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