I found a statement involving a homeomorphism $f:X\to X$ of a compact metric space $X$, with Lipshitz coefficient 1, i.e., a non-expansive map, and cannot think of an example where $f$ is not an isometry. Must it be?
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3$\begingroup$ Almost a duplicate of math.stackexchange.com/questions/12285/… $\endgroup$– Andrej BauerCommented Feb 25, 2023 at 22:18
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$\begingroup$ You are right, thanks! I can leave the question anyways in case someone else searches it using this phrasing $\endgroup$– Saúl RMCommented Feb 25, 2023 at 22:53
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1 Answer
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It follows from 1.6.15(1) in "A Course in Metric Geometry" by Burago, Burago, and Ivanov and 1.6.15(2) is a more general statement:
Any distance-noncontracting map from a compact metric space to itself is an isometry.
This statement is needed in the proof that Gromov--Hausdorff metric is a metric. Also check the solution of 1.12 in my "Pure metric geometry" it is based on a different idea.
answered Feb 26, 2023 at 21:53