I need to understand a way to prove this statement
$f$ is uniformly continuous on $D$ $\iff$ (when $x,y \in D$ then $|x-y| \rightarrow 0 \implies |f(x)-f(y)| \rightarrow 0)$
I know $\delta$ can depend on $\epsilon$ and for all the examples I have done, $\delta$ always can be set to $\delta = k \epsilon$ which would allow me to prove the rightward implication quite easily, but how can I mathematically prove this?
I am also not at all sure how to prove the leftward implication
EDIT: I have been told I was vague. I will try to reformulate.
Let $D$ be any interval of $\mathbb{R}$. Prove that a continuous function $f:D \rightarrow \mathbb{R}$ is uniformly continuous on $D$ if and only if whenever $x_n, y_n \in D$ are such that if $|x_n – y_n| \rightarrow 0$, then $|f(x_n) – f(y_n)| \rightarrow 0$
Best Answer
Your formulation is vague, and if interpreted a certain way, is false, as Lorenzo Vanni's answer shows. However, you can formulate it more precisely in a way that makes it true. Namely, $f$ is uniformly continuous if and only if for all sequences $x_n\in D$, $y_n\in D$ with $\lim_{n\to\infty}|x_n-y_n|= 0$ we have $\lim_{n\to\infty}|f(x_n)-f(y_n)|=0$.
As you observed, the rightward implication follows in a straightforward way from the $\epsilon$-$\delta$ definition of uniform continuity. [Update - I misread the question initially, see the update for an explanation of the rightward implication.]
To prove the leftward implication, note that if $f$ fails to be uniformly continuous, then there is some $\epsilon>0$ for which all $\delta$ fail to satisfy the definition of uniform continuity, meaning if we let $\delta=\frac{1}{n}$ we have some $x_n,y_n\in D$ with $|x_n-y_n|\leq \frac{1}{n}$ and $|f(x_n)-f(y_n)|> \epsilon$. Therefore letting $n\to \infty$ we have $|x_n-y_n|\to 0$, yet $|f(x_n)-f(y_n)|> \epsilon$ for all $n$, so we do not have $|f(x_n)-f(y_n)|\to 0$.
Update
Reading your question more closely, I'm not sure what you mean by $\delta=k\epsilon$. This requires something stronger than uniform continuity - it requires Lipschitz continuity.
Nonetheless, the usual $\epsilon$-$\delta$ definition of uniform continuity is enough to prove the rightward implication. Simply argue that given $\epsilon>0$, choose $\delta$ from the definition of uniform continuity, and then from the definition of convergence, eventually $|x_n-y_n|\leq \delta$ for large enough $n$, which implies from the definition of uniform continuity that $|f(x_n)-f(y_n)|\leq\epsilon$ for large enough $n$. Since $\epsilon$ was arbitrary, this in turn implies $|f(x_n)-f(y_n)|\to 0$.