Isometries between metric spaces I have three questions about when you can show there is an isometry between metric spaces. 
(1) If there is an injective non-expanding map from $X$ to $Y$ and an injective non-expanding map from $Y$ to $X$, are $X$ and $Y$ isometric?
I think the answer must be no, just let $X=[0,1]$ and $Y=[0,1/2]$ with the Euclidean metric on each  and let the morphisms just shrink each of the intervals by a 1/2.  But $X$ and $Y$ are not isometric as metric spaces.  The only reason I ask is that this question seems to imply that this is true for compact metric spaces.  So maybe I am just missing something.
(2) If there is an isometric embedding from $X$ to $Y$ and an isometric embedding from $Y$ to $X$ is it true that $X$ and $Y$ are isometric?
Here by an isometric embedding I mean a map that preserves the metric. 
(3) If the answer to (2) is yes, is there something to be said about which concrete categories this result holds for, with respect to embeddings?
Here I am taking the definition of concrete categories and embeddings from Adámek, Herrlich, Strecker.
I know this question sounds a lot like  this question, but unless I am confused, they are talking about injective maps (monomorphisms) which make sense in any category, whereas I am talking about embeddings which are only defined for concrete categories.
EDIT: Edited to remove jargon and make clearer.
Thanks very much for any information.
 A: As already observed by James, the answer to (2) is negative in general. However, the answer is positive if we assume that $X$ (or $Y$) is compact. I will show in a moment how this can be deduced from the following claim:
Let $f\colon X\to X$ be a distance-preserving map of a metric space into itself. If $X$ is compact, then $f$ is an isometry (i.e. it is surjective, its injectivity being obvious).
Let me first prove the claim: set $Y=f(X)$ and suppose that $Y\neq X$. Pick a point $x_0\in X\setminus Y$. Since $X$ is compact, $Y$ is also compact, so $x_0$ has positive distance, say $\epsilon$, from $Y$. Now let $x_n=f^n(x)$. For every $i\leq j$ we have 
$d(x_i,x_j)=d(f^i(x_0),f^i(f^{j-i}(x_0)))=d(x_0,f^{j-i}(x_0))\geq d(x_0,Y)=\epsilon$. Therefore, no subsequence of $\{x_n\}$ can be a Cauchy sequence, and this contradicts the compactness of $X$, thus proving the claim.
Coming back to question (2) (and in some sense (3)), if $h\colon X\to Y$ and $g\colon Y\to X$ are embeddings, and $X$ is compact, then by the claim $g\circ h$ is an isomorphism of $X$. In particular, $g$ is surjective, so it is an isomorphism between $Y$ and $X$.
Therefore, even if (2) does not hold in general, it holds in the category of compact metric spaces (this gives a partial answer to (3)).
A: Let me present a counter-example to question (2).
Let $\ X:=\mathbb Z_{\ge 0}\ $ be the set of all non-negative integers in their natural metrics. Let $\ Y:=X\setminus\{1\}.\ $ Then inclusion $\ Y\subseteq X\ $ is one isometric embedding, and $\ x\mapsto x+2\ $ is an isometric embedding in the opposite direction.
Spaces $\ X\ $ and $\ Y\ $ are not isometric because $\ Y\ $ has a point, namely $\ 0\in Y,\ $ which is not distant from any other point of $\ Y\ $ by $\ 1,\ $ while $\ X\ $ does not have any such point (each point
$\ x\in X\ $ is distant form $\ x+1\in X\ $ by $\ 1).$ 
A: The answer to (2) is "no". Take, for example, $X$ to consist of a countable number of copies of $[0,1]^2$ with the standard metric, where every two points in different components have distance 1000 from one another.
Then take $Y$ to be $X$, but with some stuff cut out of one of the components so it looks like an annulus or a letter of the alphabet or some other cute shape.
$X$ and $Y$ are clearly not isomorphic, because $Y$ has a weird component and $X$ doesn't. But there's clearly embeddings from each to the other, as you can embed the weird component in a normal one, and the cardinality of components is the same in $X$ and $Y$, whether you count the weird component or not.
