Help with Arzela-Ascoli theorem proof
Our teacher gaves a few days ago the standard proof of this theorem which can be found in many text books, dealing with finding a dense numerable set of the set in which the functions are defined, diagonilization, etc.
The thing is in this proof I can not understand why equicontinuity is necessary and not just simple uniform continuity for each function of the diagonal.
Any help would be appreciated.
The thing is that I don't see why the same delta is needed
Your absolutely right, if the same were needed then then inf of the deltas could be zero.
But the thing is I don't see why the same delta is needed. i have the following proof ( which I'll summarize in some parts to avoid a long text):
The proof follows much on the lines of the proof given in Elements of Real Analysis, Bartle
Given ( the set of continuous function from K to where K is compact and a subset of we want to show that if is a subset of )and is bounded and equicontinuous then there exists a subsequence for every sequence of functions in that converges.
We can find a set which is numerable and dense in K, then we can find a sequence of functions which i'll lable which for every point in A the sequence converges in .
The thing is that then from equicontinuity we can find a which makes every function , if and then take open balls with center in each and radius the union of these balls complete covers K, and because K is compact we can find a finite open subcover such that it covers K,using the fact that these functions are Cauchy convergent in . we find that for ,
Why can't we take open balls with different radius delta (i.e., use uniform continuity)?
most probably but I don't see it.
Continuing with the proof:
Then such that K is a subset of the open balls with radius delta and center in each of these points. Then we can find an N (because there is a finite number of points) such that the functions are smaller (using the norm of ) than epsilon / 3 then you just add up things using the triangle inequality and you get the result.