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I am trying to prove the following property, which seems fairly intuitive, at least in $\mathbb{R}^n$.

Let $(X,d)$ be a compact metric space where $Y \subseteq X$ arbitrary. Prove that if there exists an isometry $f: \, X \rightarrow Y$, then $X=Y$.

My idea: Prove that $X \subseteq Y$.

My issue at the moment is that I am not completely sure if the existence of an isometry would help us deduce that $X \subseteq Y$.

Another idea that I had was to perhaps use proof by contradiction. Isometries are 'distance preserving' and as $Y$ is a subset of $X$, then there must be two points in $X$ that are 'far enough' so that their images in $Y$ cannot 'match' the distance. However, this only really makes sense when $X$ and $Y$ are finite, which may not necessarily be the case.

Trogdor
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    The existence of an isometry alone is not enough. Consider, for example, $X = [0,\infty)$ and $Y=[1,\infty)$. So the compactness hypothesis is crucial. – Lee Mosher May 25 '16 at 14:39
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    See this and its links for other ideas. – David Mitra May 25 '16 at 14:42
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    "However, this only really makes sense when $X$ and $Y$ are finite, which may not necessarily be the case." Compact is a thing exactly because compact sets retain a lot of the properties that finite sets have. In analysis and topology, the concept of a compact set is often a better generalization to finite than countable is. Of course, in this case, the real trick is to figure out how to exploit this generalization. – Arthur May 25 '16 at 14:44
  • Actually the link in the comment of @DavidMitra does exactly what this question asks: it proves that $X \subset f(X)=Y$. – Lee Mosher May 25 '16 at 15:10

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If $f[X]$ is proper, pick $x_0 \in X, x_0 \notin f[X]$. Then $d(x_0, f[X]) = r > 0$, as $f[X]$ is also compact, thus closed.

Then define $x_{n+1}= f(x_n)$ for $n\ge 0$, then (as all other $x_n$ besides $x_0$ are $f$-images) we have $d(x_0, x_n) \ge d(x_0, f[X]) \ge r$ for all $n \ge 0$.

Hence, as $f$ is an isometry, $d(x_k, x_{n+k}) = d(x_0, x_n) \ge r$ for all $n \ge 0, k \ge 0$, which can be seen by induction on $k$, fixing $n$.

It follows that for all $n \neq m$, $d(x_n, x_m) \ge r$ which means that $(x_n)$ can have no convergent subsequence, contradicting compactness. So $f[X] = X$.

Henno Brandsma
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  • I understand everything except $d(x_0, f[X]) = r$. If $f[X]$ is the image of $X$ under $f$, then what does $d(x_0, f[X])$ mean? How can we have a distance between a point and a set? – Trogdor May 25 '16 at 15:34
  • @Trogdor $d(x,A) = \inf {d(x,y): y \in A }$ for any subset of $X$, and if $x \notin \overline{A}$, $d(x,A) > 0$. – Henno Brandsma May 25 '16 at 15:50
  • @Trogdor The distance between a point and a set is exactly what it sounds like: It is the smallest distance between $x_0$ and any point in $f[X]$ (have you never been asked to find the distance between a point and a line, or anything like that?). Technically, it's the infimum of all such distances, but since $f[X]$ is closed, the minimum works. – Arthur May 25 '16 at 15:55
  • I had suspected that might have been the case (in actual fact your example of the distance between a point and a line was what led me to suspect so), but was not completely sure, hence my query. Thanks for clearing that up. – Trogdor May 25 '16 at 15:57
  • @Arthur because it is compact, it is a minimum, it need not be a minimum for closed sets. – Henno Brandsma May 25 '16 at 16:16
  • @HennoBrandsma You're right. I'm thinking too much about Euclidean space. – Arthur May 25 '16 at 16:23