Let $(E, d)$ be a metric space. I'm trying to prove below result which is used in this question. Could you have a check on my attempt?
Theorem: $E$ is totally bounded if and only if every sequence in $E$ has a Cauchy subsequence.
My attempt:
- Let's prove direction $\implies$.
Let $(x_n) \subset E$. For each $n \in \mathbb N^*$, there exists a finite cover $\{B \left (c_{ni}, 1/n \right) \}_{i = 1}^{\varphi (n)}$. We define an increasing map $\psi:\mathbb N^* \to \mathbb N$ and a decreasing sequence $(I_n)$ recursively as follows.
- Let $I_0 := \mathbb N$.
- There exists $i \in \{1, \ldots \varphi (n)\}$ such that the set $J :=\left \{m \in I_{n-1} \,\middle\vert\, x_m \in B \left (c_{ni}, 1/n \right ) \right\}$ is infinite. Let $\psi(n) := \min J$ and $I_n := \{m \in I_{n-1} \mid m > \min J\}$.
Indeed, $(x_{\psi(n)})$ is a Cauchy sequence. Given $\varepsilon >0$, there is $N \in \mathbb N$ such that $1/N < \varepsilon/2$. On the other hand, $x_{\psi(n)} \in I_N$ for all $n > N$. So $$d \left (x_{\psi(n)}, x_{\psi(n')} \right) \le d \left (x_{\psi(n)}, c_{ni} \right) + d \left (x_{\psi(n')}, c_{ni} \right) < 2N^{-1}, \quad \forall n,n' > N.$$
- Let's prove direction $\impliedby$.
Assume $E$ is not totally bounded. We pick $x_0 \in E$. Then there is $x_1 \notin B(x_0, \varepsilon)$. There is $x_2 \notin \bigcup_{i=0}^1 B(x_i, \varepsilon)$. Recursively, there is $$ x_{n+1} \notin \bigcup_{i=0}^n B(x_i, \varepsilon). $$
It's clear that $(x_n)$ has no Cauchy subsequence. This completes the proof.