#  Characterizations of Urysohn universal ultrametric spaces

Yoshito Ishiki  Photonics Control Technology Team RIKEN Center for Advanced
Photonics 2-1 Hirasawa, Wako, Saitama 351-0198, Japan<EMAIL_ADDRESS>
###### Abstract.

In this paper, using the existence of equidistant subsets in closed balls, we
characterize the injectivity of ultrametric spaces for finite ultrametric
spaces, which also characterize Urysohn universal ultrametric spaces. As
applications, we first observe that the countable power of the countable
discrete space equipped with a canonical metric (the Baire space) is the
Urysohn universal ultrametric space. Next we prove that the hyperspace
consisting of all non-empty compact subsets of the (standard) Urysohn
universal ultrametric spaces $(X,d)$ is isometric to $(X,d)$. We also
establish that all complete ultrametric spaces injective for all finite
ultrametric space contain the Urysohn universal ultrametric space.

###### Key words and phrases:

Urysohn universal ultrametric spaces

###### 2020 Mathematics Subject Classification:

Primary 54E35, Secondary 51F99

##  1\. Introduction

For a class $\mathscr{C}$ of metric spaces, we say that a metric space $(X,d)$
is $\mathscr{C}$-injective if for all pair $(Y,e)$ and $(Z,h)$ of metric
spaces in $\mathscr{C}$ and isometric embeddings $\phi\colon Y\to Z$ and
$f\colon Y\to X$, there exists an isometric embedding $F\colon Z\to X$ such
that $F\circ\phi=f$. We denote by $\mathscr{F}$ the class of all finite metric
spaces. Urysohn constructed a separable complete $\mathscr{F}$-injective
metric space and proved its uniqueness up to isometry, which space is today
called the Urysohn universal (metric) space. Let $(\mathbb{U},\rho)$ denote
the Urysohn universal metric space. Our main subjects of this paper are non-
Archimedean analogues of $(\mathbb{U},\rho)$. For more information of
$(\mathbb{U},\rho)$ and related topics, we refer the readers to, for instance,
[1] [10], [4], [7], [8], [14], [11] and [12].

To explain our main results and backgrounds, we prepare some notations and
notions. We say that a set $R$ is a _range set_ if it is a subset of
$[0,\infty)$ and contains $0$. A metric on a set $X$ is said to be an
_ultrametric_ if it satisfies the so-called strong triangle inequality
$d(x,y)\leq d(x,z)\lor d(z,y)$ for all $x,y,z\in X$, where $\lor$ stands for
the maximal operator on $\mathbb{R}$. For a range set $R$, an ultrametric
space $(X,d)$ is $R$-valued if $d(x,y)\in R$ for all $x,y\in R$. For a range
set $R$, we denote by $\mathscr{N}(R)$ the class of all finite $R$-valued
ultrametric spaces.

For a countable range set $R$, we say that an ultrametric $(X,d)$ is the _$R$
-Urysohn universal ultrametric space_ if it is a separable complete
$\mathscr{N}(R)$-injective $R$-valued ultrametric space. As is the case of
$(\mathbb{U},\rho)$, we can prove the uniqueness of the $R$-Urysohn universal
ultrametric space up to isometry (see [3]). Since for an uncountable range set
$R$ all $\mathscr{N}(R)$-injective ultrametric space are non-separable,
mathematicians often treated only the case where $R$ is countable to guarantee
the separability. In [6], for an uncountable range set $R$, the author
introduced the $R$-petaloid ultrametric space and proved its uniqueness up to
isometry. For the definition of the petaloid spaces, see Section 2. Based on
this result, even when $R$ is uncountable, we also say that $(X,d)$ is the
$R$-Urysohn universal ultrametric space if it is $R$-petaloid ultrametric
space. Namely, in this paper, we define the $R$-Urysohn universal ultrametric
space as follows:

######  Definition 1.1.

Let $R$ be a range set. When $R$ is countable, the _$R$ -Urysohn universal
ultrametric space_ is a (unique) separable complete $\mathscr{N}(R)$-injective
$R$-valued ultrametric space. When $R$ is uncountable, the _$R$ -Urysohn
universal ultrametric space_ is the $R$-petaloid ultrametric space.

We now explain our main results. In this paper, we first characterize the
$\mathscr{N}(R)$-injectivity using the existence of equidistant sets in closed
balls (see Theorem 3.2). We next explain applications of Theorem. We denote by
$\omega_{0}$ the set of all non-negative integers. Remark that
$\omega_{0}=\mathbb{Z}_{\geq 0}$ as a set. We also denote by
${}^{\omega_{0}}\omega_{0}$ the set of all maps from $\omega_{0}$ into
$\omega_{0}$ (in this paper, we often use set-theorists’ notations). Using a
strictly decreasing sequence convergent to $0$, we can define an ultrametric
on ${}^{\omega_{0}}\omega_{0}$ as follows:

######  Definition 1.2.

Let $\alpha\colon\omega_{0}\to(0,\infty)$ be a map such that
$\lim_{n\to\infty}\alpha(n)$ and $\alpha(n+1)<\alpha(n)$ for all
$n\in\omega_{0}$. We define an ultrametric $\sigma_{\alpha}$ by
$\sigma_{\alpha}(x,y)=\alpha(v(x,y))$, where
$v(x,y)=\min\\{\,n\in\omega_{0}\mid x(n)\neq y(n)\,\\}$.

Remark that $d_{\alpha}$ certainly generates the product topology on
${}^{\omega_{0}}\omega_{0}$. This metric space is known as the
($0$-dimensional) Baire space and the case of $\alpha(n)=(n+1)^{-1}$ is
famous. Using Theorem 3.2, we observe that the space
$({}^{\omega_{0}}\omega_{0},\sigma_{\alpha})$ is $\mathscr{N}(R)$-injective,
where $R=\\{0\\}\cup\\{\,\alpha(n)\mid n\in\omega_{0}\,\\}$. Namely,
$({}^{\omega_{0}}\omega_{0},\sigma_{\alpha})$ is the $R$-Urysohn universal
ultrametric space (see Theorem 3.3).

For a metric space $(X,d)$, we denote by $\mathscr{K}(X)$ the set of all non-
empty compact subsets of $X$. We also denote by $\mathcal{HD}_{d}$ the
Hausdorff distance on $\mathscr{K}(X)$ induced from $d$ (for the precise
definition, see Section 4). As a main result, we show that for every range set
$R$, and for the $R$-Urysohn universal ultrametric space $(W,h)$, the
hyperspace $(\mathscr{K}(W),\mathcal{HD}_{h})$ is also the $R$-Urysohn
universal ultrametric space (see Theorem 4.5).

We also establish that for every range set $R$, every complete
$\mathscr{N}(R)$-injective ultrametric space contains a subspace isometric to
the $R$-Urysohn universal ultrametric space (see Theorem 5.1). This result
confirms the validity of the petaloid spaces.

The organization of this paper is as follows: In Section 2, we prepare basic
notations and notions. We shall explain the petaloid ultrametric space. In
Section 3, the we characterize the $\mathscr{N}(R)$-injectivity using the
existence of an equidistant subsets of closed balls (see Theorem 3.2). We also
prove Theorem 3.3. In Section 4, we prove that the hyper space of compact
subsets of the $R$-Urysohn universal ultrametric space $(X,d)$ is isometric to
$(X,d)$ (see Theorem 4.5). Section 5 is devoted to proving that all complete
$\mathscr{N}(R)$-injective ultrametric spaces contain the $R$-Urysohn
universal ultrametric space (see Theorem 5.1).

##  2\. Preliminaries

$\kappa\in\omega_{0}$ or $\kappa=\omega_{0}$. For a metric space $(X,d)$, for
$a\in X$, and for $r\in(0,\infty)$, we denote by $B(a,r)$ the closed ball
centered at $a$ of radius $r$.

The proofs of next three lemmas are presented in [13, Propositions 18.2, 18.4
and 18.5].

######  Lemma 2.1.

Let $(X,d)$ be an ultrametric space. Then the following are true:

  1. (1)

for all $x,y,z\in X$, the inequality $d(x,z)<d(z,y)$ implies $d(z,y)=d(x,y)$.

  2. (2)

For all $a\in X$, $r\in(0,\infty)$, and for all $q\in B(a,r)$ we have
$B(a,r)=B(q,r)$.

  3. (3)

For all $a,b\in X$, and for $r,l\in(0,\infty)$, if $B(a,r)\cap
B(b,l)\neq\emptyset$, then we have $B(a,r)\subseteq B(b,l)$ or
$B(b,r)\subseteq B(a,r)$.

A subset $E$ of $[0,\infty)$ is said to be _semi-sporadic_ if there exists a
strictly decreasing sequence $\\{a_{i}\\}_{i\in\mathbb{Z}_{\geq 0}}$ in
$(0,\infty)$ such that $\lim_{i\to\infty}a_{i}=0$ and
$E=\\{0\\}\cup\\{\,a_{i}\mid i\in\mathbb{Z}_{\geq 0}\,\\}$. A subset of
$[0,\infty)$ is said to be _tenuous_ if it is finite or semi-sporadic (see
[5]). For a range set $R$, we denote by $\mathbf{TEN}(R)$ the set of all
tenuous range subsets of $R$. In this paper, we often represent a restricted
metric as the same symbol to the ambient one. We now provide the definition of
the petaloid ultrametric space that is introduced in [6] by the author.

######  Definition 2.1.

Let $R$ be an uncountable range set. We say that a metric space $(X,d)$ is
_$R$ -petaloid_ if it is an $R$-valued ultrametric space and there exists a
family $\\{\Pi(X,S)\\}_{S\in\mathbf{TEN}(R)}$ of subspaces of $X$ satisfying
the following properties:

  1. (P1)

The set $\Pi(X,\\{0\\})$ is a singleton.

  2. (P2)

If $S\in\mathbf{TEN}(R)$ satisfies $S\neq\\{0\\}$, then $(\Pi(X,S),d)$ is a
separable complete $\mathscr{N}(S)$-injective $S$-ultrametric space.

  3. (P3)

We have $\bigcup_{S\in\mathbf{TEN}(R)}\Pi(X,S)=X$.

  4. (P4)

If $S,T\in\mathbf{TEN}(R)$, then $\Pi(X,S)\cap\Pi(X,T)=\Pi(X,S\cap T)$.

  5. (P5)

If $S,T\in\mathbf{TEN}(R)$ and $x\in\Pi(X,T)$, then $d(x,\Pi(X,S))$ belongs to
$(T\setminus S)\cup\\{0\\}$.

We call the family $\\{\Pi(X,S)\\}_{S\in\mathbf{TEN}(R)}$ an _$R$ -petal of
$X$_, and call $\Pi(X,S)$ the $S$-piece of the $R$-petal
$\\{\Pi(X,S)\\}_{S\in\mathbf{TEN}(R)}$. We simply write $\Pi(S)=\Pi(X,S)$ when
the whole space is clear by the context.

######  Lemma 2.2.

Let $R$ be a range set, and $(X,d)$ be the $R$-petaloid ultrametric space.
Then for all $S\in\mathbf{TEN}(R)$, and for all $x\in X$, there exists
$p\in\Pi(X,S)$ such that $d(x,\Pi(X,S))=d(x,p)$.

###### Proof.

The lemma is reduced from the property that $\Pi(X,S)$ is the $S$-Urysohn
universal ultrametric space (the property (P2)) and [13, Propositions 20.2 and
21.1]. ∎

######  Remark 2.1.

In the preprint of [6], as the property (P6), the author assumed the statement
of Lemma 2.2. However, as shown above, it is reduced from the property (P2).

##  3\. Characterizations

We write $n<\omega_{0}$ if $n\in\omega_{0}$. Thus, $\kappa\leq\omega_{0}$
means that

######  Definition 3.1.

For a cardinal $\kappa$, and for a range set $R$, we denote by
$\mathscr{N}(R,\kappa)$ the class of all $R$-valued ultrametric spaces $(X,d)$
such that $\operatorname{Card}(X)<\kappa$. Remark that in the case of
$\kappa=\omega_{0}$, then $\mathscr{N}(R,\kappa)$ coincides with the class
$\mathscr{N}(R)$.

For a metric space $(X,d)$, and $r\in(0,\infty)$, a subset $A$ of $X$ is said
to be _$r$ -equidistant_ if for all distinct $x,y\in A$ we have $d(x,y)=r$.

######  Definition 3.2.

Let $\kappa\leq\omega_{0}$ and $R$ be a range set. An $R$-valued metric space
$(X,d)$ is said to be _$(R,\kappa)$ -haloed_ if for all $a\in X$ and $r\in
R\setminus\\{0\\}$, there exists an $r$-equidistant subset $A$ of $B(a,r)$
such that $\kappa\leq\operatorname{Card}(A)$.

######  Definition 3.3.

Let $\kappa\leq\omega_{0}$ and $R$ be a range set. An $R$-valued ultrametric
space $(X,d)$ is _$(R,\kappa)$ -avoidant_ if for all $a\in A$, and $r\in
R\setminus\\{0\\}$, and for all subset $A$ of $B(a,r)$ with
$\operatorname{Card}(A)<\kappa$, there exists $p\in B(a,r)$ such that
$d(x,p)=r$ for all $x\in A$.

By the definitions, we obtain the next equivalences:

######  Lemma 3.1.

Let $(X,d)$ be an ultrametric space and $R$ be a range set. Then the following
statements are true.

  1. (1)

The ultrametric space $(X,d)$ is $(R,\omega_{0})$-haloed if and only if
$(X,d)$ is $(R,n)$-haloed for all $n\in\omega_{0}$.

  2. (2)

The ultrametric space $(X,d)$ is $(R,\omega_{0})$-avoidant if and only if
$(X,d)$ is $(R,n)$-avoidant for all $n\in\omega_{0}$.

  3. (3)

The ultrametric space $(X,d)$ is $\mathscr{N}(R,\omega_{0})$-injective if and
only if $(X,d)$ is $\mathscr{N}(R,n)$-injective for all $n\in\omega_{0}$.

######  Theorem 3.2.

Let $n\in\omega_{0}$ and $R$ be a range set. Let $(X,d)$ be an $R$-valued
ultrametric space. Then the following statements are equivalent to each other.

  1. (A1)

The space $(X,d)$ is $(R,n)$-haloed;

  2. (A2)

The space $(X,d)$ is $(R,n)$-avoidant

  3. (A3)

The space $(X,d)$ is $\mathscr{N}(R,n+1)$-injective.

As a consequence, we also observe that the following statements are
equivalent:

  1. (B1)

The space $(X,d)$ is $(R,\omega_{0})$-haloed;

  2. (B2)

The space $(X,d)$ is $(R,\omega_{0})$-avoidant

  3. (B3)

The space $(X,d)$ is $\mathscr{N}(R,\omega_{0})$-injective.

###### Proof.

Let $n\in\omega_{0}$ and $R$ be a range set. Let $(X,d)$ be an ultrametric
space. As a strategy, we prove
$\ref{item:18:main11}\Longrightarrow\ref{item:18:main12}$,
$\ref{item:18:main12}\Longrightarrow\ref{item:18:main13}$, and
$\ref{item:18:main13}\Longrightarrow\ref{item:18:main11}$.

First assume that $(X,d)$ is $(R,\kappa)$-haloed. To show that $(X,d)$ is
$(R,\kappa)$-avoidant, take $a\in X$, $r\in R$ and take a subset $A$ of
$B(a,r)$ with $\operatorname{Card}(A)<\kappa$. Since $(X,d)$ is
$(R,\kappa)$-haloed, we can find a subset $H$ of $B(a,r)$ such that
$\operatorname{Card}(H)=\kappa$ and $d(x,y)=r$ for all $x,y\in H$. Since $d$
is an ultrametric, and since $\operatorname{Card}(A)<\operatorname{Card}(H)$,
there exists $p\in H$ such that $p\not\in U(x,r)$ for all $x\in A$. Then the
point $p$ satisfies $d(x,p)=r$ for all $x\in A$.

Next assume that $(X,d)$ is $(R,\kappa)$-avoidant. Take an $R$-valued
ultrametric space $(Y\sqcup\\{\theta\\},e)$ with
$\operatorname{Card}(Y\sqcup\\{\theta\\})<n+1$, and take an isometric map
$\phi\colon Y\to X$. To prove the $\mathscr{N}(R,n+1)$-injectivity, we need to
prove that there exists an isometry $\Phi\colon Y\sqcup\\{\theta\\}\to X$. Put
$r=\min_{y\in Y}e(y,\theta)$ and take $q\in Y$ such that $r=e(q,\theta)$. In
this situation, we have $r\in R$. Put $A=\phi(Y)\cap B(\phi(q),r)$. Since
$(X,d)$ is $(R,n+1)$-avoidant and since $\operatorname{Card}(A)<n$, there
exists $t\in B(\phi(q),r)$ such that $d(a,t)=r$ for all $a\in A$. We now prove
that $d(\phi(y),t)=e(y,\theta)$ for all $y\in Y$. In what follows, we use (1)
in Lemma 2.1 without citation. If $e(y,q)\leq r$, then by the minimality of
$r$, we have $e(y,\theta)=r$. By the definition of $t$, we have
$d(\phi(y),t)=r$. Thus $d(\phi(y),t)=e(y,\theta)$. If $r<e(y,q)$, then we have
$e(y,\theta)=e(y,q)$. We also have $\phi(y)\not\in B(\phi(q),r)=B(t,r)$ (see
(2) in Lemma 2.1). Thus we obtain $r<d(\phi(y),t)$. Hence, using
$r=d(\phi(q),t)$, we also obtain
$d(\phi(y),t)=d(\phi(y),\phi(q))=e(y,q)=e(y,\theta)$. Namely,
$d(\phi(y),t)=e(y,\theta)$ We define an isometry $\Phi\colon
Y\sqcup\\{\theta\\}\to X$ by $\Phi(x)=x$ for all $x\in Y$ and
$\Phi(\theta)=t$. Therefore the space $(X,d)$ is
$\mathscr{N}(R,\kappa)$-injective.

We now prove the implication
$\ref{item:18:main13}\Longrightarrow\ref{item:18:main11}$. Take $a\in X$, and
$r\in R\setminus\\{0\\}$. By induction, we construct $\\{A_{i}\\}_{i<\kappa}$
such that

  1. (1)

we have $A_{i}\subseteq B(a,r)$ for all $i<n+1$;

  2. (2)

for all $i<n+1$, we have $\operatorname{Card}(A_{i})=i+1$;

  3. (3)

for all $i<n$, we have $A_{i}\subseteq A_{i+1}$.

Take $s\in B(a,r)$ and put $A_{0}=\\{s\\}$. Fix $l<n+1$ and we assume that we
have already constructed $\\{A_{i}\\}_{i<l}$. Define an ultrametric $e$ on
$A\sqcup\\{\theta\\}$ by $e|_{A^{2}}=d$ and $d(x,\theta)=r$ for all $x\in A$.
We also define an isometric map $\phi\colon A\to X$ by $\phi(a)=a$. Using the
$\mathscr{N}(R,\kappa)$-injectivity, we obtain $p\in B(a,r)$ such that
$d(\phi(x),p)=e(x,\theta)=r$. Put $A_{l}=A_{l-1}\sqcup\\{p\\}$. Then $A_{n}$
is $r$-equidistant, and hence $(X,d)$ is $(R,\kappa)$-haloed.

The latter part can be deduced from the former part and Lemma 3.1. This
finishes the proof. ∎

######  Theorem 3.3.

Let $\alpha$ satisfies that $\lim_{n\to\infty}\alpha(n)=0$ and
$\alpha(n+1)<\alpha(n)$ for all $n\in\omega_{0}$ and put
$R=\\{0\\}\cup\\{\,\alpha(n)\mid n\in\omega_{0}\\}$. Then the space
$({}^{\omega_{0}}\omega_{0},\sigma_{\alpha})$ is
$\mathscr{N}(R,\kappa)$-injective. Namely, it is isometric to the $R$-Urysohn
universal ultrametric space.

###### Proof.

Since $R$ is countable, according to Theorem 3.2, it suffices to show that the
space $({}^{\omega_{0}}\omega_{0},\sigma_{\alpha})$ is
$(R,\omega_{0})$-haloed. Take
$a=(a(i))_{i\in\omega_{0}}\in{}^{\omega_{0}}\omega_{0}$ and $k\in\omega_{0}$.
For each $n\in\omega_{0}$, we define $b(n)=(b(n,i))_{i\in\omega_{0}}$ by

$b(n,i)=\begin{cases}a(i)&\text{$k\leq i$;}\\\ n&\text{otherwise.}\end{cases}$

Then we have $b(n)\in B(a,r_{k})$ and $d(b(n),b(m))=r_{k}$ for all distinct
$n,m\in\omega_{0}$. Therefore we conclude that the space is
$(R,\omega_{0})$-haloed. ∎

##  4\. Hyperspaces

For a metric space $(X,d)$, and for a subset $A$ of $X$, we write
$d(x,A)=\inf_{a\in A}d(x,a)$. For a metric space $(X,d)$, we denote by
$\mathscr{K}(X)$ the set of all non-empty compact subsets of $X$. We also
denote by $\mathcal{HD}_{d}$ the Hausdorff distance on $\mathscr{K}(X)$
induced from $d$. More precisely, we define
$\mathcal{HD}_{d}(A,B)=\max\\{\sup_{a\in A}d(a,B),\sup_{b\in B}d(b,A)\\}$.

Let $(X,d)$ be an ultrametric space, and $r\in(0,\infty)$. For
$A\in\mathscr{K}(X)$ we denote by $\langle A\rangle_{\mathfrak{c}(r)}$ the set
$\\{\,B(a,r)\mid a\in A\,\\}$. Remark that it is can happen that
$B(a,r)\in\langle A\rangle_{\mathfrak{c}(r)}$ contains some elements of
$X\setminus A$. By (3) in Lemma 2.1 and the compactness of $A$, the set
$\langle A\rangle_{\mathfrak{c}(r)}$ is finite. The next theorem is an
analogue of [9, Theorem 5.1] and [5, Corollary 2.30].

######  Proposition 4.1.

Let $(X,d)$ be an ultrametric space. Then for all subsets $A$ and $B$ of $X$,
the value $\mathcal{HD}_{d}(A,B)$ is equal to the minimum $r\in R$ such that
$\langle A\rangle_{\mathfrak{c}(r)}=\langle B\rangle_{\mathfrak{c}(r)}$.

###### Proof.

Let $u$ be the infimum of $r\in R$ such that $\langle
A\rangle_{\mathfrak{c}(r)}=\langle B\rangle_{\mathfrak{c}(r)}$. Put
$S=d(A^{2})\cup d(B^{2})$. Then $S$ is tenuous (see [5, Corollary 2.28]). If
$u\not\in S$, then we can take $a,b\in S$ such that $u\in(a,b)$ and $(a,b)\cap
S=\emptyset$. In this setting, we have $B(x,r)=B(x,a)$ for all $x\in A\cup B$
and for all $r\in(u,b)$. Thus, we have $\langle
A\rangle_{\mathfrak{c}(a)}=\langle B\rangle_{\mathfrak{c}(a)}$, which is a
contradiction. Therefore $u\in S$, and hence $u\in R$. We also notice that $u$
is the minimum. By the definition of the Hausdorff distance, we have
$\mathcal{HD}_{d}(A,B)\leq u$. To obtain the opposite inequality, take
arbitrary $l$ with $\mathcal{HD}_{d}(A,B)<l$. Then we see that $\sup_{a\in
A}d(a,B)\leq l$ and $\sup_{b\in B}d(b,A)\leq l$. Then $\langle
A\rangle_{\mathfrak{c}(l)}=\langle B\rangle_{\mathfrak{c}(l)}$. Hence
$u\leq\mathcal{HD}_{d}(A,B)$. This completes the proof. ∎

######  Corollary 4.2.

Let $R$ be a range set and $(X,d)$ be an ultrametric. The the following two
statements are true:

  1. (1)

The metric $\mathcal{HD}_{d}$ is an ultrametric on $\mathscr{K}(X)$.

  2. (2)

If $(X,d)$ is $R$-valued, then so is $(\mathscr{K}(X),\mathcal{HD}_{d})$.

######  Proposition 4.3.

Let $n\in\omega_{0}$, $R$ be a range set, and $(X,d)$ be an $(R,n)$-haloed
ultrametric space. Then the space $(\mathscr{K}(X),\mathcal{HD}_{d})$ is
$(R,2^{n}-1)$-haloed.

###### Proof.

Put $M_{n}=\\{0,\dots,n-1\\}$ and define an ultrametric $e$ on $M_{n}$ by
$e(a,b)=r$ for all distinct $a,b\in M_{n}$. First we see that
$\operatorname{Card}(\mathscr{K}(M_{n}))=2^{n}-1$. Hence we only need to prove
that $(\mathscr{K}(M_{n}),\mathcal{HD}_{e})$ is $r$-equidistant. Take
$A,B\in\mathscr{K}(M_{n})$ with $A\neq B$. It is clear that $d(A,B)\leq r$. We
may assume that there exists $k\in A$ with $k\not\in B$. Then for all $l<r$,
we have $k\not\in\bigcup_{b\in B}B(b,l)$. Thus Proposition 4.1 implies that
that $d(A,B)=r$. ∎

######  Corollary 4.4.

If a range set $R$ is at most countable and $(X,d)$ is the $R$-Urysohn
universal ultrametric space, then $(\mathscr{K}(X),\mathcal{HD}_{d})$ is the
$R$-Urysohn universal ultrametric space. Namely, it is isometric to $(X,d)$.

###### Proof.

Due to Corollary 4.2 and Proposition 4.3, the space
$(\mathscr{K}(X),\mathcal{HD}_{d})$ is an $R$-valued
$\mathscr{N}(R,\omega_{0})$-injective ultrametric space. Similarly to [2,
Proposition 7.3.7], we see that $(\mathscr{K}(X),\mathcal{HD}_{d})$ is
complete. To prove the separability, take a countable dense subset $A$ of $X$.
Then the set of all non-empty subset of $A$ is dense in
$(\mathscr{K}(X),\mathcal{HD}_{d})$. Thus, it is separable. Therefore
$(\mathscr{K}(X),\mathcal{HD}_{d})$ is the $R$-Urysohn universal ultrametric
space and isometric to $(X,d)$. ∎

For a metric space $(X,d)$ and for $r\in(0,\infty)$, a subset of $A$ is called
an _$r$ -net_ if for all distinct $x,y\in A$ we have $r<d(x,y)$. An $r$-net is
_maximal_ if it is maximal with respect to the inclusion $\subseteq$.

######  Theorem 4.5.

Let $R$ be a range set, and let $(X,d)$ be the $R$-Urysohn universal
ultrametric space. Then the space $(\mathscr{K}(X),\mathcal{HD}_{d})$ is the
$R$-Urysohn universal ultrametric space and isometric to $(X,d)$.

###### Proof.

Corollary 4.2 implies that $(\mathscr{K}(X),\mathcal{HD}_{d})$ is $R$-valued.
The case where $R$ is at most countable is proven in Corollary 4.4.

We now treat the case where $R$ is uncountable. Remark that in this case,
$(X,d)$ is the $R$-petaloid space (see Section 1). We define a petal on
$(\mathscr{K}(X),\mathcal{HD}_{d})$ as follows: For every
$S\in\mathbf{TEN}(R)$, we define

$\Pi(\mathscr{K}(X),S)=\\{\,A\in\mathscr{K}(X)\mid A\subseteq\Pi(X,S)\,\\}.$

By this definition, the properties (P1) and (P4) are satisfied.

Since each $\Pi(X,S)$ is the $S$-Urysohn universal ultrametric space and $S$
is at most countable, according to Corollary 4.4, the property (P2) is true.
Due to [6, Proposition 2.8.], the property (P3) is satisfied.

We now show that the property (P5) is true for
$(\mathscr{K}(X),\mathcal{HD}_{d})$. Take $S,T\in\mathbf{TEN}(R)$ with $S\neq
T$ and $B\in\Pi(\mathscr{K}(X),T)$. Let $L$ stand for the distance between $B$
and $\Pi(\mathscr{K}(X),S)$. We shall show that $L$ belongs to
$\\{0\\}\cup(T\setminus S)$. We may assume that
$B\not\in\Pi(\mathscr{K}(X),S)$. Then $B\subseteq\Pi(X,T)$ and
$B\not\subseteq\Pi(X,S)$. We put $H=\\{\,d(y,\Pi(X,S))\mid y\in B\,\\}$. Due
to (P5) for $(X,d)$, we see that $H\subseteq\\{0\\}\cup(T\setminus S)$. Put
$h=\max H$ and take $z\in B$ such that $d(z,\Pi(X,S))=h$. The existence of $h$
is guaranteed by the fact that $H$ is a subset of the tenuous set $T$. Note
that $h\in T\setminus S$. We also take a maximal finite $h$-net $Q$ of $B$
with $z\in Q$. For all $p\in Q$, take $v(p)\in\Pi(X,S)$ with
$d(p,v(p))=d(p,\Pi(X,S))$ (see Lemma 2.2). Put $V=\\{\,v(p)\mid p\in Q\,\\}$.
Then $\mathcal{HD}_{d}(B,V)\leq h$. For the sake of contradiction, suppose
that there exists $F\in\Pi(X,S)$ with $\mathcal{HD}_{d}(B,F)<h$. Put
$l=\mathcal{HD}_{d}(B,F)$. Then Proposition 4.1 implies that $\langle
B\rangle_{\mathfrak{c}(l)}=\langle F\rangle_{\mathfrak{c}(l)}$. Hence there
exists $f\in F$ such that $d(z,f)\leq l<h$ (see (2) in Lemma (2)). This is a
contradiction to the fact that $h=\inf_{x\in\Pi(X,S)}d(z,x)$. Thus, for all
$F\in\Pi(X,S)$, we have $h\leq\mathcal{HD}_{d}(B,F)$ and
$\mathcal{HD}_{d}(B,V)=h$. This means that $L=h$ and $h\in T\setminus S$. This
finishes the proof. ∎

##  5\. Existence theorem

This section is devotes to the proof of the following theorem:

######  Theorem 5.1.

Let $R$ be a range set. Let $(X,d)$ be a complete $R$-valued
$\mathscr{N}(R,\omega_{0})$-injective ultrametric space. Then $(X,d)$ contains
a subspace $F$ isometric to the $R$-Urysohn universal ultrametric space.
Moreover, for every compact subset $K$ of $X$ satisfying that
$d(K^{2})\subseteq R$, we can take $F$ so that $K\subseteq F$.

In what follows, the symbols $R$, $(X,d)$ and $K$ are the same objects as the
statement in Theorem 5.1.

######  Definition 5.1.

We say that a family $\\{\Omega[a,r]\\}_{a\in X,r\in R\setminus\\{0\\}}$ of
subsets of $X$ is an _$R$ -seed of $X$_ if the following conditions are
satisfied for all $a\in X$, and $r\in R\setminus\\{0\\}$:

  1. (1)

We have $a\in\Omega[a,r]$;

  2. (2)

the set $\Omega[a,r]$ is $r$-equidistant;

  3. (3)

we have $\operatorname{Card}(\Omega[a,r])=\aleph_{0}$.

Remark that since $(X,d)$ is $(R,\omega_{0})$-haloed, there exists an $R$-seed
(see Theorem 3.2).

######  Definition 5.2.

Fix an $R$-seed $\\{\Omega[a,r]\\}_{a\in X,r\in R\setminus\\{0\\}}$ and a
point $P\in X$. For $S\in\mathbf{TEN}(R)$, we say that a point $x\in X$ is an
_$S$ -heir of $P$_ if there exists sequences $\\{v_{i}\\}_{i=0}^{m}$ in $X$
and $\\{r_{i}\\}_{i=0}^{m-1}$ in $S$ such taht

  1. (1)

$v_{0}=P$;

  2. (2)

$v_{m}=x$;

  3. (3)

$v_{i+1}\in\Omega[v_{i},r_{i}]$ for all $i\in\\{0,\dots,m-1\\}$;

  4. (4)

$v_{i}\neq v_{i+1}$ for all $i\in\\{0,\dots,m-1\\}$;

  5. (5)

if $2\leq m$, then $r_{i+1}<r_{i}$ for all $i\in\\{0,\dots,m-2\\}$.

In this case, the pair $\\{v_{i}\\}_{i=0}^{m}$ and $\\{r\\}_{i=0}^{m-1}$ is
called an _$R$ -inheritance of $p$ from $o$ with length $n+1$_. If $m=0$, then
we consider that $\\{r_{i}\\}_{i=0}^{m-1}$ be the empty sequence.

Remark that the point $P$ is the unique $S$-heir from $P$ that has the
$S$-inheritance of length $1$.

######  Definition 5.3.

For an $R$-seed $\\{\Omega[a,r]\\}_{a\in X,r\in R\setminus\\{0\\}}$, and a
point $P\in X$, we denote by $\mathbb{A}(S)$ the set of all $S$-heirs of $P$.
We also denote by $\mathbb{E}(S)$ the closure of $\mathbb{A}(S)$ in $X$.

Notice that since $X$ is complete, $\mathbb{E}(S)$ is nothing but the
completion of $\mathbb{A}(S)$ for all $S\in\mathbf{TEN}(R)$.

######  Proposition 5.2.

Let $S,T\in\mathbf{TEN}(R)$, and $x$ and $y$ be $S$-heir and $T$-heir of $P$,
respectively. Let $(\\{v_{i}\\}_{i=0}^{m},\\{r_{i}\\}_{i=0}^{m-1})$ and
$(\\{w_{i}\\}_{i=0}^{n},\\{l_{i}\\}_{i=0}^{n-1})$ be $S$-inheritance and
$T$-inheritance of $x$ and $y$, respectively. In this situation, the next
three statements are true:

  1. (1)

If there exists $k\in\omega_{0}$ with $k\leq\min\\{m,n\\}$ and $x_{i}=y_{i}$
for all $i\leq k$, then $r_{i}=l_{i}$ for all $i<k$.

  2. (2)

If $k\in\omega_{0}$ satisfies that

    1. (a)

$v_{i}=w_{i}$ for all $i\leq k$;

    2. (b)

$v_{k+1}\neq w_{k+1}$,

then we have $\max\\{r_{k},l_{k}\\}=d(x,y)$.

  3. (3)

If $m<n$ and $v_{i}=w_{i}$ for all $i\leq m$, then we have $d(x,y)=l_{m}$.

###### Proof.

Since $r_{i}=d(v_{i},v_{i+1})$ and $l_{i}=d(w_{i},w_{i+1})$, under the
assumption of (1) we see that $r_{i}=l_{i}$ for all $i<k$ This finishes the
proof of (1).

We next prove (2). By the definition of heirs and the strong triangle
inequality, we have $d(v_{k+1},x)\leq d(v_{k},v_{k+1})\lor\dots\lor
d(v_{m-1},v_{m})\leq\max_{i=k+1,\dots,m-1}r_{i}<r_{k}$. Similarly, we also
have $d(w_{k+1},y)<l_{k}$. Since $d(v_{k},v_{k+1})=r_{k}$ and
$d(w_{k},w_{k+1})=l_{k}$, we have $d(x,y)=\max\\{r_{k},l_{k}\\}$. Thus the
strong triangle inequality implies that $d(x,y)=\max\\{r_{k},l_{k}\\}$. This
proves the statement (2).

Under the assumption of (3), we obtain $d(w_{m},w_{m+1})=l_{m}$ and
$d(w_{m+1},y)<l_{m}$. Hence $d(x,y)=d(w_{m},y)=l_{k}$. This finishes the
proof. ∎

######  Lemma 5.3.

For each $S\in\mathbf{TEN}(R)$, the space $(\mathbb{E}(S),d)$ is the
$S$-Urysohn universal ultrametric space.

###### Proof.

Since $\mathbb{A}(S)$ is countable, the space $\mathbb{E}(S)$ is separable.
From (2) and (3) in Proposition 5.2, it follows that $(\mathbb{E}(S),d)$ is
$R$-valued. We now show that $(\mathbb{E}(S),d)$ is $(S,\omega_{0})$-haloed.
It suffices to show that for all $a\in\mathbb{A}(S)$ and for all $r\in
S\setminus\\{0\\}$, there exists an $r$-equidistant subset of $B(a,r)$. Let
$(\\{v_{i}\\}_{i=0}^{m},\\{r_{i}\\}_{i=0}^{m-1})$ be an $S$-inheritance of
$a$. we divide the proof into three cases.

Case 1. [$r>r_{0}$]: In this case, the set $\Omega[P,r]$ is contained in
$\mathbb{E}(S)$. Indeed, for all $q\in\Omega[P,r]$, Define
$\\{w_{i}\\}_{i=0}^{1}$ and $\\{l_{i}\\}_{i=0}^{0}$ by $w_{0}=P$, $w_{1}=q$,
and $l_{0}=r$. Then $(\\{w_{i}\\}_{i=0}^{1},\\{l_{i}\\}_{i=0}^{0})$ is an
$S$-inheritance of $q$ from $P$. Hence $\Omega[P,r]\subseteq\mathbb{E}(S)$. By
the statement (2) in Proposition 5.2, we obtain $\Omega[P,r]\subseteq B(a,r)$.
Thus $\Omega[P,r]$ is a desired $r$-equidistant subset.

Case 2. [There exists $k$ such that $r_{k}>r>r_{k+1}$]: To prove
$\Omega[v_{k},r]\subseteq\mathbb{E}(S)$, for each $q\in\Omega[v_{k},r]$, we
define $(\\{w_{i}\\}_{i=0}^{k+1},\\{l_{i}\\}_{i=0}^{k})$ by

$w_{i}=\begin{cases}v_{i}&\text{if $i\leq k$;}\\\ q&\text{if
$i=k+1$}.\end{cases}$

and

$l_{i}=\begin{cases}r_{i}&\text{if $i\leq k$;}\\\ r&\text{if
$i=k+1$}.\end{cases}$

Then $(\\{w_{i}\\}_{i=0}^{k+1},\\{l_{i}\\}_{i=0}^{k})$ is an $S$-inheritance
of $q$ from $P$, and hence $\Omega[v_{k},r]\subseteq\mathbb{E}(S)$. By the
statement (2) in Proposition 5.2, we obtain $\Omega[v_{k},r]\subseteq B(a,r)$.
Thus $\Omega[v_{k},r]$ is a desired $r$-equidistant subset.

Case 3. [$r_{i}>r$ for all $i=0,\dots,m-1$]: In a similar way to the cases
explained above, we observe that $\Omega[a,r]\subseteq B(a,r)$.

Thus, we conclude that $(\mathbb{E}(S),d)$ is $(S,\omega_{0})$-haloed. Due to
Theorem 3.2, the space $(\mathbb{E}(S),d)$ is
$\mathscr{N}(S,\omega_{0})$-injective, and hence it is the $S$-Urysohn
universal ultrametric space. ∎

######  Lemma 5.4.

Let $S,T\in\mathbf{TEN}(R)$. Let $x$ be a $T$-heir of $P$, and
$(\\{v_{i}\\}_{i=0}^{m},\\{r_{i}\\}_{i=0}^{m-1})$ be a $T$-inheritance of $x$.
If $x\not\in\mathbb{E}(S)$, then there exists $i$ such that $r_{i}\in
T\setminus S$.

###### Proof.

If all $r_{i}$ were in $S$, we would have $x\in\mathbb{E}(S)$. ∎

######  Lemma 5.5.

Let $S,T\in\mathbf{TEN}(R)$ with $S\neq T$, and $x\in\mathbb{E}(T)$. Then the
value $d(x,\mathbb{E}(S))$ belongs to $\\{0\\}\cup(T\setminus S)$.

###### Proof.

We may assume that $x\not\in\mathbb{E}(S)$. Put $L=d(x,\mathbb{E}(S))$. Take a
sufficient close point $z\in\mathbb{A}(T)$ to $x$ so that
$L=d(z,\mathbb{E}(S))$ and $z\in\mathbb{E}(T)\setminus\mathbb{E}(S)$. Let
$(\\{v_{i}\\}_{i=0}^{m},\\{r_{i}\\}_{i=0}^{m-1})$ be a $T$-inheritance of $z$
from $P$. Note that since $L>0$, we especially have $z\neq P$. Thus $0<m$. Let
$k$ be the minimal number such that $r_{k}\in T\setminus S$. The existence of
$k$ is guaranteed by Lemma 5.4. From (2) in Proposition 5.2, it follows that
for all $y\in\mathbb{A}(S)$, we have $d(z,y)\geq r_{k}$. If $k=0$, we define
$q=P$; otherwise $q=v_{k-1}$. Then Proposition 5.2 implies that
$d(z,q)=r_{k}$. Therefore we conclude that $L=r_{k}\in T\setminus S$. ∎

We now provide the proof of Theorem 5.1.

###### Proof of Theorem 5.1.

We only need to prove the latter part of Theorem 5.1 in the case where $R$ is
uncountable. Choose a point $P$ such that $P\in K$ and choose an $R$-seed
$\\{\Omega[a,r]\\}_{a\in X,r\in R\setminus\\{0\\}}$ so that

  1. (A)

If $a\in K$, then for all $r\in R\setminus\\{0\\}$ the set $K\cap\Omega[a,r]$
is a maximal $r$-net of $K$.

This is possible since $K$ is compact and $d(K^{2})\subseteq R$. We define
$F=\bigcup_{S\in\mathbf{TEN}(R)}\mathbb{E}(S)$. We also define a petal of $F$
by $\Pi(F,S)=\mathbb{E}(S)$. By this definition, we confirm that the
properties (P1), (P3), and (P4) are satisfied. Using Lemma 5.3, the property
(P2) is true. Lemma 5.5 proves the property (P5). Therefore $(F,d)$ is the
$R$-petaloid ultrametric space. Since $F$ is complete (see [6, Proposition
2.9]), due to the assumption (A), we see that $K\subseteq F$. This completes
the proof. ∎

######  Question 5.6.

Let $R$ be a range set, and $(X,d)$ be the $R$-Urysohn universal ultrametric
space. Is there exist a natural isometry between $(X,d)$ and
$(\mathscr{K}(X),\mathscr{K}(d))$?

######  Question 5.7.

Is $(\mathscr{K}(\mathbb{U}),\mathcal{HD}_{\rho})$ isometric to
$(\mathbb{U},\rho)$?

## References

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# A dichotomy phenomenon for
Bad minus normed Dirichlet

Dmitry Kleinbock  and  Anurag Rao  Brandeis University, Waltham
<EMAIL_ADDRESS>Israel Institute of Technology, Haifa
<EMAIL_ADDRESS>
###### Abstract.

Given a norm $\nu$ on $\mathbb{R}^{2}$, the set of $\nu$-Dirichlet improvable
numbers $\mathbf{DI}_{\nu}$ was defined and studied in the papers [AD, KR1].
When $\nu$ is the supremum norm,
$\mathbf{DI}_{\nu}=\mathbf{BA}\cup{\mathbb{Q}}$, where $\mathbf{BA}$ is the
set of badly approximable numbers. Each of the sets $\mathbf{DI}_{\nu}$, like
$\mathbf{BA}$, is of measure zero and satisfies the winning property of
Schmidt. Hence for every norm $\nu$, $\mathbf{BA}\cap\mathbf{DI}_{\nu}$ is
winning and thus has full Hausdorff dimension. In the present article we prove
the following dichotomy phenomenon: either
$\mathbf{BA}\subset\mathbf{DI}_{\nu}$ or else
$\mathbf{BA}\smallsetminus\mathbf{DI}_{\nu}$ has full Hausdorff dimension. We
give several examples for each of the two cases. The dichotomy is based on
whether the critical locus of $\nu$ intersects a precompact $g_{t}$-orbit,
where $\\{g_{t}\\}$ is the one-parameter diagonal subgroup of
$\operatorname{SL}_{2}({\mathbb{R}})$ acting on the space $X$ of unimodular
lattices in $\mathbb{R}^{2}$. Thus the aforementioned dichotomy follows from
the following dynamical statement: for a lattice $\Lambda\in X$, either
$g_{\mathbb{R}}\Lambda$ is unbounded (and then any precompact
$g_{\mathbb{R}_{>0}}$-orbit must eventually avoid a neighborhood of
$\Lambda$), or not, in which case the set of lattices in $X$ whose
$g_{\mathbb{R}_{>0}}$-trajectories are precompact and contain $\Lambda$ in
their closure has full Hausdorff dimension.

The first-named author was supported by NSF grant DMS-1900560. This material
is based upon work supported by a grant from the Institute for Advanced Study
School of Mathematics. The second-named author received funding from the
European Research Council (ERC) under the European Union’s Horizon 2020
Research and Innovation Program, Grant agreement no. 754475.

##  1\. Introduction

Dirichlet’s theorem ([S2, Theorem 1A]) states that for every real number
$\alpha$ and every $T>1$, the set of inequalities

$\begin{cases}\langle\alpha q\rangle&\leq T^{-1},\\\ \ \ |q|&<T\end{cases}$

has a solution $q\in{\mathbb{N}}$. Here $\langle\cdot\rangle$ denotes the
distance of a real number to a nearest integer. The question of improving
Dirichlet’s theorem was initiated by Davenport–Schmidt in [DS]. A real number
$\alpha$ is said to be Dirichlet improvable, written
$\alpha\in\mathbf{DI}_{\infty}$, if there is a $c<1$ such that the system

$\begin{cases}\langle\alpha q\rangle&<cT^{-1},\\\ \ \ |q|&<T\end{cases}$ (1.1)

can be solved in $q\in{\mathbb{N}}$ for all sufficiently large $T$. [DS,
Theorem 1] shows that $\mathbf{DI}_{\infty}$ is exactly the set
$\mathbf{BA}\cup{\mathbb{Q}}$. In fact, this was known as far back as the
paper of Morimoto [Mo]. We recall that badly approximable numbers are those
real numbers whose continued fraction digits are bounded from above.

A normed variant of Dirichlet’s theorem (also referred to as Minkowski’s
approximation theorem) was studied in [AD, KR1]. The starting point is the
observation that the defining condition in (1.1) can be restated as

$\Lambda_{\alpha}\cap\left[{\begin{array}[]{cc}cT^{-1}&0\\\
0&T\end{array}}\right]B_{\infty}(1)\neq\\{0\\}$

where $\Lambda_{\alpha}$ denotes the unimodular lattice
$\left[{\begin{array}[]{cc}1&\alpha\\\ 0&1\end{array}}\right]{\mathbb{Z}}^{2}$
and $B_{\infty}(1)$ denotes the supremum norm ball in ${\mathbb{R}}^{2}$
centered at the origin and with radius $1$.

Now, let $X$ denote the set of unimodular lattices in ${\mathbb{R}}^{2}$, fix
a norm $\nu$ on ${\mathbb{R}}^{2}$ and define the critical radius of $\nu$:

$r_{\nu}:=\sup\big{\\{}r\in{\mathbb{R}}:\text{there exists }\Lambda\in X\text{
with }\Lambda\cap B_{\nu}(r)=\\{0\\}\big{\\}}.$ (1.2)

By Minkowski’s convex body theorem, $r_{\nu}$ is finite. Let $B_{\nu}(r)$
denote the $\nu$-norm ball with radius $r$ centered at the origin. We say that
a real number $\alpha$ is $\nu$-Dirichlet improvable, written
$\alpha\in\mathbf{DI}_{\nu}$, if there is some $c<1$ with

$\Lambda_{\alpha}\cap\left[{\begin{array}[]{cc}cT^{-1}&0\\\
0&T\end{array}}\right]B_{\nu}\left(r_{\nu}\right)\neq\\{0\\}$

for all sufficiently large $T$. When $\nu$ is the supremum norm (denoted by
the subscript $\infty)$, we have that $r_{\infty}=1$ and so recover the
definition of Davenport–Schmidt. When $\nu$ is the Euclidean norm (denoted by
the subscript $2$), we see that $\alpha\in\mathbf{DI}_{2}$ if and only if
there is some $c<1$ such that

$\left(\frac{T\langle\alpha
q\rangle}{c}\right)^{2}+\left(\frac{q}{T}\right)^{2}<\frac{2}{\sqrt{3}}$

is solvable in $q\in{\mathbb{N}}$ for all sufficiently large $T$.

Consider the following preliminary properties:

######  Theorem 1.1 ([KR1], Theorems 3.1 and 1.3).

For each norm $\nu$ on ${\mathbb{R}}^{2}$, the set $\mathbf{DI}_{\nu}$ is of
measure zero but winning in the sense of Schmidt. In particular,
$\mathbf{DI}_{\nu}$ has full Hausdorff dimension.

See [S1] and a paragraph before Proposition 2.2 for a discussion of Schmidt
games and winning. The fact that $\mathbf{BA}$ has full Hausdorff dimension
goes back to Jarnik in 1928 (see [BRV, Theorem 7.1] for an elementary proof).
Thus, natural questions that come up in the study of Theorem 1.1 are

  1. (a)

Do there exist norms for which $\mathbf{BA}\not\subset\mathbf{DI}_{\nu}$?

  2. (b)

If yes, can one characterize the norms for which
$\mathbf{BA}\not\subset\mathbf{DI}_{\nu}$?

The existence of norms which satisfy the above non-containment is a
consequence of the equality

${\mathbb{Q}}=\bigcap_{\nu}\mathbf{DI}_{\nu}$

(see [KR2, Theorem 1.7]). In the present paper we address Question (b) by
giving a convenient dynamical condition on $\nu$ which completely determines
whether or not $\mathbf{BA}$ is a subset of $\mathbf{DI}_{\nu}$. In fact, our
main results are completely dynamical in nature and this paper is a twofold
study of limit points of bounded diagonal orbits in $X$ and criteria to detect
whether certain submanifolds intersect precompact orbits.

We briefly explain how dynamics comes into play; the idea essentially comes
from the work of Davenport and Schmidt [DS]. Each norm $\nu$ gives rise to the
following compact subset of $X$:

$\mathcal{L}_{\nu}:=\big{\\{}\Lambda\in X:\Lambda\cap
B_{\nu}\left(r_{\nu}\right)=\\{0\\}\big{\\}}.$

$\mathcal{L}_{\nu}$ is referred to as the critical locus of the norm. It
follows from the definition of $r_{\nu}$ and from Mahler’s compactness
criterion that $\mathcal{L}_{\nu}$ is a non-empty compact subset of $X$.
Equivalently, lattices in $\mathcal{L}_{\nu}$ give the densest lattice
packings of ${\mathbb{R}}^{2}$ by the domain
$B_{\nu}\left(\frac{r_{\nu}}{2}\right)$. We have the following notation for
present and future use:

${g_{t}:=\left[{\begin{array}[]{cc}e^{t}&0\\\
0&e^{-t}\end{array}}\right],\text{ where }t\in{\mathbb{R}},\ \text{ and
}g_{E}:=\left\\{g_{t}:t\in E\right\\}\text{ for a subset $E$ of
${\mathbb{R}}$}.}$ (1.3)

######  Proposition 1.2 ([D], Theorem 2.20 and [KR1], Theorem 2.1).

We have the following dynamical reformulations:

  1. (i)

A real number $\alpha$ belongs to $\mathbf{BA}$ if and only if

$g_{{\mathbb{R}}_{>0}}\Lambda_{\alpha}\text{ is precompact in }X.$

  2. (ii)

Given a norm $\nu$ on ${\mathbb{R}}^{2}$, we have that
$\alpha\in\mathbf{DI}_{\nu}$ if and only if there exists $t_{0}>0$ and a
neighborhood $U\supset\mathcal{L}_{\nu}$ suth that

$g_{{\mathbb{R}}_{>t_{0}}}\Lambda_{\alpha}\cap U=\varnothing.$

Thus initiated, we state our main theorem:

######  Theorem 1.3.

Fix a lattice $\Lambda\in X$. If $g_{\mathbb{R}}\Lambda$ is precompact in $X$,
then the set

$\mathcal{S}_{\Lambda}:=\left\\{\alpha\in\mathbf{BA}:\Lambda\in\overline{g_{{\mathbb{R}}_{>0}}\Lambda_{\alpha}}\right\\}$

has Hausdorff dimension $1$. Conversely, if $g_{\mathbb{R}}\Lambda$ is not
precompact in $X$, then for every $\alpha\in\mathbf{BA}$ there exists
$t_{0}>0$ and a neighbourhood $U\supset\Lambda$ such that

$g_{{\mathbb{R}}_{>t_{0}}}\Lambda_{\alpha}\cap U=\varnothing.$

######  Remark 1.4.

The converse is elementary and left to the reader as an exercise, see [KR2,
Proposition 4.1]. Thus the main content of the theorem is the first part. Its
proof below can be easily modified to show that the intersection of
$\mathcal{S}_{\Lambda}$ with any open interval has Hausdorff dimension $1$.
However, $\mathcal{S}_{\Lambda}$ is not winning as follows from [AGK, Theorem
2.8]. (This last theorem was used to prove Theorem 1.1.)

As a corollary of Proposition 1.2 and Theorem 1.3, we have our dichotomy
phenomenon:

######  Theorem 1.5.

Fix a norm $\nu$ on ${\mathbb{R}}^{2}$. If $\mathcal{L}_{\nu}$ contains a
lattice $\Lambda$ such that $g_{\mathbb{R}}\Lambda$ is precompact in $X$, then
$\mathbf{BA}\smallsetminus\mathbf{DI}_{\nu}$ has full Hausdorff dimension.
Conversely, if no such lattice exists, then
$\mathbf{BA}\subset\mathbf{DI}_{\nu}$.

######  Remark 1.6.

The converse has already been proved in the more general multidimensional
setting of weighted approximation of systems of linear forms, see [KR2,
Proposition 4.1].

It can be shown that the Euclidean critical locus $\mathcal{L}_{2}$ intersects
a precompact ${\mathbb{R}}$-orbit (in fact, quite a few of them). Moreover,
there is a very large class of norms which also satisfy this condition. Let us
say that a norm $\nu$ on ${\mathbb{R}}^{2}$ is irreducible if whenever
$\eta\neq\nu$ is a norm on ${\mathbb{R}}^{2}$ with $\eta({\bf v})\geq\nu({\bf
v})$ for any ${\bf v}\in{\mathbb{R}}^{2}$, we have $r_{\eta}>r_{\nu}$. Mahler
introduced this distinguished set of norms (or rather, in his terminology, of
convex bounded symmetric domains) in [Ma1], and they turned out to be central
to proving Theorem 1.1 among other results. Examples of irreducible norms are
those whose unit balls are ellipses, parallelograms and Reinhardt’s
curvilinear octagon (see [KRS, Examples 2.2, 2.3, 3.3]). The critical locus
$\mathcal{L}_{\nu}$ of an irreducible norm $\nu$ which does not come from a
parallelogram is necessarily a one-dimensional $\mathcal{C}^{1}$-submanifold
of $X$ (see [Ma2, Theorem 3]). Considerations in the tangent bundle of $X$
then allow us to conclude that $\mathcal{L}_{\nu}$ must intersect some
precompact orbit $g_{\mathbb{R}}\Lambda$ (see Proposition 2.5 below). Thus, we
obtain as a corollary:

######  Theorem 1.7.

If $\nu$ is an irreducible norm on ${\mathbb{R}}^{2}$ whose unit ball is not a
parallelogram, then $\mathbf{BA}\smallsetminus\mathbf{DI}_{\nu}$ has full
Hausdorff dimension. In particular, $\mathbf{BA}\smallsetminus\mathbf{DI}_{2}$
has full Hausdorff dimension.

We also mention that, since the critical locus of a hexagonal norm is a
singleton (see [KRS, Example 2.4]), one can easily construct norms $\nu$ which
are not irreducible and for which $\mathbf{BA}\smallsetminus\mathbf{DI}_{\nu}$
has full dimension. For example, if $\nu$ is a norm with
$\mathcal{L}_{\nu}=\\{\Lambda_{0}\\}$, one can choose an element
$g\in\operatorname{SL}_{2}({\mathbb{R}})$ which makes
$g_{\mathbb{R}}g\Lambda_{0}$ precompact, and then notice that
$\mathcal{L}_{\nu\circ g^{-1}}=g\mathcal{L}_{\nu}$.

We organize the paper as follows. In §2 we explore the structure of $X$ more
carefully to establish, under the assumption that $\nu$ is a non-parallelogram
irreducible norm, the existence of lattices in $\mathcal{L}_{\nu}$ with
precompact ${\mathbb{R}}$-orbits. With this out of the way, all that remains
is to prove Theorem 1.3 in the case when the mentioned lattice $\Lambda$ has a
precompact ${\mathbb{R}}$-orbit. To prove it we use a symbolic representation
of the diagonal flow. In §3 we study a specific Poincare section for $X$ which
arises in the exposition of [C]. From this it becomes apparent that the set
$\mathcal{S}_{\Lambda}$ from Theorem 1.3 consists of badly approximable
numbers having certain predetermined blocks of digits appearing in their
continued fraction expansion. We then apply a dimension estimate from [U] in
§4 to show this set has full Hausdorff dimension. Some remarks on possible
generalizations of the set-up of this paper are made in the last section.

##  2\. Critical loci which intersect precompact orbits

We denote $G:=\operatorname{SL}_{2}({\mathbb{R}})$ and
$\Gamma:=\operatorname{SL}_{2}({\mathbb{Z}})$. Identify $X$ with $G/\Gamma$
via the map $g\mapsto g{\mathbb{Z}}^{2}$. In addition to the notation of
(1.3), for $x,y,\in{\mathbb{R}}$ denote

$u_{x}:=\left[{\begin{array}[]{cc}1&x\\\ 0&1\end{array}}\right]\text{ and
}v_{y}:=\left[{\begin{array}[]{cc}1&0\\\ y&1\end{array}}\right].$

We next have the stability of precompact orbits under the action of certain
subgroups of $G$:

######  Proposition 2.1.

For any $\Lambda\in X$ there exists $\varepsilon>0$ such that the maps

$(-\varepsilon,\varepsilon)^{3}\to X;\ (x,t,y)\mapsto v_{y}g_{t}u_{x}\Lambda\
\text{ and }\ (x,t,y)\mapsto u_{x}g_{t}v_{y}\Lambda$ (2.1)

are diffeomorphisms. Moreover, in these coordinates, we have the equivalences:

$g_{{\mathbb{R}}_{>0}}\left(v_{y}g_{t}u_{x}\Lambda\right)\text{ is precompact
in }X\iff g_{{\mathbb{R}}_{>0}}\left(u_{x}\Lambda\right)\text{ is precompact
in }X.$

And

$g_{{\mathbb{R}}_{<0}}\left(u_{x}g_{t}v_{y}\Lambda\right)\text{ is precompact
in }X\iff g_{{\mathbb{R}}_{<0}}\left(v_{y}\Lambda\right)\text{ is precompact
in }X.$ (2.2)

###### Proof.

The first assertion follows from [V, Theorem 2.10.1]. For the second, note
that

$g_{t}v_{y}=v_{ye^{-2t}}g_{t}\ \text{ and }\ g_{t}u_{x}=u_{xe^{2t}}g_{t}.$
(2.3)

Alternatively, see [D, Proposition 2.12]. ∎

Our main tool in establishing the existence of precompact orbits passing
through critical loci is the following proposition which constructs a winning
set of precompact orbits along the $u_{x}$ and $v_{y}$ directions about any
fixed lattice. It is a well known application of Schmidt’s results in [S1]; we
give an elementary proof here for convenience (see also [KW, Theorem 3.7] and
[Mc, Theorem 1.3]).

For the benefit of the reader we recall the rules of Schmidt’s game introduced
in [S1]. It involves two parameters $\alpha,\beta\in(0,1)$ and is played by
two players Alice and Bob on a complete metric space (which we shall take to
be the set of real numbers) with a target set $S$. Bob starts the game by
choosing a closed ball $B_{0}=B(x_{0},r_{0})$ in ${\mathbb{R}}$ with center
$x_{0}$ and radius $r_{0}$. After Bob chooses a closed ball
$B_{i}=B(x_{i},{r}_{i})$, Alice chooses
$A_{i}=B(x_{i}^{\prime},{r}_{i}^{\prime})\subset B_{i}$ with
${r}^{\prime}_{i}=\alpha{r}_{i}$, and then Bob chooses
$B_{i+1}=B(x_{i+1},{r}_{i+1})\subset A_{i}$ with
${r}_{i+1}=\beta{r}^{\prime}_{i}$, etc. Alice wins the game if the unique
point $\bigcap_{i=0}^{\infty}A_{i}=\bigcap_{i=0}^{\infty}B_{i}$ belongs to
$S$, and Bob wins otherwise. The set $S$ is $(\alpha,\beta)$-winning if Alice
has a winning strategy, and is winning if it is $(\alpha,\beta)$-winning for
some $\alpha>0$ and all $\beta\in(0,1)$.

######  Proposition 2.2.

For any $\Lambda\in X$, the sets

$\left\\{x\in{\mathbb{R}}:g_{{\mathbb{R}}_{>0}}u_{x}\Lambda\text{ is
precompact in }X\right\\}\ \text{ and }\
\left\\{y\in{\mathbb{R}}:g_{{\mathbb{R}}_{<0}}v_{y}\Lambda\text{ is precompact
in }X\right\\}$

are winning.

###### Proof.

Recall that we are given $\Lambda\in X$. Let us first consider the case when

$\Lambda=\left[{\begin{array}[]{cc}a&0\\\
c&a^{-1}\end{array}}\right]\left[{\begin{array}[]{cc}1&b\\\
0&1\end{array}}\right]{\mathbb{Z}}^{2}$ (2.4)

for some $a\in{\mathbb{R}}\smallsetminus\\{0\\}$ and $b,c\in{\mathbb{R}}$.
Take $\alpha<1/2$. After making the first move Alice can ensure that all $x$
in the ball of her choice satisfy $a+cx\neq 0$. For
$\phi(x):=a^{-1}x(a+cx)^{-1}$, we have

$u_{x}\Lambda=\left[{\begin{array}[]{cc}a+cx&0\\\
c&(a+cx)^{-1}\end{array}}\right]\ \left[{\begin{array}[]{cc}1&\phi(x)+b\\\
0&1\end{array}}\right]{\mathbb{Z}}^{2}.$

Using the commutator argument of Proposition 2.1 and the characterization of
Proposition 1.2(i), we see that

$\phi(x)+b\in\mathbf{BA}{\ \Longleftrightarrow\
}g_{{\mathbb{R}}_{>0}}u_{x}\Lambda\text{ is precompact in }X.$

Since $\phi^{-1}$ is Lipschitz, the winning property of $\mathbf{BA}$ and [S1,
Theorem 1] establishes the winning property for the first set in the
proposition.

Next we consider the case when $\Lambda$ cannot be written in the form (2.4).
This happens if and only if $\Lambda=\left[{\begin{array}[]{cc}0&a\\\
b&c\end{array}}\right]{\mathbb{Z}}^{2}$ for some $a,b,c\in{\mathbb{R}}$. A
consequence is that, for any $x\neq 0$, $u_{x}\Lambda$ can be written in the
form (2.4); thus one can replace $\Lambda$ with $u_{x}\Lambda$ and reduce the
problem to the previously considered case. Hence in both cases we have the
winning property for the first set in the proposition; the second set, in view
of an elementary observation that

${g_{{\mathbb{R}}_{>0}}u_{\alpha}{\mathbb{Z}}^{2}\text{ is precompact in }X\
\Longleftrightarrow\ \text{so is
}g_{{\mathbb{R}}_{<0}}v_{\alpha}{\mathbb{Z}}^{2},}$ (2.5)

can be handled similarly. ∎

We have the notation for the following subgroups of $G$:

$F:=\\{g_{t}:t\in{\mathbb{R}}\\},\ H^{+}:=\\{u_{x}:x\in{\mathbb{R}}\\},\
H^{-}:=\\{v_{y}:y\in{\mathbb{R}}\\}.$

######  Proposition 2.3.

Let $Z\subset X$ be a compact $\mathcal{C}^{1}$-submanifold satisfying, for
some $z\in Z$,

$T_{z}(X)=T_{z}(Z)+T_{z}(Fz)+T_{z}(H^{+}z)$ (2.6)

and

$T_{z}(X)=T_{z}(Z)+T_{z}(Fz)+T_{z}(H^{-}z).$ (2.7)

Then $Z$ contains a lattice $\Lambda$ for which $g_{\mathbb{R}}\Lambda$ is
precompact in $X$.

######  Remark 2.4.

We use the notation $T_{m}(M)$ to denote the tangent space of a
($\mathcal{C}^{1}$ or smoother) manifold $M$ at a point $m$. We also use the
prefix $d$ to indicate the derivative of a smooth map. $T_{z}(Fz)$ is
understood to be the image, in $T_{z}(X)$, of the tangent space $T_{e}(F)$ via
the derivative of the map $f\mapsto fz$. The same interpretation is used for
the other subgroups of $G$.

###### Proof of Proposition 2.3.

Let

$\phi:(-\varepsilon,\varepsilon)\to X$

be a local parameterization of $Z$ with $\phi(0)=z$. Using the first set of
local coordinates at $z$ in (2.1), and the tangent space decomposition of
(2.7), we can write

$\phi(s)=v_{y(s)}g_{t(s)}u_{x(s)}z$

where $x(\cdot)$, ${t}(\cdot)$, $y(\cdot)$ are now $\mathcal{C}^{1}$-functions
and $x^{\prime}(0)\neq 0$. Thus the inverse function $x^{-1}$ exists in a
neigborhood of $0$ and is Lipschitz. Using Proposition 2.2, [S1, Theorem 1]
and the equivalence (2.1), we see that

$\left\\{s\in(-\varepsilon,\varepsilon):g_{{\mathbb{R}}_{>0}}\phi(s)\text{ is
precompact in }X\right\\}$

is winning. An entirely analogous argument using (2.6) shows that

$\left\\{s\in(-\varepsilon,\varepsilon):g_{{\mathbb{R}}_{<0}}\phi(s)\text{ is
precompact in }X\right\\}$

is winning. Since the intersection of winning sets is also winning [S1,
Theorem 2], we arrive at the desired result. ∎

######  Proposition 2.5.

If $\nu$ is an irreducible norm on ${\mathbb{R}}^{2}$ whose unit ball is not a
parallelogram, then $\mathcal{L}_{\nu}$ is a compact
$\mathcal{C}^{1}$-submanifold of $X$ satisfying conditions (2.6) and (2.7) at
every point $z\in\mathcal{L}_{\nu}$. Consequently, $\mathcal{L}_{\nu}$ must
intersect some precompact ${\mathbb{R}}$-orbit.

###### Proof.

The first condition (2.6) is proved in [KR1, Theorem 3.11]. For the second
condition, consider the linear automorphism of ${\mathbb{R}}^{2}$ given by

${\bf v}\mapsto p{\bf v},\text{ where }p:=\left[{\begin{array}[]{cc}0&1\\\
1&0\end{array}}\right].$

Of course, it also induces a map on subsets of ${\mathbb{R}}^{2}$ which gives
a diffeomorphism of $X$

$\Lambda\mapsto p\Lambda.$

One the level of critical loci and norms, we have that

$p\mathcal{L}_{\nu}=\mathcal{L}_{\nu\circ p}.$

Moreover, if $\nu$ is irreducible with unit ball not a parallelogram, then the
same holds true for $\nu\circ p$. On the level of tangent spaces we have that,
for $z\in X$,

$(dp)T_{z}(Fz)=T_{pz}(Fpz),\ (dp)T_{z}(H^{+}z)=T_{pz}(H^{-}pz),\
(dp)T_{z}(H^{-}z)=T_{pz}(H^{+}pz).$

The upshot now is that since $\mathcal{L}_{\nu\circ p}$ satisfies (2.6) at
$pz\in p\mathcal{L}_{\nu}$, it follows that $\mathcal{L}_{\nu}$ satisfies
(2.7) at $z$. This shows that Proposition 2.3 can be applied, giving the final
assertion in the present proposition. ∎

##  3\. Minimal vectors and a Poincare section for $X$

We study a well known correspondence between diagonal orbits in $X$ and orbits
in an invertible extension of the Gauss map. We follow the exposition of [C,
§3.1] and presume the reader is familiar with the theory of continued
fractions at the level of [K].

If $a,b$ are real numbers, we define the rectangle $R(a,b)$ to be

$R(a,b):=\left\\{(x_{1},x_{2})\in{\mathbb{R}}^{2}:|x_{1}|\leq a\text{ and
}|x_{2}|\leq b\right\\}.$

######  Definition 3.1.

(Minimal vectors) Let $\Lambda\in X$. A vector ${\bf
r}=(r_{1},r_{2})\in\Lambda$ is called a minimal vector if it is nonzero, and
if

${\bf s}=(s_{1},s_{2})\in
R\left(|r_{1}|,|r_{2}|\right)\cap\Lambda\smallsetminus\\{0\\}\implies|s_{1}|=|r_{1}|\text{
and }|s_{2}|=|r_{2}|.$

######  Definition 3.2.

(Consecutive minimal vectors) An ordered pair of minimal vectors ${\bf
r}=(r_{1},r_{2})$ and ${\bf s}=(s_{1},s_{2})$ in $\Lambda$ is said to be the
pair of consecutive minimal vectors if $|r_{2}|<|s_{2}|$, and if there are no
minimal vectors ${\bf w}=(w_{1},w_{2})\in\Lambda$ with
$|r_{2}|<|w_{2}|<|s_{2}|$.

######  Lemma 3.3 ([C], Lemma 2 and Proposition 3).

If ${\bf r},{\bf s}$ are consecutive minimal vectors in $\Lambda$ then we have

$\operatorname{Interior}(R\left(|r_{1}|,|s_{2}|\right))\cap\Lambda=\\{0\\}.$

Moreover, ${\bf r}$ and ${\bf s}$ form a basis of $\Lambda$.

If $x\in{\mathbb{R}}$, let $\lfloor x\rfloor$ and $\\{x\\}$ denote the integer
and fractional parts of $x$. Consider the set

$\mathcal{U}=(0,1)^{2}\cup\left(\left[0,\frac{1}{2}\right]\times\\{0\\}\right)\cup\left(\\{0\\}\times\left[0,\frac{1}{2}\right]\right).$
$(0,0)$Here is a beautiful picture of $\mathcal{U}$.

Consider also the maps

######  Definition 3.4.

$T:(0,1)^{2}\cup\left(\left(0,\frac{1}{2}\right]\times\\{0\\}\right)\to(0,1)^{2}\cup\left(\\{0\\}\times\left(0,\frac{1}{2}\right]\right)$
given by

$T(x,y)=\left(\left\\{x^{-1}\right\\},\frac{1}{\left\lfloor
x^{-1}\right\rfloor+y}\right)$ (3.1)

and
$S:(0,1)^{2}\cup\left(\\{0\\}\times\left(0,\frac{1}{2}\right]\right)\to(0,1)^{2}\cup\left(\left(0,\frac{1}{2}\right]\times\\{0\\}\right)$
given by

$S(a,b)=\left(\frac{1}{a+\left\lfloor
b^{-1}\right\rfloor},\left\\{b^{-1}\right\\}\right).$

It is then straightforward to check that

######  Lemma 3.5.

The compositions $S\circ T$ and $T\circ S$ give the identity on the domains of
$T$ and $S$ respectively.

We now discuss a mapping from pairs of minimal vectors of a lattice onto
$\mathcal{U}\times\left\\{\pm 1\right\\}$ that, in a sense made precise below,
intertwines the diagonal action on $X$ with the $T$-action on $\mathcal{U}$.

${\bf r}$${\bf s}$${\bf w}={\bf r}+3{\bf s}$${\bf r}+{\bf s}$${\bf r}+2{\bf
s}$The Euclidean algorithm for finding the minimal vector following ${\bf s}$.

######  Proposition 3.6 (Proposition 6 in [C]).

Let $\Lambda\in X$ and let ${\bf r},{\bf s}$ be a pair of consecutive minimal
vectors in $\Lambda$ with

$0\leq r_{2}<s_{2}.$

If we happen to have $r_{2}=0$ and $r_{1}s_{1}>0$, replace ${\bf r}$ by $-{\bf
r}$. For such pairs of minimal vectors, define the functions

$x({\bf r},{\bf s}):=-\frac{s_{1}}{r_{1}},\ y({\bf r},{\bf
s}):=\frac{r_{2}}{s_{2}},\ \varepsilon({\bf r},{\bf
s}):=\frac{r_{1}}{|r_{1}|}.$

We then have that $(x,y)\in\mathcal{U}.$ Moreover, if $s_{1}\neq 0$,
$n:=\left\lfloor\frac{1}{x}\right\rfloor$ and ${\bf w}:={\bf r}+n{\bf s},$
then ${\bf s},{\bf w}$ is a pair of consecutive minimal vectors for $\Lambda$.
Further, we have that

$\big{(}x({\bf s},{\bf w}),y({\bf s},{\bf w}),\varepsilon({\bf s},{\bf
w})\big{)}=\big{(}T\left(x({\bf r},{\bf s}),y({\bf r},{\bf
s})\big{)},-\varepsilon({\bf r},{\bf s})\right),$ (3.2)

where $T$ is as in Definition 3.4.

######  Remark 3.7.

Note that $r_{1}$ cannot be zero by minimal vector considerations. In
subsequent notation, we often drop the dependence of $x,y,\varepsilon$ on the
minimal vectors when the context is clear. We repeat the proof from [C] in
order to make use of equation (3.4) below.

###### Proof of Proposition 3.6.

We first show that $(x,y)\in\mathcal{U}$. It is clear that $0\leq y<1$.

If $y=0$, then we have $r_{2}=0$ and, by definition, either $s_{1}=0$ or
$r_{1}s_{1}<0$. This shows that $0\leq x$. Moreover, if $s_{1}=0$, then
$(x,y)=(0,0)\in\mathcal{U}$ as desired. So, assume we are in the case when
$s_{2}=0$ and $r_{1}s_{1}<0$. Consider the lattice vector ${\bf r}+{\bf
s}=(r_{1}+s_{1},s_{2})$. Since ${\bf s}$ is minimal, we must have

$|r_{1}+s_{1}|\geq|s_{1}|.$ (3.3)

When $r_{1}>0$, since $|s_{1}|<r_{1}$ and $r_{1}s_{1}<0$, this implies

$r_{1}+s_{1}\geq-s_{1}$

which shows that $x\leq 1/2$. When $r_{1}<0$, equation (3.3) leads to

$-(r_{1}+s_{1})\geq s_{1},$

which again shows that $x\leq 1/2$.

Now consider $x$. By minimality, $|s_{1}|<|r_{1}|$ so that $|x|<1$. If $x=0$
so that $s_{1}=0$, we consider the lattice vector ${\bf r}-{\bf
s}=(r_{1},r_{2}-s_{2})$. Since ${\bf r}$ is minimal, we must have

$|r_{2}-s_{2}|=s_{2}-r_{2}\geq r_{2}$

which leads to $y\leq 1/2$. Taking stock so far, we have shown that $(x,y)$
always lies in

$[0,1/2]\times\\{0\\}\cup(0,1)\times(0,1)\cup(-1,0)\times(0,1)\cup\\{0\\}\times[0,1/2].$

Thus, we are left with proving that $x>0$ when $y>0$. If $y>0$, consider the
vector ${\bf r}-{\bf s}$ which is equal to $(r_{1}-s_{1},r_{2}-s_{2})$. If
$r_{1}$ and $s_{1}$ had the same sign, this vector would belong to the
interior of the rectangle $B(|r_{1}|,|s_{2}|)$ which contradicts Lemma 3.3.
Thus, $x>0$ and this concludes the proof that $(x,y)\in\mathcal{U}$.

Moreover, we can write ${\bf r},{\bf s}$ in a matrix as

$\left[{\begin{array}[]{cc}{\bf r}&{\bf
s}\end{array}}\right]=\left[{\begin{array}[]{cc}\varepsilon|r_{1}|&-\varepsilon|r_{1}|x\\\
s_{2}y&s_{2}\end{array}}\right].$

Or rather, noting that $\Lambda$ is a covolume one lattice, we can solve for
$s_{2}$ and write:

$\left[{\begin{array}[]{cc}{\bf r}&{\bf
s}\end{array}}\right]=\left[{\begin{array}[]{cc}|r_{1}|&0\\\
0&|r_{1}|^{-1}\end{array}}\right]\left[{\begin{array}[]{cc}\varepsilon&-\varepsilon
x\\\ \frac{y}{1+xy}&\frac{1}{1+xy}\end{array}}\right].$ (3.4)

Now we study the case when $s_{1}\neq 0$. In this case, Minkowski’s convex
body theorem shows that there is a nonzero lattice vector in the strip

$\left\\{(x_{1},x_{2})\in{\mathbb{R}}^{2}:x_{2}>0\text{ and
}|x_{1}|<|s_{1}|\right\\}.$

But using the discreteness of $\Lambda$, we get a minimal lattice vector ${\bf
w}=(w_{1},w_{2})$ in this region such that ${\bf s},{\bf w}$ form a
consecutive pair of minimal vectors. By Lemma 3.3, we see that ${\bf r},{\bf
s}$ and ${\bf s},{\bf w}$ form bases of $\Lambda$. The orientations of each
basis are distinct by what we have proved above; namely, if $r_{1}>0$, then
$\det({\bf r},{\bf s})=1$, and if $r_{1}<0$, then $\det({\bf r},{\bf s})=-1$.
And further, we always have the condition that $r_{1}s_{1}<0$. Thus we can
write

${\bf w}={\bf r}+k{\bf
s}=\left(\varepsilon|r_{1}|x\left(\frac{1}{x}-k\right),s_{2}(y+k)\right)\text{
for some }k\in\mathbb{N}.$

(Note that $x\neq 0$ since $s_{1}\neq 0$.) Since ${\bf w}$ is minimal, we must
have

$\left|\frac{1}{x}-k\right|<1,$

so that $k=n$ or $n+1$. On the other hand the presence of the lattice vector

${\bf r}+n{\bf
s}=\left(\varepsilon|r_{1}|x\Big{(}\frac{1}{x}-n\Big{)},s_{2}(y+n)\right),$

and the fact that ${\bf s},{\bf w}$ are consecutive shows that we indeed have
${\bf w}={\bf r}+n{\bf s}$. Also we have

$x({\bf s},{\bf w})=-\frac{w_{1}}{s_{1}}=\left\\{\frac{1}{x({\bf r},{\bf
s})}\right\\},\ y({\bf s},{\bf w})=\frac{s_{2}}{w_{2}}=\frac{1}{y({\bf r},{\bf
s})+n},\ \varepsilon({\bf s},{\bf w})=-\varepsilon({\bf r},{\bf s})$

thereby completing the proof. ∎

######  Remark 3.8.

Henceforth, whenever we refer to a minimal vector ${\bf r}=(r_{1},r_{2})$, we
always make the tacit assumption that $r_{2}\geq 0$.

We also note the following characterization of precompact orbits:

######  Proposition 3.9.

The orbit $g_{\mathbb{R}}\Lambda$ is precompact if and only if, for any choice
of consecutive minimal vectors ${\bf r},{\bf s}\in\Lambda$, we have that both
$x({\bf r},{\bf s})$ and $y({\bf r},{\bf s})$ are badly approximable real
numbers.

###### Proof.

Say $g_{\mathbb{R}}\Lambda$ is precompact in $X$. Let ${\bf
r}=(r_{1},r_{2}),{\bf s}=(s_{1},s_{2})$ be any pair of consecutive minimal
vectors. By Lemma 3.3, they form a basis of $\Lambda$ and we can write

$\Lambda=\left[{\begin{array}[]{cc}r_{1}&s_{1}\\\
r_{2}&s_{2}\end{array}}\right]{\mathbb{Z}}^{2}.$

Moreover, precompactness shows that $\Lambda$ does not contain any vector on
the coordinate axes, and so we have

$\left[{\begin{array}[]{cc}r_{1}&s_{1}\\\
r_{2}&s_{2}\end{array}}\right]=\left[{\begin{array}[]{cc}r_{1}&0\\\
r_{2}&s_{2}-\frac{s_{1}r_{2}}{r_{1}}\end{array}}\right]\left[{\begin{array}[]{cc}1&\frac{s_{1}}{r_{1}}\\\
0&1\end{array}}\right]=\left[{\begin{array}[]{cc}r_{1}&0\\\
r_{2}&s_{2}-\frac{s_{1}r_{2}}{r_{1}}\end{array}}\right]u_{-x({\bf r},{\bf
s})},$

so that, from the commutator relation (2.3) and the equivalence of Proposition
1.2(i), $-x({\bf r},{\bf s})$ is badly approximable, which, of course, is
equivalent to $x({\bf r},{\bf s})$ being badly approximable.

Similarly, we also have the decomposition

$\left[{\begin{array}[]{cc}r_{1}&s_{1}\\\
r_{2}&s_{2}\end{array}}\right]=\left[{\begin{array}[]{cc}r_{1}-\frac{s_{1}r_{2}}{s_{2}}&s_{1}\\\
0&s_{2}\end{array}}\right]\left[{\begin{array}[]{cc}1&0\\\
\frac{r_{2}}{s_{2}}&1\end{array}}\right]=\left[{\begin{array}[]{cc}r_{1}-\frac{s_{1}r_{2}}{s_{2}}&s_{1}\\\
0&s_{2}\end{array}}\right]v_{y({\bf r},{\bf s})}.$

Using (2.3) and (2.5), we see that $y({\bf r},{\bf s})$ is badly approximable.
We remark that, in the above proof, it does not matter if the matrix formed by
the ordered pair ${\bf r},{\bf s}$ has determinant $-1$.

For the converse, we simply note that we can reverse the order in the argument
just given. ∎

##  4\. An abundance of precompact orbits with a prescribed limit point

We set out some further notation involving continued fractions. The symbols

$a_{1},a_{2},\dots,a_{n},\dots$

denote the countable family of measurable functions giving the continued
fraction digits of a number in $[0,1]$. Of course, $a_{n}$ is not defined on
certain rationals. Other letters such as $b_{n},c_{n},d_{n}$ are, as before,
used to denote continued fraction digits of specific numbers. We also have the
associated convergent functions

$\frac{p_{n}}{q_{n}}:=[0;a_{1},\dots,a_{n}],$

and the remainder functions defined by

$\alpha=[0;a_{1}(\alpha),\dots,a_{n-1}(\alpha),\rho_{n}(\alpha)].$

Finally, we denote the the successive iterates of the Gauss map on $[0,1)$ by

$z_{n}(\alpha):=\rho_{n}(\alpha)-a_{n}(\alpha)\in[0,1).$

Given a multi-index ${\bf k}=(k_{1},\dots,k_{n})$ of natural numbers, we
define the $n$th order cylindrical interval

$I({\bf k}):=\left\\{\alpha\in[0,1]:a_{i}(\alpha)=k_{i}\text{ for
}i=1,\dots,n\right\\}.$

It is well known that these intervals are given by

$I({\bf
k})=\begin{cases}\left[\frac{p_{n}}{q_{n}},\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}\right)&\text{
if $n$ is even,}\\\
\left(\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}},\frac{p_{n}}{q_{n}}\right]&\text{ if
$n$ is odd}.\end{cases}$ (4.1)

Here, the convergents are associated to ${\bf k}$. The following estimate on
conditional probability is essential for our computation of dimension. Here
and hereafter, $|\cdot|$ denotes Lebesgue measure.

######  Proposition 4.1 (cf. [IK], Corollary 1.2.6).

If ${\bf k}=(k_{1},\dots,k_{n})$ is a multi-index of natural numbers and
$x\in[0,1]$, then we have

$\frac{\left|\left\\{\alpha\in I({\bf k}):z_{n}(\alpha)\geq
x\right\\}\right|}{\left|I({\bf k})\right|}=\frac{1-x}{\sigma_{n}x+1}$

where $\sigma_{n}:=q_{n-1}q_{n}^{-1}$ ($\sigma_{n}$ is dependent on ${\bf k}$
only).

As a corollary, we obtain

######  Corollary 4.2.

If $0<x<y<1$, we have

$\frac{\left|\\{\alpha\in I({\bf k}):x\leq z_{n}(\alpha)\leq
y\\}\right|}{|I({\bf
k})|}=\frac{(y-x)(\sigma_{n}+1)}{(\sigma_{n}x+1)(\sigma_{n}y+1)}.$

We also have the following consequence of the mass distribution principle, see
[U, Lemma 2.1]. For each $m\in{\mathbb{N}}\cup\\{0\\}$ let
$\mathcal{E}_{m}\subset 2^{[0,1]}$ be a finite collection of nondegenerate
compact intervals in $[0,1]$. By abuse of notation we write

$\cup\mathcal{E}_{m}:=\bigcup_{I\in\mathcal{E}_{m}}I.$

Define for each $I\in\mathcal{E}_{m}$,

$\text{density}(\mathcal{E}_{m+1},I):=\frac{|\cup\mathcal{E}_{m+1}\cap
I|}{|I|}.$

Assume we have the following properties:

  1. (a)

$\cup\mathcal{E}_{0}=[0,1]$.

  2. (b)

For $I\neq J$ in $\mathcal{E}_{m}$, we have $|I\cap J|=0$.

  3. (c)

For every $I\in\mathcal{E}_{m+1}$, there is a unique $J\in\mathcal{E}_{m}$
with $I\subset J$.

  4. (d)

We have that

$\Theta_{m}:=\inf\left\\{\text{density}(\mathcal{E}_{m+1},I):I\in\mathcal{E}_{m}\right\\}>0.$

  5. (e)

If $\text{diam}_{m}$ is the supremum of $\text{diameter}(I)$ over all
$I\in\mathcal{E}_{m}$, then we have

$\lim_{m\to\infty}\text{diam}_{m}=0.$

######  Theorem 4.3 (Theorem 2.1, [U]).

Given the conditions above, let
${E}=\bigcap_{m\in{\mathbb{N}}}\cup\mathcal{E}_{m}$. Then we have that

$1-\dim{{E}}\leq\limsup_{m\to\infty}\frac{\sum_{j=1}^{m-1}\log\Theta_{j}}{\log\operatorname{diam}_{m}}.$

###### Proof of Theorem 1.3.

We have that $g_{\mathbb{R}}\Lambda$ is precompact. Choose consecutive minimal
vectors ${\bf r},{\bf s}\in\Lambda$. By Proposition 3.9, we have that
$x=x({\bf r},{\bf s})$ and $y=y({\bf r},{\bf s})$ are badly approximable. From
Proposition 3.6, we can assume without loss of generality, that
$\varepsilon=\varepsilon({\bf r},{\bf s})=1$. From equation (3.4), ${\bf
r},{\bf s}$ have coordinates given by

$\left[{\begin{array}[]{cc}{\bf r}&{\bf
s}\end{array}}\right]=\left[{\begin{array}[]{cc}|r_{1}|&0\\\
0&|r_{1}|^{-1}\end{array}}\right]\left[{\begin{array}[]{cc}1&-x\\\
\frac{y}{1+xy}&\frac{1}{1+xy}\end{array}}\right].$

And by Proposition 3.3, ${\bf r}$ and ${\bf s}$ form a basis of $\Lambda$.

Write out the continued fraction expansions

$x=[0;b_{1},\dots,b_{k},\dots]\text{ and }y=[0;c_{1},\dots,c_{k},\dots]$

and assume we have $M\in{\mathbb{R}}$ as a uniform upper bound for
$b_{k},c_{k}$. For each $m\in{\mathbb{N}}$, let $B_{m}$ be the block of $2m$
digits

$B_{m}=(c_{m},\dots,c_{1},b_{1},\dots,b_{m}).$

Let $\alpha\in(0,1)$ be any real number which has infinite continued fraction
expansion

$\alpha=[0;d_{1},\dots,d_{k},\dots].$

Assume that, for each $m\in{\mathbb{N}}$, there is a digit $d_{k_{m}}$ with
$k_{m}$ even and such that

$\left(d_{k_{m}-(m-1)}\dots,d_{k_{m}},\dots,d_{k_{m}+m}\right)=B_{m}.$ (4.2)

Moreover, assume that $d_{1}=2$. (This condition on the first digit is not
necessary but taken for convenience in writing out the minimal vectors of
$\Lambda_{\alpha}$.) Of course, there exist badly approximable $\alpha$ with
these properties.

We then claim that
$\Lambda\in\overline{g_{{\mathbb{R}}_{>0}}\Lambda_{\alpha}}$. To prove the
claim, it suffices to show that

$g_{-\log(|r_{1}|)}\Lambda=\left[{\begin{array}[]{cc}1&-x\\\
\frac{y}{1+xy}&\frac{1}{1+xy}\end{array}}\right]\cdot{\mathbb{Z}}^{2}\text{
belongs to }\overline{g_{{\mathbb{R}}_{>0}}\Lambda_{\alpha}}.$ (4.3)

Consider the following consecutive minimal vectors in $\Lambda_{\alpha}$:

${\bf w}_{1}=\left[{\begin{array}[]{c}-1\\\ 0\end{array}}\right]\text{ and
}{\bf w}_{2}=\left[{\begin{array}[]{c}\alpha\\\ 1\end{array}}\right].$

Here we have used that the first digit of $\alpha$ is $2$. According to
Proposition 3.6, the intrinsic coordinates of ${\bf w}_{1},{\bf w}_{2}$ are
given by

$z:=x({\bf w}_{1},{\bf w}_{2})=\alpha,\ w:=y({\bf w}_{1},{\bf w}_{2})=0,\text{
and }\varepsilon:=\varepsilon({\bf w}_{1},{\bf w}_{2})=-1.$

By induction, one sees that it is possible to continue choosing minimal
vectors ${\bf w}_{n}$ for each $n\in{\mathbb{N}}$ such that ${\bf w}_{n},{\bf
w}_{n+1}$ form consecutive pairs and that

$x({\bf w}_{n},{\bf w}_{n+1})=[0;d_{n+1},\dots,d_{n+k},\dots],\ y({\bf
w}_{n},{\bf w}_{n+1})=[0;d_{n},\dots,d_{1}],\ \varepsilon({\bf w}_{n},{\bf
w}_{n+1})=(-1)^{n}.$ (4.4)

Write $x_{n},y_{n},\varepsilon_{n}$ for these coordinates. Again using (3.4)
and the fact that ${\bf w}_{n},{\bf w}_{n+1}$ form a basis of
$\Lambda_{\alpha}$, we write, for some $t_{n}>0$, that

$\Lambda_{\alpha}=\left[{\begin{array}[]{cc}e^{-t_{n}}&0\\\
0&e^{t_{n}}\end{array}}\right]\left[{\begin{array}[]{cc}\varepsilon_{n}&-\varepsilon_{n}x_{n}\\\
\frac{y_{n}}{1+x_{n}y_{n}}&\frac{1}{1+x_{n}y_{n}}\end{array}}\right]{\mathbb{Z}}^{2}.$

Or rather,

$g_{t_{n}}\Lambda_{\alpha}=\left[{\begin{array}[]{cc}\varepsilon_{n}&-\varepsilon_{n}x_{n}\\\
\frac{y_{n}}{1+x_{n}y_{n}}&\frac{1}{1+x_{n}y_{n}}\end{array}}\right]{\mathbb{Z}}^{2}.$

If we choose $n=k_{m}$, the choices in (4.2) and the formulas in (4.4) show
that

$x_{k_{m}}=[0;d_{k_{m}+1},\dots]=[0;b_{1},\dots,b_{m},d_{k_{m}+m+1},\dots],$
$y_{k_{m}}=[0;d_{k_{m}},\dots,d_{k_{m}-(m-1)},\dots,d_{1}]=[0;c_{1},\dots,c_{m},d_{k_{m}-m},\dots,d_{1}]$

and that

$\varepsilon_{k_{m}}=(-1)^{k_{m}}=1.$

Hence the matrices

$\left[{\begin{array}[]{cc}1&-x\\\
\frac{y}{1+xy}&\frac{1}{1+xy}\end{array}}\right]\text{ and
}\left[{\begin{array}[]{cc}\varepsilon_{n}&-\varepsilon_{n}x_{n}\\\
\frac{y_{n}}{1+x_{n}y_{n}}&\frac{1}{1+x_{n}y_{n}}\end{array}}\right]$

can be made arbitrarily close, which establishes the claim in (4.3). Thus we
have shown that $\mathcal{S}_{\Lambda}$ is nonempty, and it remains to
estimate its Hausdorff dimension.

We will work with a convenient subset of $\mathcal{S}_{\Lambda}$. What we have
so far is that the lattice $\Lambda$ with precompact orbit gives sequences of
digits $(b_{n}),(c_{n})$ and a bound $M\in{\mathbb{N}}$ from which we get the
blocks

$B_{m}=\left(c_{m},\dots,c_{1},b_{1},\dots,b_{m}\right)\ \text{ with }1\leq
c_{j},b_{j}\leq M.$

The set we are concerned with is

$\widetilde{\mathcal{S}}_{\Lambda}:=\bigcap_{k}\bigcup_{m\text{ is
even}}\left\\{\alpha\in\mathbf{BA}:a_{1}(\alpha)=2\text{ and
}\big{(}a_{i}(\alpha)\big{)}^{m+k}_{m-(k-1)}=B_{k}\right\\},$

which we have just shown to be a subset of $\mathcal{S}_{\Lambda}$. We are
left to show $\dim\widetilde{\mathcal{S}}_{\Lambda}=1$. Let $L>M$ be a natural
number. Choose a sequence of even numbers $(m_{k})_{k\in{\mathbb{N}}}$ sparse
enough so that

  1. (a)

For each $k>1$, $m_{k}-m_{k-1}>2k$.

  2. (b)

If we define

$\text{density}(\big{(}m_{k}),n\big{)}:=\frac{\sum_{m_{k}<n}2k}{n},\ \text{
then }\lim_{n\to\infty}\text{density}((m_{k}),n)=0.$ (4.5)

We define the following subset of $\widetilde{\mathcal{S}}_{\Lambda}$:

$\mathcal{S}\big{(}L,(m_{k})\big{)}:=\bigcap_{k}\left\\{\alpha\in\tilde{\mathcal{S}}_{\Lambda}:a_{n}(\alpha)\leq
L\text{ for each }n,\text{ and
}\big{(}a_{n}(\alpha)\big{)}^{m_{k}+k}_{m_{k}-k+1}=B_{k}\right\\}.$

This is nothing but the set of badly approximable numbers with $a_{1}=2$
having upper bound $L$ on their digits and having the block $B_{k}$ appear at
the $m_{k}$ position. The upper bound on the entries of $B_{k}$ and condition
(a) above guarantee that this set is nonempty.

We now use the Cantor set structure of $\mathcal{S}\big{(}L,(m_{k})\big{)}$ to
estimate its Hausdorff dimension. According to the notation in Theorem 4.3, we
first define

$\mathcal{E}_{0}:=\\{[0,1]\\}\text{ and
}\mathcal{E}_{1}:=\left\\{\overline{\\{\alpha\in[0,1]:a_{1}(\alpha)=2\\}}\right\\}{=\left\\{[1/3,1/2]\right\\}}.$

Now, assuming we have defined $\mathcal{E}_{m}$ as a collection of closed
cylindrical intervals $\\{I\\}$ as in (4.1), we define the family
$\mathcal{E}_{m+1}$ as follows; $\mathcal{E}_{m+1}:=$

$\begin{cases}\bigcup\limits_{I\in\mathcal{E}_{m}}\bigcup\limits_{i\leq
L}\left\\{\overline{\\{\alpha\in I:a_{m+1}(\alpha)=i\\}}\right\\}&\text{if for
all }k,\ m+1\notin\\{m_{k}-(k-1),\dots,m_{k}+k\\},\\\
\bigcup\limits_{I\in\mathcal{E}_{m}}\left\\{\overline{\\{\alpha\in
I:a_{m+1}(\alpha)=c_{i+1}\\}}\right\\}&\text{if }m+1=m_{k}-i\text{ where
}0\leq i\leq k-1,\\\
\bigcup\limits_{I\in\mathcal{E}_{m}}\left\\{\overline{\\{\alpha\in
I:a_{m+1}(\alpha)=b_{i}\\}}\right\\}&\text{if }m+1=m_{k}+i\text{ where }1\leq
i\leq k.\end{cases}$ (4.6)

The cylindrical intervals of (4.1) are either disjoint or satisfy a
containment relation. Each $I$ in $\mathcal{E}_{m}$ is a set where the first
$m$ digits of the continued fraction are defined. The diameter of these sets
converge to $0$ as $m\to\infty$. Moreover for each $I\in\mathcal{E}_{m}$, we
have

$|\cup\mathcal{E}_{m+1}\cap I|>0.$

Thus the conditions for Theorem 4.3 are satisfied. We also have that

$\mathcal{S}\big{(}L,(m_{k})\big{)}=\bigcap_{m}\cup\mathcal{E}_{m}.$

We now turn to computing $\Theta_{m}$. Assume $\mathcal{E}_{m+1}$ is defined
according to the first case in equation (4.6). If $I\in\mathcal{E}_{m}$, then,
using Proposition 4.1, we can wirte

$\begin{split}|\cup\mathcal{E}_{m+1}\cap I|&=\left|\left\\{\alpha\in
I:a_{m+1}(\alpha)\leq L\right\\}\right|\\\ &=\left|\left\\{\alpha\in
I:z_{m}(\alpha)\geq(L+1)^{-1}\right\\}\right|\\\
&=\frac{1-(L+1)^{-1}}{\sigma_{m}(L+1)^{-1}+1}\cdot|I|.\end{split}$

Here, $\sigma_{m}$ depends on the defining digits for $I$. But since
$\sigma_{m}<1$ always, we get that

$\Theta_{m}\geq\frac{L}{L+2}.$ (4.7)

Now if $\mathcal{E}_{m+1}$ is defined according to the second or third case in
equation (4.6), using Corollary 4.2 and the bound $M$, we have

$\Theta_{m}\geq\frac{1}{4(M+1)^{2}}.$ (4.8)

In order to differentiate these two ways in which $\mathcal{E}_{m+1}$ can be
defined, we introduce, for each $m\in{\mathbb{N}}$, the index sets

$\mathcal{I}_{m}=\bigcap\limits_{k=1}^{\infty}\left\\{j\in{\mathbb{N}}_{<m}:j+1\notin\\{m_{k}-(k-1),\dots,m_{k}+k\\}\right\\}$

and

$\mathcal{J}_{m}=\bigcup\limits_{k=1}^{\infty}\left\\{j\in{\mathbb{N}}_{<m}:j+1\in\\{m_{k}-(k-1),\dots,m_{k}+k\\}\right\\}.$

In considering $\text{diam}_{m}$, we note by induction and Corollary 4.2 that

$\text{diam}_{m}\leq 2^{-m}.$ (4.9)

Thus, we can finally estimate, using (4.7), (4.8), (4.9),

$\begin{split}\frac{-\sum_{j=1}^{m-1}\log\Theta_{j}}{-\log\text{diam}_{m}}&=\frac{1}{-\log\text{diam}_{m}}\left(-\sum\limits_{j\in\mathcal{I}_{m}}\log\Theta_{j}-\sum\limits_{j\in\mathcal{J}_{m}}\log\Theta_{j}\right)\\\
&\leq\frac{1}{m\log
2}\left(-\sum\limits_{j\in\mathcal{I}_{m}}\log\frac{L}{L+2}-\sum\limits_{j\in\mathcal{J}_{m}}\log\frac{1}{4(M+1)^{2}}\right)\\\
&=\frac{-1}{\log
2}\left(\frac{\\#\mathcal{I}_{m}}{m}\log\frac{L}{L+2}+\frac{\\#\mathcal{J}_{m}}{m}\log\frac{1}{4(M+1)^{2}}\right).\end{split}$

Using Theorem 4.3 and (4.5) on the above estimate, we get

$1-\dim\mathcal{S}\big{(}L,(m_{k})\big{)}\leq\limsup_{m\to\infty}\frac{-\sum_{j=1}^{m-1}\log\Theta_{j}}{-\log\text{diam}_{m}}\leq\frac{1}{\log
2}\cdot\log\frac{L+2}{L}.$

Since $\mathcal{S}\big{(}L,(m_{k})\big{)}\subset\mathcal{S}_{\Lambda}$, taking
$L\to\infty$ gives the result. ∎

##  5\. Further questions on Dirichlet improvability and precompact orbits

Some amusing problems on $\nu$-Dirichlet numbers which we omitted are:

  1. (c)

In the case of the Euclidean norm, does
$\mathbf{DI}_{2}\smallsetminus\mathbf{BA}$ have full Hausdorff dimension?

  2. (d)

In general, for two norms $\nu_{1},\nu_{2}$ on ${\mathbb{R}}^{2}$, what can be
said about $\mathbf{DI}_{\nu_{1}}\smallsetminus\mathbf{DI}_{\nu_{2}}$?

We expect (c) to have an affirmative answer, although constructing the
required orbits in $X$ will require a more delicate inductive procedure than
the once produced here.

Dirichlet-improvability can be studied in a variety of different settings. And
each setting relates, via the Dani correspondence, to diagonal orbits on a
homogeneous space avoiding a critical locus. The setting of diagonal flows on
$\operatorname{SL}_{d}({\mathbb{R}})/\operatorname{SL}_{d}({\mathbb{Z}})$,
described in [KR2], is as follows. Fix positive integers $m,n$ with $m+n=d$
and a set of weights

${{\boldsymbol{\omega}}}=({\boldsymbol{\alpha}},{\boldsymbol{\beta}})\in{\mathbb{R}}^{m}_{+}\times{\mathbb{R}}^{n}_{+}\text{
with }\sum_{i}\alpha_{i}=\sum_{j}\beta_{j}=1.$

Fix also a norm $\nu$ on ${\mathbb{R}}^{d}$ and define the critical radius
$r_{\nu}$ in analogy with (1.2). A matrix $A\in M_{m,n}({\mathbb{R}})$ is said
to be $(\nu,{\boldsymbol{\omega}})$-Dirichlet improvable (written
$A\in\mathbf{DI}_{\nu,{\boldsymbol{\omega}}}$) if there is a constant $c<1$
such that

$\Lambda_{A}\cap\left[{\begin{array}[]{cc}\left(ct^{-1}\right)^{\boldsymbol{\alpha}}&0\\\
0&(t)^{\boldsymbol{\beta}}\end{array}}\right]B_{\nu}(r_{\nu})\neq\\{0\\}$

for all $t$ sufficiently large. Here, we define a positive real number $x$
raised to a vector power ${\bf a}\in{\mathbb{R}}^{k}$ as the $k\times k$
matrix

$(x)^{\bf a}:=\operatorname{diag}(x^{a_{1}},\dots,x^{a_{k}}),$

and use the notation

$\Lambda_{A}:=\left[{\begin{array}[]{cc}I_{m}&A\\\
0&I_{n}\end{array}}\right]{\mathbb{Z}}^{d}.$

We also have the set of weighted badly approximable matrices:

$\mathbf{BA}_{\boldsymbol{\omega}}:=\left\\{\text{$A\in
M_{m,n}$}({\mathbb{R}}):\inf_{{\bf p}\in{\mathbb{Z}}^{m},\,{\bf
q}\in{\mathbb{Z}}^{n}\smallsetminus\\{{\bf 0}\\}}\|A{\bf q}-{\bf
p}\|_{\boldsymbol{\alpha}}\|{\bf q}\|_{\boldsymbol{\beta}}>0\right\\}$

where we have used the quasi-norms

$\|{\bf
x}\|_{{\boldsymbol{\alpha}}}:=\max_{i}|x_{i}|^{1/\alpha_{i}}\quad\textrm{and}\quad\|{\bf
y}\|_{{\boldsymbol{\beta}}}:=\max_{j}|y_{j}|^{1/\beta_{j}}.$

Now one can naturally ask for a comparison study between
$\mathbf{DI}_{\nu,{\boldsymbol{\omega}}}$ and
$\mathbf{BA}_{\boldsymbol{\omega}}$. We predict that the same dichotomy
phenomenon as in Theorem 1.5 holds for this case. (See [KR2] for some
preliminary theorems about these sets.)

An approach to this problem via symbolic dynamics is currently unavailable.
However, one can try and modify the construction (via equidistibution of
expanding horocycles) in [KM] to force the resulting precompact orbits to have
prescribed limit points. (Of course, it is necessary that these limit points
themselves have precompact ${\mathbb{R}}$-orbits.) We mention that, although
this technique via equidistribution was available to us, we chose to use
continued fractions in the case of
$\operatorname{SL}_{2}({\mathbb{R}})/\operatorname{SL}_{2}({\mathbb{Z}})$ for
the simplicity of their heuristics and as an exercise in understanding
Poincare sections.

### Acknowledgements

The authors thank Barak Weiss for helpful questions and comments on an earlier
version of the paper. The second-named author thanks S. G. Dani and Anish
Ghosh for their patience and helpful discussions.

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