Pointwise convergent subsequence of a sequence of bounded functions on a countable domain

Question

I was studying analysis on Abbot's understanding analysis, and I came across this exercise:

Let $A = \{x_1, x_2, x_3, \cdots\}$ be a countable set. For each $n \in\mathbb N$, let $f_n$ be defined on $A$ and assume there exists an $M > 0$ such that |$f_n(x)| \le M$ for all $n \in\mathbb N$ and $x\in A$. Follow these steps to show that there exists a subsequence of $(f_n)$ that converges on $A$.

The textbook gave me some hints which are:

6 (a) Why does the sequence of real numbers $f_n(x_1)$ necessarily contain a convergent subsequence $(f_{n_k} )$? To indicate that the subsequence of functions $(f_{n_k} ) $ is generated by considering the values of the functions at $x_1$, we will use the notation $f_{n_k} = f_{1,k}$.

(b) Now, explain why the sequence $f_{1,k}(x_2)$ contains a convergent subsequence.

(c) Carefully construct a nested family of subsequences $(f_{m,k})$, and show how this can be used to produce a single subsequence of $(f_n)$ that converges at every point of $A$.

I didn't have any problem with the first two (even if I still don't understand how in (b) he can tell the "form" of $(f_{n_k}(x))$, since we only know it's form if we evaluate it at $x=x_1$). On the third one though I didn't really understand how I was supposed to construct this particular sequence, so after playing around some time without getting any result I decide to take a look at the solution the author gives in the solution manual and I found this:

Keep in mind that if $m′ > m$ then $(f_{m′,k})$ is a subsequence of $(f_{m,k})$.

I honestly don't get why this should be true, since as I understand it, it should hold only for the specific sequences of real numbers $(f_{m',k}(x_m))$ and $(f_{m,k}(x_m))$.

$f_{n_k} = f_{k,k }= (f_{1,1}, f_{2,2}, f_{3,3},\cdots)$. The nested quality shows that $(f_{k,k})$ is a subsequence of $f_{1,k}$ and thus $f_{k,k}(x_1)$ converges. But what about $f_{k,k}(x_m)$ for an arbitrary $x_m\in A$? Well, after the first $m$ terms, we see that $f_{k,k}$ becomes a proper subsequence of $f_{m,k} (i.e., $f_{k,k}$ is eventually in $f_{m,k}$)

I also don't get why this true.

And it follows that $f_{k,k}(x_m)$ converges. This shows $f_{k,k}$ converges pointwisely on $A$

I would really appreciate if somebody could help me understand this proof, or even furnish me a different proof of this, since I was not able to find an alternative proof anywhere else.

Answer 1

Your comment about "not knowing the form of $(f_{n_k}(x))$" is confusing. What "form" are you referring to? It is just some subsequence of $f_n$ for which $f_{n_k}(x_1)$ converges. What more is there to know?

Let me fill various details of the construction. This is an inductive definition.

Start by setting $f_{0, n} = f_n$ for all $n$. We are going to define for each $m > 0$ a sequence of functions $(f_{m,n})_n$ and a value $y_m$ such that

• $(f_{m,n})_n$ is a subsequence of $(f_{m-1,n})_n$ (and therefore a subsequence of $(f_{M,n})_n$ for every $M < m$).
• for every $0 < i \le m, (f_{m,n}(x_i))_n \to y_i$.

When $m = 1$, we form the sequence of values $(u_n)_n = (f_{0, n}(x_1))_n$. This is a bounded sequence, so it has some convergent subsequence $(u_{n_j})_j$. Define $y_1 = \lim_j u_{n_j}$. Note that we indicate which elements of the original sequence are in the subsequence by a sequence of indices: $(n_j)_j$. We can use those indices to pick a corresponding sequence of functions: $$f_{1, j} = f_{0, n_j}$$ so for all $j$, $$f_{1, j}(x_1) = f_{0, n_j}(x_1) = u_{n_j} \overset j \to y_1$$

Now suppose that $((f_{m,n})_n)_{m=0}^{N-1}$ and $(y_m)_{m=1}^{N-1}$ have been defined so that the two conditions above are satisfied for all $m < N$. We do the same thing as for the start: Define a new sequence $(u_n)$ by setting $u_n = f_{N-1,n}(x_N)$. Our assumptions for convergence of $f_{N-1,n}$ say nothing about $x_N$. But we do know that this sequence is bounded. Therefore once again we have some convergent subsequence $(u_{n_j})_j$. Define $$y_N = \lim_j u_{n_j}\\(f_{N,j})_j = (f_{N-1,n_j})_j$$

Then $(f_{N,j})_j$ is by definition a subsequence of $(f_{N-1,n})_n$. So for every $i < N, (f_{N,j}(x_i))_j$ is a subsequence of $(f_{N-1,n}(x_i))_n$. But that is a convergent sequence, so the subsequence must also converge, and to the same limit. That is, for $i < N, f_{N,j}(x_i) \overset j \to y_i$. And for $i = N$, $$(f_{N,j}(x_N))_j = f_{N-1,n_j}(x_N) = u_{n_j} \overset j \to y_N$$

So in all cases, $(f_{N, j})_j$ is a subsequence of $(f_{N-1, n})_n$, and for all $i \le N, (f_{N, j}(x_i))_j \to y_i$. meeting the induction requirements.

By induction, this defines $f_{m,n}$ for all $m, n$ with the desired properties.