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This is theorem 7.23 in Baby Rudin.

If $\{ f_n \}$ is pointwise bounded sequence of complex functions on a countable set $E$, then $\{ f_n \}$ has a subsequence $\{ f_{n_k} \}$ such that $\{ f_{n_k}(x) \}$ convverges for every $x \in E$ (i.e. it has a pointwise convergent subsequence).

The proof says that $\{ f_{n}(x_1 \}$ is bounded so there exists a subsequence $\{ f_{1,k} \}$ such that $\{ f_{1,k}(x_1 \}$ converges as $k \to \infty$. I understand that. And I also understand the next part which says that to repeat this process and create a "matrix" with rows of these sequences. But then comes the part that I don't understand. Rudin then writes that we need to go down the diagonal of this "matrix" and choose $\{ f_{1,1}, f_{2,2}, f_{3,3}, \dots \}$ because this sequence will be our pointwise convergent subsequence. Why do we need to go along the diagonal and why can't we, for example, take the first "column", i.e. $\{ f_{1,1}, f_{2,1}, f_{3,1}, \dots \}$? It feels like the first column would still fulfill all the conditions that he sets, but obviosuly I am wrong but I can't see why

user5744148
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  • All elements of the first column could each be $ f_1$; you need to insure you wind up with a proper subsequence of $(f_n)$ (and have the required convergence properties). – David Mitra Aug 23 '21 at 10:50
  • @DavidMitra Okay, so it is here then the property that the order of which "the functions appear is the same in each sequence" so that "functions may only move to the left but never to the right"? – user5744148 Aug 23 '21 at 11:06
  • We cannot take the first column for it wouldn't be convergent as you see ${ f_{1,k} }$ means ${ f_{1,k}(x_1 }$ converges as $k \to \infty$ and ${ f_{2,k} }$ means ${ f_{2,k}(x_2 }$ converges as $k \to \infty$ and so on. If you take the first column i.e., ${ f_{1,1}, f_{2,1}, f_{3,1}, \dots }$ it could be the sequence ${f_{k,1}}$ that does not converge at all for in this sequence you only collect the first term of each convergent subsequence. Only then we go down diagonally, we can touch every point of E and get the convergence. – Myo Nyunt Sep 06 '22 at 10:42
  • This answer https://math.stackexchange.com/a/4525960/828003 may explain what you want to know completely. – Myo Nyunt Sep 06 '22 at 13:50

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