Despite of being aware of the duplicates of this question, I am posting it just to have my approach verified.

Let $f:X \to X$ be an isometry on $X$. By definition, when $f(a)=f(b)$, then $ d_X(f(a),f(b))=0=d(a,b) \implies a=b$, i.e. it is injective.

The set $X$ is compact, hence it is closed and bounded. Suppose, $X-f(X)=Y \neq \emptyset$. Taking $p\in Y$, we take the minimum distance from $f(X)$ to be $\delta$.

We now take an open cover of $X$, in such a manner, that in each $G_{X_i}$, $\displaystyle\sup\{d_X(s,t): s, t \in G_{X_i}\}=\delta$. Now, let $\displaystyle\sup\{d_X(s,t): s, t \in X\}=\lambda$. We need at most $\lceil \frac{\lambda}{\delta} \rceil $ open sets. Again, by isometry, we get corresponding $\sup{d_X(f(s),f(t))}=\delta$. But, by Pigeon Hole Principle, we see that if not surjective, then the function fails to be injective due to overlapping.

I know that this "proof" is very vague and inappropriate, but I need some direction.

Another doubt: How do I prove that $f(X)$ is compact?

Attempt: $X \subset G_1\cup G_2 \cup...\cup G_n \implies f(X) \subset f(G_1) \cup ...\cup f(G_n)$. For an arbitrary $g\in G_k$, $d_X(z,g)<\beta \implies z\in G_k$ for some $\beta >0$. Now, $f(g) \in f(G_k)$ and $d_X(f(z),f(g))<\beta \implies f(z) \in f(G_k)$. We repeat this for every possible open cover.

Does this work?

  • $\begingroup$ The image of a compact set under a continuous function is always again compact. And $f$, being an isometry, is in particular continuous (pre-image of open sets are open). $\endgroup$ – Berni Waterman Mar 14 at 22:06
  • $\begingroup$ I can sense some sort of continuity, but can't prove it. Does my proof work at all (compactness). ? $\endgroup$ – Subhasis Biswas Mar 14 at 22:09
  • $\begingroup$ In order to show that $f(X)$ is compact, you would want to start with an arbitrary open cover of $f(X)$ and not of $X$, as you seem to do in your sketch. $\endgroup$ – Berni Waterman Mar 14 at 22:10
  • $\begingroup$ Let me try to prove the continuity here in the comments, do check it. $\endgroup$ – Subhasis Biswas Mar 14 at 22:12
  • 1
    $\begingroup$ It is an essential part of the intuition behind the proof (also because all the compact sets that we can visualize are bounded in some sense), but it is not actually needed in the formal proof. $\endgroup$ – Berni Waterman Mar 15 at 8:29

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