# Isometry from Sequentially Compact Metric Space to Subset

I am working through an old analysis qualifier question, and I was hoping someone could help me verify or correct my proof. I would also appreciate any feedback for how I could strengthen my proof, and for alternative ways of approaching the problem. Thank you in advance for any help!

Theorem: Let $X$ be a sequentially compact metric space, and let $Y \subset X$. Let $f : X \to Y$ be an isometry between $X$ and $Y$. Recall that this means that $f$ is a distance preserving homeomorphism from $X$ onto $Y$. Thus, for each $a, b \in X$, $d(a, b) = d(f(a), f(b))$. Show that $Y = X$.

Proof: Suppose to the contrary that $X \neq Y$. As $f$ is an isometry, it is continuous. So as $f$ is continuous and $X$ is compact, it follows that $f(X)$ is connected and compact. In particular, $f(X)$ is closed, so $X \setminus f(X)$ is open. Let $a \in X \setminus f(X)$ s.t. there exists $y \in f(X)$ s.t. $d(a, y) > \text{diam}(f(X))$, and let $r > 0$ s.t. $B_{r}(a) \subset (X \setminus f(X))$ (such an $r$ exists as $X \setminus f(X)$ is open).

In particular, we have $\overline{B_{r/2}(a)} \subset B_{r}(a)$. As $\overline{B_{r/2}(a)}$ is a closed subset of a compact space, it is compact. So by the Weierstrass Extreme Value Theorem, let:

$$m := \max_{x \in \overline{B_{r/2}(a)}, y \in f(X)} d(x, y)$$

By construction, $m > \text{diam}(f(X))$. Let $x \in \overline{B_{r/2}(a)}, y \in f(X)$ achieving $d(x, y) = m$. By isometry, $d(f(x), f(y)) = m$. So $m \leq \text{diam}(f(X))$, a contradiction. QED.

In my proof, am I correct in selecting:

Let $a \in X \setminus f(X)$ s.t. there exists $y \in f(X)$ s.t. $d(a, y) > \text{diam}(f(X))$

If so, how can I justify such an $a$ exists? If not, is there a way I can correct this hole (as well as the others) in my proof?

Thank you in advance for any help!

• Why is $f(X)$ connected? Is $X$ connected? – ajotatxe Dec 28 '16 at 21:28
• The existence of $a$ is just wishful thinking on your part, there is no argument leading in that direction. What you probably want is this kind of argument. Also, sequentially compact = compact for metric spaces, so it's unnecessary to say "sequential". – user357151 Dec 28 '16 at 21:30
• @ajotatxe- That's a good point. There was no assumption that $X$ was connected. The finite union of disjoint intervals in $\mathbb{R}$ would be a good counterexample here. zaq- I realize. The problem statement included sequential compactness though. Thanks for the link! I'll check it out. – ml0105 Dec 28 '16 at 21:30
• Very short argument at math.stackexchange.com/questions/170989/… – user399601 Dec 28 '16 at 21:39
• I'm not trying to show that $f$ is onto- this is given. I'm trying to show $X = Y$. Edit: That is a nice argument! I see now how it applies here. Thank you for sharing it! – ml0105 Dec 28 '16 at 21:40

1. Since you have not assumed $X$ is connected, it does not follow from continuity that $f(X)$ is connected.
2. The way you choose your point $a\in X\setminus f(X)$ needs clarification (and probably no such $a$ can be proved to exist). It is not obvious that you can choose $a$ such that there is $y\in f(X)$ with $d(a,y) > \text{diam}f(X)$. Imagine, for example, that somehow you had $X = [0,3]$ and $f(X) = [0,1]\cup [2,3]$. Then $\text{diam} f(X) = 3$. But for any $a\notin [0,1] \cup [2,3]$, the largest distance to any point in $[0,1]\cup[2,3]$ is $2$. Since at this point you have not provided any further properties of $f(X)$, it is unclear how to conclude this magical $a$ exists.
3. You really are being asked to prove that $f$ is surjective as a map $X\to X$. For if $f$ is surjective as a map $X\to X$, then $Y = f(X) = X$, and conversely if $Y = X$, then $f(X) = X$ so $f:X\to X$ is surjective. user399601's link provides a clean proof.