# If $X,X'$ have the same $\varepsilon$-packing numbers and $f:X \to X'$ surjective $1$-Lipschitz, then $f$ is an isometry

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!

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.