Let $f: X \subset \mathbb{R}^n\to $. Then, $f$ is uniformly continuous if, and only if, for every sequence, $(x_n)_{n\in\mathbb{N}}$, $(y_n)_{n\in\mathbb{N}}$ such that $d(x_n,y_n) \to 0$, then $d(f(x_n),f(y_n)) \to 0.$ Even more, if $f$ is uniformly continuous then for every $a \in \overline{X}$, the limit $\lim_{x\to a}f(x)$ exist.
I am thinking very hard about this problem. It seems that every moment I will solve it, but it never happens.
I imagine that my problem resides in does not understand well the meaning of uniform continuity, and principally, I am sure that I can not apply the definition to this case.
Please, I appreciate any elucidate solution with an explanation about where to use, and how to use uniform continuity and in what part to use.
Thanks a lot.
Hints:
Assume $f$ is uniformly continuous. Let $x_n,y_n\in X$ with $d(x_n, y_n)\to 0$.
Let $\epsilon > 0$. Then, there exists a $\delta > 0$ with $ d(f(x), f(y)) < \epsilon $ for every $x,y\in X$ with $d(x,y) < \delta$. For this $\delta$ there exists a $N\ge 1$ with $d(x_n, y_n) < \delta$ for every $n\ge N$. Thus, for every $n\ge N$ it follows $d(f(x_n), f(y_n)) < \epsilon$.
Assume $f$ is not uniformly continuous. Then there exists a $\epsilon > 0$ such that for every $\delta > 0$ there exists some $x,y\in X$ with $d(x,y) < \delta$ and $d(f(x), f(y)) \ge \epsilon$.
As you said that $Y$ is complete: Let $a\in \overline X$. Then, there exists a sequence $\{x_n\}$ in $X$ with $x_n\to a$. Now, show that $f(x_n)$ is Cauchy. Does the limit depend on the sequence $\{x_n\}$?