Proof of Theorem 1.1

We first recall some notation in §1. Let M be a locally symmetric space of non-compact type with finite Riemannian volume. Then, M can be identified with $Y=K\backslash G/\Gamma $ , where G is a semisimple Lie group without compact factors and with finite center, K is a maximal compact subgroup of G, and $\Gamma $ is a lattice in G. Let $X=G/\Gamma $ , $\mathfrak {g}=\mathrm {Lie}(G),\mathfrak {k}=\mathrm {Lie}(K)$ , and $\mathfrak {p}\subset \mathfrak {g}$ be the orthogonal complement of $\mathfrak {k}$ . Then, it suffices to show that: for any point $y=[Kg\Gamma ]\in Y$ , the set

$$ \begin{align*} \{\mathbf{v}\in\mathfrak{p}^1: \{\exp(t\mathbf{v})g\Gamma: t\geq 0\} \text{ is bounded in }X\} \end{align*} $$

is thick in $\mathfrak {p}^1$ .

Let us choose a maximal abelian subspace $\mathfrak {a}\subset \mathfrak {p}$ and an open Weyl chamber $\mathfrak {a}_+\subset \mathfrak {a}$ . Let $L=Z_K(\mathfrak {a})$ and $\mathfrak {a}^1_+=\mathfrak {a}_+\cap \mathfrak {p}^1$ . Then, the map

$$ \begin{align*} \theta: \mathfrak{a}^1_+\times L\backslash K\to \mathfrak{p}^1, \quad (\mathbf{v}, Lk)\mapsto (\mathrm{Ad} (k^{-1}))\mathbf{v} \end{align*} $$

is a diffeomorphism onto an open dense subset of $\mathfrak {p}^1$ . Then, it suffices to prove that the set

(3.1) $$ \begin{align} \{(\mathbf{v},Lk)\in \mathfrak{a}^1_+\times L\backslash K:\{\exp(t\mathbf{v})kg\Gamma: t\geq 0\} \text{ is bounded in }X\} \end{align} $$

is thick in $\mathfrak {a}^1_+\times L\backslash K$ . Moreover, by Marstrand’s slicing and projection theorems [Reference Marstrand30], we only need to prove that: for any $\mathbf {v}\in \mathfrak {a}^1_+$ , the set

(3.2) $$ \begin{align} E_K^+([g\Gamma],\infty):=\{k\in K: \{\exp(t\mathbf{v})kg\Gamma: t\geq 0\} \text{ is bounded in }X\} \end{align} $$

is thick in K, since the set (3.1) can be decomposed into a union of slices, each of which is a quotient of a set of the form (3.2).

Let $F_{\mathbf {v}}^+=\{\exp (t\mathbf {v}): t\geq 0\}$ be the one-parameter subsemigroup of G, and $H=H(F_{\mathbf {v}}^+)$ (respectively $H^0=H^0(F_{\mathbf {v}}^+),H^-=H^-(F_{\mathbf {v}}^+)$ ) be the unstable (respectively neutral, stable) horospherical subgroup with respect to $F_{\mathbf {v}}^+$ . Let $B=H^0H$ and $B^-=H^-H^0$ , and consider the action of the subgroup K on the real flag variety $G/B$ by left translation. By Iwasawa decomposition, this action is transitive. It follows that for any $g\in G$ , the map

$$ \begin{align*} \phi_g: K\rightarrow G/B,\quad k\mapsto kgB \end{align*} $$

is a projection map with compact fiber. Write $K'=K\cap B^-Bg^{-1}=K\cap B^-Hg^{-1}$ . Then, $K\setminus K'=\phi _g^{-1}((G\setminus B^{-1}B)/B)$ is a finite union of manifolds of strict lower dimensions. So, it suffices to show that $E_K^+([g\Gamma ],\infty )\cap K'$ is thick in $K'$ .

For $k\in K'$ , we set $\sigma _1(k)\in B^-$ and $\sigma _2(k)\in H$ to be the unique pair satisfying ${kg=\sigma _1(k)\sigma _2(k)}$ . It is easy to check that the map $\sigma _2$ is a projection map. However,

$$ \begin{align*} F_{\mathbf{v}}^+ kg\Gamma \text{ is bounded } \Longleftrightarrow F_{\mathbf{v}}^+ \sigma_2(k)\Gamma \text{ is bounded}. \end{align*} $$

Consequently,

$$ \begin{align*}E_K^+([g\Gamma],\infty)\cap K'=\sigma_2^{-1}(\{h\in H:F_{\mathbf{v}}^+h\Gamma \text{ is bounded in }X\})=\sigma_2^{-1}(E_H^+([1_G\Gamma],\infty)).\end{align*} $$

In view of Theorem 3.1, we see that $E_H^+([1_G\Gamma ],\infty )$ is thick in H and, hence, by Marstrand’s slicing theorem, $E_K^+([g\Gamma ],\infty )\cap K'$ is thick in $K'$ . This completes the proof.

Proof of Theorem 1.5

We begin with a simple observation, that is, it suffices to prove Theorem 1.7 for $x=[1_G\Gamma ]\in X$ . Indeed, for any $x'=[g\Gamma ]\in X$ and one-parameter subgroup $F\subset G$ , $F^+x'$ is bounded if and only if $g^{-1}F^+gx$ is bounded, which implies

$$ \begin{align*} E^+(x', \infty)=\mathrm{Ad}(g)(E^+(x,\infty)). \end{align*} $$

Thus, it suffices to prove Theorem 1.5 for $x=[1_G\Gamma ]\in X$ .

Let $p:\mathfrak {g}\setminus \{0\}\to \mathbf {P}^+(\mathfrak {g})$ denote the projection map. By Marstrand’s projection and slicing theorems, to prove Theorem 1.5 is equivalent to show that the set

$$ \begin{align*} \{\mathbf{v}\in\mathfrak{g}: F_{\mathbf{v}}^+\Gamma\text{ is bounded in }X\} \end{align*} $$

is thick in $\mathfrak {g}$ . We claim that we may reduce to the case when G is a connected semisimple Lie group with trivial center and without compact factors for the following reasons.

First, let $G^0$ denote the identity component of G. It is clear that $\mathrm {Lie}(G^0)=\mathfrak {g}$ and that $X^0=G^0/(G^0\cap \Gamma )$ is a connected component of X containing $[1_G\Gamma ]$ . In particular,

$$ \begin{align*} F_{\mathbf{v}}^+\Gamma\text{ is bounded in }X\Longleftrightarrow F_{\mathbf{v}}^+\Gamma\text{ is bounded in }X^0. \end{align*} $$

So, we may assume that $G=G^0$ is connected without loss of generality.

Second, let $\mathfrak {g}=\mathfrak {r}\rtimes \mathfrak {l}$ be the Levi decomposition, where $\mathfrak {r}$ is the solvable radical of $\mathfrak {g}$ and $\mathfrak {l}$ is a Levi subalgebra of $\mathfrak {g}$ . Let $\mathfrak {l}=\mathfrak {l}_c\oplus \mathfrak {l}_{nc}$ be the decomposition into simple ideals where $\mathfrak {l}_c$ is the direct sum of all compact simple ones. The projection map $\mathrm {d}q:\mathfrak {g}\to \mathfrak {l}_{nc}$ is a Lie algebra homomorphism, which induces a Lie group homomorphism $q:G\to L_{nc}$ and a projection map $X=G/\Gamma \to X_{nc}:=L_{nc}/q(\Gamma )$ . Note that $q(\Gamma )$ is a lattice in $L_{nc}$ , and the fiber of $X\to X_{nc}$ is a compact homogeneous space $RL_c/(RL_c\cap \Gamma )$ (see [Reference Dani19, Lemma 9.1]). Then, for $\mathbf {v}\in \mathfrak {g}$ ,

$$ \begin{align*} F_{\mathbf{v}}^+\Gamma\text{ is bounded in }X\Longleftrightarrow q(F_{\mathbf{v}})^+q(\Gamma)=F_{\mathrm{d}q(\mathbf{v})}^+q(\Gamma) \text{ is bounded in }X_{nc}. \end{align*} $$

It follows that

$$ \begin{align*} \{\mathbf{v}\in\mathfrak{g}: F_{\mathbf{v}}^+\Gamma\text{ is bounded in }X\} =(\mathrm{d}q)^{-1}\{\mathbf{w}\in\mathfrak{l}_{nc}: F_{\mathbf{w}}^+q(\Gamma)\text{ is bounded in }X_{nc}\}. \end{align*} $$

By Marstrand’s slicing theorem, it suffices to show that

$$ \begin{align*} \{\mathbf{w}\in\mathfrak{l}_{nc}: F_{\mathbf{w}}^+q(\Gamma)\text{ is bounded in }X_{nc}\} \end{align*} $$

is thick in $\mathfrak {l}_{nc}$ . So, we may assume that G is a semisimple Lie group without compact factors.

Finally, since $X=G/\Gamma $ is a finite cover of $G/(Z(G)\cdot \Gamma )$ , we may replace G by $G/Z(G)$ and $\Gamma $ by $(\Gamma \cdot Z(G))/Z(G)$ without loss of generality. These verify the claim.

Now, let G be a connected semisimple Lie group with trivial center and without compact factors. Let $G=K\cdot \exp (\mathfrak {p})$ be the global Cartan decomposition, $\mathfrak {a}\subset \mathfrak {p}$ be a maximal abelian subspace, and $L=Z_K(\mathfrak {a})$ . We choose an open Weyl chamber $\mathfrak {a}_+\subset \mathfrak {a}$ , and for any $\mathbf {v}\in \mathfrak {a}_+$ , let $H=H(F_{\mathbf {v}}^+)$ (respectively $H^0=H^0(F_{\mathbf {v}}^+),H^-=H^-(F_{\mathbf {v}}^+)$ ) be the unstable (respectively neutral, stable) horospherical subgroup with respect to $F_{\mathbf {v}}^+$ . Note that the subgroups $H,H^0,H^-$ only depend on the choice of the open Weyl chamber $\mathfrak {a}_+$ .

Consider the map

$$ \begin{align*} \theta_0: \mathfrak{a}_+\times Z_G(\mathfrak{a}_+)\backslash G \rightarrow \mathfrak{g}, \quad (\mathbf{v},Z_G(\mathfrak{a}_+)g)\mapsto \mathrm{Ad}(g^{-1})\mathbf{v}. \end{align*} $$

It is easy to check that $\theta $ is a diffeomorphism onto $\mathfrak {g}$ . Using Bruhat decomposition, there is a diffeomorphism onto its image:

$$ \begin{align*} H\times L\times H^-\to Z_G(\mathfrak{a}_+)\backslash G, \quad (h_1,h_2,h_3)\mapsto Z_G(\mathfrak{a}_+)h_3h_2h_1. \end{align*} $$

It follows that the composition map

$$ \begin{align*} \theta: \mathfrak{a}_+\times H\times L\times H^- \rightarrow \mathfrak{g}, \quad (\mathbf{v}, h_1,h_2,h_3)\mapsto\mathrm{Ad}(h_1^{-1}h_2^{-1}h_3^{-1})\mathbf{v} \end{align*} $$

is also a diffeomorphism onto its image. Moreover, we see that p is a projection map when restricted to $\mathrm {Im}(\theta )$ and that $\mathfrak {g}\setminus \mathrm {Im}\ \theta $ is a finite union of submanifolds of strict lower dimensions. Thus, it suffices to prove that

$$ \begin{align*} \{(\mathbf{v},h_1,h_2,h_3)\in\mathfrak{a}_+\times H\times L\times H^-: F_{\mathbf{v}}^+h_3h_2h_1\Gamma\text{ is bounded in }X\} \end{align*} $$

is thick on $\mathfrak {a}_+\times H\times L\times H^-$ . However, we have for any $\mathbf {v}\in \mathfrak {a}_+$ ,

$$ \begin{align*} F_{\mathbf{v}}^+h_3h_2h_1\Gamma\text{ is bounded in }X\Longleftrightarrow F_{\mathbf{v}}^+h_1\Gamma\text{ is bounded in }X. \end{align*} $$

So, the conclusion follows from Marstrand’s slicing theorem and Theorem 3.1. This completes the proof.

Proof of Theorem 1.3 modulo Theorem 4.1

We first recall basic facts about geodesic flows on Y from [Reference Mautner31]. Let $\mathfrak {k}=\mathrm {Lie}(K)$ and $\mathfrak {p}\subset \mathfrak {g}$ be its orthogonal complement. Then, for any $y=[Kg\Gamma ]\in Y$ , $S_y(Y)$ can be identified with $\mathfrak {p}^1=\{\mathbf {v}\in \mathfrak {p}: \|\mathbf {v}\|=1\}$ . Moreover, for any $y=[Kg\Gamma ]\in Y$ and $\mathbf {v}\in \mathfrak {p}^1$ , the associated geodesic ray is given by

$$ \begin{align*} \gamma(y,\mathbf{v})=\{K\exp(t\mathbf{v})g\Gamma: t\geq 0\}.\end{align*} $$

Since K is compact, we have

$$ \begin{align*}E^+(y,\infty)=\{\mathbf{v}\in \mathfrak{p}^1: F_{\mathbf{v}}^+g\Gamma \text{ is bounded in }X\}.\end{align*} $$

The proof of Theorem 1.3 will be given in two steps. To begin, we consider the map

$$ \begin{align*} \theta: \Pi\times K\rightarrow \mathfrak{p}^1, \quad \theta(\unicode{x3bb}, k)=\mathrm{Ad}(k^{-1})\mathbf{v}_\unicode{x3bb}'\quad \text{where } \mathbf{v}_\unicode{x3bb}'=\|\mathbf{v}_\unicode{x3bb}\|^{-1}\mathbf{v}_\unicode{x3bb}\in \mathfrak{p}^1. \end{align*} $$

It is easy to check that $\mathfrak {p}^1\setminus (\mathrm {Im}\ \theta \cup -\mathrm {Im}\ \theta )$ is a finite union of manifolds of strict lower dimension. Hence, in view of Lemma 2.3, it suffices to prove that ${E^+(y,\infty )\cap \pm \mathrm {Im}\ \theta }$ is HAW on $\pm \mathrm {Im}\ \theta $ . By applying $\overline {\tau }:X\rightarrow X$ induced by the outer automorphism ${\tau (g)=(g^T)^{-1}}$ defined on G, we get

$$ \begin{align*}E^+(y,\infty)\cap -\mathrm{Im}\ \theta=\mathrm{d}\tau(E^+(y',\infty)\cap \mathrm{Im}\ \theta) \quad\text{where }y'=(g^T)^{-1}\Gamma.\end{align*} $$

Thus, it suffices to show $E^+(y,\infty )\cap \mathrm {Im}\ \theta $ is HAW on $\mathrm {Im}\ \theta $ for any $y\in Y$ . However, it is also easy to check that $\theta $ is a cover map. Hence, in view of Lemma 2.2(iii), it suffices to show that

$$ \begin{align*}E':=\theta^{-1}(E^+(y,\infty))=\{(\unicode{x3bb},k)\in \Pi\times K: F_\unicode{x3bb}^+ kg\Gamma \text{ is bounded in }X\}\end{align*} $$

is HAW on $\Pi \times K$ .

Let $ B\subset G$ (respectively $B^- \subset G$ ) be the subgroup of upper triangular (respectively lower triangular) matrices, and consider the action of the subgroup K on the real flag variety $G/B$ by left translations. By Iwasawa’s decomposition, this action is transitive and has finite stabilizer. It follows that for any $g\in G$ , the map $\phi _g: K\rightarrow G/B$ defined by $\phi _g(k)= kgB$ is a cover map. Write $K'=K\cap B^-Bg^{-1}=K\cap B^-Ug^{-1}$ . Then, $K\setminus K'=\phi _g^{-1}((G\setminus B^{-1}B)/B)$ is a finite union of manifolds of strict lower dimensions. Using Lemma 2.3 again, it suffices to show that $E":=E'\cap (\Pi \times K')$ is HAW on $\Pi \times K'$ . For $k\in K'$ , we set $\sigma _1(k)\in B^-$ and $\sigma _2(k)\in U$ to be the unique pair satisfying $kg=\sigma _1(k)\sigma _2(k)$ . It is easy to check that the map $\sigma _2$ is a cover map. However,

$$ \begin{align*} F_\unicode{x3bb}^+ kg\Gamma \text{ is bounded in }X \Longleftrightarrow F_\unicode{x3bb}^+ \sigma_2(k)\Gamma \text{ is bounded in }X. \end{align*} $$

Consequently, $E"=(\mathrm {Id}\times \sigma _2)^{-1}S$ . Thus, in view of Lemma 2.2(iii) and Theorem 4.1, we see that $E"$ is HAW on $\Pi \times K'$ . This completes the proof.

Proof of Theorem 1.7 modulo Theorem 4.1

We begin with a simple observation, that is, it suffices to prove Theorem 1.7 for $x=[1_G\Gamma ]\in X$ . Indeed, for any $x'=[g\Gamma ]\in X$ and one-parameter subgroup $F\subset G$ , $Fx'$ is bounded if and only if $g^{-1}Fgx$ is bounded, which implies

$$ \begin{align*} E^+(x', \infty)=\mathrm{Ad}(g)(E^+(x,\infty)). \end{align*} $$

Hence, in view of Lemma 2.2(iii), it suffices to prove Theorem 1.7 for $x=[1_G\Gamma ]\in X$ .

Consider the map

$$ \begin{align*} \theta: \Pi\times U\times U^- \rightarrow \mathfrak{g}, \quad (\unicode{x3bb}, u_1, u_2)\mapsto\mathrm{Ad}(u_1^{-1}u_2^{-1})\mathbf{v}_\unicode{x3bb}. \end{align*} $$

It is straightforward to check that the map $\theta $ is a diffeomorphism onto its image. Let p be the projection from $\mathfrak {g}\setminus \{0\}$ to $\mathbf {P}^+(\mathfrak {g})$ . Then, it can be checked that p is a diffeomorphism onto its image when restricted on $\mathrm {Im}\ \theta \cup -\mathrm {Im}\ \theta$ and $\mathbf {P}^+(\mathfrak {g})\setminus (p(\mathrm {Im}\ \theta )\cup -p(\mathrm {Im}\ \theta ))$ is a finite union of submanifolds of strict lower dimensions. In view of Lemmas 2.2(iv) and 2.3, it suffices to show that $E^+([1_G\Gamma ],\infty )\cap \pm p(\mathrm {Im}\ \theta )$ is HAW on $\pm p(\mathrm {Im}\ \theta )$ , respectively. Set

$$ \begin{align*} E^+(\infty)=\{\mathbf{v}\in \mathfrak{g}: F_{\mathbf{v}}^+\Gamma \text{ is bounded in } X \}. \end{align*} $$

Then, $E^+([1_G\Gamma ],\infty )=p(E^+(\infty )\setminus \{0\})$ . We claim that Theorem 1.7 can be deduced from the following statement:

(4.2) $$ \begin{align} E^+(\infty)\cap \mathrm{Im}\ \theta \text{ is HAW on } \mathrm{Im}\ \theta. \end{align} $$

Indeed, on one hand, it follows from Lemma 2.2(iii) and (4.2) that

$$ \begin{align*} E^+([1_G\Gamma],\infty)\cap p(\mathrm{Im}\ \theta)=p(E^+(\infty)\cap\mathrm{Im}\ \theta) \text{ is HAW on } p(\mathrm{Im}\ \theta). \end{align*} $$

On the other hand, by applying the map $\overline {\tau }:X\to X$ induced by the outer automorphism $\tau (g)=(g^T)^{-1}$ defined on G, we see that

$$ \begin{align*} E^+([1_G\Gamma],\infty)\cap -p(\mathrm{Im}\ \theta)= (\mathrm{d}\tau\circ p)(E^+(\infty)\cap\mathrm{Im}\ \theta). \end{align*} $$

It follows again from Lemma 2.2(iii) and (4.2) that

$$ \begin{align*} E^+([1_G\Gamma],\infty)\cap -p(\mathrm{Im}\ \theta) \text{ is HAW on } (\mathrm{d}\tau\circ p)(\mathrm{Im}\ \theta)=-p(\mathrm{Im}\ \theta). \end{align*} $$

These verify our claim.

Finally, to prove (4.2), it suffices to establish the HAW property of $\theta ^{-1} (E^+(\infty ))$ on $\Pi \times U\times U^-$ . However, we have

$$ \begin{align*} (\unicode{x3bb}, u_1,u_2)\in \theta^{-1} (E^+(\infty)) &\Longleftrightarrow F_\unicode{x3bb}^{+}u_2u_1\Gamma \text{ is bounded }\\ &\Longleftrightarrow (\unicode{x3bb}, u_1)\in S, \end{align*} $$

which means that $\theta ^{-1} (E^+(\infty ))=S\times U^-$ . So, the conclusion follows from Lemma 2.2(v) and Theorem 4.1. This completes the proof.

Proof. The first inequality is obvious. By Minkowski linear forms theorem, $\mathcal {W}(\Omega ,\mathbf {v})$ contains a vector $(a_0,b_0,c_0)$ with $|a_0|\le q^{\unicode{x3bb} _\Omega }$ and $|b_0+z_\Omega a_0|\le q^{1-\unicode{x3bb} _\Omega }$ . Thus, it follows from (5.2) that

Hence, the second inequality.

Proof. In view of (5.8), it suffices to show that if $k>[{n}/{2}]$ , then $\mathcal {V}_{\Omega ,k}=\varnothing $ . Suppose to the contrary that $\mathcal {V}_{\Omega ,k}\ne \varnothing $ for some $k>[{n}/{2}]$ . Let $(p,r,q)\in \mathcal {V}_{\Omega ,k}$ . Then,

However,

which is a contradiction.

Proof of Lemma 5.1

By Lemma 2.4, it suffices to prove that $S'$ is HPW on $\Sigma $ . Let ${\beta \in (0,1)}$ , $\gamma>0$ . We prove that Alice has a strategy to win the $(\beta ,\gamma )$ -hyperplane potential game on $\Sigma $ with target set $S'$ . In the first round of the game, Bob chooses a closed ball $\Omega _0\subset \Sigma $ . By [Reference An, Guan and Kleinbock5, Remark 2.4], we may without loss of generality assume that Bob will play so that $\rho _i:=\rho (\Omega _i)\to 0$ . Let $\kappa $ and R be positive numbers satisfying (5.4) and (5.5). We also require that

(5.10) $$ \begin{align} (R^\gamma-1)^{-1}\le (\beta^2/2)^\gamma. \end{align} $$

Let $\epsilon , \mathscr {B}_n, \mathcal {V}_\Omega , \mathcal {V}_{\Omega ,k}$ , and $\mathscr {B}^{\prime }_n$ be as above. Then, Lemmas 5.3–5.5 hold. Let Alice play according to the following strategy. Suppose that $i\ge 0$ and Bob has chosen the closed ball $\Omega _i$ . If there is $n\ge 0$ such that

(5.11) $$ \begin{align} \Omega_i\in\mathscr{B}^{\prime}_n, \text{ and } i \text{ is the smallest non-negative integer with } \Omega_i\in\mathscr{B}_n, \end{align} $$

then Alice chooses the family of neighborhoods $\{L_k(\Omega _i)^{(2R^{-(n+k)}\rho _0)}:k\in \mathbb {N}\}$ , where $L_k(\Omega _i)$ are the hyperplanes given in Lemma 5.5. Otherwise, Alice makes an arbitrary move. Since $\Omega _i\in \mathscr {B}_n$ , we have that $\rho _i>\beta R^{-n}\rho _0$ . This, together with (5.10), implies that

$$ \begin{align*}\sum_{k=1}^\infty(2R^{-(n+k)}\rho_0)^\gamma=(2R^{-n}\rho_0)^\gamma(R^\gamma-1)^{-1}\le(\beta \rho_i)^\gamma. \end{align*} $$

So, Alice’s move is legal. We prove that this guarantees a win for Alice, that is, the unique point $\mathbf {x}_\infty =(\unicode{x3bb} _\infty , \omega _\infty )$ in $\bigcap _{i=0}^\infty \Omega _i$ lies in

$$ \begin{align*} S'\cup\bigcup_{n\in\mathcal{N}}\bigcup_{k=1}^n L_k(\Omega_{i_n})^{(2R^{-(n+k)}\rho_0)}, \end{align*} $$

where

$$ \begin{align*} \mathcal{N}:=\{n\ge0:\text{ there exists } i=i_n \text{ such that (5.11) holds}\}. \end{align*} $$

There are two cases.

Case (1). Assume $\mathcal {N}=\mathbb {N}\cup \{0\}$ . For any $\mathbf {v}=(p,r,q)\in \Xi $ , there is n such that ${q^{2}\le 2H_{n+1}}$ . Then, it follows that $q^{1+\unicode{x3bb} _n}\le q^2\le 2H_{n+1}.$ Since $n\in \mathcal {N}$ , we have that $\Omega _{i_n}\in \mathscr {B}^{\prime }_n$ . Then, Lemma 5.4 implies that $\Delta (\mathbf {v},\epsilon )\cap \Omega _{i_n}=\varnothing $ . Thus, $\mathbf {x}_\infty \notin \Delta (\mathbf {v}, \epsilon )$ . It then follows from (5.1) and the arbitrariness of $\mathbf {v}$ that $\mathbf {x}_\infty \in S(\epsilon )\subset S'$ . Hence, Alice wins.

Case (2). Assume $\mathcal {N}\ne \mathbb {N}\cup \{0\}$ . Let n be the smallest non-negative integer with $n\notin \mathcal {N}$ . Since $B_0\in \mathscr {B}^{\prime }_0$ , we have $0\in \mathcal {N}$ . Hence, $n\ge 1$ . Since $\rho _i\to 0$ and $\rho _{i+1}\ge \beta \rho _i$ , there must exist $i\ge 1$ with $\Omega _i\in \mathscr {B}_n$ and $\Omega _{i-1}\notin \mathscr {B}_n$ . It follows that $\Omega _i\notin \mathscr {B}^{\prime }_n$ . Since $n-1\in \mathcal {N}$ , we have $\Omega _{i_{n-1}}\in \mathscr {B}^{\prime }_{n-1}$ . Thus, it follows from (5.9) that $\Omega _i\cap \bigcup _{\mathbf {v}\in \mathcal {V}_{\Omega _i}}\Delta (\mathbf {v},\epsilon )\ne \varnothing $ . By Lemma 5.3, there is $k\in [1,[n/2]]$ and $\mathbf {v}\in \mathcal {V}_{\Omega _i,k}$ such that ${\Omega _i\cap \Delta (\mathbf {v}, \epsilon )\ne \varnothing }$ . Applying Lemma 5.5 to $\Omega =\Omega _{i_{n-k}}$ and $\Omega '=\Omega _i$ , we obtain $\Omega _i\cap \Delta (\mathbf {v},\epsilon )\subset L_k(\Omega _{i_{n-k}})^{(R^{-n}\rho _0)}$ . So, $\Omega _i\cap L_k(\Omega _{i_{n-k}})^{(R^{-n}\rho _0)}\ne \varnothing $ . In view of $\rho _i\le R^{-n}\rho _0$ , it follows that $\mathbf {x}_\infty \in \Omega _i\subset L_k(\Omega _{i_{n-k}})^{(2R^{-n}\rho _0)}$ . Hence, Alice also wins.

This completes the proof of Theorem 4.1 modulo Lemmas 5.4 and 5.5.

Proof. The second inequality is straightforward. For the first inequality, the case $n=0$ is trivial since $H_2<1$ . Let $n\geq 1$ . Since the function $x^{-1}\log x$ is monotonically decreasing when $x>e$ , it follows that

$$ \begin{align*}2\sqrt{\rho}\log H_{2n+2}\le \sqrt{R^{-n}}\log R^{2n+2}\le \sqrt{R^{-n}}\log R^{4n}\le \sqrt{R^{-1}}\log R^4\le 1.\end{align*} $$

This completes the proof of the first inequality.

Proof of Lemma 5.4

The case $n=0$ is trivial since $2H_1<1$ . Let $n\geq 1$ . We denote ${\Omega _n=\Omega }$ and let $\Omega _n\subset \cdots \subset \Omega _1\subset \Omega _0$ be such that $\Omega _k\in \mathscr {B}^{\prime }_k$ . Write $\unicode{x3bb} _{\Omega _k}=\unicode{x3bb} _k$ and ${\rho _{\Omega _k}=\rho _k}$ for simplicity. Suppose in contrast that the conclusion of the lemma is not true. Then, there exists $\mathbf {v}=(p,r,q)\in \mathbb {Z}^2\times \mathbb {N}$ with $q^{1+\unicode{x3bb} _n}\le 2H_{n+1}$ such that ${\Delta (\mathbf {v},\epsilon )\cap \Omega _k\ne \varnothing }$ for every $1\le k\le n$ . It follows from the definition of $\mathscr {B}^{\prime }_k$ that $\mathbf {v}\notin \mathcal {V}_{\Omega _k}$ , that is,

(6.2) $$ \begin{align} H_{\Omega_k}(\mathbf{v})\notin[H_k,3H_{k+1}], \quad 1\le k\le n. \end{align} $$

Let $1\le m\le n$ be such that

(6.3)

We inductively prove that

(6.4) $$ \begin{align} H_{\Omega_k}(\mathbf{v})<H_k, \quad \bigg[\dfrac{m}{2}\bigg]\le k\le m. \end{align} $$

By letting $k=[{m}/{2}]$ in (6.4), we have

$$ \begin{align*} q\le H_{\Omega_{[{m}/{2}]}}(\mathbf{v})<H_{[{m}/{2}]}<\sqrt{H_m}<q, \end{align*} $$

which is a contradiction. So Lemma 5.4 follows from (6.4). Since

it follows from (6.2) that (6.4) holds for $k=m$ . Suppose that $[{m}/{2}]\le k\le m-1$ and (6.4) holds if k is replaced by $k+1$ . We prove that

(6.5) $$ \begin{align} H_{\Omega_k}(\mathbf{v})\le 2H_{\Omega_{k+1}}(\mathbf{v}). \end{align} $$

To prove (6.5), the first step is to establish

(6.6) $$ \begin{align} \mathbf{w}(\Omega_{k+1},\mathbf{v})\in\mathcal{W}(\Omega_k,\mathbf{v}). \end{align} $$

Write $\mathbf {w}(\Omega _k,\mathbf {v})=(a_k,b_k,c_k)$ and $z_{\Omega _k}=z_k$ . Since $\mathbf {w}(\Omega _{k+1},\mathbf {v})\in \mathcal {W}(\Omega _{k+1},\mathbf {v})$ , we have

Note that $\sqrt {\rho _{k+1}}\le \sqrt {\beta ^{-1} R^{-1}}\sqrt {\rho _k}\le ({1}/{10})\sqrt {\rho _k}$ and by Lemma 6.1,

Now, we are in position to prove (6.6). First, we have

Second, we have

Hence, (6.6) holds. It then follows from (6.6) and the definition of $\mathbf {w}(\Omega _k,\mathbf {v})$ that

$$ \begin{align*} H_{\Omega_k}(\mathbf{v})&=q\max\{|a_k|,|b_k+z_ka_k|\}\\ &\le q\max\{|a_{k+1}|,|b_{k+1}+z_ka_{k+1}|\}\\ &\le 2q\max\{|a_{k+1}|,|b_{k+1}+z_{k+1}a_{k+1}|\}\\ &=2H_{\Omega_{k+1}}(\mathbf{v}). \end{align*} $$

This proves (6.5).

It follows from (6.5) and the induction hypothesis that $H_{\Omega _k}(\mathbf {v})< 2H_{k+1}$ . By (6.2), we have $H_{\Omega _k}(\mathbf {v})<H_k$ . Thus, (6.4) is proved.

Proof. Let $(\unicode{x3bb} ,x,y,z)\in \Delta (\mathbf {v},\epsilon )\cap \Omega $ . Then,

$$ \begin{align*} \bigg|x-\frac{p}{q}-z\bigg(y-\frac{r}{q}\bigg)\bigg|<\frac{\epsilon}{q^{1+\unicode{x3bb}}}, \quad \bigg|y-\frac{r}{q}\bigg|<\frac{\epsilon}{q^{2-\unicode{x3bb}}}. \end{align*} $$

Hence, to show (7.1), it suffices to show that, for any $\unicode{x3bb} \in I_{\Omega }$ , $q^{|\unicode{x3bb} -\unicode{x3bb} _{\Omega }|}\le 3$ . According to Lemma 6.1, we have

This completes the proof.

Proof. For simplicity, we write $\rho _{\Omega }=\rho $ and $\unicode{x3bb} _{\Omega }=\unicode{x3bb} $ . We first give a bound on $q_1$ and $q_2$ in terms of $\unicode{x3bb} $ . Indeed, for any $\unicode{x3bb} '\in I_{\Omega }$ , we have

(7.3) $$ \begin{align} H_{n+k}^{|({1}/{1+\unicode{x3bb}'})-({1}/{1+\unicode{x3bb}})|}\le H_{2n}^{|\unicode{x3bb}'-\unicode{x3bb}|} \le e^{\rho\log H_{2n}}\le 3. \end{align} $$

By applying (7.3) to $\unicode{x3bb} _{\Omega _1}$ and $\unicode{x3bb} _{\Omega _2}$ , we get

(7.4) $$ \begin{align} \begin{aligned} q_1, q_2\in & [\tfrac{1}{3}H_{n+k}^{({1}/{1+\unicode{x3bb}})},\ 3H_{n+k}^{({1}/{1+\unicode{x3bb}})}R^{20}], & k=1, \\ q_1, q_2\in & [\tfrac{1}{3}H_{n+k}^{({1}/{1+\unicode{x3bb}})}R^{2k+16}, \ 3H_{n+k}^{({1}/{1+\unicode{x3bb}})}R^{2k+18}], & k\ge 2. \end{aligned} \end{align} $$

Let $(\unicode{x3bb} _j',x_j,y_j,z_j)\in \Delta (\mathbf {v}_j,\epsilon )\cap \Omega $ . Since $\Omega _j\in \mathscr {B}_{n+k}$ and $\mathbf {v}_j\in \mathcal {V}_{\Omega _j}$ , we have

$$ \begin{align*}q_j\le H_{\Omega_j}(\mathbf{v}_j)\le 3H_{n+k+1} \le 3H_{2n+1}\le H_{2n+2}.\end{align*} $$

Thus, it follows from Lemma 7.1 that $(x_j,y_j,z_j)\in \Delta _{\unicode{x3bb} }(\mathbf {v}_j,3\epsilon )$ . Hence,

$$ \begin{align*} \bigg|x_j-\frac{p_j}{q_j}-z_j\bigg(y_j-\frac{r_j}{q_j}\bigg)\bigg|<\frac{3\epsilon}{q_j^{1+\unicode{x3bb}}}, \quad \bigg|y_j-\frac{r_j}{q_j}\bigg|<\frac{3\epsilon}{q_j^{2-\unicode{x3bb}}}. \end{align*} $$

The latter inequality implies that

$$ \begin{align*} \bigg|\frac{r_j}{q_j}\bigg|\le|y_j|+\frac{3\epsilon}{q_j^{2-\unicode{x3bb}}}\le \kappa. \end{align*} $$

Thus,

$$ \begin{align*} &\ \bigg|\frac{p_1}{q_1}-\frac{p_2}{q_2}-z_{\Omega_2}\bigg(\frac{r_1}{q_1}-\frac{r_2}{q_2}\bigg)\bigg|\\& \quad=\ \bigg|-\bigg(x_1-\frac{p_1}{q_1}-z_1\bigg(y_1-\frac{r_1}{q_1}\bigg)\bigg)+\bigg(x_2-\frac{p_2}{q_2}-z_2\bigg(y_2-\frac{r_2}{q_2}\bigg)\bigg)\\& \qquad +(x_1-x_2)+\frac{r_1}{q_1}(z_1-z_{\Omega_2})+\frac{r_2}{q_2}(z_{\Omega_2}-z_2)-(y_1z_1-y_2z_2)\bigg|\\& \quad\le\ \bigg|x_1-\frac{p_1}{q_1}-z_1\bigg(y_1-\frac{r_1}{q_1}\bigg)\bigg|+\bigg|x_2-\frac{p_2}{q_2}-z_2\bigg(y_2-\frac{r_2}{q_2}\bigg)\bigg|\\& \qquad +|x_1-x_2|+\bigg|\frac{r_1}{q_1}\bigg||z_1-z_{\Omega_2}|+\bigg|\frac{r_2}{q_2}\bigg||z_{\Omega_2}-z_2|+|y_1||z_1-z_2|+|z_2||y_1-y_2|\\& \quad\le\ \frac{3\epsilon}{q_1^{1+\unicode{x3bb}}}+\frac{3\epsilon}{q_2^{1+\unicode{x3bb}}}+10\kappa \rho \end{align*} $$

and

$$ \begin{align*} \bigg|\frac{r_1}{q_1}-\frac{r_2}{q_2}\bigg|=\bigg|-\bigg(y_1-\frac{r_1}{q_1}\bigg)+\bigg(y_2-\frac{r_2}{q_2}\bigg)+(y_1-y_2)\bigg| \le\frac{3\epsilon}{q_1^{2-\unicode{x3bb}}}+\frac{3\epsilon}{q_2^{2-\unicode{x3bb}}}+2\rho. \end{align*} $$

Write $\mathbf {w}_j=(a_j,b_j,c_j)$ . Then, according to Lemma 6.1, we have

Set $d_k$ to be a function such that $d_k=20$ if $k=1$ , and $d_k=2$ otherwise. It follows that

$$ \begin{align*} |\mathbf{v}_1\cdot\mathbf{w}_2|=&\ q_1|(q_1^{-1}\mathbf{v}_1-q_2^{-1}\mathbf{v}_2)\cdot\mathbf{w}_2|\\ =&\ q_1\bigg|a_2\bigg(\frac{p_1}{q_1}-\frac{p_2}{q_2}\bigg)+b_2\bigg(\frac{r_1}{q_1}-\frac{r_2}{q_2}\bigg)\bigg| \\ =&\ q_1\bigg|a_2\bigg(\frac{p_1}{q_1}-\frac{p_2}{q_2}-z_{\Omega_2}\bigg(\frac{r_1}{q_1}-\frac{r_2}{q_2}\bigg)\bigg) +(b_2+z_{\Omega_2}a_2)\bigg(\frac{r_1}{q_1}-\frac{r_2}{q_2}\bigg)\bigg| \\ \le&\ q_1|a_2|\bigg(\frac{3\epsilon}{q_1^{1+\unicode{x3bb}}}+\frac{3\epsilon}{q_2^{1+\unicode{x3bb}}}+10\kappa \rho\bigg)+q_1|b_2+z_{\Omega_2}a_2|\bigg(\frac{3\epsilon}{q_1^{2-\unicode{x3bb}}}+\frac{3\epsilon}{q_2^{2-\unicode{x3bb}}}+2\rho\bigg) \\ \le&\ 9q_1q_2^{\unicode{x3bb}}\bigg(\frac{3\epsilon}{q_1^{1+\unicode{x3bb}}}+\frac{3\epsilon}{q_2^{1+\unicode{x3bb}}}\bigg) +9q_1q_2^{1-\unicode{x3bb}}\bigg(\frac{3\epsilon}{q_1^{2-\unicode{x3bb}}}+\frac{3\epsilon}{q_2^{2-\unicode{x3bb}}}\bigg) \\ &\ +12\kappa \rho q_1\max\{|a_2|,|b_2+z_{\Omega_2}a_2|\}\\ \le&\ 27\epsilon \bigg(\frac{q_2^{\unicode{x3bb}}}{q_1^{\unicode{x3bb}}}+\frac{q_1}{q_2}+\frac{q_2^{1-\unicode{x3bb}}}{q_1^{1-\unicode{x3bb}}} +\frac{q_1}{q_2}\bigg)+12\kappa R^{-n}\rho_0H_{\Omega_2}(\mathbf{v}_2)\frac{q_1}{q_2} \\ \le &\ 108R^{d_k}\epsilon+720\kappa^2R^{d_k+k+1}\epsilon \le R^{d_k+k+2}\epsilon. \end{align*} $$

This proves (7.2) for $|\mathbf {v}_1\cdot \mathbf {w}_2|$ . A similar argument shows the same inequality for $|\mathbf {v}_2\cdot \mathbf {w}_1|$ .

Proof. We proceed by considering two separate cases.

Case (1). Suppose $k=1$ . We consider two subcases.

Case (1.1). The linear span of the set

$$ \begin{align*} \mathcal{V}:=\bigcup_{\substack{\Omega'\in \mathscr{B}_{n+1}\\ \Omega'\subset\Omega}}\{\mathbf{v}\in \mathcal{V}_{\Omega',1}:\Delta(\mathbf{v},\epsilon)\cap \Omega\neq \varnothing\} \end{align*} $$

is of dimension $\geq 2$ . We prove that condition (i) holds. It suffices to prove that for ${j=1,2}$ , if $\Omega _j\in \mathscr {B}_{n+1}$ , $\Omega _j\subset \Omega $ , $\mathbf {v}_j\in \mathcal {V}_{\Omega _j,1}$ , $\Delta (\mathbf {v}_j, \epsilon )\cap \Omega \ne \varnothing $ , and $\mathbf {v}_1$ and $\mathbf {v}_2$ are linearly independent, then $L(\Omega _1,\mathbf {v}_1)=L(\Omega _2,\mathbf {v}_2)$ . Write $\mathbf {w}_j=\mathbf {w}(\Omega _j,\mathbf {v}_j)$ . It follows from Lemma 7.2 that

$$ \begin{align*}\max\{|\mathbf{v}_1\cdot\mathbf{w}_2|, |\mathbf{v}_2\cdot\mathbf{w}_1|\}\le R^{23}\epsilon<1.\end{align*} $$

Since both $\mathbf {v}_2\cdot \mathbf {w}_1$ and $\mathbf {v}_1\cdot \mathbf {w}_2$ are integers, they must be $0$ . Hence, both $\mathbf {w}_1$ and $\mathbf {w}_2$ are orthogonal to $\mathrm {span}\{\mathbf {v}_1,\mathbf {v}_2\}$ . Thus, $\mathbf {w}_1$ and $\mathbf {w}_2$ are linearly dependent. This means that $L(\Omega _1,\mathbf {v}_1)=L(\Omega _2,\mathbf {v}_2)$ .

Case (1.2). The linear span of $\mathcal {V}$ is of dimension $1$ . We prove that condition (ii) holds. Let $\mathbf {v}_0=(p_0,r_0,q_0)$ be any element in $\mathcal {V}$ . It follows from $\mathcal {V}\subset \Xi $ that $\mathcal {V}=\{\mathbf {v}_0\}$ . Moreover, since $\mathbf {v}_0\in \mathcal {V}_{\Omega ',1}$ for some $\Omega '\in \mathscr {B}_{n+1}$ with $\Omega '\subset \Omega $ , we have $\tfrac 13H_{n+1}^{({1}/{1+\unicode{x3bb} _{\Omega }})}\leq q_0\leq 3H_{n+2}$ . This completes the proof of Case (1).

Case (2). Suppose $k\ge 2$ . We prove that condition (i) holds. It suffices to prove that if $\Omega _j\in \mathscr {B}_{n+k}$ , $\Omega _j\subset \Omega $ , $\mathbf {v}_j\in \mathcal {V}_{\Omega _j,k}$ , $\Delta (\mathbf {v}_j, \epsilon )\cap \Omega \ne \varnothing $ , and $\mathbf {w}_j=\mathbf {w}(\Omega _j,\mathbf {v}_j)$ ( $j=1,2$ ), then $\mathbf {w}_1$ and $\mathbf {w}_2$ are linearly dependent. Let $\mathbf {v}_j=(p_j,r_j,q_j)$ , $\mathbf {w}_j=(a_j,b_j,c_j)$ . Then,

$$ \begin{align*} \max\{|a_j|,|b_j+z_{\Omega_j}a_j|\}&=q_j^{-1}H_{\Omega_j}(\mathbf{v}_j)\le(\tfrac{1}{3}H_{n+k}^{({1}/{1+\unicode{x3bb}})}R^{2k+16})^{-1}\cdot 3H_{n+k+1}\\ &=9H_{n+k}^{({\unicode{x3bb}}/{1+\unicode{x3bb}})}R^{-2k-15}\le H_{n+k}^{({\unicode{x3bb}}/{1+\unicode{x3bb}})}R^{-2k-14}. \end{align*} $$

Moreover, for any $(x,y,z)\in B_\Omega $ , we have

Let

$$ \begin{align*}\mathbf{v}_0=(p_0,r_0,q_0)=\mathbf{w}_1\times \mathbf{w}_2.\end{align*} $$

It suffices to prove that $\mathbf {v}_0=0$ . By the triple cross product expansion, we have

$$ \begin{align*}\mathbf{v}_1\times \mathbf{v}_0=\mathbf{v}_1\times(\mathbf{w}_1\times \mathbf{w}_2)=(\mathbf{v}_1\cdot\mathbf{w}_2)\mathbf{w}_1.\end{align*} $$

Comparing the first two components of the vectors on both sides, we obtain

(7.6) $$ \begin{align}\kern-3.5pt q_0\frac{r_1}{q_1}-r_0&=q_1^{-1}(\mathbf{v}_1\cdot\mathbf{w}_2)a_1, \end{align} $$

(7.7) $$ \begin{align} q_0\frac{p_1}{q_1}-p_0&=-q_1^{-1}(\mathbf{v}_1\cdot\mathbf{w}_2)b_1. \end{align} $$

Suppose in contrast that $\mathbf {v}_0\ne 0$ . There are two cases.

Case (2.1). Suppose $q_0=0$ . Then, $(p_0,r_0)\ne (0,0)$ . It then follows from (7.6) and (7.7) that

$$ \begin{align*} 1&\le\max\{|r_0|,|p_0-z_{\Omega_1}r_0|\}\\ &=q_1^{-1}|\mathbf{v}_1\cdot\mathbf{w}_2|\max\{|a_1|,|b_1+z_{\Omega_1}a_1|\}\\ &\le (\tfrac{1}{3}H_{n+k}^{({1}/{1+\unicode{x3bb}})}R^{2k+16})^{-1}\cdot R^{k+4}\epsilon \cdot H_{n+k}^{({\unicode{x3bb}}/{1+\unicode{x3bb}})}R^{-2k-14}\\ &\le 3 H_{n+k}^{({\unicode{x3bb}-1}/{1+\unicode{x3bb}})}R^{-3k-26}\epsilon<1, \end{align*} $$

which is a contradiction.

Case (2.2). Suppose $q_0\ne 0$ . Without loss of generality, we may assume that $q_0>0$ . Then,

(7.8) $$ \begin{align} q_0&=|a_1b_2-a_2b_1|=|a_1(b_2+za_2)-a_2(b_1+za_1)|\notag\\ &\le 2\cdot H_{n+k}^{({\unicode{x3bb}}/{1+\unicode{x3bb}})}R^{-2k-14}\cdot H_{n+k}^{({1-\unicode{x3bb}}/{1+\unicode{x3bb}})}R^{k+10}= 2H_{n+k}^{({1}/{1+\unicode{x3bb}})}R^{-k-4}. \end{align} $$

It follows that

$$ \begin{align*}q_0/q_1\le2H_{n+k}^{({1}/{1+\unicode{x3bb}})}R^{-k-4}\cdot 3H_{n+k}^{-({1}/{1+\unicode{x3bb}})}R^{-2k-16}=6R^{-3k-20}\le 1/6.\end{align*} $$

We prove that $\Delta (\mathbf {v}_1,\epsilon )\cap \Omega \subset \Delta (\mathbf {v}_0,\epsilon )$ . In view of (7.8) and Lemma 7.1, it suffices to show that

$$ \begin{align*} \Delta_{\unicode{x3bb}}(\mathbf{v}_1,3\epsilon)\cap B_\Omega\subset \Delta_{\unicode{x3bb}}(\mathbf{v}_0,\epsilon/3). \end{align*} $$

Let $( x,y,z)\in \Delta _{\unicode{x3bb} }(\mathbf {v}_1,3\epsilon )\cap B_\Omega $ . It follows from (7.6) and (7.7) that

$$ \begin{align*} &\ q_0^{1+\unicode{x3bb}}\bigg|x-\frac{p_0}{q_0}-z\bigg(y-\frac{r_0}{q_0}\bigg)\bigg|\\ \le &\ q_0^{1+\unicode{x3bb}}\bigg|x-\frac{p_1}{q_1}-z\bigg(y-\frac{r_1}{q_1}\bigg)\bigg| +q_0^{1+\unicode{x3bb}}\bigg|\frac{p_1}{q_1}-\frac{p_0}{q_0}-z\bigg(\frac{r_1}{q_1}-\frac{r_0}{q_0}\bigg)\bigg|\\ < &\ q_0^{1+\unicode{x3bb}}\frac{\epsilon}{q_1^{1+\unicode{x3bb}}}+q_0^{\unicode{x3bb}} q_1^{-1}|b_1+za_1||\mathbf{v}_1\cdot\mathbf{w}_2|\\ \le &\ \epsilon/6+2H_{n+k}^{({\unicode{x3bb}}/{1+\unicode{x3bb}})}\cdot 3H_{n+k}^{-({1}/{1+\unicode{x3bb}})}R^{-2k-16}\cdot H_{n+k}^{({1-\unicode{x3bb}}/{1+\unicode{x3bb}})}R^{k+10}\cdot R^{k+4}\epsilon\\ \le &\ (1/6+6R^{-2})\epsilon<\epsilon/3 \end{align*} $$

and

$$ \begin{align*} q_0^{2-\unicode{x3bb}}\bigg|y-\frac{r_0}{q_0}\bigg| \le &\ q_0^{2-\unicode{x3bb}}\bigg|y-\frac{r_1}{q_1}\bigg|+q_0^{2-\unicode{x3bb}}\bigg|\frac{r_1}{q_1}-\frac{r_0}{q_0}\bigg|\\ < &\ q_0^{2-\unicode{x3bb}}\frac{\epsilon}{q_1^{2-\unicode{x3bb}}}+q_0^{1-\unicode{x3bb}} q_1^{-1}|a_1||\mathbf{v}_1\cdot\mathbf{w}_2|\\ \le &\ \epsilon/6+2H_{n+k}^{({1-\unicode{x3bb}}/{1+\unicode{x3bb}})}\cdot 3H_{n+k}^{-({1}/{1+\unicode{x3bb}})}R^{-2k-16}\cdot H_{n+k}^{({\unicode{x3bb}}/{1+\unicode{x3bb}})}R^{-2k-14}\cdot R^{k+4}\epsilon\\ \le &\ (1/6+ 6R^{-26})\epsilon<\epsilon/3. \end{align*} $$

Thus, $(x,y,z)\in \Delta _\unicode{x3bb} (\mathbf {v}_0,\epsilon /3)$ . This proves $\Delta (\mathbf {v}_1,\epsilon )\cap \Omega \subset \Delta (\mathbf {v}_0,\epsilon )$ . It then follows from $\Delta (\mathbf {v}_1,\epsilon )\cap \Omega \ne \varnothing $ that $\Delta (\mathbf {v}_0,\epsilon )\cap \Omega \ne \varnothing $ . By Lemma 5.4, we have $q_0^{1+\unicode{x3bb} }>3H_{n+1}$ . However, in view of (7.8), we have

$$ \begin{align*}q_0^{1+\unicode{x3bb}}\le 2^{1+\unicode{x3bb}}H_{n+k}R^{-(1+\unicode{x3bb})(k+4)}\le 4R^{-4}H_{n+1}\le 3H_{n+1},\end{align*} $$

which leads to a contradiction. This completes the proof of the lemma.

Proof of Lemma 5.5

Let $n\ge 0$ , $\Omega \in \mathscr {B}^{\prime }_n$ , $k\ge 1$ . We need to prove that there is a hyperplane $L_k(\Omega )$ such that for any $\Omega '\in \mathscr {B}_{n+k}$ with $\Omega '\subset \Omega $ and any $\mathbf {v}\in \mathcal {V}_{\Omega ',k}$ , we have $\Delta (\mathbf {v},\epsilon )\cap \Omega '\subset L_k(\Omega )^{(R^{-(n+k)}\rho _0)}$ . We need only to consider the case that ${\Delta (\mathbf {v},\epsilon )\cap \Omega '\ne \varnothing }$ . By Lemma 7.3, one of the following statements holds:

1. (i) there is a plane $L_k(\Omega )$ such that for any $\Omega '\in \mathscr {B}_{n+k}$ with $\Omega '\subset \Omega $ , if $\mathbf {v}\in \mathcal {V}_{\Omega ',k}$ and $\Delta (\mathbf {v},\epsilon )\cap \Omega \ne \varnothing $ , then $L(\Omega ',\mathbf {v})=L_k(\Omega )$ ;

2. (ii) $k=1$ and there is $\mathbf {v}_0=(p_0,r_0,q_0)\in \Xi $ with $\tfrac 13H_{n+1}^{({1}/{1+\unicode{x3bb} _{\Omega }})}\leq q_0\leq 3H_{n+2}$ such that for any $\Omega '\in \mathscr {B}_{n+1}$ with $\Omega '\subset \Omega $ , if $\mathbf {v}\in \mathcal {V}_{\Omega ',1}$ and $\Delta (\mathbf {v},\epsilon )\cap \Omega \neq \varnothing $ , then $\mathbf {v}=\mathbf {v}_0$ .

Suppose that statement (i) holds. Let us fix such a $\Omega '$ and denote $\unicode{x3bb} _{\Omega '}=\unicode{x3bb} ',\;\rho _{\Omega '}=\rho '$ . By Lemma 7.1, it suffices to show that

(7.9) $$ \begin{align} (I_{\Omega'}\times\Delta_{\unicode{x3bb}'}(\mathbf{v}, 3\epsilon))\cap \Omega'\subset L(\Omega',\mathbf{v})^{(R^{-(n+k)}\rho_0)}\quad \text{ for any } \mathbf{v}=(p,r,q)\in\mathcal{V}_{\Omega'}. \end{align} $$

First, we have

$$ \begin{align*} \rho q\le R^{-(n+k)}\rho_0\cdot3H_{n+k+1}=60\epsilon \kappa R\le 1/4. \end{align*} $$

Denote $\mathbf {w}(\Omega ',\mathbf {v})=(a,b,c)$ . If $(x,y,z)\in \Delta _{\unicode{x3bb} '}(\mathbf {v}, 3\epsilon )\cap B_{\Omega '}$ , then

$$ \begin{align*} |ax+by+c|&=\bigg|a\bigg(x-\frac{p}{q}-z\bigg(y-\frac{r}{q}\bigg)\bigg)+(b+za)\bigg(y-\frac{r}{q}\bigg)\bigg|\\ &< |a|\frac{3\epsilon}{q^{1+\unicode{x3bb}'}}+|b+za|\frac{3\epsilon}{q^{2-\unicode{x3bb}'}}\\ &\le 3\epsilon q^{-1}(|a|q^{-\unicode{x3bb}'}+|b+z_{\Omega'} a|q^{-1+\unicode{x3bb}'}+|z-z_{\Omega'}||a|q^{-1+\unicode{x3bb}'})\\ &\le 3\epsilon q^{-1}(3+3+\rho' q^{-1+2\unicode{x3bb}'})\\ &\le 3\epsilon q^{-1}(6+\rho' q)\\ &\le 20\epsilon q^{-1}. \end{align*} $$

Note that

$$ \begin{align*}H_{\Omega'}(\mathbf{v})=q\max\{|a|,|b+z_{\Omega'}a|\}\le\kappa q\max\{|a|,|b|\}. \end{align*} $$

Thus,

$$ \begin{align*} \frac{|ax+by+c|}{\sqrt{a^2+b^2}}<\frac{20\epsilon}{q\max\{|a|,|b|\}}\le\frac{20\epsilon \kappa }{H_{\Omega'}(\mathbf{v})}\le\frac{20\epsilon \kappa }{H_{n+k}}=R^{-(n+k)}\rho_0. \end{align*} $$

This means that $I_{\Omega '}\times \{(x,y,z)\}\subset L(\Omega ',\mathbf {v})^{(R^{-(n+k)}\rho _0)}$ . This proves (7.9).

Suppose that statement (ii) holds. Let us write $\unicode{x3bb} _{\Omega }=\unicode{x3bb} ,\;\rho _{\Omega }=\rho $ . Consider the hyperplane

$$ \begin{align*} L_1(\Omega)=\bigg\{(\unicode{x3bb},x,y,z)\in \Sigma: x-\frac{p_0}{q_0}-z_{\Omega}\bigg(y-\frac{r_0}{q_0}\bigg)=0\bigg\}. \end{align*} $$

We proceed by showing that $\Delta (\mathbf {v}_0,\epsilon )\cap \Omega \subset L_1(\Omega )^{(R^{-n+1}\rho _0)}$ . By Lemma 7.1, it suffices to show that

(7.10) $$ \begin{align} I_{\Omega}\times \Delta_{\unicode{x3bb}}(\mathbf{v}_0,3\epsilon)\subset L_1(\Omega)^{(R^{-(n+1)\rho_0})}. \end{align} $$

If $(x,y,z)\in \Delta _{\unicode{x3bb} }(\mathbf {v}_0,3\epsilon )\cap B_{\Omega }$ , then

$$ \begin{align*} (1+z_{\Omega}^2)^{-1/2}\bigg|x-\frac{p_0}{q_0}-z_{\Omega}\bigg(y-\frac{r_0}{q_0}\bigg)\bigg| &\leq\bigg|x-\frac{p_0}{q_0}-z\bigg(y-\frac{r_0}{q_0}\bigg)\bigg|+|z-z_{\Omega}|\bigg|y-\frac{r_0}{q_0}\bigg|\\ &< \frac{3\epsilon}{q_0^{1+\unicode{x3bb}}}+\rho\frac{3\epsilon}{q_0^{2-\unicode{x3bb}}}\\ &\le 3\epsilon q_0^{-(1+\unicode{x3bb})}(1+\rho q_0)\\ &\le 27\epsilon H_{n+1}^{-1}(1+R^{-n}\rho_0\cdot 3H_{n+2})\\ &=27(20\kappa)^{-1}R^{-(n+1)}\rho_0(1+60\epsilon\kappa R^2)\\ &< R^{-(n+1)}\rho_0. \end{align*} $$

This means that $I_{\Omega }\times \{(x,y,z)\}\subset L_1(\Omega )^{(R^{-(n+1)\rho _0})}.$ This proves (7.10).