How to prove if $f$ is defined and uniformly continuous on a bounded set $E$, then $f$ is bounded on $E$? How to prove if $f$ is defined and uniformly continuous on a bounded set $E$, then $f$ is bounded on $E$?
I know how to prove it on an open or closed interval, but my problem is the question says bounded sets instead of intervals and I don't know how to deal with that.
Thanks very much!!
 A: Here is a proof that does not use extension to $\bar E$.
By uniform continuity, there exists $\delta>0$ such that
$$
|x-y|\le\delta\implies|f(x)-f(y)|\le1.
$$
Since $E$ is bounded, there exist $x_1,\dots,x_N\in E$ such that
$$
E\subset\bigcup_{k=1}^N B(x_i,\delta),
$$
where $B(x,r)$ is the open ball of center $x$ and radius $r$. Note that this holds in $\mathbb{R}^n$, but not necessarily in general metric spaces. Let $M=\max_{1\le i\le N}|f(x_i)|$. Given $x\in E$ there exists $i$, $1\le i\le N$ such that $|x-x_i|<\delta$. Then
$$
|f(x)|\le|f(x)-f(x_i)|+|f(x_i)|\le M+1.
$$
A: By uniform continuity, $f$ can be continuously extend to the closure of $E$. (OP indicated awareness of this in comments.)
By boundedness of $E$, there is a closed interval $[a,b]$ containing $E$, and $f$ can be continuously extended further from the closure of $E$ to $[a,b]$ (OP indicated awareness of continuous extensions of functions defined on closed sets in comments.)
This extension on $[a,b]$ is bounded, hence $f$ is bounded.
