This question already has an answer here:

Let $(X,d)$ be a compact metric space and $f : X \to X$ be isometric, i.e. for every $x,y \in X$ : $ d(f(x),f(y)) = d(x,y) $. How I can show the following? $f(X) = X$

My first thought was to show the surjection $ x \in X \setminus f(x) $. Is this the right approach?

Thanks in advance


marked as duplicate by Marso, Stefan Hamcke, Hanul Jeon, Dan Rust, M Turgeon Oct 6 '13 at 15:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See this and its "linked" section. $\endgroup$ – David Mitra Oct 6 '13 at 14:05