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Does there exist a bounded set $A \subset \mathbb{R}^ n$ (for some $n$) and a function $f:A\to A$ (not necessarily onto) such that $\forall x,y \in A$ then $\|x-y\| < \|f(x)-f(y)\|$?

If such a set $A$ (and a function $f$) exists, does there even exist such a set $A$ and a function $f$ such that $f$ is continuous? (Or - can it be proved that such a continuous $f$ doesn't exist?)

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    $\begingroup$ This should be moved over to/deleted and re-asked at math.stackexchange. $\endgroup$
    – Alec Rhea
    Commented Jul 18, 2018 at 5:58
  • $\begingroup$ Can it be moved automatically or can I do it somehow? Thank you $\endgroup$
    – Check drummer
    Commented Jul 20, 2018 at 23:23
  • $\begingroup$ It's possible for a moderator to move it and that would preserve the rep boost I believe, but probably faster to copy then delete it yourself and repost. No problem. $\endgroup$
    – Alec Rhea
    Commented Jul 21, 2018 at 9:35
  • $\begingroup$ Choose $x_0 \ne y_0$, and let $x_{n+1} = f(x_n)$ and $y_{n+1} = f(y_n)$. Next, choose a convergent subsequence $(x_{k_n},y_{k_n})$, and set $l_n=k_{n+1}-k_n$. Since $|x_0-x_{l_n}| \le |x_{k_n}-x_{k_{n+1}}|$, $(x_{l_n})$ converges to $x_0$. Similarly, $y_{l_n}$ converges to $y_0$. Therefore, $|x_{l_n}-y_{l_n}| \to |x_0 - y_0|$. On the other hand, $|x_{l_n}-y_{l_n}|\ge|x_1-y_1|$, so $|f(x_0)-f(y_0)| = |x_1-y_1|\le|x_0-y_0|$, a contradiction. See here. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 21, 2018 at 14:26
  • $\begingroup$ To Mateusz: but $|x_1−y_1|≤|x_0−y_0|$ is not genrally true, $\endgroup$
    – Check drummer
    Commented Jul 23, 2018 at 1:30

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