Why do we pick $n_0$ such that $\frac{1}{n_0}< \delta$? Let $f: \mathbb{R} \to \mathbb{R}$ uniformly continuous.We set $f_n(x)=f(x+\frac{1}{n})$.Show that $f_n \to f \text{ uniformly }$.
Let $\epsilon>0$.
Since $f: \mathbb{R} \to \mathbb{R}$ uniformly continuous, $\exists \delta>0 $ such that $\forall x,y \in \mathbb{R}$ with $|x-y|< \delta \Rightarrow |f(x)-f(y)|<\epsilon$.
For $y=x+\frac{1}{n}$,we have $|x-x-\frac{1}{n}|< \delta \Rightarrow |f(x)-f(x+\frac{1}{n})|< \epsilon \Rightarrow \frac{1}{n}< \delta \Rightarrow |f(x)-f(x+\frac{1}{n})|< \epsilon$
We pick $n_0$ such that $\frac{1}{n_0}< \delta$
$|f_n(x)-f(x)|=|f(x+\frac{1}{n})-f(x)| < \epsilon$
and taking the supremun,we get $\sup_{x \in \mathbb{R}} {|f_n(x)-f(x)|} < \epsilon$,so $f_n \to f \text{ uniformly }$.
Could you explain me why we pick  $n_0$ such that $\frac{1}{n_0}< \delta$ ??
 A: Since you need $1/n<\delta$, you must choose any $n_0 \in \mathbb{N}$ such that $1/n_0<\delta$. As you write, $|x-y|=1/n$, and the choice of $n$ is dictated by the uniform continuity of $f$.
A: The part that says "$\frac{1}{n}\lt\delta\Rightarrow...$", is actually saying if
"$\frac{1}{n}\lt\delta$" this implies "..."
So to continue with the desired property we must pick an $n_0$ such that $\frac{1}{n_0}\lt\delta$
It may become more apparent if we look at a real life analogue:
If I pick up an apple, this implies that I can eat it.
Which means I must pick it up to eat it. But I am free to choose when I pick it up and with which hand. This is anologous to the choice of picking an $n_0$.
A: What does it mean for a sequence $f_n$ to be uniformly convergent to $f$? It just means that for every $\varepsilon>0$ you can choose a $n_0$ so that for all $n\geq n_0$: $sup_{x \in \mathbb{R}} {|f_n(x)-f(x)|} < \epsilon$. So what's been done is, that $n_0$ is chosen according to the premise, which is possible hence f is uniformly continous.
