TITLE: If $f$ is uniformly continuous on $(a, b)$, then $f$ is bounded on $(a, b)$.
QUESTION [8 upvotes]: So I know that since $f$ is uniformly continuous on $(a,b)$, then for every 
$\varepsilon > 0$, there exists $\delta > 0$ such that for all $x$ and $y$ in $(a,b)$, 
if $\,\lvert x - y\rvert < \delta$, then $\,\lvert\,f(x) - f(y)\rvert < \varepsilon$. 
I also know that I need to show that there are some numbers $M$, $N$, so that 
$M ≤ f(x) ≤ N$, for all $x$ and $y$ in $(a,b)$. 
It makes logical sense to me because it means that for every x and y that you pick, the function values can only jump by epsilon, so there would be no way for the function suddenly approach infinity at the endpoints. However, I don't know really how to make a rigorous argument. I was thinking that maybe I could do something by contradiction, like saying that $f(y)>N$, but $f(x)<N$, and then somehow showing that this jump is now greater than 
$\varepsilon$, but I'm not sure that would really work since we only have an arbitrary function with arbitrary bounds.

REPLY [2 votes]: Let $\epsilon = 1$ in the definition of uniform continuity.
There is $\delta$ such as
$$
|x-y| < \delta \implies |f(x) - f(y)|<1
$$
Now let $N\in \Bbb N$, $N>\frac{b-a}{\delta}$. Then for every $x\in (a,b)$ one can find 
$k$ points $x_k$ with $|x_{m+1} - x_m| < \delta$, $x_1 = x$, $x_k = \frac{a+b}2$ and $k\le N$.
Using the triangle inequality,
$$
|f(x)|\le 
\left|f\left(\frac{a+b}2\right)\right| + \left|f\left(\frac{a+b}2
\right)- f(x)
\right| \le 
\left|f\left(\frac{a+b}2\right)\right| + \sum_{m=1}^{k-1}
\left|
f\left(x_{m+1}\right)- f(x_m)
\right|
\\
\le \left|f\left(\frac{a+b}2\right)\right| + k
\le \left|f\left(\frac{a+b}2\right)\right| + N 
$$
which is a universal bound.