Is $f$ an isometry and $X$ compact implies $f$ is bijective and a homeomorphism?

[Chris L T521]

Here's this week's problem.

-----

Problem: Let $(X,d)$ be a metric space. If $f:X\rightarrow X$ satisfies the condition
\[d(f(x),f(y))=d(x,y)\]
for all $x,y\in X$, then $f$ is called an isometry of $X$. Show that if $f$ is an isometry and $X$ is compact, then $f$ is bijective and hence a homeomorphism.

-----

Hint:

Spoiler

If $a\notin f(X)$, choose $\epsilon$ so that the $\epsilon$-neighborhood of $a$ is disjoint from $f(X)$. Set $x_1=a$, and $x_{n+1}=f(x_n)$ in general. Show that $d(x_n,x_m)\geq \epsilon$ for $n\neq m$.

 

[Chris L T521]

No one answered this week's problem. You can find the solution below.

Spoiler

Proof: Let $a\notin f(X)$. Since $d$ is continuous, we have that it maps $X$ (which is compact) to a compact set. Thus, $X\backslash f(X)$ is open and some $\epsilon$-neighborhood of $a$ is disjoint from $f(X)$. In that case, $d(a,f(a))=l>\epsilon$. By the isometry property, $d(f(a),f(f(a)))=l$ and so on. Hence, by iterating the application of $f$ to $a$, we generate an infinite sequence of points in $f(X)$ (that is, $f(a)$, $f(f(a))$, $f(f(f(a)))$, etc). Let's call these points $a_1$, $a_2$, $a_3$, etc. Observe that $d(a_j,a_k)$ for any $j,k$ with $k>j$ will be the same as $d(a,a_{k-j})$ by the isometry property and since $d(a,a_{k-j})>\epsilon$, it follows that all of the points in the sequence $(a_i)$ are farther than $\epsilon$ from one another. Thus, we have an infinite sequence with no convergent subsequence, which contradicts the assumption of $X$ being compact. Therefore, $f$ must be surjective. It's also clear from the isometry property that $f$ is injective, hence $f$ is bijective. Furthermore, since isometry guarantees the continuity of $f^{-1}$, we have that $f$ is a homeomorphism. $\hspace{.25in}\blacksquare$