Let $E$ be a dense subset of a metric space $X$, and let $f$ be a uniformly continuous function defined on $E$ to a complete metric space $Y$. Prove that $f$ has a continuous extension from $E$ to $X$.
Proof:Let, $l\in X$ then there exist a sequence $\{x_n\}\in E$ which converges to $l$.
Then, $\{f(x_n)\}$ becomes a cauchy sequence in $Y$, therefore it converges somewhere in $Y$ call the point $g(l)$.
Now, it's easy to see that $g$ is well defined, since if $d_X(x_n,y_n)\to 0$ then $d_Y(f(x_n),f(y_n))\to 0$.Again, it's not hard to see that $g$ is continuous on $X$.
Is it a correct approach?
Can we extend $f$ in such a way preserving uniformly continuity?
Best Answer
Your proof is correct. For uniform continuity let $\epsilon >0$ and choose $\delta >0$ such that $d(x,y) <\delta$ and $x,y \in E$ implies $d(f(x),f(y)) <\epsilon$. Now let $x,y \in X$ with $d(x,y) <\delta $ and choose $x_n, y_n$ in $E$ converging to $x$ and $y$ respectively. Then $d(x_n,y_n)\leq d(x_n,x)+d(y_n,y)+d(x,y) <\delta$ for $n$ sufficiently large. Hence $d(f(x_n),f(y_n)) <\epsilon$ for such $n$. Using continuity of the metric this implies $d(f(x),f(y)) \leq \epsilon$.