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In every Arzela Ascoli proof you see the following:

Let $S = \mathbb{Q} \cap [a,b]$, where $[a,b]$ is an interval in $\mathbb{R}$, then $S$ is a countable dense subset and there exists a finite number of points $\{x_1, \ldots, x_k\}$ such that $\forall x \in [a,b], |x - x_j| < \delta$, for at least one $j = \{1, \ldots, k\}$

It seems trivially true, since $Q$ is dense in $\mathbb{R}$. But there are some interesting details....

  1. there exists finitely number of points $\{x_1, \ldots, x_k\}$ , however $S$ is countable but not finite. What happened to the rest of the points?

  2. For at least one $j$...? Why not for all $x_j$ since for every rational we can stick a real in between that is $\delta$ close.

Can someone provide a proof to the claim and resolve the problems?

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    $\begingroup$ I guess you did not get the point of the argument. We want to cover $[a,b]$ with finitely many intervals of length $2\delta$ where the centers are rational numbers. That's it. You can choose the centers to be rational because $S$ is dense. $\endgroup$ – Friedrich Philipp Mar 4 '16 at 4:23
  • $\begingroup$ I'm not sure what $\delta$ is supposed to be here. But I think you're just using the compactness of $[a, b]$. For any fixed $\delta > 0$, put a ball of radius $\delta$ around each element of $S$. Since $S$ is dense in $[a, b]$, this is an open cover of $[a, b]$, so it can be thinned to a finite subcover. This gives you what you want. $\endgroup$ – Alex Wertheim Mar 4 '16 at 4:25
  • $\begingroup$ @FriedrichPhilipp Is Arzela Ascoli do-able if we considered the entire domain instead from the start? I don't understand why we need to first consider a countable dense set, then prove that uniform convergence on dense set holds for the entire domain $\endgroup$ – Carlos - the Mongoose - Danger Mar 4 '16 at 4:56
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Fix some $\delta > 0$. Then $\{ (q-\delta, q+\delta) \mid q \in \mathbb Q \}$ is an open covering of the compact set $[a,b]$. Therefore, there is a finite subset $\mathcal F \subseteq \{ (q-\delta, q+\delta) \mid q \in \mathbb Q \}$ that also covers $[a,b]$. This in return means that there is a finite set $\{q_1, \ldots, q_n\}$ of rational numbers such that for all $x \in [a,b]$ there is some $i \in \{1, \ldots, n\}$ with $\mid q_i - x \mid < \delta$.

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