I'm reading about Functions of Bounded Variation in Carother's Real Analysis and trying to prove the following:
Helly's Selection Theorem:
Let ($f_n$) be a uniformly bounded sequence of real-valued functions defined on a set X, and let D be any countable subset of X. Then, there is a subsequence of ($f_n)$ that converges pointwise on D.
By uniformly boundedness of $(f_{n})$ on X, we have that $(f_{n}(x_{1}))$ is bounded in $\mathbb{R}$. Therefore, we can contain $(f_{n}(x_{1}))$ in some compact interval in $\mathbb{R}$. By compactness, we know that any subsequence has a convergent subsequence. Hence, there exists a subsequence $(f_{n}^{(1)}(x_{1}))$ that converges. Similarly, by uniform boundedness of $(f_{n}^{1})$, we can pass to a subsequence $(f_{n}^{(2)}(x_{2}))$ that converges. By our way of construction $(f_{n}^{(2)})$ $\subseteq$ $(f_{n}^{(1)})$, therefore $(f_{n}^{(2)}(x_{2}))$ and $(f_{n}^{(2)}(x_{1}))$ both converge. We can continue this process to obtain a subsequence $(f_{n}^{(m)})$ $\subseteq$ $(f_{n})$ such that $(f_{n}^{(m)}(x_{i}))$ converges for $i = 1, 2, 3, ...$. We now have a subsequence $(f_{n}^{(m)})$ $\subseteq$ $(f_{n})$ that converges for all $x$ $\in$ $D$.
This is the part of the proof I get confused at...
However, instead of using this subsequence we takes the diagonal of all the constructed subsequences $(f_{n}^{(n)}(x_{k}))_{n=1}^{\infty}$ to obtain a subsequence convergent for all $x$ $\in$ $D$. Why we need to go through this diagonalization process? Furthermore, it isn't obvious to me that this "diagonal" subsequence converges for all $x$ $\in$ $D$. If someone could explain this to me that would be greatly appreciated.
Thanks for any help.
By construction a subsequence $\{f_n^{(m)}(x_i)\}_{n\ge 1}$ converges for each $i\le m$. "Diagonalization" is used to get a sequence that converges for all $i\ge 1$. For fixed $i$, the tail of the diagonal sequence $\{f_n^{(n)}(x_i)\}_{n\ge i}$ is a subsequence of $\{f_n^{(i)}(x_i)\}_{n\ge 1}$ which is convergent (by construction).