For $(X,d)$ a compact metric space and $ Y \subsetneq X$ there is no isometry $f:X \to Y$.

Question

Let $(X,d)$ be a compact metric space and $Y \subsetneq X$. Prove there does not exist an isometry $f:X \to Y$.

I will try to prove this by contradiction. As $ Y \subsetneq X$ we can take some $x_{0} \in X$ such $x_{0} \notin Y$, but $f:X \to Y$ is an isometry so for every $a,b \in X$ we have $$d(a,b)=d(f(a),f(b)).$$

So $f(x_{0}) \in Y$ and $x_{0} \neq f(x_{0})$, implying $d(x_{0},f(x_{0}))= \epsilon$ such $\epsilon >0$ but Im not able to show a contradiction from here. Can anyone help me how to end the proof by constructing a contradiction from supposing there is an isometry $f:X \to Y$? Thanks

Answer 1

We should first note that the only reason we can say $d(x_0,Y)=\varepsilon>0$ is because assuming $f$ is an isometry means it is continuous, and the continuous image of a compact set is compact.

Now consider the sequence $(f^n(x_0))$. As $X$ is compact there must exist some subsequence $(f^{n_k}(x_0))$ converging to some $x\in X$. In particular this implies that $(f^{n_k}(x_0))$ is Cauchy. However, $d(f^{n}(x_0),x_0)\geq \varepsilon$ for all $n\in\mathbb N$, because the image of $f$ lies in $Y$. As $d$ is assumed to be an isometry this implies that $d(f^{n+m}(x_0),f^m(x_0))\geq \varepsilon$ for all $n,m\in\mathbb N$. This contradicts the fact that $(f^{n_k}(x_0))$ is Cauchy.

Answer 2

Indeed, let $p \in X\setminus Y$. Then $d(p, f[X]) > 0$ because $f[X]$ is closed in $X$, being compact.

Let $0 < r < d(p,f[X])$, and note that $X$ can be covered by finitely many open sets of diameter $< r$. Let $N$ be the smallest number of open sets of diameter $<r$ that cover $X$ (which is well-defined as $\mathbb{N}$ is a well-order and the set $$\{n \in \mathbb{N}: \exists O_1, O_2, O_n \text{ open } \land (\forall_i \operatorname{diam}(O_i) < r) \land \bigcup_{i=1}^n O_i =X \}$$ is non-empty, if we want to be formal)

In any such covering by small open sets, the member that contains $p$ does not intersect $f[X]$ (as $d(p,f[X])$ is larger than than the diameter, and so $f[X]$ can be covered by $N-1$ such small open sets. But then their inverse images under $f$ (which have the same diameter, as $f$ is an isometry!) form an open cover of $X$ by $N-1$ such small open sets too, a blatant contradiction with the minimality of $N$. This contradiction shows no such $p$ can exist.

Answer 3

Suppose $y\in X\backslash f(X)$. Let $d=d(y,f(X)).$ There is a subset $% E$ having the largest number of elements such that the distance between any two points exceeds $d$. To see that such a set exists note that there is a finite set $\{x\_1,x\_2,..,x\_N\}$ such that $X$ is covered by the balls $% B(x_{k},d/2)$. Then we cannot have more than $N$ points separated by distance $d$ (because two of these points would belong to the same ball and have the distance between them is less than $d).$ Now we get a contradiction by looking at $f(E)\cup \{y\}$.