# What's the diagonal Cauchy subsequence (to show totally bounded if and only if every sequence has a Cauchy subsequence)?

I read a textbook showing a subset of a normed linear space is totally bounded if and only if every sequence in it has a Cauchy subsequence. It proves as follows:

$\Rightarrow$

Let $(x_n)$ be an infinite sequence in $K$ that is totally bounded. There is a finite set of points $\{y_{11},y_{12},...,y_{1r}\}$ in $K$ such that $K \subset \cup_{j=1}^{r} B(y_{1j},\frac{1}{2})$. Then, at least one of the balls contains an infinite subsequence $(x_{n1})$ of $(x_n)$. Again, there is a finite set of points $\{y_{21},y_{22},...,y_{2s}\}$ in $K$ such that $K \subset \cup_{j=1}^{s} B(y_{2j},\frac{1}{2^2})$. Then, at least one of the balls contains an infinite subsequence $(x_{n2})$ of $(x_{n1})$. Continuing in this way, at the $m$-th step we obtain a subsequence $(x_{nm})$ of $(x_{n(m-1)})$ which is contained in a ball of the form $B(y_{mj},\frac{1}{2^m})$. The diagonal subsequence $(x_{nn})$ of $(x_n)$ is Cauchy.

I am hoping that you may answer my following questions in details:

Q1: What does the sequence $(x_n)$ look like here? If it looks like $x_n = (x_1,x_2,x_3...)$, and for particular example $(x_n) = \{x_n:x_n=\frac{1}{n},n\in\mathbb{N}\} = (1,\frac{1}{2},\frac{1}{3},...)$, why does it have a diagonal subsequence $(x_{nn})$, sounds like a matrix?

Q2: The author denotes $(x_{nm})$ as a subsequence of $(x_{n(m-1)})$, and again a subsequence of $(x_{n(m-2)})$ and again... a subsequence of $(x_{n1})$, which is a subsequence of $(x_n)$. I don't understand this chain. Why does the original sequence have only one subscript $n$ but its subsequences all have two subscript $nm$?

Q3: Could you show me a very simple and specific example of the sequence and subsequences, i.e., $(x_n)$ and $(x_{nm})$, in form of a table or matrix?

Q4: Generally, what are the different meanings of math symbols for sequence $(x_n)$, $(x_n)_1^{\infty}$, $(x_n)_{n\in\mathbb{N}}$, $\{x_n\}$? I think $(x_n)_1^{\infty}$ and $(x_n)_{n\in\mathbb{N}}$ should be identical to each other, but I am not very sure about $(x_n)$ and $\{x_n\}$.

• Out of curiosity, does this result appear shortly after a result of the form "a subset of a totally bounded set is totally bounded"?. – Mike F Aug 19 '17 at 16:08

Suppose you have a subsequence of a subsequence of a subsequence, and so forth. We want to find a single subsequence of the original sequence that captures the properties of all of these subsequence. That's what the diagonal subsequence does. Consider a sequence $(x_{1,n})$ with subsequence $(x_{2,n})$, and $(x_{3,n})$ a subsequence of $(x_{2,n})$, and so forth. $$\begin{matrix} x_{1,1} & x_{1,2} & x_{1,3} & \cdots \\ x_{2,1} & x_{2,2} & x_{2,3} & \cdots \\ x_{3,1} & x_{3,2} & x_{3,3} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{matrix}$$ Each row represents the sequence $(x_{k,n})_n$, which is a subsequence of the previous row $(x_{k-1,n})_n$, and the entries of the rows are the sequence elements. The idea is that we want a sequence $(x_n)_n$ that captures the properties of all of these subsequences. To do this, we will take $(x_n)_n$ so that for each $k$, a tail of $(x_n)_n$ is a subsequence of $(x_{k,n})_n$. We define $x_n := x_{n,n}$, which is nothing more than the diagonal entries in the above picture. Then the tail $(x_n)_{n\geq k}$ is a subsequence of $(x_{k,n})_n$, for every $k$, so this new sequence $(x_n)_n$ behaves as if it is a subsequence of all of these subsequences.
For your fourth question, the notations are usually treated to be identical. Sometimes $\{x_n\}$ will be used to represent the set of the sequence instead of the sequence, but this is rare and will typically be mentioned by the author.