set with compact closure under isometry

Question

In a 2nd countable and hausdorff metric space, how to see that the image under an isometry of a subset with compact closure also has compact closure? Thanks.

Answer 1

If $A$ has compact closure and $T$ is an isometry then the closure of $T(A)$ is $T(\overset {-} A)$. Since $\overset {-} A$ is compact and $T$ is continuous, it follows that the closure of $T(A)$ is compact. [ $T(\overset {-} A) \subset \overset {-} T(A)$ is true for any continuous function and it is easy to prove. Suppose $y \in \overset {-} {T(A)}$. There is a sequence $(a_n)$ in $A$ such that $T(a_n) \to y$. Since $A$ has compact closure the sequence $(a_n)$ lies in a compact set, so it has a convergent subsequence $(a_{n_k})$. Let the limit of this subsequence be $a$. Then $a \in \overset {-} {T(A)}$ and $y=\lim T((a_{n_k}) =T(a)$].

Answer 2

Extend $T: X \to X$ to $T: c(X) \to c(X)$, where $c(X)$ is the completion of $X$. It's standard that you can do this for isometries.

working in $c(X)$:

$A$ has compact closure iff $A$ is totally bounded and $T$ preserves totally bounded subsets, being uniformly continuous.