Helly's Selection Theorem Problem (From Rudin):
Assume that $\{f_n\}$ is a sequence of monotonically increasing functions on $\mathbb{R}^1$ with $0 \le f_n(x) \le 1$ for all $x$ and all $n$.
Prove that there is a function $f$ and a sequence $\{n_k\}$ such that $f(x) = lim_{n \to \infty} f_{n_k}(x)$ for every $x \in \mathbb{R}^1$.
Why does this proof not work:
For each $x \in \mathbb{R}^1$, the range of $f_n(x)$ is $[0,1]$ and so bounded and closed, and since it is in Euclidean space, the range is compact. By the Bolzano-Weierstrass Theorem, $f_n(x)$ thus has a convergent subsequence (call it $f_{n_k}$). Since this is true of all $x$, we can define a limit function $f$ such that $f(x) = lim_{n \to \infty} f_{n_k}(x)$ for all $x$.
I haven't seen anywhere saying that this proof is incorrect, but I notice that all proofs I can find online are much longer and more involved. What am I missing?
thus has a convergent subsequence...This is subtly wrong, in that you've found a convergent subsequence for a fixed value of $x$, but there's no guarantee this will converge for other values of $x$. E.g. it's not obvious that $\lim_{k\rightarrow\infty} f_{n_k}(y)$ even exists for $y\neq x$.