A question about compact sets: how to prove $g$ must be an isometry

Asked Nov 27 '14 at 20:26

Active Dec 08 '14 at 17:31

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Let $(X,p)$ be a compact metric space. Suppose that $g:X\rightarrow X$ is a function such that for all $x_1,x_2\in X$ we have $p(g(x_1),g(x_2))\geq p(x_1,x_2)$. Prove that, in fact, $g$ must be an isometry ($p(g(x_1),g(x_2))=p(x_1,x_2)$ for all $x_1,x_2\in X$) and that g is bijective.

I am stuck here, and I have tried many times, but still have no idea how to solve the question. Can someone tell me how to solve this question?

edited Nov 27 '14 at 20:33

Jonathan Gleason

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asked Nov 27 '14 at 20:26

python3

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is $f$ onto function? – mesel Nov 27 '14 at 20:42
see Theorem 1.4 https://people.math.osu.edu/memoli.2/courses/cse-topics-2014/background.pdf – Mirko Nov 27 '14 at 20:44
@mesel You don't need to assume it to be onto. Once you show that this is an isometry, it follows that $g$ must be onto (actually bijective, but injectivity is easy).. – Jonathan Gleason Nov 27 '14 at 20:56
we need to prove f is onto function – python3 Nov 27 '14 at 20:59
see the link in my previous comment, Theorem 1.3 shows the map is onto – Mirko Nov 27 '14 at 21:00
See also: http://math.stackexchange.com/questions/1201072/let-m-d-be-a-compact-metric-space-and-fm-to-m-such-that-dfx-fy-g – Martin Sleziak Mar 30 '15 at 13:14

A question about compact sets: how to prove $g$ must be an isometry

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