Let $E \subset X$ which is a dense subset of $X$.The function $f$ has its range in $\mathbb{R}$.We need to show that there exists a continuous extension from $E$ to $X$.
My attempt :
let $x \in X$ and we choose a cauchy sequence {$e_n$} $\to x$.Now, since $\{e_n\}$ is a convergent sequence so it is a cauchy sequence. As $f$ is a uniformly continuous function, then $\{f(e_n)\}$ is also a cauchy sequence and so it is convergent to a point,we define that to be $f(x)$[As $\mathbb{R}$ is a complete metric space].
Now, we choose another sequence $\{(e_n)'\} \to x$.We see that
$d(e_n,x)<\frac{\delta}{2}$ for all $n \ge K_1$ and $d((e_n)',x)) < \frac{\delta}{2}$ for all $n \ge K_2$
Then, we see that $d_y(f((e_n)'),f(e_n)) \le \frac{\epsilon }{2}$ when $d(e_n,(e_n)')\le \delta$ for all $n \ge max\{K_1,K_2\}$.Using this equation we can show that $f((e_n)') \to f(x)$.
To prove:$f$ is uniformly continuous. We choose $d(x_m,x_n) < \frac{\delta}{3}$.
Let $\{(e_n^m)\} \to x_m$ and $\{(e_n^k)\} \to x_k$.Now, $\{f(e_n^m)\} \to f(x_m)$ and $\{f(e_n^k)\} \to f(x_k)$.So,
$d_y(f(e_n^k),f(x_k)) < \frac{\epsilon}{3}$ for all $n \ge K_1$ and
$d_y(f(e_n^m),f(x_m)) < \frac{\epsilon}{3}$ for all $n \ge K_2$.
Now by some manipulations we can find a $N$ such that when $n_1,n_2 \ge N$ we have $d((e_{n_1}^m), (e_{n_2}^k)) < \delta $ and by choosing such a delta we can have $d_y(f(e_{n_1}^k),f(e_{n_2}^m) )< \frac{\epsilon}{3}$ for all $n_1,n_2\ge N$.Also we see that,
$d_y(f(x_m),f(x_k)) \le d_y(f(e_n^k),f(x_k)) + d_y(f(e_n^m),f(x_m)) + d_y(f(e_n^k),f(e_n^m))< \epsilon$ for all $n \ge max\{K_1,K_2,N\}$
This has been an attempt. Can someone go through my proof?I am not sure about the part where I am proving uniform continuity.
Best Answer
Your proof is mainly correct, although the notation might seem a bit confusing at times. I will try to write a cleaner version of what you did in the last paragraph, where you proved the uniform continuity of the extension of $f$, which is denoted exactly the same. Of course, I shall assume that $(X, d)$ is a complete metric space and by abuse of notation, I shall also denote the standard euclidian distance on $\mathbb{R}$ by $d.$
Let $\varepsilon > 0.$ Since $f$ is uniformly continuous on the dense set $E,$ there is $\delta_\varepsilon > 0$ such that $d(f(e_1), f(e_2)) < \frac{\varepsilon}{3}$ whenever $d(e_1, e_2) < \delta_\varepsilon$ and $e_1, e_2 \in E.$ Take now $x, y \in X$ such that $d(x, y) < \frac{\delta_\varepsilon}{3}.$ From the definition of the extension (thus implicitly the density of $E$), we infer that there are $e_x, e_y \in E$ such that $d(e_x, x) < \frac{\delta_\varepsilon}{3} > d(y, e_y)$ and $d(f(x), f(e_x)) < \frac{\varepsilon}{3} > d(f(e_y), f(y)).$ In particular, it follows from the triangle inequality that we also have that $d(e_x, e_y) < \delta_\varepsilon.$ Hence we have that $d(f(e_x), f(e_y)) < \frac{\varepsilon}{3}.$ Finally, applying once again the triangle inequality, we deduce that $d(f(x), f(y)) \leq d(f(x), f(e_x)) + d(f(e_x), f(e_y)) + d(f(e_y), f(y)) < \varepsilon.$ This proves the claim that the extension of $f$ is uniformly continuous.