$(M,d)$ is a compact metric space and $f:M \to M$ is injective such that $d(f(x),f(y)) \le d(x,y) , \forall x,y \in M$ , then is $f$ an isometry?

Question

$(M,d)$ is a compact metric space and $f:M \to M$ is an injective function such that $d(f(x),f(y)) \le d(x,y) , \forall x,y \in M$ , then is $f$ an isometry i.e. $d(f(x),f(y)) = d(x,y) , \forall x,y \in M$ ?

It looks like you changed the question, invalidating user10000100_u's (originally correct) answer. I think a better solution would be to revert this to its original form and make a new question, or edit this in such a way that you're asking both questions, so that user10000100_u's answer is still ok. — Jason DeVito - on hiatus, Mar 23 '15 at 16:01
See also: http://math.stackexchange.com/questions/12285/isometry-in-compact-metric-spaces — Martin Sleziak, Mar 30 '15 at 13:12
I'm going to revert this to the original since 1) answer is for injective, 2) bijective already has answer elsewhere, 3) posts should ideally not change their contents — Jakobian, Apr 22 '25 at 07:29

score 4 · Answer 1 · answered Mar 22 '15 at 15:41

4

I guess that $x\mapsto \frac{1}{2} x$ from $[0,1]$ into intself is injective, but is not an isometry. The answer is no.

answered Mar 22 '15 at 15:41

Olórin

12,448

$(M,d)$ is a compact metric space and $f:M \to M$ is injective such that $d(f(x),f(y)) \le d(x,y) , \forall x,y \in M$ , then is $f$ an isometry?

1 Answers1