Let $(X, d)$ be a compact metric space.

  • We say that $\{x_1, \cdots, x_n\} \subseteq X$ is an $\varepsilon$-covering of $X$ if for any $x \in X$, there exists $i \in \{1, \ldots, n\}$ such that $d(x, x_i) \leq \varepsilon$. Let $$ \operatorname{Cov} (X, \varepsilon) := \min \{n: \exists \varepsilon \text {-covering of } X \text { with size } n\} $$ be the $\varepsilon$-covering number of $X$.

  • We say that $\{x_1, \cdots, x_n\} \subseteq X$ is an $\varepsilon$-packing of $X$ if $d(x_i, x_j)\ge\varepsilon$ for all distinct $i, j$. Let $$ \operatorname{Pack} (X, \varepsilon) := \max \{n: \exists \varepsilon \text {-packing of } A \text { with size } n\} $$ be the $\varepsilon$-packing number of $A$.

The metric spaces $(X, d)$ and $(X', d')$ are said to be isometric (denoted by $X \cong X'$) if there is a bijective isometry between them. Then we have a theorem.

Theorem Let $(X, d)$ be a compact metric space and $f:X \to X$ be $1$-Lipschitz. Then $f$ is an isometry if and only if $f$ is surjective.

I have found the proof of one direction from here and the other one from here, i.e.,

  • $\implies$ Let $f$ be an isometry. Assume the contrary that there is $y \in X$ such that $y \notin Y:= f(X)$. Then there is $\varepsilon>0$ such that $d(y, Y)\ge \varepsilon$. Let $n:= \operatorname{Cov} (X, \varepsilon/2)$. Because $X \cong Y$, we get $n = \operatorname{Cov} (Y, \varepsilon/2)$. Let $C:=\{x_1, \ldots, x_n\}$ be an $\varepsilon/2$-covering of $X$. It follows that $y \in C$. Then $C \setminus \{y\}$ is an $\varepsilon/2$-covering of $Y$. Hence $\operatorname{Cov} (Y, \varepsilon/2) \le n-1$. This is a contradiction.

  • $\impliedby$ Let $f$ be surjective. Fix $x, y\in X$ such that $x \neq y$. Fix $\varepsilon>0$ such that $\delta := d(x,y) - \varepsilon/2 > 0$. Let $S$ be an $\varepsilon/4$-covering of $X$ that minimizes the quantity $$ \mathcal N(S) := \operatorname{card} (\{ (s_1, s_2) \in S^2 : d(s_1, s_2) \ge \delta\}). $$ Because $f$ is $1$-Lipschitz and surjective, $f(S)$ is also an $\varepsilon/4$-covering of $X$. Hence $\mathcal N(f(S)) \ge \mathcal N(S)$. This implies if $s_1, s_2 \in S$ with $d(s_1, s_2) \ge \delta$, then $d(f(s_1), f(s_2)) \ge \delta$. Now we pick $s_1, s_2 \in S$ with $d(s_1,x)\le \varepsilon/4$ and $d(s_2,y)\le \varepsilon/4$. Then $$ d(s_1, s_2) \ge d(x, y)-d(s_1, x)-d(s_3, y) \ge d(x, y)- \varepsilon/2= \delta. $$ So $d(f(s_1), f(s_2)) \ge \delta= d(x,y) - \varepsilon/2$. The claim then follows by taking the limit $\varepsilon \to 0^+$.

Now let $(X, d)$ and $(X', d')$ be compact metric spaces and $f:X \to X'$ be $1$-Lipschitz.

  • Just as in the proof of direction $\implies$ above, if $f$ is an isometry and $\operatorname{Cov} (X, \varepsilon)= \operatorname{Cov} (X', \varepsilon)$ for all $\varepsilon>0$, then $f$ is surjective.
  • The proof of direction $\impliedby$ above uses the quantity $\mathcal N(S)$ which looks related to $\operatorname{Pack} (S, \delta)$.

I would like to ask if below statement is true, i.e.,

If $f$ is surjective and $\operatorname{Pack} (X, \varepsilon)= \operatorname{Pack} (X', \varepsilon)$ for all $\varepsilon>0$, then $f$ is an isometry.

Thank you so much for your elaboration!


1 Answer 1


Consider two metrics on $\{x,x',y\}$ defined by $$|x-y|_1=|x'-y|_1=|x-y|_2=3, \quad |x-x'|_1=|x-x'|_2=1, \quad |x'-y|_2=2.$$ Denote by $X_1$ and $X_2$ the corresponding metric spaces.

Note that $\mathrm{pack}_\varepsilon X_1\equiv \mathrm{pack}_\varepsilon X_2$. Indeed, for both spaces we have

  • $\mathrm{pack}_\varepsilon=1$ if $\varepsilon>3$,
  • $\mathrm{pack}_\varepsilon=2$ if $3\geqslant \varepsilon>1$, and
  • $\mathrm{pack}_\varepsilon=3$ if $1\geqslant \varepsilon>0$.

The identity map on the set $\{x,x',y\}$ defines an onto short map $X_1\to X_2$ which is not an isometry.

  • $\begingroup$ Thank you so much for your help! $\endgroup$
    – Akira
    Mar 12, 2023 at 20:19

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