Sequence of uniformly bounded functions over the positive integers has a certain convergent subsequence This is Theorem 1.24 from Reed and Simon's Methods of Modern Mathematical Physics, Vol 1: Functional Analysis

Let $f_n(m)$ be a sequence of functions on the positive integers which is uniformly bounded, i.e. $|f_n(m)| \le C$ for all $n,m$. Then there is a subsequence $\{f_{\hat n(i)}(m)\}^\infty_{i = 1}$ so that for each fixed $m$,  the sequence $f_{\hat n(i)}(m)$ converges as $i \to \infty$.

Their proof:

Consider the sequence $f_n(1)$. It is a bounded set of numbers, so we can find a subsequence $f_{n(i)}$ such that $f_{n_1(i)}(1) \to f_\infty(1)$, for some number $f_\infty(1)$. Now consider the sequence $f_{n_1(i)}(2)$. We can find a subsequence $f_{n_2(i)}(2) \to f_\infty(2)$ as $i \to \infty$. Proceeding inductively, we find successive subsequences $f_{n_k(i)}$ so that $f_{n_{k+1}(i)}$ is a subsequence of $f_{n_k(i)}$, and $f_{n_k(i)} \to f_\infty(k)$ as $i \to \infty$. Thus, in particular, $f_{n_k(i)}(j) \to f_\infty(j)$ as $i \to \infty$ for $j = 1,2,\dots,k$. To get a subsequence $f_{\hat n(i)}$ converging for each $j$, one is tempted to try and take the limit of the horizontal sequence but that won't work! (for it may happen $n_k(1) \rightarrow \infty$). The simple way out is to take the diagonal sequence $\hat n(k) = n_k(k)$. Then $f_{\hat n(k)}, f_{\hat n(k+1)}, \dots$ is a subsequence of $f_{n_k(i)}$ so $f_{\hat n(i)}(k) \to f_\infty(k)$ as $i \to \infty$ for each $k$.

I have gone over their proof a number of times but I still do not fully understand it. My understanding is that since $f_n(1)$ is bounded for all $n$ (by assumption of uniform boundedness) there exists a subsequence $f_{n_1(i)}$ such that $f_{n_1(i)}(1)$ converges, whose existence is given by the Bolzano-Weierstrass theorem. This process is repeated inductively so that

*

*$f_{n_{k+1}(i)}$ is a subsequence of $f_{n_k(i)}$

*$f_{n_k(i)} \to f_\infty(k)$ as $i \to \infty$.

What confuses me is their claim that $f_{n_k(i)}(j) \to f_\infty(j)$ as $i \to \infty$ for $j = 1,2,\dots,k$. How do we know, for example, that $f_{n_k(i)}(k-1) \to f_\infty(k-1)$ as $i \to \infty$. In other words, how do we know that the subsubsequence when evaluated at $k-1$ converges to the same limit as the subsequence also evaluated at $k-1$.
Secondly, they claim that taking the limit of the horizontal sequence won't work due to the possible case where $n_k(1) \rightarrow \infty$, and that is why they use their "diagonal sequence trick". Why does taking the limit fail? I do not see why their example is a counterexample.
 A: 
How do we know that the subsubsequence when evaluated at $k−1$ converges to the same limit as the subsequence also evaluated at $k−1$?

In fact, that is obvious since any subsequence of a sequence $S$ that converges also converges to the same limit. A subsequence will converge only "sooner".
In fact, not only a subsequence converges to the same limit, but any sequence that is a multi-subset of $S$ as a multi-set also converges to the same limit (hence the order of elements is not important). We do not need this more general fact for the proof in the question; hopefully it may help understand similar situations.


They claim that taking the limit of the horizontal sequence won't work due to the possible case where $n_k(1)\to\infty$.

By "taking the limit of the horizontal sequence" they mean the strategy to let $\hat n(i)=\lim_{k\to\infty}n_k(i)$ for all $i$. However, it won't work necessarily.
For example, suppose for all $k$ the sequence $n_{k+1}$ is the subsequence of all elements of $n_k$ at even indices and $n_1(i)=2i$, i.e., $n_k(i)=2^ki$. Then $n_k(1)=2^k$, which goes to $\infty$.

Exercise.  Let $\{f_n(\cdot)\}^\infty_{i=1}$ be a sequence of functions on the positive integers which is uniformly bounded, i.e. $|f_n(m)| \le C$ for all $n,m$. Then there is a sequence of indices $\hat n=\{\hat n(i)\}^\infty_{i = 1}$ so that  the sequences $\{f_{\hat n(i)}(m)\}^{\infty}_{i=1}$ with $i \to\infty$ for all positive integer $m$ converge uniformly. (Some notations here are slightly different from but clearer than the notations in the proposition in the question.)
