathtt{i}_k|_{n_k}),\overline{\vartheta}_1(\mathtt{i}_k)) < \frac{\varepsilon}{2} \end{align*} $$

for all $k\geqslant k_0$ . By the assumption, by making $k_0$ larger if needed, we also have

$$ \begin{align*} \sphericalangle(\vartheta_1(\mathtt{i}_k|_{n_k}),W) < \frac{\varepsilon}{2}, \end{align*} $$

for all $k\geqslant k_0$ . Thus, by the triangle inequality, we have

$$ \begin{align*} \sphericalangle(\overline{\vartheta}_1(\mathtt{i}_k),W)\leqslant \sphericalangle(\overline{\vartheta}_1(\mathtt{i}_k),\vartheta_1(\mathtt{i}_k|_{n_k}))+\sphericalangle(\vartheta_1(\mathtt{i}_k|_{n_k}),W) < \varepsilon \end{align*} $$

and therefore, $\lim _{k\to \infty }\overline {\vartheta }_1(\mathtt {i}_k)=W$ .

Proof. Let $(\mathtt {i}_k)_{k \in \mathbb {N}}$ be a sequence of infinite words in $\Sigma $ and $(r_k)_{k \in \mathbb {N}}$ be a sequence of positive real numbers converging to zero such that

$$ \begin{align*} M_{\pi\mathtt{i}_k,r_k}(X)\cap B(0,1)\to T. \end{align*} $$

Recall from the proof of Lemma 3.2 that there exist a sequence $(n_k)_{k \in \mathbb {N}}$ of integers and, for each $i \in I$ , finite words $\mathtt {j}_{n_k}^i \in \Sigma _*$ and sets $T_i$ such that

$$ \begin{align*} M_{\pi\mathtt{i}_{n_k},r_{n_k}} \circ \varphi_{\mathtt{j}_{n_k}^i}(X) \cap B(0,1) \to T_i \end{align*} $$

for all $i \in I$ in Hausdorff distance. Fix $y\in T^{\circ }$ and choose $i\in I$ such that $y\in T_i^{\circ }$ . Since $y\not \in \partial B(0,1)$ , there is $\delta>0$ depending only on y such that $B(y,2\delta )\subset B(0,1)$ . Therefore, there are infinite words $\mathtt {j}_k\in [\mathtt {j}_{n_k}^i]$ such that $M_{\pi \mathtt {i}_{n_k},r_{n_k}}(\pi (\mathtt {j}_k))\to y$ and

(4.1) $$ \begin{align} M_{\pi\mathtt{i}_{n_k},r_{n_k}}(X\cap B(\pi\mathtt{j}_k,\delta r_{n_k}))\subset M_{\pi\mathtt{i}_{n_k},r_{n_k}}\circ \varphi_{\mathtt{j}_{n_k}^{i}}(X)\cap B(0,1) \end{align} $$

for all large enough $k\in \mathbb {N}$ . Let $m_k\geqslant n_k$ be the unique integer which satisfies

(4.2) $$ \begin{align} \alpha_1(A_{\mathtt{j}_k|_{m_k}})\leqslant \delta r_{n_k}< \alpha_1(A_{\mathtt{j}_k|_{m_k-1}}). \end{align} $$

By again passing to a subsequence if necessary, there exists an affine map $P_y$ such that

$$ \begin{align*} M_{\pi\mathtt{i}_{n_k},r_{n_k}}\circ\varphi_{\mathtt{j}_k|_{m_k}}\to P_y \end{align*} $$

in the uniform convergence in X. By compactness and (4.1), we have $y \in P_y(X) \subset T_i$ and, by domination, we have

$$ \begin{align*} \frac{\alpha_2(A_{\mathtt{j}_k|_{m_k}})}{\alpha_1(A_{\mathtt{j}_k|_{m_k}})}\leqslant C\tau^{m_k}, \end{align*} $$

so in particular $\det (r_{n_k}^{-1}A_{\mathtt {j}_k|_{m_k}})\to 0$ as $k\to \infty $ . Also, by (4.2) and [Reference Bárány, Käenmäki and Morris8, Corollary 2.4], there exists a constant $C>0$ such that $\|r_{n_k}^{-1}A_{\mathtt {j}_k|_{m_k}}\|\geqslant C\delta $ for all $k\in \mathbb {N}$ . Therefore, we see that ${\mathrm {rank}}(P_y)=1$ . Let $W_y={\mathrm {im}}(P_y)$ and note that by Lemma 2.3, $W_y\in Y_F$ . Recall that $P_y(X)$ and ${\mathrm {proj}}_{\ker (P_y)^{\perp }}(X)$ are bi-Lipschitz equivalent, so by the assumption,

(4.3) $$ \begin{align} {\mathrm{dim}_{\mathrm{H}}}(T\cap (W_y+y))&\geqslant {\mathrm{dim}_{\mathrm{H}}}(P_y(X)\cap (W_y+y)) \nonumber\\ &\geqslant {\mathrm{dim}_{\mathrm{H}}}({\mathrm{proj}}_{\ker(P_y)^{\perp}}(X)). \end{align} $$

Let us show that $\ker (P_y)\in X_F$ . Observe that the linear part of the map $M_{\pi \mathtt {i}_{n_k},r_{n_k}}\circ \varphi _{\mathtt {j}_k|_{m_k}}$ is $r_{n_k}^{-1}A_{\mathtt {j}_k|_{m_k}}$ and notice that this sequence converges to the linear part of $P_y$ in the uniform convergence in X. Denote by $A_y$ the linear part of $P_y$ and let v be a unit vector in the kernel of $A_y$ . Since the eigenspaces of $(r_{n_k}^{-1}A_{\mathtt {j}_k|_{m_k}})^T(r_{n_k}^{-1}A_{\mathtt {j}_k|_{m_k}})$ converge to the eigenspaces of $A_y^TA_y$ and since $\ker (A_y)=\ker (A_y^TA_y)$ is the eigenspace corresponding to the singular value $0$ , we see that there is a sequence of unit vectors $v_k\to v$ , such that

$$ \begin{align*} A_{\mathtt{j}_k|_{m_k}}^TA_{\mathtt{j}_k|_{m_k}}v_k=\alpha_2(A_{\mathtt{j}_k|_{m_k}})^2v_k \end{align*} $$

for all $k\in \mathbb {N}$ . Let us define $w_k= \alpha _2(A_{\mathtt {j}_k|_{m_k}})^{-1}A_{\mathtt {j}_k|_{m_k}}v_k$ . By the foregoing, we have

$$ \begin{align*} \|w_k\|&=\alpha_2(A_{\mathtt{j}_k|_{m_k}})^{-1}\|A_{\mathtt{j}_k|_{m_k}}v_k\|=\alpha_2(A_{\mathtt{j}_k|_{m_k}})^{-1}(A_{\mathtt{j}_k|_{m_k}}v_k\,\cdot\,A_{\mathtt{j}_k|_{m_k}}v_k)^{{1}/{2}}\\ &=\alpha_2(A_{\mathtt{j}_k|_{m_k}})^{-1}\langle A_{\mathtt{j}_k|_{m_k}}^TA_{\mathtt{j}_k|_{m_k}}v_k\,|\,v_k\rangle^{{1}/{2}}=\alpha_2(A_{\mathtt{j}_k|_{m_k}})^{-1}(\alpha_2(A_{\mathtt{j}_k|_{m_k}})^{-1}v_k\,\cdot\,v_k)^{{1}/{2}}\\ &=(v_k\,\cdot\,v_k)^{{1}/{2}}=\|v_k\|=1, \end{align*} $$

where $\cdot $ is the standard inner product on $\mathbb {R}^2$ . Therefore, by possibly passing to a subsequence, we may assume that $w_k$ converges to some unit vector w and that

$$ \begin{align*} \frac{(A_{\mathtt{j}_k|_{m_k}})^{-1}}{\|(A_{\mathtt{j}_k|_{m_k}})^{-1}\|}\to B_y, \end{align*} $$

for some $2\times 2$ matrix $B_y$ . Since $\mathsf {A}^{-1}=(A_1^{-1},\ldots ,A_N^{-1})$ is dominated, we see as before that $\det (\|(A_{\mathtt {j}_k|_{m_k}})^{-1}\|^{-1}(A_{\mathtt {j}_k|_{m_k}})^{-1})\to 0$ , so $B_y$ has rank at most one. Since

$$ \begin{align*} \frac{(A_{\mathtt{j}_k|_{m_k}})^{-1}}{\|(A_{\mathtt{j}_k|_{m_k}})^{-1}\|}\cdot\frac{A_{\mathtt{j}_k|_{m_k}}v_k}{\alpha_2(A_{\mathtt{j}_k|_{m_k}})}=v_k, \end{align*} $$

by taking limits, we have that $B_y(w)=v$ , so v is a unit vector in the image of $B_y$ . Therefore, $B_y$ is a rank-one matrix and $\ker (A_y)=\langle v\rangle ={\mathrm {im}}(B_y)\in X_F$ . By the assumption and (4.3), we have that ${\mathrm {dim}_{\mathrm {H}}}(T\cap (W_y+y))\geqslant 1$ . The upper bound is trivial, since $T\cap (W_y+y)$ is contained in a line.

Finally, let us show that $W_y$ is constant in $T_i^{\circ }$ . By passing to a subsequence if necessary, we may assume that $\lim _{k\to \infty }\vartheta _1(\mathtt {j}_{n_k}^i)=W_i$ for some $W_i\in Y_F$ . Notice that $W_i$ does not depend on the choice of $y\in T_i^{\circ }$ . By Lemma 2.3 (or rather its proof), it is easy to see that

$$ \begin{align*} W_y={\mathrm{im}}(P_y)=\lim_{k\to\infty}\vartheta_1(\mathtt{j}_k|_{m_k}). \end{align*} $$

Therefore, by Lemma 4.2, we have ${\mathrm {im}}(P_y)=\lim _{k\to \infty }\overline {\vartheta }_1(\mathtt {j}_k)$ . Noting that $\mathtt {j}_{n_k}^i=\mathtt {j}_k|_{|\mathtt {j}_{n_k}^i|}$ and, applying Lemma 4.2 again, we see that

$$ \begin{align*} W_y=\lim_{k\to\infty} \overline{\vartheta}_1(\mathtt{j}_k)=\lim_{k\to\infty} \vartheta_1(\mathtt{j}_{n_k}^i)=W_i \end{align*} $$

finishing the proof.

Proof. Let $V\in \mathbb {RP}^1$ and $x\in X$ and, by recalling Lemma 2.1, let $T_{\max }\in {\mathrm {Tan}}(X\cap (V+x))$ be a weak tangent, which satisfies ${\mathrm {dim}_{\mathrm {H}}}(T_{\max })={\mathrm {dim}_{\mathrm {A}}}(X\cap (V+x))$ . Let $(x_k)_{k \in \mathbb {N}}$ be a sequence of points in $X\cap (V+x)$ and $(r_k)_{k \in \mathbb {N}}$ be a sequence of positive real numbers converging to zero such that

$$ \begin{align*} M_{x_k,r_k}(X\cap(V+x))\cap B(0,1)\to T_{\max} \end{align*} $$

in Hausdorff distance. Since $x_k\in V+x$ for all $k\in \mathbb {N}$ , and each $M_{x_k,r_k}$ is a similarity, we have $M_{x_k,r_k}(V+x)\to V$ . Let T be an accumulation point of the sequence $M_{x_k,r_k}(X)\cap B(0,1)$ . Then $T\in {\mathrm {Tan}}(X)$ and, by compactness, $T_{\max }\subset T\cap V$ . Therefore, we have

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}(X\cap (V+x))\leqslant{\mathrm{dim}_{\mathrm{A}}}(X\cap (V+x))={\mathrm{dim}_{\mathrm{H}}}(T_{\max})\leqslant{\mathrm{dim}_{\mathrm{H}}}(T\cap V) \end{align*} $$

as required.

Proof of Theorem 4.1

By Lemma 2.1, we have

(4.4) $$ \begin{align} {\mathrm{dim}_{\mathrm{A}}}(X)=\max_{T\in{\mathrm{Tan}}(X)}{\mathrm{dim}_{\mathrm{H}}}(T), \end{align} $$

so we may choose $T_{\max }\in {\mathrm {Tan}}(X)$ such that ${\mathrm {dim}_{\mathrm {A}}}(X)={\mathrm {dim}_{\mathrm {H}}}(T_{\max })$ . Recalling Proposition 3.1, there are $x\in X$ and $V\in X_F\subset \mathbb {RP}^1\setminus Y_F$ such that

(4.5) $$ \begin{align} {\mathrm{dim}_{\mathrm{A}}}(X) = {\mathrm{dim}_{\mathrm{H}}}(T_{\max}) \leqslant 1+ {\mathrm{dim}_{\mathrm{H}}}(X\cap (V+x)). \end{align} $$

By Lemma 4.4, there exists a tangent set T such that

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}(T\cap V) \geqslant {\mathrm{dim}_{\mathrm{A}}}(X\cap(V+x)) \geqslant {\mathrm{dim}_{\mathrm{H}}}(X\cap(V+x)). \end{align*} $$

If ${\mathrm {dim}_{\mathrm {H}}}(T\cap V)=0$ , then trivially ${\mathrm {dim}_{\mathrm {H}}}(T)\geqslant 1+{\mathrm {dim}_{\mathrm {H}}}(X\cap (V+x))=1$ , by Lemma 4.3. Therefore, we may assume that ${\mathrm {dim}_{\mathrm {H}}}(T\cap V)>0$ . Notice that $T\cap V\cap \partial B(0,1)$ consists of at most two points, so ${\mathrm {dim}_{\mathrm {H}}}(T^{\circ }\cap V)={\mathrm {dim}_{\mathrm {H}}}(T\cap V)$ . Let $0<s<{\mathrm {dim}_{\mathrm {H}}}(T\cap V)$ and let $\mu $ be a Frostman measure on $T^{\circ }\cap V$ ; see [Reference Mattila36, Theorem 8.8]. Let $\{T_i\}_{i\in I}$ be a tangent decomposition of T given by Lemma 3.2. Since $T=\bigcup _{i\in I}T_i$ , at least one of the sets $T_i^{\circ }\cap V$ has positive $\mu $ -measure. Let $T_i^{\circ }$ be such a set and let $W_i\in Y_F$ be the line given by Lemma 4.3. Since $V\not \in Y_F$ , we have $V\ne W_i$ and so, by Marstrand’s slicing theorem [Reference Bishop and Peres11, Theorem 3.3.1],

$$ \begin{align*} 1={\mathrm{dim}_{\mathrm{H}}}(T\cap (W_i+y))\leqslant {\mathrm{dim}_{\mathrm{H}}}(T)-s \end{align*} $$

for $\mu $ -almost every $y\in T_i^{\circ }\cap V$ . In particular, since $\mu (T_i^{\circ }\cap V)>0$ , such a point y exists. By letting $s\uparrow {\mathrm {dim}_{\mathrm {H}}}(T\cap V)$ , we get

$$ \begin{align*} {\mathrm{dim}_{\mathrm{A}}}(X) &\geqslant{\mathrm{dim}_{\mathrm{H}}}(T)\geqslant 1+{\mathrm{dim}_{\mathrm{H}}}(T\cap V) \\ &\geqslant 1+{\mathrm{dim}_{\mathrm{A}}}(X\cap (V+x)) \geqslant 1+{\mathrm{dim}_{\mathrm{H}}}(X\cap (V+x)). \end{align*} $$

Combining this with (4.5), we get

$$ \begin{align*} {\mathrm{dim}_{\mathrm{A}}}(X) = 1+{\mathrm{dim}_{\mathrm{A}}}(X\cap (V+x)) = 1+{\mathrm{dim}_{\mathrm{H}}}(X\cap(V+x)) \end{align*} $$

as claimed.

It remains to show that ${\mathrm {dim}_{\mathrm {A}}}(X\cap (W+y))\leqslant {\mathrm {dim}_{\mathrm {H}}}(X\cap (V+x))$ for all $y\in X$ and $W\in \mathbb {RP}^1\setminus Y_F$ . By repeating the above proof for such y and W, we find a tangent set T such that

$$ \begin{align*} 1+{\mathrm{dim}_{\mathrm{A}}}(X\cap(W+y))\leqslant {\mathrm{dim}_{\mathrm{H}}}(T)\leqslant {\mathrm{dim}_{\mathrm{H}}}(T_{\max})\leqslant1+ {\mathrm{dim}_{\mathrm{H}}}(X\cap(V+x)), \end{align*} $$

where we used (4.4) in the middle inequality. This finishes the proof.

Proof. Since X satisfies the SOSC, [Reference Bárány, Käenmäki and Yu10, Theorem 2.18] shows that

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}({\mathrm{proj}}_{V^\bot}(X)) = 1 \end{align*} $$

for all $V \in \mathbb {RP}^1 \setminus \mathcal {I}$ , where $\mathcal {I} = \{W \in \mathbb {RP}^1 \colon W = A_iW \text { for all } i \in \{1,\ldots ,N\} \}$ contains at most one element. If $\mathcal {I}=\emptyset $ , then the claim follows from Theorem 4.1. Bárány, Käenmäki and Yu [Reference Bárány, Käenmäki and Yu10, Lemma 2.11] show that if $X_F$ is not a singleton, then $\mathcal {I}$ is non-empty if and only if the matrices $A_i$ are of the form

$$ \begin{align*} A_i=\begin{pmatrix} a_i&b_i\\ 0&d_i \end{pmatrix}\!, \end{align*} $$

possibly after a change of basis, where $0<|d_i|<|a_i|<1$ , and the matrices are not simultaneously diagonalizable, and clearly in this case, $\mathcal {I}=Y_F$ . Since $X_F\subset \mathbb {RP}^1\setminus Y_F$ , Theorem 4.1 gives the claim.

Proof. Let $\mathtt {i}\in \Sigma _*$ and recall that $\alpha _1(A_{\mathtt {i}})=\|A_{\mathtt {i}}\|$ . The lower bound for $\alpha _1(A_{\mathtt {i}})$ follows from the fact that $\unicode{x3bb} ^{|\mathtt {i}|}$ is an eigenvalue of $A_{\mathtt {i}}$ . Similarly, since $\alpha _2(A_{\mathtt {i}})=\|A_{\mathtt {i}}^{-1}\|^{-1}$ , the upper bound for $\alpha _2(A_{\mathtt {i}})$ is trivial as $2^{|\mathtt {i}|}$ is an eigenvalue of $A_{\mathtt {i}}^{-1}$ . We prove the upper bound for $\alpha _1(A_{\mathtt {i}})$ and only remark that the proof of the lower bound for $\alpha _2(A_{\mathtt {i}})$ follows similarly. Let $s(\mathtt {i})=\sum _{k=1}^{|\mathtt {i}|}(-1)^{i_{|\mathtt {i}|-k}+1}2^{-k}\unicode{x3bb} ^{|\mathtt {i}|-k}$ and notice that

$$ \begin{align*} |s(\mathtt{i})|\leqslant \sum_{k=1}^{|\mathtt{i}|}2^{-k}\unicode{x3bb}^{|\mathtt{i}|-k}\leqslant\unicode{x3bb}^{|\mathtt{i}|}\sum_{k=1}^{\infty}2^{-k}\unicode{x3bb}^{-k}=K_{\unicode{x3bb}}\unicode{x3bb}^{|\mathtt{i}|} \end{align*} $$

for all $\mathtt {i} = i_1 \cdots i_{|\mathtt {i}|} \in \Sigma _*$ . Writing $y=(y_1,y_2)\in S^1$ , we see that

$$ \begin{align*} \|A_{\mathtt{i}}y\|^2&=|(2^{-|\mathtt{i}|}y_1)^2+(s(\mathtt{i})y_1+\unicode{x3bb}^{|\mathtt{i}|}y_2)^2| \\ &= |2^{-2|\mathtt{i}|}y_1^2+s(\mathtt{i})^2y_1^2+2s(\mathtt{i})\unicode{x3bb}^{|\mathtt{i}|}y_1y_2+\unicode{x3bb}^{2|\mathtt{i}|}y_2^2|\\ &\leqslant(2^{-2|\mathtt{i}|}+|s(\mathtt{i})|^2)y_1^2+2|s(\mathtt{i})|\unicode{x3bb}^{|\mathtt{i}|}|y_1y_2|+\unicode{x3bb}^{2|\mathtt{i}|}y_2^2 \\ &\leqslant\unicode{x3bb}^{2|\mathtt{i}|}+K_{\unicode{x3bb}}^2\unicode{x3bb}^{2|\mathtt{i}|}+2K_{\unicode{x3bb}}\unicode{x3bb}^{2|\mathtt{i}|}+\unicode{x3bb}^{2|\mathtt{i}|}\\ &=(K_{\unicode{x3bb}}^2+2K_{\unicode{x3bb}}+2)\unicode{x3bb}^{2|\mathtt{i}|}, \end{align*} $$

so the claim holds with $C=\sqrt {(K_{\unicode{x3bb} }+1)^2+1}>1$ . Finally, since

$$ \begin{align*} \alpha_2(A_{\mathtt{i}})\leqslant \Bigl(\frac{1}{2\unicode{x3bb}}\Bigr)^{|\mathtt{i}|}\alpha_1(A_{\mathtt{i}}), \end{align*} $$

we see that $T_{\unicode{x3bb} }$ is dominated.

Proof. Let $\mathtt {i} \in \Sigma $ and observe that, by (5.1), we may define

$$ \begin{align*} B_{\mathtt{i}}=\lim_{n\to\infty}2^{-n}A_{\overleftarrow{\mathtt{i}|_n}}^{-1}= \begin{pmatrix} 1&0\\ \sum_{k=1}^{\infty}(-1)^{i_k}2^{-k}\unicode{x3bb}^{-k} & 0 \end{pmatrix}\!. \end{align*} $$

Since $T_\unicode{x3bb} $ is dominated by Lemma 5.3 and $\lim _{n \to \infty }\eta _1(A_{\overleftarrow {\mathtt {i}|_n}}^{-1}) = (1,0)$ , it follows from Lemma 2.3 that the word $\mathtt {i}$ determines an element $V_{\mathtt {i}} = {\mathrm {im}}(B_{\mathtt {i}})$ of the set $X_F$ by

$$ \begin{align*} V_{\mathtt{i}}=\langle B_{\mathtt{i}}(1,0)\rangle=\bigg\langle 1,\sum_{k=1}^{\infty}(-1)^{i_k}2^{-k}\unicode{x3bb}^{-k}\bigg\rangle. \end{align*} $$

Hence, it is clear that for any $V\in \{V_t \in \mathbb {RP}^1 \colon t \in [-K_\unicode{x3bb} ,K_\unicode{x3bb} ]\}$ , there exists $\mathtt {i}\in \Sigma $ such that $V=V_{\mathtt {i}}\in X_F$ .

For the other inclusion, let $V\in X_F$ . By the definition of $X_F$ , we find a sequence $(c_n)_{n \in \mathbb {N}}$ of positive real numbers, a sequence $(\mathtt {i}_n)_{n \in \mathbb {N}}$ of finite words $\mathtt {i}_n = i^n_1 \cdots i^n_{|\mathtt {i}_n|} \in \Sigma _*$ , and a linear map A of rank one such that $A = \lim _{n \to \infty }c_nA^{-1}_{\overleftarrow {\mathtt {i}_n}}$ and $V={\mathrm {im}}(A)$ . Passing through a subsequence if necessary, we see that

$$ \begin{align*} \lim_{n\to\infty}2^{-|\mathtt{i}_n|}A^{-1}_{\overleftarrow{\mathtt{i}_n}}=\begin{pmatrix} 1&0\\ \lim_{n\to\infty}\sum_{k=1}^{n}(-1)^{i^n_k}2^{-k}\unicode{x3bb}^{-k} & 0 \end{pmatrix}\!, \end{align*} $$

where the limit $\lim _{n\to \infty }\sum _{k=1}^{n}(-1)^{i^n_k+1}2^{-k}\unicode{x3bb} ^{-k}$ exists. In particular, ${c_n2^{-|\mathtt {i}_n|}\kern1.3pt{\to}\kern1.3pt c\kern1.3pt{\in}\kern1.3pt \mathbb {R}\kern1.3pt{\setminus}\kern1.3pt \{0\}}$ and A is a constant multiple of the above matrix, which finishes the proof.

For $Y_F$ , let $\mathtt {i}\in \Sigma $ and, using (5.1), define

$$ \begin{align*} B_{\mathtt{i}}=\lim_{n\to\infty} \unicode{x3bb}^{-n} A_{\mathtt{i}|_n}= \begin{pmatrix} 0&0\\ \sum_{k=1}^{n}(-1)^{i_{n-k+1}+1}2^{-k}\unicode{x3bb}^{-k} & 1 \end{pmatrix}\!. \end{align*} $$

Since ${\mathrm {im}}(B_{\mathtt {i}})=V_{\infty }$ , we have $\{V_{\infty }\}\subset Y_F$ . For the other inclusion, as before, if $V\in Y_F$ , we find a sequence $(c_n)_{n \in \mathbb {N}}$ of positive real numbers, a sequence $(\mathtt {i}_n)_{n \in \mathbb {N}}$ of finite words $\mathtt {i}_n = i^n_1 \cdots i^n_{|\mathtt {i}_n|} \in \Sigma _*$ , and a linear map A of rank one such that $A = \lim _{n \to \infty }c_nA_{\mathtt {i}_n}$ and $V={\mathrm {im}}(A)$ . Passing through a subsequence if necessary, we see that

$$ \begin{align*} \lim_{n\to\infty}\unicode{x3bb}^{-|\mathtt{i}_n|}A_{\mathtt{i}_n}=\begin{pmatrix} 0&0\\ \sum_{k=1}^{n}(-1)^{i^n_{n-k+1}+1}2^{-k}\unicode{x3bb}^{-k} & 1 \end{pmatrix}\!, \end{align*} $$

and since ${\mathrm {im}}(A)=V_{\infty }$ , the proof is finished.

Proof. Let $C> 1$ be the constant of Lemma 5.3. Fix $x\in T_{\unicode{x3bb} }$ and $0<r<C^{-1}$ , and let $k\in \mathbb {N}$ be the smallest natural number satisfying

$$ \begin{align*} 2^kC^{-1}\geqslant 1. \end{align*} $$

Define

$$ \begin{align*} \Phi_n(x,r)=\{\varphi_{\mathtt{i}}\colon 2^nC^{-1}2^{-|\mathtt{i}|}\leqslant r < 2^nC^{-1}2^{-|\mathtt{i}|+1} \text{ and } \varphi_{\mathtt{i}}(X)\cap B(x,r)\ne\emptyset\} \end{align*} $$

for all $n \in \mathbb {N}$ . It follows from Lemma 5.3 that $\Phi (x,r)\subset \bigcup _{n=0}^k\Phi _n(x,r)$ , so it suffices to show that the cardinality of $\Phi _n(x,r)$ is uniformly bounded for all $n \in \{0,\ldots ,k\}$ . For each $n\in \{0,\ldots ,k\}$ , let $m_n$ be the unique integer satisfying

(5.2) $$ \begin{align} 2^nC^{-1}2^{-m_n}\leqslant r < 2^{n}C^{-1}2^{-m_n+1}. \end{align} $$

Then clearly $\Phi _n(x,r)=\{\varphi _{\mathtt {i}}\colon \mathtt {i}\in \Sigma _{m_n}$ and $\varphi _{\mathtt {i}}(X)\cap B(x,r)\ne \emptyset \}$ . We observe from the construction that each $\varphi _{\mathtt {i}}$ with $\mathtt {i}\in \Sigma _{m_n}$ maps $T_\unicode{x3bb} $ inside a unique set of the form $[{k}/{2^{m_n}},({k+1})/{2^{m_n}}]\times \mathbb {R}$ , where k is an integer satisfying $0\leqslant k < 2^{m_n}$ . Therefore, (5.2) implies that $B(x,r)$ can intersect at most

$$ \begin{align*} 2^{m_n}r\leqslant 2^{m_n}2^nC^{-1}2^{-m_n+1}\leqslant C^{-1}2^{k+1}, \end{align*} $$

of the sets $\varphi _{\mathtt {i}}(X)$ with $\mathtt {i}\in \Sigma _{m_n}$ . Since the upper bound is independent of x and r, the set $T_{\unicode{x3bb} }$ satisfies the BNC.

Proof of Theorem 5.1

It follows from Lemmas 5.3, 5.4, and 5.5 that the Takagi function $T_\unicode{x3bb} $ is a dominated self-affine set satisfying the BNC such that $X_F$ is a non-trivial projective interval. Furthermore, since the Takagi function is continuous, $T_\unicode{x3bb} (0)=0=T_\unicode{x3bb} (1)$ and, by [Reference Mishura and Schied37] (see also Lemma 6.7), $M_\unicode{x3bb} = \max _{x \in [0,1]}T_\unicode{x3bb} (X) = {1}/{3(1-\unicode{x3bb} )}> 0$ , we see that ${\mathrm {dim}_{\mathrm {H}}}({\mathrm {proj}}_{V^{\perp }}(T_\unicode{x3bb} ))=1$ for all $V\in \mathbb {RP}^1$ . Therefore, by Theorem 4.1, we have

$$ \begin{align*} {\mathrm{dim}_{\mathrm{A}}}(T_{\unicode{x3bb}})&=1+\max_{\atop{x\in T_\unicode{x3bb}}{V \in X_F}}{\mathrm{dim}_{\mathrm{H}}}(T_{\unicode{x3bb}}\cap (V+x))\\ &=1+\max_{\atop{x\in T_\unicode{x3bb}}{V\in\mathbb{RP}^1\setminus Y_F}}{\mathrm{dim}_{\mathrm{A}}}(T_{\unicode{x3bb}}\cap (V+x)). \end{align*} $$

Furthermore, by Lemma 5.4, $Y_F$ is a singleton containing only the y-axis $V_{\infty }$ . Since $T_{\unicode{x3bb} }$ is a graph of a function, we have ${\mathrm {dim}_{\mathrm {A}}}(T_{\unicode{x3bb} }\cap (V_{\infty }+x))={\mathrm {dim}_{\mathrm {A}}}(\{x\})=0$ for all $x\in T_{\unicode{x3bb} }$ , which concludes the proof.

Proof of Theorem 5.2

Since $T_\unicode{x3bb} $ is a graph of a function, it suffices to show that

(5.3) $$ \begin{align} {\mathrm{dim}_{\mathrm{H}}}(T_{\unicode{x3bb}}\cap(V+x))\leqslant \frac{ \log(2^{n_{\unicode{x3bb}}}-1)}{\log 2^{n_{\unicode{x3bb}}}} \end{align} $$

for all $x \in T_\unicode{x3bb} $ and $V \in \mathbb {RP}^1\setminus \{V_\infty \}$ . We write

$$ \begin{align*} \Sigma_n(A) = \{\mathtt{i}\in\Sigma_n\colon \varphi_{\mathtt{i}}(T_{\unicode{x3bb}})\cap A\ne\emptyset\} \end{align*} $$

for all $A \subset \mathbb {R}^2$ and $n \in \mathbb {N}$ . Let us first show by induction that

(5.4) $$ \begin{align} \#\Sigma_{kn_{\unicode{x3bb}}}(V+x)\leqslant (2^{n_{\unicode{x3bb}}}-1)^k, \end{align} $$

for all $k\in \mathbb {N}$ , $x\in T_\unicode{x3bb} $ and $V\in \mathbb {RP}^1\setminus \{V_\infty \}$ .

First let $k=1$ , $x\in T_\unicode{x3bb} $ , and $V\in \mathbb {RP}^1\setminus \{0\}$ . By symmetry, we may assume that $V=V_t$ , with $t\geqslant 0$ , and without loss of generality we may also assume that $x=(0,y)$ for some $y\in \mathbb {R}$ so that

$$ \begin{align*} V_t+x=\{(s,ts+y)\colon s\in\mathbb{R}\}. \end{align*} $$

Since $\#\Sigma _{n_{\unicode{x3bb} }}=2^{n_{\unicode{x3bb} }}$ , it suffices to show that there exists at least one $\mathtt {i}\in \Sigma _{n_{\unicode{x3bb} }}$ such that $\varphi _{\mathtt {i}}(T_{\unicode{x3bb} })\cap (V_t+x)=\emptyset $ . Assume without loss of generality that $\varphi _{\mathtt {i}_1}(T_{\unicode{x3bb} })\cap (V_t+x)\ne \emptyset $ , where $\mathtt {i}_1=\overline {1}|_{n_{\unicode{x3bb} }}$ . From (5.1), we may deduce that $\varphi _{\mathtt {i}_1}(T_{\unicode{x3bb} })$ is contained in the rectangle $ [0,2^{-n_{\unicode{x3bb} }}]\times [0,(K_{\unicode{x3bb} }+M_{\unicode{x3bb} })\unicode{x3bb} ^{n_{\unicode{x3bb} }}]$ , and since $\varphi _{\mathtt {i}_1}(T_{\unicode{x3bb} })\cap (V_t+x)\ne \emptyset $ , we have $y\leqslant (K_\unicode{x3bb} +M_{\unicode{x3bb} })\unicode{x3bb} ^{n_{\unicode{x3bb} }}$ . Write $x_{1/2}=(\tfrac 12,\tfrac 12)$ and let $\mathtt {i}=1\overline {2}$ . Since $T_{\unicode{x3bb} }(\tfrac 12)=\tfrac 12$ , we have $x_{{1}/{2}}\in T_{\unicode{x3bb} }$ . In fact,

$$ \begin{align*} \varphi_{\mathtt{i}|_{n_{\unicode{x3bb}}}}(1)=\big(\tfrac{1}{2},\tfrac{1}{2}\big). \end{align*} $$

It is also a simple exercise to show that $\varphi _{\mathtt {i}|_{n_{\unicode{x3bb} }}}(0)\geqslant \tfrac 12$ . Therefore, if $\tfrac {1}{2}t+y< \tfrac 12$ , then $\varphi _{\mathtt {i}|_{n_{\unicode{x3bb} }}}(X)\cap (V_t+x)=\emptyset $ . On the other hand, if $t\tfrac {1}{2}+y\geqslant \tfrac 12$ , then

$$ \begin{align*} t\geqslant 1-2y\geqslant 1-2(K_{\unicode{x3bb}}+M_{\unicode{x3bb}})\unicode{x3bb}^{n_{\unicode{x3bb}}}>0 \end{align*} $$

by the choice of $n_{\unicode{x3bb} }$ . This means that the line $V_t+x$ has a positive slope. Since $\varphi _{\mathtt {i}_2}(T_{\unicode{x3bb} })\subset [1-2^{-n_{\unicode{x3bb} }},1]\times [0,(K_{\unicode{x3bb} }+M_{\unicode{x3bb} })\unicode{x3bb} ^{n_{\unicode{x3bb} }}]$ by symmetry, where $\mathtt {i}_2=\overline {2}|_{n_{\unicode{x3bb} }}$ , and since $t(1-2^{n_{\unicode{x3bb} }})+y>\tfrac 12\geqslant (K_{\unicode{x3bb} }+M_{\unicode{x3bb} })\unicode{x3bb} ^{n_{\unicode{x3bb} }}$ , we have $\varphi _{\mathtt {i}_2}(T_{\unicode{x3bb} })\cap (V_t+x)=\emptyset $ , which finishes the proof for $k=1$ .

Let us then assume that (5.4) holds for $k\in \mathbb {N}$ . Let $x\in T_\unicode{x3bb} $ and $V\in \mathbb {RP}^1\setminus \{0\}$ . To finish the proof of (5.4), we have to show that $\#\Sigma _{(k+1)n_{\unicode{x3bb} }}(V+x)\leqslant (2^{n_{\unicode{x3bb} }}-1)^{k+1}$ . Notice that trivially

(5.5) $$ \begin{align} \Sigma_{(k+1)n_{\unicode{x3bb}}}=\{\mathtt{i}\mathtt{j}\in\Sigma_{(k+1)n_{\unicode{x3bb}}}\colon \mathtt{i}\in\Sigma_{kn_{\unicode{x3bb}}} \text{ and } \mathtt{j}\in\Sigma_{n_{\unicode{x3bb}}}\}. \end{align} $$

Let $\mathtt {i}\in \Sigma _{kn_{\unicode{x3bb} }}$ . If $\mathtt {i}\not \in \Sigma _{kn_{\unicode{x3bb} }}(V+x)$ , then $\mathtt {i}\mathtt {j}\not \in \Sigma _{(k+1)n_{\unicode{x3bb} }}(V+x)$ for all $\mathtt {j}\in \Sigma _{n_{\unicode{x3bb} }}$ , so we may assume that $\mathtt {i}\in \Sigma _{kn_{\unicode{x3bb} }}(V+x)$ . Since $\varphi _{\mathtt {i}}$ is a bijection, we have for any $\mathtt {j}\in \Sigma _{n_{\unicode{x3bb} }}$ , that $\varphi _{\mathtt {i}\mathtt {j}}(T_{\unicode{x3bb} })\cap (V+x)\ne \emptyset $ if and only if $\varphi _{\mathtt {j}}(T_{\unicode{x3bb} }\cap \varphi _{\mathtt {i}}^{-1}(V+x))\ne \emptyset $ . Since $\varphi _{\mathtt {i}}^{-1}$ is affine, there exist $x_{\mathtt {i}}\in T_\unicode{x3bb} $ and $V_{\mathtt {i}}\in \mathbb {RP}^1\setminus \{V_\infty \}$ such that $\varphi _{\mathtt {i}}^{-1}(V+x)=V_{\mathtt {i}}+x_{\mathtt {i}}$ , and thus $\varphi _{\mathtt {i}\mathtt {j}}(T_{\unicode{x3bb} })\cap (V+x)\ne \emptyset $ if and only if $\mathtt {j}\in \Sigma _{n_{\unicode{x3bb} }}(V_{\mathtt {i}}+x_{\mathtt {i}})$ . Since this is true for all $\mathtt {i}\in \Sigma _{kn_{\unicode{x3bb} }}$ , we have, by (5.5) and the fact that the claim holds for $\Sigma _{n_{\unicode{x3bb} }}(V_{\mathtt {i}}+x_{\mathtt {i}})$ and $\Sigma _{kn_{\unicode{x3bb} }}(V+x)$ , that

$$ \begin{align*} \#\Sigma_{(k+1)n_{\unicode{x3bb}}}(V+x)&=\sum_{\mathtt{i}\in\Sigma_{kn_{\unicode{x3bb}}}(V+x)}\#\Sigma_{n_{\unicode{x3bb}}}(V_{\mathtt{i}}+x_{\mathtt{i}}) \\ &\leqslant\#\Sigma_{kn_{\unicode{x3bb}}}(V+x)\cdot (2^{n_{\unicode{x3bb}}}-1) \leqslant (2^{n_{\unicode{x3bb}}}-1)^{k+1}. \end{align*} $$

This concludes the proof of (5.4).

Let us then show (5.3) by relying on (5.4). It follows from the construction that for any $k\in \mathbb {N}$ and $\mathtt {i}\in \Sigma _{kn_{\unicode{x3bb} }}$ , the image $\varphi _{\mathtt {i}}(T_{\unicode{x3bb} })$ is contained in a vertical strip of width $2^{-kn_{\unicode{x3bb} }}$ . For any $V=V_t\in \mathbb {RP}^1\setminus \{V_{\infty }\}$ , we therefore have

(5.6) $$ \begin{align} {\mathrm{diam}}(\varphi_{\mathtt{i}}(T_{\unicode{x3bb}})\cap (V_t+x))\leqslant 2^{-kn_{\unicode{x3bb}}}\sqrt{t^2+1}. \end{align} $$

Thus $\{\varphi _{\mathtt {i}}(T_{\unicode{x3bb} })\cap (V_t+x)\}_{\mathtt {i}\in \Sigma _{kn_{\unicode{x3bb} }}(V_t+x)}$ is a $2^{-kn_{\unicode{x3bb} }}\sqrt {t^2+1}$ -cover of $T_{\unicode{x3bb} }\cap (V_t+x)$ which together with (5.4) shows that

$$ \begin{align*} N_{2^{-kn_{\unicode{x3bb}}}\sqrt{t^2+1}}(T_{\unicode{x3bb}}\cap (V_t+x))\leqslant\#\Sigma_{kn_{\unicode{x3bb}}}(V_t+x)\leqslant (2^{n_{\unicode{x3bb}}}-1)^k. \end{align*} $$

Taking logarithms, dividing by $-\log (2^{-kn_{\unicode{x3bb} }}\sqrt {t^2+1})$ , and letting $k\to \infty $ yields

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}(T_{\unicode{x3bb}}\cap (V_t+x))\leqslant{\underline{\mathrm{dim}}_{\mathrm{M}}}(T_{\unicode{x3bb}}\cap (V_t+x))\leqslant \frac{\log (2^{n_{\unicode{x3bb}}}-1)}{\log 2^{n_{\unicode{x3bb}}}} \end{align*} $$

as claimed.

Proof. Let $x \in T_\unicode{x3bb} $ and $V \in X_F$ . By Lemma 5.4, there is $t \in [-K_\unicode{x3bb} ,K_\unicode{x3bb} ]$ such that $V = V_t = \langle (1,t)\rangle $ . Similarly as in (5.6), we have

$$ \begin{align*} {\mathrm{diam}}(\varphi_{\mathtt{i}}(T_{\unicode{x3bb}})\cap (V+x)) \leqslant 2^{-n}\sqrt{t^2+1} \end{align*} $$

for all $\mathtt {i}\in \Sigma _{n}(V+x)$ . Therefore, the collection $\{\varphi _{\mathtt {i}}(T_{\unicode{x3bb} })\cap (V+x)\}_{\mathtt {i}\in \Sigma _n(V+x)}$ is a $2^{-n}\sqrt {t^2+1}$ -cover of $T_{\unicode{x3bb} }\cap (V+x)$ , which proves the upper bounds. The lower bounds follow by observing that if $\{U_i\}$ is any $2^{-n}$ -cover of $T_{\unicode{x3bb} }\cap (V+x)$ , then every $U_i$ intersects at most two of the sets in $\{\varphi _{\mathtt {i}}(T_{\unicode{x3bb} })\cap (V+x)\}_{\mathtt {i}\in \Sigma _n(V+x)}$ .

Proof. Let $x\in T_{\unicode{x3bb} }$ and $V\in \mathbb {RP}^1\setminus \{V_{\infty }\}$ , and note that

$$ \begin{align*} ({\mathrm{proj}}_{V^{\perp}})_*\nu(B({\mathrm{proj}}_{V^{\perp}}(x),r))=\nu([V+x]_r) \end{align*} $$

for all $r> 0$ . Write $c = \sqrt {2}(K_{\unicode{x3bb} }+M_{\unicode{x3bb} })$ . It is easy to see that for every $\mathtt {i}\in \Sigma _n$ we have

$$ \begin{align*} {\mathrm{diam}}(\varphi_{\mathtt{i}}(T_{\unicode{x3bb}}))\leqslant c\unicode{x3bb}^n, \end{align*} $$

so for any $\mathtt {i}\in \Sigma _n(V+x)$ , the cylinder $\varphi _{\mathtt {i}}(T_{\unicode{x3bb} })$ is contained in $[V+x]_{c\unicode{x3bb} ^{n}}$ . Since the $\nu $ -measure of $\varphi _{\mathtt {i}}(T_{\unicode{x3bb} })$ is $2^{-n}$ , we have

$$ \begin{align*} \nu([V+x]_{c\unicode{x3bb}^{n}}) \geqslant \sum_{\mathtt{i}\in\Sigma_n(V+x)} \nu(\varphi_{\mathtt{i}}(T_{\unicode{x3bb}}))\geqslant 2^{-n}\cdot\#\Sigma_n(V+x). \end{align*} $$

Therefore,

$$ \begin{align*} \frac{\log\nu([V+x]_{c\unicode{x3bb}^{n}})}{\log c\unicode{x3bb}^n}&\leqslant\frac{-n\log2}{\log c\unicode{x3bb}^n}+\frac{\log\#\Sigma_n(V+x)}{\log c\unicode{x3bb}^n}\\ &=\frac{n\log1/2}{\log c\unicode{x3bb}^n}-\frac{n\log 1/2}{\log c\unicode{x3bb}^n}\frac{\log\#\Sigma_{n}(V+x)}{n\log2}, \end{align*} $$

and taking the limit superior or the limit inferior, Lemma 6.2 yields the respective upper bounds.

The lower bound is more subtle and for that we apply Lemma 6.3. Let $c,K>0$ be as in Lemma 6.3. Since the collection $\{\varphi _{\mathtt {i}}(X)\colon \mathtt {i}\in \Sigma _n(V+x)\cup {\mathrm {Bad}}_{n,c}\}$ covers $T_\unicode{x3bb} \cap [V+x]_{c\unicode{x3bb} ^n}$ , Lemma 6.3 shows that

$$ \begin{align*} \nu([V+x]_{c\unicode{x3bb}^n})&\leqslant\sum_{\mathtt{i}\in\Sigma_n(V+x)}\nu(\varphi_{\mathtt{i}}(T_\unicode{x3bb}))+\sum_{\mathtt{i}\in{\mathrm{Bad}}_{n,c}}\nu(\varphi_{\mathtt{i}}(T_\unicode{x3bb}))\\ &=2^{-n}\cdot\#\Sigma_n(V+x)+2^{-n}\cdot\#{\mathrm{Bad}}_{n,c}\\ &=2^{-n}\cdot\#\Sigma_n(V+x)+2^{-n}\sum_{k=1}^{n}\#{\mathrm{Bad}}_{n,c}^k\\ &\leqslant2^{-n}\cdot\#\Sigma_n(V+x)+2^{-n}K\sum_{k=1}^{n}\#\Sigma_k(V+x)\\ &\leqslant2^{-n}(Kn+1)\cdot\#\Sigma_n(V+x). \end{align*} $$

Taking logarithms, dividing by $c\unicode{x3bb} ^n$ , and taking the limits then gives the desired lower bounds.

Proof of Theorem 6.1

Let us first assume that

$$ \begin{align*} {\overline{\mathrm{dim}}_{\mathrm{loc}}}(({\mathrm{proj}}_{V^{\perp}})_*\nu,{\mathrm{proj}}_{V^{\perp}}(x))\geqslant 1 \end{align*} $$

for all $x \in T_\unicode{x3bb} $ and $V\in X_F$ . Then Proposition 6.4 and (1.3) give us

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}(T_{\unicode{x3bb}}\cap(V+x))&\leqslant{\underline{\mathrm{dim}}_{\mathrm{M}}}(T_{\unicode{x3bb}}\cap(V+x))\\ &=1-\frac{\log\unicode{x3bb}}{\log1/2}{\overline{\mathrm{dim}}_{\mathrm{loc}}}(({\mathrm{proj}}_{V^{\perp}})_*\nu,{\mathrm{proj}}_{V^{\perp}}(x))\\ &\leqslant 1+\frac{\log\unicode{x3bb}}{\log 2}={\mathrm{dim}_{\mathrm{H}}}(T_{\unicode{x3bb}})-1, \end{align*} $$

for all $x\in T_{\unicode{x3bb} }$ and $V \in X_F$ . Since $T_\unicode{x3bb} $ is a graph of a function and recalling the proof of Theorem 4.1, this estimate extends to all $V \in \mathbb {RP}^1$ . Therefore, by Theorem 5.1,

$$ \begin{align*} {\mathrm{dim}_{\mathrm{H}}}(T_{\unicode{x3bb}})-1\leqslant {\mathrm{dim}_{\mathrm{A}}}(T_{\unicode{x3bb}})-1=\max_{\atop{x\in T_\unicode{x3bb}}{V\in \mathbb{RP}^1}}{\mathrm{dim}_{\mathrm{H}}}(T_\unicode{x3bb} \cap (V+x))\leqslant {\mathrm{dim}_{\mathrm{H}}}(T_{\unicode{x3bb}})-1 \end{align*} $$

and the claim follows.

Let us then assume that

(6.2) $$ \begin{align} \max_{\atop{x\in T_\unicode{x3bb}}{V\in\mathbb{RP}^1}}{\mathrm{dim}_{\mathrm{H}}}(T_\unicode{x3bb} \cap (V+x))= {\mathrm{dim}_{\mathrm{H}}}(T_\unicode{x3bb})-1. \end{align} $$

If $x\in T_\unicode{x3bb} $ and $V\in \mathbb {RP}^1\setminus \{V_{\infty }\}$ are such that

$$ \begin{align*} {\underline{\mathrm{dim}}_{\mathrm{loc}}}(({\mathrm{proj}}_{V^{\perp}})_*\nu,{\mathrm{proj}}_{V^{\perp}}(x))<1, \end{align*} $$

then, by Proposition 6.4, (1.3), (6.2), and Theorem 5.1, we have

$$ \begin{align*} {\overline{\mathrm{dim}}_{\mathrm{M}}}(T_{\unicode{x3bb}}\cap(V+x))&>1-\frac{\log\unicode{x3bb}}{\log1/2}=1+\frac{\log\unicode{x3bb}}{\log 2}={\mathrm{dim}_{\mathrm{H}}}(T_\unicode{x3bb})-1\\ &=\max_{\atop{x\in T_\unicode{x3bb}}{V\in\mathbb{RP}^1}}{\mathrm{dim}_{\mathrm{H}}}(T_\unicode{x3bb} \cap (V+x))=\max_{\atop{x\in T_\unicode{x3bb}}{V\in\mathbb{RP}^1}}{\mathrm{dim}_{\mathrm{A}}}(T_\unicode{x3bb} \cap (V+x)) \end{align*} $$

which is a contradiction.