Why diagonal $S$ is a subsequence of $S_n?$

real-analysis

Rudin theorem $7.23$

If $(f_n)_{n \in \mathbb{N}}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $(f_n)_{n \in \mathbb{N}}$ has a subsequence $(f_{n_k})_{k \in \mathbb{N}}$ such that $(f_{n_k}(x))_{k \in \mathbb{N}}$ converges for each $x \in E$.

\begin{matrix}
S_1: & f_{1, 1} & f_{1, 2} & \ldots \\
S_2: & f_{2, 1} & f_{2, 2} & \ldots \\
\vdots \\
S_n: & f_{n, 1} & f_{n, 2} & \ldots \\
\end{matrix}

In the theorem of the proof Rudin say that diagonal $S: f_{1,1},f_{2,2},……f_{n,n}…$ is a subsequence of $S_n$

My thinking : Here $S_n$ is a subsequence of $S_{n-1}$. we are starting from sequence $S_1$ so by bolzano Wierstrass theorem we can derived another sequence .similarly we can derived $S_2$ from $S_1$,….$S_n$ from $S_{n-1}$

Therefore by Bolzano Wierstrass theorem we can derived $S_n$ from $S \implies S_n$ is a subsequence of $S$

My confusion :why diagonal $S$ is a subsequence of $S_n?$

Best Answer

Presumably, we enumerate $E=\{e_1,e_2,\dots\}.$ Then:

  1. Define $S_1=\{f_{1,j}\}_{j=1}^\infty$ as a subsequence of $f_i$ such that $\lim_{j\to\infty}f_{1,j}(e_1)$ exists.
  2. Given the $S_i$ defined, define $S_{i+1}=\{f_{i+1,j}\}_{j=1}^\infty$ as a subsequence of $S_i$ such that $\lim_{j\to\infty}f_{i+1,j}(e_{i+1})$ exists.

Then certainly, $\{f_{1,1},f_{2,2},\dots,\}$ is a sequence of functions from our original sequence, $\{f_1,f_2,\dots\}.$ The real question is whether we can get repeated $f_i$ or disordered $f_i$ in this sequence.

Let $m(i,j)$ be the index in the original series corresponding to $f_{i,j}.$ Then, since $S_{i+1}$ is a subsequence of $S_i,$ you get that $m(i+1,j)\geq m(i,j).$

You also have that $m(i,j+1)>m(i,j).$

To prove these two inequalities rigorously, you need to have a fairly rigorous definition of “subsequence,” but they are examples of results that are hopefully “obvious,” if you spend a moment thinking of them.

So $m(1,1),m(2,2),m(3,3),\dots$ is strictly increasing.

Then $f_{i,i}=f_{m(i,i)}$ is a subsequence of $f_1,f_2,\dots.$

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