# The Hilbert cube contains a minimal subshift
of full mean dimension

Lei Jin  Lei Jin: Center for Mathematical Modeling, University of Chile and
UMI 2807 - CNRS<EMAIL_ADDRESS>and  Yixiao Qiao  Yixiao Qiao (Corresponding
author): School of Mathematical Sciences, South China Normal University,
Guangzhou, Guangdong 510631, China<EMAIL_ADDRESS>
###### Abstract.

We construct a minimal dynamical system of mean dimension equal to $1$, which
can be embedded in the shift action on the Hilbert cube $[0,1]^{\mathbb{Z}}$.
Our result clarifies a seemingly plausible impression (about dynamical
embedding possibility in relation to mean dimension) and finally enables us to
have a full understanding of (a pair of) the exact ranges of all possible
values of mean dimension, within which there will always be a minimal
dynamical system that can be (resp. cannot be) embedded in the shift action on
the Hilbert cube.

###### Key words and phrases:

Mean dimension; Embedding; Minimal dynamical system.

###### 2010 Mathematics Subject Classification:

37B99; 54F45.

##  1\. Main result

Mean dimension is a topological invariant of dynamical systems, which
originates with Misha Gromov [Gro99] around 1999. It is closely connected
with, and has been deeply applied to the embedding problem, in particular,
deciding if a minimal dynamical system can be embedded in the shift action on
the Hilbert cube $[0,1]^{\mathbb{Z}}$. As follows is a brief review of the
latest progress in this direction. With the necessary historical background we
shall state our main theorem very quickly. All the precise notions and
notations related to the statements can be found in Section 2, while a
constructive proof of the main theorem is located in Section 3.

We denote by $([0,1]^{\mathbb{Z}},\sigma)$ the shift action on the Hilbert
cube $[0,1]^{\mathbb{Z}}$ (note that its mean dimension is equal to $1$). Let
$(X,T)$ be a minimal dynamical system, whose mean dimension is denoted by
$\operatorname{\mathrm{mdim}}(X,T)$ which takes values within $[0,+\infty]$.
The following statements are classically known:

  * •

If $1<\operatorname{\mathrm{mdim}}(X,T)\leq+\infty$ then $(X,T)$ cannot be
embedded in $([0,1]^{\mathbb{Z}},\sigma)$.

  * •

If $0\leq\operatorname{\mathrm{mdim}}(X,T)<1/2$ then $(X,T)$ can be embedded
in $([0,1]^{\mathbb{Z}},\sigma)$.

  * •

If $1/2\leq\operatorname{\mathrm{mdim}}(X,T)<1$, then it is possible for
$(X,T)$ to be embedded in $([0,1]^{\mathbb{Z}},\sigma)$, whereas it is also
possible that $(X,T)$ cannot be embedded in $([0,1]^{\mathbb{Z}},\sigma)$.

  * •

Suppose $\operatorname{\mathrm{mdim}}(X,T)=1$. It may happen that $(X,T)$
cannot be embedded in $([0,1]^{\mathbb{Z}},\sigma)$.

Here let us make some remarks on the above assertions. The first implication
follows directly from definition. The second significant result is due to
Gutman and Tsukamoto [GT20] (for a corresponding result of
$\mathbb{Z}^{k}$-actions we refer to [GQT19]). In relation to the third and
fourth cases, embeddable and non-embeddable examples were constructed by
Lindenstrauss–Weiss [LW00] and Lindenstrauss–Tsukamoto [LT14], respectively.
To be precise, Lindenstrauss and Tsukamoto [LT14] showed that for any given
$r\geq 1/2$ there is a minimal dynamical system of mean dimension equal to
$r$, which cannot be embedded in $([0,1]^{\mathbb{Z}},\sigma)$; Lindenstrauss
and Weiss [LW00] proved that for any given $0\leq r<1$ there exists a minimal
dynamical system of mean dimension equal to $r$, which can be embedded in
$([0,1]^{\mathbb{Z}},\sigma)$.

A careful reader may observe that in order to have a complete picture of the
possibility of embedding minimal dynamical systems in the Hilbert cube, there
is only one issue that remains open, namely, the embeddable situation for the
critical value $1$ (with mean dimension) is not clear to us. Formally, we
study the problem as follows:

  * •

Assume that a minimal dynamical system can be embedded in
$([0,1]^{\mathbb{Z}},\sigma)$. Does this imply that it has mean dimension
strictly less than $1$?

In other words, this problem asks if there exists a minimal dynamical system
of mean dimension equal to $1$, which can be embedded in
$([0,1]^{\mathbb{Z}},\sigma)$. The aim of this paper is to solve this problem.

In fact, we observe the following three aspects:

  1. (1)

With a view towards mean dimension theory: The approach of any previously
mentioned results [LW00, LT14] reveals that if we try to construct a minimal
dynamical system of mean dimension equal to $1$, then a construction within
the framework of $[0,1]^{\mathbb{Z}}$ is not adequate for our purpose, and
instead, we need consider, with the same method, the shift action on
$([0,1]^{2})^{\mathbb{Z}}$, which has mean dimension equal to $2$, rather than
consider the shift action on $[0,1]^{\mathbb{Z}}$.

  2. (2)

With a view towards entropy theory: We note moreover that a highly similar
circumstance took place in topological entropy in connection with the shift
action over any alphabet of finite cardinality; strictly speaking, such an
“analogue” asserts that any proper subsystem of the shift action on
$\\{k\in\mathbb{Z}:1\leq k\leq N\\}^{\mathbb{Z}}$ (where $N$ is a positive
integer) does not have full topological entropy (i.e. its topological entropy
must be strictly less than $\log N$).

  3. (3)

With a view towards dimension theory: A celebrated theorem in dimension theory
states that if a compact metrizable space of (topological) dimension $D$
(where $D$ is a positive integer) can be (topologically) embedded in
$[0,1]^{D}$, then it must contain (a copy of) $[0,1]^{D}$ (i.e. $[0,1]^{D}$
can be embedded in it).

Apparently, all of these three notable phenomena lead naturally to a seemingly
plausible impression, i.e. it would be true that minimal dynamical systems
satisfying the embeddable property are not able to possess full mean
dimension. However, this impression turns out to be false. Somewhat
surprisingly, we do construct a minimal dynamical system successfully, which
satisfies both embeddable and full mean dimensional conditions. This is our
main result.

######  Theorem 1.1 (Main theorem).

There exists a minimal dynamical system of mean dimension equal to $1$, which
can be embedded in the shift action on the Hilbert cube $[0,1]^{\mathbb{Z}}$.

In contrast to the seemingly reasonable observation as stated above, Theorem
1.1 enables a different (and unexpected) behaviour to become clarified.
Furthermore, our main theorem, together with the previous remarkable results,
eventually allows the embeddability of minimal dynamical systems in the
Hilbert cube to be fully understood with a view towards mean dimension. More
precisely, we have now gotten a full understanding of (a pair of) the exact
ranges (I) (i.e. $[0,1]$) and (II) (i.e. $[1/2,+\infty]$) of all possible
values of mean dimension, such that for any value assigned within (I) (resp.
within (II)) there always exists a minimal dynamical system of mean dimension
equal to this appointed value, which can be (resp. cannot be) embedded in the
shift action on the Hilbert cube. Notice that with Theorem 1.1 the range (I)
finally becomes exact, whereas the range (II) is exact because of [LT14] for
the existence of such a system of mean dimension within $[1/2,+\infty]$ and
because of [GT20] for the non-existence of such a system of mean dimension
within $[0,1/2)$.

The strategy we shall adopt generally follows the technical framework of the
“block-type” induction. Nevertheless, we have to make each step in our
construction sufficiently delicate. The key ingredient of our idea is to
produce a dense subset of the alphabet $[0,1]$ “more gently” (i.e. with a
sequence of closed intervals instead of points).

######  Remark 1.2.

The same statement (as in Theorem 1.1) also applies to the alphabet
$[0,1]^{D}$; namely, the following assertion is true.

  * •

Let $D$ be a positive integer (possibly $+\infty$). There exists a minimal
dynamical system of mean dimension equal to $D$, which can be embedded in the
shift action on $([0,1]^{D})^{\mathbb{Z}}$.

Meanwhile, the alphabet $[0,1]^{D}$ of topological dimension $D$ may be
replaced by some other compact metrizable spaces provided they have some nice
structure, e.g. polyhedrons $P$ of topological dimension $\dim(P)$. Besides,
we notice that it is straightforward to generalize our main theorem to actions
of infinite (countable discrete) amenable groups (with almost no additional
effort while using material on tilings of amenable groups).

### Acknowledgements

The initial proof we provided in a previous version of this paper contained a
gap. Professor Masaki Tsukamoto pointed out this to us, and meanwhile, he also
kindly explained to us how to fix it. We would like to thank him for his warm
help. We are also grateful to the anonymous referee for his/her insightful
comments and helpful suggestions which improve this paper greatly. L. Jin was
supported by Basal Funding AFB 170001 and Fondecyt Grant No. 3190127. Y. Qiao
was supported by NNSF of China No. 11901206.

##  2\. Terminologies

This section is devoted to a brief review of all the fundamental definitions
appearing in the previous section. By a (topological) dynamical system we
shall understand a pair $(X,T)$, where $X$ is a compact metrizable space and
$T:X\to X$ is a homeomorphism. An important class of dynamical systems are
minimal systems. A dynamical system $(X,T)$ is said to be minimal if for every
point $x\in X$ the set $\\{T^{n}x:n\in\mathbb{Z}\\}$ is dense in $X$. Among
other typical examples of dynamical systems, probably the most canonical ones
are the shift action on the Hilbert cubes $([0,1]^{D})^{\mathbb{Z}}$ (where
$D$ is a positive integer or $+\infty$), which we denote by
$(([0,1]^{D})^{\mathbb{Z}},\sigma)$, defined as follows:

$\sigma:([0,1]^{D})^{\mathbb{Z}}\to([0,1]^{D})^{\mathbb{Z}},\quad(x_{n})_{n\in\mathbb{Z}}\mapsto(x_{n+1})_{n\in\mathbb{Z}}.$

For simplicity we assume $D=1$. We say that a dynamical system $(X,T)$ can be
embedded in $([0,1]^{\mathbb{Z}},\sigma)$ if there is an equivariant
topological embedding $f:X\to[0,1]^{\mathbb{Z}}$, namely, a homeomorphism of
$X$ into $[0,1]^{\mathbb{Z}}$ satisfying $f\circ T=\sigma\circ f$. Such a
mapping $f$ is usually called an embedding of the dynamical system $(X,T)$ in
$([0,1]^{\mathbb{Z}},\sigma)$. As we have seen in Section 1, this paper mainly
focuses on the embeddability of minimal dynamical systems in
$([0,1]^{\mathbb{Z}},\sigma)$.

We denote by $\dim(P)$ the topological dimension (i.e. the Lebesgue covering
dimension) of a compact metrizable space $P$ (which is always assumed to be
nonempty). Let $X$ and $P$ be two compact metrizable spaces and $\rho$ a
compatible metric on $X$. For $\epsilon>0$ a continuous mapping $f:X\to P$ is
called an $\epsilon$-embedding with respect to $\rho$ if $f(x)=f(x^{\prime})$
implies $\rho(x,x^{\prime})<\epsilon$, for all $x,x^{\prime}\in X$. Let
$\operatorname{\mathrm{Widim}}_{\epsilon}(X,\rho)$ be the minimum topological
dimension $\dim(P)$ of a compact metrizable space $P$ which admits an
$\epsilon$-embedding $f:X\to P$ with respect to $\rho$.

######  Remark 2.1.

We may verify that the topological dimension of $X$ may be recovered by
$\dim(X)=\lim_{\epsilon\to
0}\operatorname{\mathrm{Widim}}_{\epsilon}(X,\rho)$.

Let $(X,T)$ be a dynamical system with a compatible metric $\rho$ on $X$. For
every positive integer $n$ we define on $X$ a compatible metric $\rho_{n}$ as
follows:

$\rho_{n}(x,x^{\prime})=\max_{0\leq
i<n}\rho(T^{i}x,T^{i}x^{\prime}),\quad\forall\,x,x^{\prime}\in X.$

The mean dimension of $(X,T)$ is defined by

$\operatorname{\mathrm{mdim}}(X,T)=\lim_{\epsilon\to
0}\lim_{n\to+\infty}\frac{\operatorname{\mathrm{Widim}}_{\epsilon}(X,\rho_{n})}{n}.$

It is well known that the limits in the above definition always exist, and the
value $\operatorname{\mathrm{mdim}}(X,T)$ is independent of the choices of a
compatible metric $\rho$ on $X$ and a Følner sequence
$\\{F_{n}\\}_{n=1}^{+\infty}$ (instead of $\\{k\in\mathbb{Z}:0\leq
k<n\\}_{n=1}^{+\infty}$) of $\mathbb{Z}$.

We note that the mean dimension of $(([0,1]^{D})^{\mathbb{Z}},\sigma)$ is
equal to $D$, where $D$ is a positive integer or $+\infty$. For details we
refer the reader to [LW00]. Clearly, if a dynamical system can be embedded in
$(([0,1]^{D})^{\mathbb{Z}},\sigma)$, then its mean dimension must be less than
or equal to $D$.

##  3\. Proof of Theorem 1.1

###  3.1. Construction of $(X,\sigma)$

We are going to construct a minimal dynamical system $(X,\sigma)$ such that
$X$ is a closed and shift-invariant subset of the Hilbert cube
$[0,1]^{\mathbb{Z}}$ and that the mean dimension of $(X,\sigma)$ is equal to
$1$.

First of all, we fix a compatible metric $d$ on $[0,1]^{\mathbb{Z}}$ as
follows:

$d(x,x^{\prime})=\sum_{n\in\mathbb{Z}}\frac{|x_{n}-x^{\prime}_{n}|}{2^{|n|}},\quad\left(x=(x_{n})_{n\in\mathbb{Z}},x^{\prime}=(x^{\prime}_{n})_{n\in\mathbb{Z}}\in[0,1]^{\mathbb{Z}}\right).$

For $x=(x_{n})_{n\in\mathbb{Z}}\in[0,1]^{\mathbb{Z}}$ and two integers $t\leq
t^{\prime}$ we set $x|_{t}^{t^{\prime}}=(x_{n})_{t\leq n\leq t^{\prime}}$. For
a closed interval $I$ we denote its length by $|I|$.

The construction will be fulfilled by induction. To start with, let us explain
the intuitive meaning of our notations shortly. For a nonnegative integer $k$
we build in the $k$-th step a closed shift-invariant subset $X_{k}$ of
$[0,1]^{\mathbb{Z}}$, which is generated from a “block” $B_{k}$ of the form
$\prod_{j=1}^{b_{k}}I_{j}^{(k)}$, where each $I_{j}^{(k)}$ is a closed
subinterval of $[0,1]$ and where the positive integer $b_{k}$ indicates the
“length” of the block $B_{k}$. Roughly speaking, employing a family of closed
intervals in our construction is actually to increase dimension. But this is
not enough for its mean dimension to be dominated from below. Therefore we are
going through the following approach: When we deal with the next block
$B_{k+1}$ in the $(k+1)$-th step we have to “copy” the block $B_{k}$
sufficiently many times in order to occupy a rather large proportion of
positions of the block $B_{k+1}$. The positive integer $r_{k}$ is to describe
the number of copies of the block $B_{k}$ in the block $B_{k+1}$, whereas for
an integer $0\leq s<k+1$, the number $\eta(s,k+1)>0$ is to dominate the
“density” of those intervals of length greater than or equal to
$1/2^{s(s+1)/2}$ in the block $B_{k+1}$, which will strictly decrease as $k$
increases, while the number $\eta(s)>0$ is to guarantee that the lower bound
for the value $\eta(s,k+1)$ is always under control as $k+1$ ranges over all
those integers greater than $s$, which will finally converge to $1$ as $s$
goes to $+\infty$.

Step $-\infty$. We take a two-parameter sequence $\\{\eta(s,k):0\leq
s<k;s,k\in\mathbb{Z}\\}$ of real numbers and a one-parameter sequence
$\\{\eta(s)\\}_{s=0}^{+\infty}$ of real numbers satisfying all the following
conditions (the existence of such two sequences is obvious):

  1. (1)

For any two integers $0\leq s<k$: $\;0<\eta(s,k)<1$; $\,0<\eta(s)<1$.

  2. (2)

For any nonnegative integer $s$ the one-parameter sequence
$\\{\eta(s,k)\\}_{k=s+1}^{+\infty}$ is strictly decreasing, namely

$\eta(s,k+1)<\eta(s,k),\;\quad\forall\,0\leq s<k\quad(s,k\in\mathbb{Z}).$

  3. (3)

For any nonnegative integer $s$ the one-parameter sequence
$\\{\eta(s,k)\\}_{k=s+1}^{+\infty}$ is bounded by $\eta(s)$ from below, i.e.

$\eta(s,k)>\eta(s),\;\quad\forall\,0\leq s<k\quad(s,k\in\mathbb{Z}).$

  4. (4)

$\eta(s)\to 1$ as $s\to+\infty$.

We fix them throughout this section. We would like to remind the reader that
we shall use these conditions implicitly in the sequel.

Step $0$. We take $b_{0}=1$, $B_{0}=I_{1}^{(0)}=[0,1]$ and
$X_{0}=[0,1]^{\mathbb{Z}}$.

Step $1$. We divide $I_{1}^{(0)}=[0,1]$ equally into $2$ closed subintervals
of length $1/2$ as follows:

$I_{1,1}^{(1)}=[0,1/2],\quad\;I_{1,2}^{(1)}=[1/2,1].$

We take a positive integer $r_{0}$ with

$\frac{r_{0}}{r_{0}+2}\geq\eta(0,1).$

We put $b_{1}=r_{0}+2$ and set

$B_{1}=(B_{0})^{r_{0}}\times I_{1,1}^{(1)}\times
I_{1,2}^{(1)}\subset[0,1]^{b_{1}}.$

We rename the $b_{1}$ closed subintervals of $[0,1]$ appearing in the above
product with indices, and rewrite $B_{1}$ as follows:
$B_{1}=\prod_{j=1}^{b_{1}}I_{j}^{(1)}$. More precisely, we notice here (in
relation to the indices) that $B_{0}=I_{1}^{(1)}$ and that the sets
$I_{b_{1}-1}^{(1)}$ and $I_{b_{1}}^{(1)}$ form a cover of $B_{0}$. Let $X_{1}$
be the set of all those $x=(x_{n})_{n\in\mathbb{Z}}\in[0,1]^{\mathbb{Z}}$
satisfying the following condition:

  * •

There exists some integer $0\leq l\leq b_{1}-1$ such that
$x|_{l+b_{1}m+1}^{l+b_{1}(m+1)}\in B_{1}$ for all $m\in\mathbb{Z}$.

Obviously, $X_{1}$ is a nonempty closed and shift-invariant subset of
$X_{0}=[0,1]^{\mathbb{Z}}$.

To proceed, we assume for a positive integer $k$ that $b_{k-1}$, $B_{k-1}$ and
$X_{k-1}$ have already been generated in Step $(k-1)$. We now generate
$b_{k}$, $B_{k}$ and $X_{k}$.

Step $k$. We write $B_{k-1}=\prod_{j=1}^{b_{k-1}}I_{j}^{(k-1)}$, where every
$I_{j}^{(k-1)}$ is a closed subinterval of $[0,1]$. For each $1\leq j\leq
b_{k-1}$ we divide $I_{j}^{(k-1)}$ equally into $2^{k}$ closed subintervals of
length equal to $|I_{j}^{(k-1)}|/2^{k}$, which are denoted by $I_{j,i}^{(k)}$
with $1\leq i\leq 2^{k}$. We take a positive integer $r_{k-1}$ sufficiently
large such that for every integer $0\leq s\leq k-1$

$\frac{\left(\\#\\{1\leq j\leq b_{k-1}:|I_{j}^{(k-1)}|\geq
1/2^{s(s+1)/2}\\}\right)\cdot
r_{k-1}}{b_{k-1}\cdot(r_{k-1}+2^{kb_{k-1}})}\geq\eta(s,k),$

where the symbol $\\#$ records the cardinality of a set. We put
$b_{k}=b_{k-1}\cdot(r_{k-1}+2^{kb_{k-1}})$ and set

$B_{k}\,=\,(B_{k-1})^{r_{k-1}}\,\times\,\prod_{i_{1},\dots,i_{b_{k-1}}\in\\{1,\dots,2^{k}\\}}\;\prod_{j=1}^{b_{k-1}}\,I_{j,i_{j}}^{(k)}\;\,\subset\,(B_{k-1})^{r_{k-1}+2^{kb_{k-1}}}\,.$

We note that $B_{k}$ is a closed subset of $[0,1]^{b_{k}}$. We rewrite $B_{k}$
as $B_{k}=\prod_{j=1}^{b_{k}}I_{j}^{(k)}$, where
$I_{1}^{(k)},\dots,I_{b_{k}}^{(k)}$ are closed subintervals of $[0,1]$. To be
precise here we remark that in relation to the indices we may ensure in
addition:

$\bigcup_{m=r_{k-1}}^{r_{k-1}+2^{kb_{k-1}}-1}\,\prod_{j=b_{k-1}m+1}^{b_{k-1}(m+1)}I_{j}^{(k)}\,=\,B_{k-1}\,=\,\prod_{j=1}^{b_{k-1}}I_{j}^{(k)}.$

We let $X_{k}$ be the set consisting of all the points
$x=(x_{n})_{n\in\mathbb{Z}}\in[0,1]^{\mathbb{Z}}$ satisfying the following
condition:

  * •

There is some integer $0\leq l\leq b_{k}-1$ such that
$x|_{l+b_{k}m+1}^{l+b_{k}(m+1)}\in B_{k}$ for all $m\in\mathbb{Z}$.

It follows that $X_{k}$ is a (nonempty) closed and shift-invariant subset of
$X_{k-1}\subset[0,1]^{\mathbb{Z}}$.

Step $+\infty$. The induction has now been completed, as we have already
generated $b_{k}$, $B_{k}$ and $X_{k}$ for all nonnegative integers $k$. To
end the construction, we finally take the intersection as follows:

$X=\bigcap_{k=0}^{+\infty}X_{k}.$

Since $\\{X_{k}\\}_{k=0}^{+\infty}$ is a decreasing sequence of nonempty
closed shift-invariant subsets of $[0,1]^{\mathbb{Z}}$, $X$ is a nonempty
closed shift-invariant subset of $[0,1]^{\mathbb{Z}}$ as well. Thus,
$(X,\sigma)$ becomes a dynamical system. In what follows we need verify that
it does satisfy the required conditions.

###  3.2. Minimality of $(X,\sigma)$

We will show that the dynamical system $(X,\sigma)$ is minimal. In fact, it
suffices to show that for any $x,y\in X$ and any $\epsilon>0$ there exists
some $M\in\mathbb{Z}$ such that $d(\sigma^{M}(x),y)<\epsilon$. Let us fix
$x=(x_{n})_{n\in\mathbb{Z}},y=(y_{n})_{n\in\mathbb{Z}}\in X$ and $\epsilon>0$.

For each positive integer $k$ we set

$M_{k}=\max\left\\{|I_{j}^{(k)}|:b_{k-1}\cdot r_{k-1}+1\leq j\leq
b_{k}\right\\}.$

This sequence $\\{M_{k}\\}_{k=1}^{+\infty}$ is deceasing and converges to $0$
as $k$ goes to $+\infty$. Actually, it follows from the construction that the
value $M_{k}$ is equal to $1/2^{k}$.

We choose a positive integer $L$ depending only on $\epsilon>0$ such that if
two points $x^{\prime}=(x^{\prime}_{n})_{n\in\mathbb{Z}}$ and
$x^{\prime\prime}=(x^{\prime\prime}_{n})_{n\in\mathbb{Z}}$ coming from
$[0,1]^{\mathbb{Z}}$ satisfy that
$|x^{\prime}_{n}-x^{\prime\prime}_{n}|<\epsilon/2$ for any $-L\leq n\leq L$,
then they will satisfy $d(x^{\prime},x^{\prime\prime})<\epsilon$.

We take an integer $N>2L+1$ with $1/2^{N}<\epsilon/2$. Let us look at the
$N$-th, $(N+1)$-th and $(N-1)$-th steps of the construction.

Since $x,y\in X\subset X_{N}$, there must exist two integers $0\leq p,q\leq
b_{N}-1$ such that

$x|_{p+b_{N}m+1}^{p+b_{N}(m+1)},y|_{q+b_{N}m+1}^{q+b_{N}(m+1)}\in
B_{N},\quad\forall m\in\mathbb{Z}.$

There are two integers $L_{0}\in[-L,L]$ and $m_{0}\in\\{-1,0\\}$, both of
which are uniquely determined, satisfying that

$[-L,L_{0}-1]\subset[q+b_{N}(m_{0}-1)+1,q+b_{N}m_{0}],$
$[L_{0},L]\subset[q+b_{N}m_{0}+1,q+b_{N}(m_{0}+1)].$

Here we assume by convention $[-L,-L-1]=\emptyset$.

Let us make it clearer with a short remark. Strictly speaking, the general
case is where $L_{0}\in[-L+1,L]$ (and hence where we will have
$L_{0}=q+b_{N}m_{0}+1$), with an exception (which turns out to be simpler than
the general case) if such an integer $L_{0}\in[-L+1,L]$ does not exist, for
which we set $L_{0}=-L$. As we will see in a moment, the exceptional case is
contained in (a part of) the general case. Therefore we assume without loss of
generality $-L+1\leq L_{0}\leq L$.

Since

$y|_{q+b_{N}m_{0}+1}^{q+b_{N}(m_{0}+1)}\in
B_{N}=\bigcup_{m=r_{N}}^{r_{N}+2^{(N+1)b_{N}}-1}\prod_{j=b_{N}m+1}^{b_{N}(m+1)}I_{j}^{(N+1)},$

there is an integer $r_{N}\leq m_{1}\leq r_{N}+2^{(N+1)b_{N}}-1$ with

$y|_{q+b_{N}m_{0}+1}^{q+b_{N}(m_{0}+1)}\in\prod_{j=b_{N}m_{1}+1}^{b_{N}(m_{1}+1)}I_{j}^{(N+1)}.$

Since $x\in X\subset X_{N+1}$ and since

$B_{N+1}=(B_{N})^{r_{N}}\times\prod_{j=b_{N}r_{N}+1}^{b_{N+1}}I_{j}^{(N+1)},$

there is some integer $m_{2}$ depending on $m_{1}$ such that

$x|_{p+b_{N}m_{2}+1}^{p+b_{N}(m_{2}+1)}\in\prod_{j=b_{N}m_{1}+1}^{b_{N}(m_{1}+1)}I_{j}^{(N+1)}.$

Thus, for every integer $1\leq i\leq b_{N}$

$|x_{p+b_{N}m_{2}+i}-y_{q+b_{N}m_{0}+i}|\leq M_{N+1}.$

This implies that for every integer $n\in[q+b_{N}m_{0}+1,q+b_{N}(m_{0}+1)]$,
in particular, for every integer $n\in[L_{0},L]$

$|\sigma^{b_{N}(m_{2}-m_{0})+p-q}(x)_{n}-y_{n}|\leq
M_{N+1}=1/2^{N+1}<\epsilon/2.$

Since

$\sigma^{b_{N}(m_{2}-m_{0})+p-q}(x)|_{q+b_{N}(m_{0}-1)+1}^{q+b_{N}m_{0}},y|_{q+b_{N}(m_{0}-1)+1}^{q+b_{N}m_{0}}\in
B_{N}$

and since

$2L+1\leq N-1\leq b_{N-1},$

we deduce that for every integer $n\in[-L,L_{0}-1]$

$|\sigma^{b_{N}(m_{2}-m_{0})+p-q}(x)_{n}-y_{n}|\leq M_{N}=1/2^{N}<\epsilon/2.$

Thus, we conclude that for all integers $n\in[-L,L]$

$|\sigma^{b_{N}(m_{2}-m_{0})+p-q}(x)_{n}-y_{n}|<\epsilon/2,$

which implies

$d(\sigma^{b_{N}(m_{2}-m_{0})+p-q}(x),y)<\epsilon.$

###  3.3. Mean dimension of $(X,\sigma)$

We shall show that $\operatorname{\mathrm{mdim}}(X,\sigma)=1$. Since it is
obvious that the dynamical system $(X,\sigma)$ has been embedded in
$([0,1]^{\mathbb{Z}},\sigma)$ already, its mean dimension does not exceed $1$.
Thus, it suffices to prove $\operatorname{\mathrm{mdim}}(X,\sigma)\geq 1$.
This will end the paper.

We borrow a practical lemma [Gro99, Lemma 1.1.1][LT14, Example 2.1] as
follows. The point which will be crucial to our argument is that the equality
presented in this lemma is true for all positive integers $n$ and all
sufficiently small $\delta>0$ (which should be strictly less than $\tau>0$).

######  Lemma 3.1.

For any $0<\delta<\tau$ and any positive integer $n$

$\operatorname{\mathrm{Widim}}_{\delta}\left([0,\tau]^{n},d_{l^{\infty}}\right)=n.$

Here $d_{l^{\infty}}$ is the compatible metric on $[0,\tau]^{n}$ defined by

$d_{l^{\infty}}\left((x_{i})_{i=1}^{n},(x^{\prime}_{i})_{i=1}^{n}\right)=\max_{1\leq
i\leq
n}|x_{i}-x^{\prime}_{i}|,\;\quad\forall\,(x_{i})_{i=1}^{n},(x^{\prime}_{i})_{i=1}^{n}\in[0,\tau]^{n}.$

We fix a point $z\in X$ satisfying that $z|_{b_{k}m}^{b_{k}(m+1)-1}\in B_{k}$
for any nonnegative integer $k$ and any $m\in\mathbb{Z}$. For each nonnegative
integer $k$ we define a mapping as follows:

$F_{k}:B_{k}\to X,\,\quad
x=(x_{n})_{n=0}^{b_{k}-1}\,\mapsto\,F_{k}(x)=(F_{k}(x)_{n})_{n\in\mathbb{Z}},$
$F_{k}(x)_{n}=\begin{cases}x_{n},\quad&0\leq n\leq b_{k}-1\\\
z_{n},\quad&n\in\mathbb{Z}\setminus[0,b_{k}-1]\end{cases}.$

It is clear that $F_{k}(x)$ (where $x\in B_{k}$) is indeed in $X$. Moreover,
the mapping $F_{k}:B_{k}\to X$ is continuous and distance-increasing with
respect to the compatible metric $d_{l^{\infty}}$ on $B_{k}$ and the
compatible metric $d_{b_{k}}$ on $X$, i.e. it satisfies that

$d_{l^{\infty}}(x,x^{\prime})\leq
d_{b_{k}}(F_{k}(x),F_{k}(x^{\prime})),\quad\forall\,x,x^{\prime}\in B_{k}.$

It follows that for every $\epsilon>0$ and every nonnegative integer $k$

$\operatorname{\mathrm{Widim}}_{\epsilon}(B_{k},d_{l^{\infty}})\leq\operatorname{\mathrm{Widim}}_{\epsilon}(X,d_{b_{k}}).$

Now let us take $\epsilon>0$ arbitrarily. We choose a nonnegative integer
$s=s(\epsilon)$ depending only on $\epsilon>0$ such that

$1/2^{(s+1)(s+2)/2}\leq\epsilon<1/2^{s(s+1)/2}.$

Note that for all integers $k>s$

$\frac{\\#\\{1\leq j\leq b_{k}:|I_{j}^{(k)}|\geq
1/2^{s(s+1)/2}\\}}{b_{k}}\geq\eta(s,k).$

By Lemma 3.1 this implies that for all integers $k>s$

$\frac{\operatorname{\mathrm{Widim}}_{\epsilon}(B_{k},d_{l^{\infty}})}{b_{k}}\geq\eta(s,k)\geq\eta(s).$

Thus, for all integers $k>s$

$\frac{\operatorname{\mathrm{Widim}}_{\epsilon}(X,d_{b_{k}})}{b_{k}}\geq\eta(s).$

Since $\epsilon>0$ is arbitrary and since $s\to+\infty$ as $\epsilon\to 0$, we
conclude with

$\operatorname{\mathrm{mdim}}(X,\sigma)=\lim_{\epsilon\to
0}\lim_{k\to+\infty}\frac{\operatorname{\mathrm{Widim}}_{\epsilon}(X,d_{b_{k}})}{b_{k}}\geq\lim_{\epsilon\to
0}\eta(s)=1.$

## References

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