Proof. Let $f:X\to X$ be an $N$-expansive homeomorphism of a compact metric space. We shall prove that every $N$-expansivity constant $e$ of $f$ is a uniform $N$-expansivity constant of $f$.

Suppose not. Then, there is $\epsilon >0$ such that no $M\in \mathbb {N}$ satisfies the conclusion in definition 3. From this, we obtain $N+1$ sequences

\[ (x^M_1)_{M\in\mathbb{N}},\cdots,(x^M_{N+1})_{M\in\mathbb{N}} \]

in $X$ such that

(1.1)\begin{equation} {\rm d}(f^i(x^M_l),f^i(x^M_s))\leq e,\,\forall -M\leq i\leq M, \forall l,s\in\{1,\cdots, N+1\} \end{equation}

but

(1.2)\begin{equation} {\rm d}(x^M_l,x^M_s)>\epsilon,\quad\forall M\in \mathbb{N}, \,\forall\text{ distinct }l,s\in\{1,\cdots, N+1\}. \end{equation}

Since $X$ is compact, we can assume that $\forall l\in \{1,\cdots, N+1\}$ $\exists x_l\in X$ such that $x^M_l\to x_l$ as $M\to \infty$, $\forall l\in \{1,\cdots, N+1\}$.

Since $f$ is continuous, we can obtain the inequalities below by fixing $i\in \mathbb {Z}$ and letting $M\to \infty$ in (1.1):

\[ {\rm d}(f^i(x_l),f^i(x_r))\leq e,\quad\forall i\in\mathbb{Z}, \,\forall l,s\in\{1,\cdots, N+1\}. \]

This proves

\[ \{x_1,\cdots, x_{N+1}\}\subset\Gamma_e(x_1). \]

But $e$ is an $N$-expansivity constant so $\Gamma _e(x_1)$ has at most $N$ elements thus the above inclusion implies

\[ x_l=x_s,\quad\text{for some distinct }l,s\in\{1,\cdots, N+1\}. \]

However, by letting $M\to \infty$ in (1.2), one gets ${\rm d}(x_l,x_s)\geq \epsilon$ hence

\[ x_l\neq x_s,\quad \forall\text{ distinct }l,s\in \{1,\cdots, N+1\}. \]

This is a contradiction which completes the proof.

Proof. It suffices to prove that every uniformly $N$-expansive constant $e$ of a bijective map of a metric space $f:X\to X$ is a weak asymptotic $N$-expansivity constant.

Suppose not. More precisely, that $e$ is not a weak asymptotic $N$-expansivity constant. Then, by definition 4, there are $x_1,\cdots, x_{N+1}\in X$ satisfying

(1.3)\begin{equation} {\rm d}(f^j(x_l),f^j(x_s))\leq e,\quad\forall j\geq0, \ l,s\in\{1,\cdots, N+1\}, \end{equation}

but

\[ \lim_{i\to\infty}\inf\{{\rm d}(f^i(x_l),f^i(x_s)):l,s\in \{1,\cdots N+1\}\text{ are distinct}\}\neq0. \]

From this limit, we get $\epsilon >0$ and a sequence

(1.4)\begin{equation} (i_k)_{k\in\mathbb{N}}\to \infty\quad \text{(as } k\to\infty) \end{equation}

such that

\[ \inf\{{\rm d}(f^{i_k}(x_l),f^{i_k}(x_s)):l,s\in \{1,\cdots N+1\}\text{ are distinct}\}>\epsilon,\quad\forall k\in\mathbb{N}. \]

Then,

(1.5)\begin{equation} {\rm d}(f^{i_k}(x_l)f^{i_k}(x_s))>\epsilon, \end{equation}

for all $k\in \mathbb {N}$ and all distinct $l,s\in \{1,\cdots, N+1\}$.

Now, recall that $e$ is a uniformly $N$-expansivity constant of $f$ so we can apply definition 3. Then, by taking

\[ \epsilon_*=\frac{\epsilon}2 \]

in this definition we obtain $M\in \mathbb {N}$ such that if $x_1^*,\cdots, x_{N+1}^*\in X$ and

(1.6)\begin{equation} {\rm d}(x_l^*,x_s^*)>\epsilon_*,\quad\forall \text{ distinct }l,s\in\{1,\cdots, N+1\}, \end{equation}

then

\[ {\rm d}(f^i(x_l^*),f^i(x_s^*))> e, \]

for some $-M\leq i\leq M$ and some $l,s\in \{1,\cdots, N+1\}$.

By (1.4) we can fix $k\in \mathbb {N}$ such that

\[ i_k\geq M \]

and, by (1.5),

\[ {\rm d}(f^{i_k}(x_l),f^{i_k}(x_s))>\epsilon_*,\quad\forall\text{ distinct }l,s\in\{1,\cdots, N+1\}. \]

Then, defining $x_l^*=f^{i_k}(x_l)$ for $1\leq l\leq N+1$ we obtain

\[ x_1^*,\cdots,x^*_{N+1}\in X \]

satisfying (1.6). So, the choice of $M$ implies that there are

\[{-}M\leq i\leq M \]

and two (necessarily distinct) indexes $l,s\in \{1,\cdots, N+1\}$ such that

\[ {\rm d}(f^i(f^{i_k}(x_l)),f^i(f^{i_k}(x_s)))>e. \]

Therefore,

\[ {\rm d}(f^j(x_l),f^j(x_s))>e\quad\text{where}\quad j=i_k+i. \]

Since $j=i_k+i\geq i_k-M\geq 0,$ we contradict (1.3) completing the proof.

Question 2 Is every weak asymptotically $N$-expansive homeomorphism of a compact metric space asymptotically $N$-expansive?

Proof. Consider a nonchaotic weak asymptotically $N$-expansive map of a metric space $f:X\to X$. Let $e$ be a weak asymptotic $N$-expansivity constant of $f$. Suppose that $\frac {e}2$ is not an asymptotic $N$-expansive constant of $f$. Then, there are $x_1,\cdots, x_{N+1}\in X$ such that $d(f^i(x_l),f^i(x_r))\leq \frac {e}2$ for all $i\geq 0$, $l,r\in \{1,\cdots, N+1\}$ but

\[ \lim_{i\to\infty}\,{\rm d}(f^i(x_l),f^i(x_r))\neq0, \quad \forall\text{ distinct }l,r\in\{1,\cdots, N+1\}. \]

So, there is $\epsilon >0$ such that $\forall$ distinct $l,r\in \{1,\cdots, N+1\}$ there is a sequence

(1.7)\begin{equation} (i_k(l,r))_{k\in\mathbb{N}}\to \infty\quad \text{(as }k\to\infty) \end{equation}

such that

(1.8)\begin{equation} {\rm d}(f^{i_k(l,r)}(x_l),f^{i_k(l,r)}(x_r))>\epsilon, \end{equation}

for all $k\in \mathbb {N}$ and all distinct $l,r\in \{1,\cdots, N+1\}$.

On the other hand,

\[ {\rm d}(f^j(x_l),f^j(x_r))\leq e,\quad\forall j\geq0,\,\forall l,r\in\{1,\cdots, N+1\}. \]

Since $e$ is an asymptotic $N$-expansivity constant,

\[ \lim_{j\to\infty}\inf\{d(f^j(x_l),f^j(x_r)):l,r\in \{1,\cdots, N+1\}\text{ are distinct}\}=0. \]

Then, there are sequences $l_j\neq r_j\in \{1,\cdots, N+1\}$ such that

\[ \lim_{j\to\infty}\,{\rm d}(f^j(x_{l_j}),f^j(x_{r_j}))=0. \]

By the Pigeon Principle, there is a sequence $(j_q)_{q\in \mathbb {N}}\to \infty$ such that $l_{j_q}=l$ and $r_{j_q}=r$ are constant. Then,

\[ \lim_{q\to \infty}\,{\rm d}(f^{j_q}(x_l),f^{j_q}(x_r))=0. \]

This implies

\[ \liminf_{j\to\infty}\,{\rm d}(f^j(x_l),f^j(x_r))=0. \]

Moreover, (1.8) implies

\[ \limsup_{j\to\infty}\,{\rm d}(f^j(x_l),f^j(x_r))\geq \epsilon>0. \]

Then, $(x_l,x_r)$ is a Li–Yorke pair of $f$ which is a contradiction. This completes the proof.

Proof of the theorem Let $f:X\to X$ be a nonchaotic $N$-expansive homeomorphism of a compact metric space. Since $X$ is compact, lemma 1 implies that $f$ is uniformly $N$-expansive and so weak asymptotically $N$-expansive by lemma 2. Since $f$ is nonchaotic, $f$ is asymptotically $N$-expansive by lemma 3.

Question 3 If $f$ is an $N$-expansive homeomorphism defined in a compact metric space $X$, then there exists $\epsilon >0$ such that

(2.1)\begin{equation} n(x,\epsilon)\leq N,\quad\forall x\in X. \end{equation}

Proof. Suppose that there is $\epsilon >0$ satisfying the conclusion of the lemma. Let $x_1,\cdots, x_{N+1}\in X$ such that

\[ {\rm d}(f^i(x_l),f^i(x_r))\leq\frac{\epsilon}2 \quad\forall i\geq0,\, l,r\in\{1,\cdots, N+1\}. \]

Then, the triangle inequality implies $x_1,\cdots, x_{N+1}\in \Phi _\epsilon (x_1)$. It follows from the definition of $n(\Phi _\epsilon (x_1))$ that $x_l\in W^s(x_r)$ for some distinct $l,r\in \{1,\cdots, N+1\}$. Therefore,

\[ \lim_{i\to\infty}d(f^i(x_l),f^i(x_r))=0\quad\text{ for some distinct }l,r\in\{1,\cdots, N+1\} \]

and so $e=\frac {\epsilon }2$ is an asymptotic $N$-expansivity constant of $f$.

Conversely, let $e$ be an asymptotic $N$-expansivity constant of $f$. Suppose that $x\in X$ and $x_1, \cdots, x_{N+1}\in \Phi _{\frac {e}2}(x)$. Then, $d(f^i(x_l),f^i(x_r))\leq e$ for all $i\geq 0$, $l,r\in \{1,\cdots, N+1\}$ so $\lim _{i\to \infty }{\rm d}(f^i(x_l),f^i(x_r))=0$ (i.e. $x_l\in W^s(x_r)$) for some distinct $l,s\in \{1,\dots, N+1\}$, hence $n(\Phi _\epsilon (x_1))\leq N$ proving the result.

Proof. Our theorem implies that $f$ is asymptotically $N$-expansive and so the desired $\epsilon$ exists by corollary 1.