Show that if $f$ is uniformly continuous and if $|p_n-q_n|\to 0$, then $|f(p_n)-f(q_n)|\to 0$.

Question

If $f$ is uniformly continuous on a set $D$, show that it has the property that if $p_n,q_n\in D$ and $|p_n-q_n|\to 0$, then $|f(p_n)-f(q_n)|\to 0$.

This question is a suggested problem from my professor but I couldn't do it. Since $f$ is uniformly continuous, for each $\epsilon>0$, we can find $\delta>0$ such that for every $p_n,q_n\in D$ , $|p_n-q_n|<\delta$ implies $|f(p_n)-f(q_n)|<\epsilon$. We need to show as $\delta \to 0$, we have $\epsilon \to 0$. But can we change our delta randomly for the same epsilon? I'm confused at this point because delta is dependent to epsilon. Any help is much appreciated.

Answer 1

Let $\epsilon>0$ given.

$f $ is uniformly continuous at $D $

thus

$$ \exists \eta>0 \; : \;\forall x,y\in D $$

$$|x-y|<\eta\implies |f (x)-f (y)|<\epsilon $$

on the other hand

$$\exists N\in\mathbb N :$$ $$n>N \implies |p_n-q_n|<\eta $$

thus

$$n>N\implies |f (p_n)-f (q_n)|<\epsilon $$

which means that

$$\lim_{n\to\infty}(f (p_n)-f (q_n))=0.$$