Originally Posted by

**Diego** Your absolutely right, if the same $\displaystyle \delta$ 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 $\displaystyle C(K,\mathbb{R}^{q}) $( the set of continuous function from K to $\displaystyle \mathbb{R}^{q}), $ where K is compact and a subset of $\displaystyle \mathbb{R}^{p} $ we want to show that if $\displaystyle \Im$ is a subset of $\displaystyle C(K,\mathbb{R}^{q}$)and $\displaystyle \Im$ is bounded and equicontinuous then every subsequence of a sequence of functions in $\displaystyle \Im$.

We can find a set $\displaystyle A := \left\lbrace x_{1},x_{2},...\right\rbrace$ which is numerable and dense in K, then we can find a sequence of functions which i'll lable $\displaystyle \left\lbrace g_{n}\right\rbrace $ which for every point in A the sequence $\displaystyle \left\lbrace g_{n}(x_{i})\right\rbrace$ converges in $\displaystyle \mathbb{R}^{q}$.

The thing is that then from equicontinuity we can find a $\displaystyle \delta$ which makes every function $\displaystyle \|g_{n}(x)-g_{n}(y)\|<\epsilon/3$, if $\displaystyle \|x-y\|<\delta$ and then take open balls with center in each $\displaystyle x_{i}$ and radius $\displaystyle \delta$ the union of these balls complete covers K, and because K is compact we can find an open subcover such that it covers K,using the fact that these functions are Cauchy convergent in $\displaystyle \mathbb{R}^{q}$. we find that for [tex]x \in K[\math], $\displaystyle \|g_{n}(x)-g_{m}(x)\|<\epsilon$

Why can't we take open balls with different radius delta (i.e., use uniform continuity)?