# All closed balls are compact each isometry is bijective

Let $(X,d)$ be a metric space in which all closed balls are compact and such that for any two points $x,y \in X$ there exists a function $u \in Iso(X,d)$ such that $u(x)=y$.

Prove that then each isometry $u: X \rightarrow X$ is bijective.

How can I prove this?

I know that if $X$ is compact, then each isometry is bijective.

Could you help me wih that?

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Let $u$ be an isometry of $X$, and $x$ some point of $X$. It is enough to show that $u$ is surjective.

Taking $u'$ an isometry such that $u'(u(x)) = x$ given by your hypothesis, then $u' \circ u$ is an isometry that fixes $x$. Therefore, it stabilizes the balls centered at $x$, and you can show that $u' \circ u$ is surjective because the balls are compact. The surjectivity of $u$ follows directly :

If $y \in X$, then $\exists x \in X$ such that $(u' \circ u) (x) = u'(y)$ by surjectivity of $u' \circ u$. Then $u(x) = y$ by injectivity of $u'$.

EDIT : If you remove one of the hypothesis, the assumption fails :

Take $\mathbb{N}$ with the classical metric. Then $n \mapsto n+1$ is an isometry which isn't surjective. Its closed balls are finite and therefore compact ; but the isometries group of $\mathbb{N}$ is trivial, so its action can't be transitive.

And in $l^2$, take the shift operator $e_i \mapsto e_{i+1}$. It is not surjective either. The action of its isometries group is transitive because it contains the translations ; but it is known that its closed balls aren't compact.

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Thanks. Is it possible for a metric space $(X, d)$ in which each closed ball is compact to have a non-surjective isometric mapping $X\rightarrow X$? –  Sandy May 21 '13 at 19:16