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

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. • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Commented Aug 1, 2021 at 20:15 • This is called "taking a diagonal subsequence". Commented Aug 1, 2021 at 20:51 • This doesn't help me that much to understand the reasoning of why this works :( – Evan Commented Aug 1, 2021 at 21:35 • Try to prove the case$\{ x_1,x_2\}=A$first, see how that works. Then try to prove the case of finite$A\$ by induction. I think this might be an issue of trying to do too much at once. Commented Aug 2, 2021 at 6:52
• I think that ignoring what I didn't understand in the proof and just taking it for granted I managed to probe the case where A is finite (if I did it correctly it should be equal to the proof for the case where A has only 2 elements), but now my question is: in the proof I sketched I make use of the argument from the solution which states that after taking {f_n(x)} and evaluating it at x=x1, we obtain a convergent subsequence.
– Evan
Commented Aug 2, 2021 at 10:14

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.

• It took me a while to even get an idea of what you wrote, since I'm not very experienced, but please let me know if I got this right: There are infinitely many subsequences of f_n(x), and at least one of them convergences for x_1, which we denote by f_1,k. By the same reasoning we construct f_2,k and so on, so that for each m In the naturals f_m,k convergences for x_1, x_2,...x_m, and for n<m, f_m,k is a subsequence of f_n,k. Is this correct?
– Evan
Commented Aug 3, 2021 at 9:17