# Arzela-Ascoli theorem on Wikipedia

The following is a question about the Arzela-Ascoli theorem. The theorem states that if $f_n : [0,1]\to \mathbb R$ are uniformly bounded and equicontinuous then there exists an uniformly convergent subsequence.

Assume we have constructed the limit function by taking the pointwise limit on the rationals. The proof on Wikipedia chooses $N$ to be the largest of all $N_k$ for the finite set of rational points $x_k$. The $x_k$ here are each in one of the $\delta$ balls of the finite cover of $[0,1]$.

Wouldn't it also work to argue as follows:

Let $x \in [0,1]$ be arbitrary. Then $x$ is in one of the $\delta$ balls. One of the $x_k$ is in the same $\delta$ ball. Now choose $N$ to be $N_k$ (instead of choosing $N$ to be $\max_k N_k$).

I think this choice of $N$ works, too. Can you tell me what I'm missing and why it is necessary to choose $N = \max_k N_k$?

• But for another $x'\in [0,1]$, it lies in another $\delta$-ball which contains another $x_{k'}$. Then for this $x'$ you need to choose $N_{k'}$. After all you need to take the biggest one. – user99914 Mar 13 '14 at 9:23
• @John But $x$ is chosen at the start. And then the proof makes the difference $f_n - f_m$ at that particular $x$ smaller than $\varepsilon$. Since $x$ was an arbitrary point the bound should hold for all $x$ in $[0,1]$. Why can one not argue like this? – newb Mar 13 '14 at 10:21
• Because you want that $\epsilon$ to be chosen independent of $x$. They need the uniform convergence of $f_n$ to the limit to show that the limit is a continuous function. – user99914 Mar 13 '14 at 10:40