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I have been dealing with this problem for some days without any progress:

"Let $f:K \rightarrow K$ be a weakly shrinking function defined on a compact metric space $K$ (i.e.: $\forall x,y \in K, x \neq y \implies d(f(x), f(y)) \le d(x,y)$); then, prove that $f$ is surjective if and only if $\forall x,y \in K, x \neq y \implies d(f(x), f(y)) = d(x,y)$ (that is, the equality always holds in the definition of weakly shrinking function)"

My Attempt

So far I have only been able to prove that $f$ is uniformly continuous. Also, in trying to show the direction (surjectivity $\implies$ equality) I had the idea to separate the points on which the equality holds from those on which the strict inequality holds into two sets A and B respectively; then, I thought of reaching a contradiction in the two possible cases: B finite (in which case I should use finiteness somehow) and B infinite (in which case I would probably use the existence of a limit point by the characterization of compact sets). However I haven't made any progress in this direction yet.

Any hint on how to proceed or any comment on my idea are highly appreciated as always!

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  • $\begingroup$ Related: math.stackexchange.com/q/3955652/42969 $\endgroup$
    – Martin R
    Mar 15, 2022 at 9:23
  • $\begingroup$ @MartinR Thank you for the link! However, that question deals with only one direction of the equivalence and it does so with a stronger condition. $\endgroup$ Mar 15, 2022 at 9:52

1 Answer 1

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Claim: Given a function $f:K \to K$ which does not increase distances on a compact metric space $(K,d)$, surjectivity is equivalent to preserving distances.

Proof: (a) First suppose $f$ preserves distances but is not surjective, so there exists $z_0 \in K \setminus f(K)$. Let $K_0=K$. Since $K_1:=f(K)$ is compact, we have $$\delta:=d(z_0,K_1)>0 \,.$$ For $n \ge 1$, define inductively $K_n=f(K_{n-1})$ and $z_n=f(z_{n-1}) \in K_n$. Then $d(z_{n-1},K_n)=\delta$ for all $n \ge 1$, and for positive integers $\ell<n$ we have $z_n \in K_{\ell+1}$, so $$ d(z_\ell,z_n) \ge \delta \,.$$ Thus the sequence $\{z_n\}$ does not have a convergent subsequence, contradicting compactness.

(b) Suppose that $f$ is surjective, and $x_0,y_0 \in K$ are distinct. Given $\epsilon>0$, there is a finite set $S \subset K$ such that $\cup_{s \in S} B(s,\epsilon) =K$, where $B(s,\epsilon)$ is the open ball of radius $\epsilon$ centered at $s$. For $n \ge 1$, define inductively preimages $x_n,y_n$ such that $f(x_n)=x_{n-1}$ and $f(y_n)=y_{n-1}$. For each $n$, there exist $\tilde{x}_n, \tilde{y}_n \in S$ such that $d(\tilde{x}_n,x_n)<\epsilon$ and $d(\tilde{y}_n,y_n)<\epsilon$. In the infinite sequence $\{(\tilde{x}_n, \tilde{y}_n) \}_{n \ge 1}$, some element of the finite set $ S \times S$ must repeat. Thus there exist $\ell>k>0$ such that $ (\tilde{x}_k, \tilde{y}_k)=(\tilde{x}_\ell, \tilde{y}_\ell) \,,$ so $$d(x_k,x_\ell)<2\epsilon \quad \mbox{ and} \quad d(y_k,y_\ell)<2\epsilon \,. $$ Applying $f^{k+1}$, we infer that $$d(f(x_0),x_{\ell-k-1})<2\epsilon \quad \mbox{ and} \quad d(f(y_0),y_{\ell-k-1})<2\epsilon \,. $$ Therefore, by the triangle inequality, $$d(x_{\ell-k-1}, y_{\ell-k-1}) < d(f(x_0),f(y_0))+4\epsilon \,.$$ Applying $f^{\ell-k-1}$ yields that $$d(x_{0}, y_{0}) < d(f(x_0),f(y_0))+4\epsilon \,.$$ Since $\epsilon>0$ was arbitrary, this proves that $$d(x_{0}, y_{0}) = d(f(x_0),f(y_0)) \,,$$ so $f$ is distance preserving.

Edit: According to [1], The equivalence in the claim is due to [2].

[1] Hu, Thakyin, and W. A. Kirk. "On local isometries and isometries in metric spaces." In Colloquium Mathematicum, vol. 44, no. 1, pp. 53-57. Institute of Mathematics Polish Academy of Sciences, 1981.

[2] Freudenthal, Hans, and Witold Hurewicz. "Dehnungen, verkürzungen, isometrien." Fundamenta Mathematicae 26, no. 1 (1936): 120-122.

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  • $\begingroup$ Thank you very much for the precise and clear answer! $\endgroup$ Mar 20, 2022 at 8:35
  • $\begingroup$ @Matteo Menghini: How did this problem arise? I really enjoyed thinking about it. Do you know any reference where this question was discussed? $\endgroup$ Mar 20, 2022 at 22:57
  • $\begingroup$ I searched some more, one direction is also proved in math.stackexchange.com/questions/12285/… $\endgroup$ Mar 20, 2022 at 23:15
  • $\begingroup$ @Yuvale Peres actually, it was taken from a supplement to Rudin’s “Principle of Mathematical Analysis” which contains many problems as thought-provoking as this one, I’ll leave the link here: math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.pdf $\endgroup$ Mar 20, 2022 at 23:17
  • $\begingroup$ Apparently this equivalence was proved in 1936. I added a reference to my answer. $\endgroup$ Mar 20, 2022 at 23:21

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