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I am having trouble understanding the mechanics of this theorem's proof.

I don't see why uniform continuity is required, and I don't see why $n> \cfrac{1}{\delta}$ is needed.

So far I see this $\left|\cfrac{k}{n}-s\right|<\delta \iff\cfrac{|k-sn|}{|n|}<\delta \iff\cfrac{|k-sn|}{\delta}<n$ and I am stuck

Any help would be greatly appreciated. I copied part of this question from another user "MathLover"

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Kam
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From uniform continuity, for given $\epsilon >0$ we can find $\delta>0$ such that for any $s',s \in (\frac {k-1}{n}, \frac k{n})$ $$| s'-s|<\frac 1{n}<\delta \implies |F(z,s')-F(z,s)|<\epsilon \tag{for $n>1/ \delta$} $$

Now and taking $s'=\frac k{n}$ we would have $$|\frac k{n}-s|<\delta \tag{$\forall s\in (\frac {k-1}{n}, \frac k{n})$}$$

And this would imply

$$|F(z,\frac k{n})-F(z,s)|<\epsilon \tag{$\forall s\in (\frac {k-1}{n}, \frac k{n})$} $$

Math_user
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  • Thank you! I did not realize we were picking $s,s^{\prime}$ from that interval. We can define such an interval because $[0,1]$ is compact, and because it is compact and $F$ is continuous, it is uniformly continuous, so this allows us to pick any two points in such an interval. Still have a question, if we couldn't pick any two points in this defined interval, then why would this fail? In other words, why was uniform continuity required? – Kam Apr 30 '20 at 15:38
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    If we try to prove last inequality in my answer assuming point wise continuity at each $k/n$ then we would find different "delta's" for each $s=k/n$, whereas uniform continuity gives single delta which works for all – Math_user Apr 30 '20 at 16:01
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    Without uniform continuity can be done ,if we take $1/n<\delta =min{\delta_k }$,where $\delta_k $ is such that $|k/n-s|<\delta _k \implies |F(z,k/n)-F(z,s)|<\epsilon$ – Math_user Apr 30 '20 at 16:14
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    FANTASTIC, Thank YOU:) – Kam Apr 30 '20 at 16:16