Isometries between metric spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:54:14Z http://mathoverflow.net/feeds/question/73309 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73309/isometries-between-metric-spaces Isometries between metric spaces Paul Tupper 2011-08-21T04:11:59Z 2011-08-21T14:44:28Z <p>I have three questions about when you can show there is an isometry between metric spaces. </p> <p>(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?</p> <p>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 <a href="http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold" rel="nofollow">this question</a> seems to imply that this is true for compact metric spaces. So maybe I am just missing something.</p> <p>(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?</p> <p>Here by an isometric embedding I mean a map that preserves the metric. </p> <p>(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?</p> <p>Here I am taking the definition of concrete categories and embeddings from <a href="http://katmat.math.uni-bremen.de/acc/acc.pdf" rel="nofollow">Adámek, Herrlich, Strecker</a>.</p> <p>I know this question sounds a lot like <a href="http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold" rel="nofollow">this question</a>, 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.</p> <p>EDIT: Edited to remove jargon and make clearer.</p> <p>Thanks very much for any information.</p> http://mathoverflow.net/questions/73309/isometries-between-metric-spaces/73318#73318 Answer by James Cranch for Isometries between metric spaces James Cranch 2011-08-21T10:14:14Z 2011-08-21T10:14:14Z <p>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.</p> <p>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.</p> <p>$X$ 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.</p> http://mathoverflow.net/questions/73309/isometries-between-metric-spaces/73331#73331 Answer by Roberto Frigerio for Isometries between metric spaces Roberto Frigerio 2011-08-21T14:43:30Z 2011-08-21T14:43:30Z <p>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:</p> <p>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).</p> <p>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.</p> <p>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 $X$.</p> <p>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)).</p>