Rudin's proof theorem 7.23

It is entirely possible to create such sequences. As to the reason, here it is.

Consider the sequence

$f_{1,1}(x_2),f_{1,2}(x_2),\ldots$

This is a bounded sequence since the functions are pointwise bounded.

Thus, it contains a convergent subsequence, $f_{1,n_i}(x_2)$

This subsequence of functions, we denote by $f_{2,i}$.

You can go through such a process repeatedly to get $S_n\subset S_{n-1}$ with the required properties.

I see this is an old post. But if I am still looking at it, then probably someone else is, too. I'm not a fan of Rudin's notation in this one. I think it adds confusion, not clarity. However, I really like Daniel Fischer's comment. Do I understand correctly (I too am only an undergraduate) that the proof is as follows?

Proof

Let $A$ be a countably infinite set. Let $(f_n)$ be a sequence of mappings $A\to\mathbb{C}.$ Thus, by definition of countably infinite, consider an enumeration $a_1,a_2,\ldots$ of $A$. Assume that $(f_n)$ is point-wise bounded (i.e., the sequence $f_1(a_n), f_2(a_n),\ldots$ in $\mathbb{C}$ is bounded for each given $n\in\mathbb{Z}^+$). In particular, the sequence of complex numbers $(f_n(a_1))$ is bounded. Hence, by the Bolzano-Weierstrass theorem, there exists a convergent sub-sequence. $(f_{n_k}(a_1))$. Consider this sub-sequence. Now, we have a sequence of functions $(f_{n_k})$ which is convergent when "evaluated at" $a_1$. However, it may still be evaluated at other points. In particular, it is reasonably clear that a sub-sequence of a point-wise bounded sequence of functions is, itself, a point-wise bounded sequence of functions. Thus, the sequence of complex numbers $(f_{n_k}(a_2))$ is bounded, so that there again exists a convergent sub-sequence $(f_{n_{k_m}}(a_2))$. Consider this sub-sequence.

The crucial logical step is that not only does the sequence of complex numbers $(f_{n_{k_m}}(a_2))$ converge, but also the sequence $(f_{n_{k_m}}(a_1))$ converges since it is a sub-sequence of the convergent sequence $(f_{n_k}(a_1))$! So, at this point in our informal construction, we have already identified a sub-sequence $(f_{n_{k_m}})$ of $(f_n)$ which converges evaluated at, not one, but two points $a_1,a_2 \in A$. The idea is, thus, to (implicitly, I think, use the axiom of choice to) extend this iterative construction ad infinitum.

The technical details of infinite recursive constructions are a subject in their own right. However, we (following Rudin) are typically not shy about taking for granted that these constructions are non-problematic. So, let us simply "continue in this way" without end, and take the sub-sequence $$ f_{ n_{ \scriptstyle{k}_{ \scriptstyle{\ell}_{ ._{._{._{.}}} } } } } $$ which results from this construction as evidence of the asserted existence statement.