Not Cauchy Implies No Convergent Subsequence?

Question

Show that if $f : X \to X$ is an isometry and $X$ is compact metric space, then $f$ is bijective and hence a homeomorphism Hint: If $a\in X$ \ $f(X)$, choose so that the neighborhood of a is disjoint from $f(X)$. Set $x_1 = a$, and $x_{n+1} = f(x_n)$. Show that $d(x_n, x_m) \ge \epsilon$ for $n \neq m$.

Okay. So this question was addressed here. I was able to show that $d(x_n, x_m) \ge \epsilon$ for $n \neq m$, but was unable to discern the contradiction from this. As the answer in the link suggests, this evidently implies that $(x_n)$ cannot have a convergent subsequence, which contradicts sequential compactness. I haven't been able to find a proof of this, so I provide my own and hope that someone could critique it:

Suppose that there exists an $\epsilon > 0$ such that $d(x_n, x_m) \ge \epsilon$ for $n \neq m$, but, by way of contradiction, suppose that there exists a subsequence $x_{n_k}$ converging to $L$. Given $\frac{\epsilon}{2} > 0$, there exists a natural number $N$ such that $d(x_{n_k},L) < \frac{\epsilon}{2}$ for every $k \ge N$. Choosing $k=N$ and $k=N+1$, we get $d(x_{n_N},L) < \frac{\epsilon}{2}$ and $d(x_{n_{N+1}},L) < \frac{\epsilon}{2}$, and adding the two gives us $$d(x_{n_N},x_{n_{N+1}}) \le d(x_{n_N},L) + d(x_{n_{N+1}},L) < \epsilon,$$

contradicting the hypothesis.

How does this sound?

Answer 1

Your proof is correct, but as stated in the comments, your title is wrong.

By negating the definition of a Cauchy sequence, we obtain that a sequence is not Cauchy if there exists a positive number $\epsilon$ such that for every natural number N, there exists natural numbers $n,m\geq N$ such that $d(x_n,x_m)\geq\epsilon$.

Now your sequence satisfies $d(x_n,x_m)\geq\epsilon$ for $n\ne m$. Yes, this sequence is not Cauchy. However it is much more than that because the "there exists natural numbers $n,m\geq N$" can be replaced by "for every distinct pair of natural numbers $n,m$". So not only is your sequence not Cauchy, but in fact every subsequence of your sequence is not Cauchy. Since a convergent sequence must be Cauchy, we see that every subsequence is not convergent.

Answer 2

Cauchy-ness nor sequential compactness are necessary here:

Any compact space (metric or not) has the property that every countable set $A$ has an $\omega$-accumulation point $p \in X$: every neighbourhood of $p$ contains countably many points from $A$.

The proof is quite easy: suppose not, then every $x \in X$ has an open neighbourhood $O_x$ that contains at most finitely members of $A$. This defines an open cover of $X$, that has a finite subcover $\{O_{x_i}: i = 1\ldots n\}$; and then:

$$A = A \cap X = A \cap (\cup_{i=1}^n O_{x_i}) =\cup_{i=1}^n (O_{x_i} \cap A)$$

which would mean that $A$ is finite, as a finite union of finite sets; contradiction, so such a $p$ must exist.

Apply this to the set $\{x_n: n \in \mathbb{N}\}$, and we get $p\in X$ as above. In particular we have two $x_n \neq x_m$ both in $B(p, \frac{\varepsilon}{2})$, which by the triangle inequality implies $d(x_n, x_m) < \varepsilon$, contradiction. Quite similar to the sequence idea, without actually needing sequential notions like Cauchy-ness or subsequential limits etc.