I am confused about a remark for the following proof.
Statement: Let $X$ be a totally bounded metric space. If $(x_n)_{n\geq 1}$ is a sequence in $X$ then it possesses a Cauchy subsequence.
Sketch Proof:
Suppose $(x_n)$ is a sequence in $X.$
Since $X$ is totally bounded, for each $n\in\mathbb{N},$ we set the radius of the balls to be $2^{-n}$ (with the intention that as $n\to\infty$, we have the case where the radius is tending to $0$), such that $X$ is the union of finite balls with radius $2^{-n}$.
For instance, to start of, set $n=0$ then we can find a finite list of open balls, say $\{O_1^0,..,O_n^0\}$, such that $X =\cup_{i=1}^{n} O_i^0.$ By Pigeon Hole Principle, if you like, there must be some $k$ such that $O_k^0$ contains infinitely many $x_n$ terms. In other words, define $S_0$ such that $S_0:=\{n\in\mathbb{N}:x_0\in O^0_k\}$ has infinite cardinality.
Doing this inductively, we have that $S_0\supseteq S_1\supseteq...\supseteq S_k$, each a infinite subset of $\mathbb{N}.$ (Basically at each $i$, we can always find such open ball that contain infinitely many $x_n$ terms , which happens to also lie in the open ball from the previous step and $S_i$ is just a set that records the index of $x_n$ in such open ball at step $i$.) Now, order the elements in $S_i$ such that $S_i=\{n_1^i<n_2^i<...\}.$ Consider the subsequence induced by $n_i^i$ and this subsequence is a Cauchy Subsequence.
Question:
This proof is followed by this comment
We needed a infinite sequence nested subsequences and hence we need a Cantor's diagonal method.
Now I understand why this may be an issue but how does Cantor's Diagonal Method resolve this issue? At least, it appeals to me that two things are quite unrelated.
Thank you for reading this far and m any thanks in advance!
We have a descending sequence of infinite subsets $S_k$ of $\mathbb N$. We do not know much about this sequence, but we have to pick $n_k \in S_k$ such that the sequence $(n_k)$ is strictly increasing. Then $(x_{n_k})$ is a subsequence of $(x_n)$ and it is easy to show that $(x_{n_k})$ is a Cauchy sequence.
The simplest way is to take $n_k$ to be the $k$-th element of $S_k$ (with respect to its natural order as a subset of $\mathbb N$). This is precisely Cantor's diagonal method.
However, we may also proceed inductively and and choose some $n_{k+1} \in S_{k+1}$ such that $n_{k+1} > n_k$. This $n_{k+1}$ may be among the first $k-1$ element of $S_k$ or come after the $k$-th element. It is even possible that $n_k$ is always the first element of $S_k$.
Anyway, it seems that there is no benefit in comparison to Cantor's diagonal method.