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A metric space $(X,d)$ is said to be bounded if there is $r\in\mathbb{R}$ such that for all $x,y\in X$ we have $d(x,y) \leq r$.

A self-isometry is a map $\iota:X\to X$ such that for all $x,y\in X$ we have $d(x,y) = d(\iota(x), \iota(y))$.

Does there exist a bounded metric space with a non-surjective self-isometry?

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  • $\begingroup$ Let $t$ be an irrational. Define $X_t=\{e^{int}:n\ge 0\}$; this is a subset of the unit circle. Then the rotation $z\mapsto e^{it}z$ maps $X_t$ into a proper subset of itself; with the Euclidean distance this is a self-isometry of $X_t$. Unlike the Hilbert shift example, it's not a complete metric space. $\endgroup$
    – YCor
    Commented Jul 29, 2015 at 7:44
  • $\begingroup$ Can you post this as an answer? $\endgroup$
    – Dominic van der Zypen
    Commented Jul 29, 2015 at 7:53

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Let $B$ be the unit ball in the sequence space $\ell^2$. Send $(x_0, x_1,\dots)$ to $(0,x_0,x_1,\dots)$.

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