Uniformly continuous on a bounded set How do I prove that If $f$ is uniformly continuous on a bounded set $D \subset \mathbb{R}$, then $f[D]$ is bounded in $\mathbb{R}$. "
Thank you 
 A: I am assuming $f:D\to \mathbb{R}$ is uniformly continuous and $D$ is totally bounded.


*

*Prove that $\overline{D}$ is bounded : Since $D$ is totally bounded, so is $\overline{D}$. Hence $\overline{D}$ is compact

*Extend $f$ to a continuous function $g:\overline{D} \to \mathbb{R}$ as follows: 
a) For a $x\in \overline{D}$, choose a sequence $x_n \in D$ such that $x_n \to x$, then $(x_n)$ is Cauchy. Now prove that $f(x_n)$ is Cauchy.
b) Since $\mathbb{R}$ is complete, $f(x_n)$ converges to a point $y\in R$. Define $g(x) := y$, and show that this definition is independent of your choice of sequence $(x_n)$.
c) Now check that $g$ is continuous.
Hence, $g:\overline{D} \to \mathbb{R}$ is a continuous function defined on a compact set. So, $g(\overline{D})$ is compact in $\mathbb{R}$. In particular, $f(D) \subset g(\overline{D})$ is bounded.
A: Since $f$ is uniformly continuous, you can find a $\delta$ such that whenever we have $|x-y|<\delta$, then $|f(x)-f(y)|<1$.
Take $M$ big enough so that $D\subset [-M,M]$. By the archimedean property, we can cover this interval with a finite number of intervals of length $\delta$. Take those who intersect $D$ and call them $I_1,\ldots,I_n$.
For each $i$, pick an $x_i\in I_i\cap D$ and let $M=\max\{|f(x_1)|,\ldots, |f(x_n)|\}$.  Let's see that $1+M$ is an upper bound for $f$ in $D$:
For every $y\in D$, there's a $1\leq j\leq n$ such that $y\in I_j$. Since $x_j\in I_j$ and the length of $I_j$ is less than $\delta$, then $|f(y)-f(x_j)|<1$. It follows that
$|f(y)|<1+|f(x_j)|\leq 1+M$.
