Equivalence of two definitions for uniformly continuous functions in metric spaces

Question

Given metric spaces $(X,d_{X})$ and $(Y,d_{Y})$; a function $f:X \to Y$ is called uniformly continuous if $$ \forall \, \varepsilon>0 \quad \exists \, \delta>0 \quad \forall \, x,y\in X \quad (d_{X}(x,y)<\delta \quad \Rightarrow \quad d_{Y}(f(x),f(y))<\varepsilon). $$ How to prove that $f:X \to Y$ is uniformly continuous if and only if for every pair of sequences $\{x_{n}\}, \{y_{n}\}\subset X$ such that $$ \lim\limits_{n\to\infty} d_{X}(x_{n},y_{n})=0 $$ we have $$ \lim\limits_{n\to\infty} d_{Y}(f(x_{n}),f(y_{n}))=0 \, ? $$

Answer 1

HINT: Proving that the first statement implies the second is very straightforward: you do the most natural thing, and it works. That is, assume that $\lim_{n\to\infty}d_X(x_n,y_n)=0$, let $\epsilon>0$ be arbitrary, and show that there is an $m_\epsilon\in\Bbb N$ such that $d_Y\big(f(x_n),f(y_n)\big)<\epsilon$ whenever $n\ge m_\epsilon$. You’ll use both the first statement, which tells you that getting $x_n$ close enough to $y_n$ gets $f(x_n)$ as close to $f(y_n)$ as you want no matter where $x_n$ and $y_n$ are, and the hypothesis that $\lim_{n\to\infty}d_X(x_n,y_n)=0$.

For the other direction I suggest that you try proving the contrapositive. Suppose that there is some $\epsilon>0$ such that for each $\delta>0$ there are $x_\delta,y_\delta\in X$ such that $$d_X(x_\delta,y_\delta)<\delta\;,\quad\text{ but }\quad d_Y\big(f(x_\delta),f(y_\delta)\big)\ge\epsilon\;.$$ Use this assumption to produce two sequences that violate the second statement.