How can it be proved that a continuous function is bounded? Say $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$, what's the most concise way you know of to show that it's bounded?
I was thinking let $A=\{u : f(x) \text{ is bounded on }x<u\}$
Is there a way to show that $\sup(A)=b$?
 A: Here is a "from scratch" attempt. Suppose $f$ were to be unbounded on $[a,b]$ yet continuous there.  Then $f$ is unbounded on one of $[a,(a + b)/2]$ or $[(a + b)/2, b]$.  Denote this interval by $I_1$.  Keep subdividing in this fashion to obtain a sequence of intervals $I_n$ so that $I_{n+1}\subseteq I_n$ for all $n$ and so that the length of $I_n$ is $$(b-a)/2^n.$$
Now write $I_n = [a_n, b_n]$ for each $n$.  The sequence $a_n$ is increasing and bounded by $b_1$ so it converges to a limit $l$.  Since the lengths of the $I_n$ converge to zero, we have $b_n\rightarrow l$.  By continuity, $f(l) = \lim f(a_n) = \lim f(b_n).$  By continuity, we can choose $\delta > 0$ so that $f(x) < f(l) + 1$ for $l - \delta < x < l + \delta$.  
Pick $n$ so that $I_n \subseteq (l - \delta, l + \delta)$.  The function $f$ must be bounded on $I_n$, a contradiction of our construction.  
The arabesque executed here has a feel very similar to that of the proof of Heine-Borel theorem.
A: Yes, this is a good idea. Since $A \subseteq [a,b]$, it has a supremem, say $c = \sup A$. If $c<b$, then since $f$ is continuous at $c$, there is $\delta>0$ such that $|f(x)-f(c)|< \frac{1}{2}$ whenever $x \in (c, c+\delta)$. Then $|f(x)| < |f(c)| + \frac{1}{2}$ for every whenever $x \in (c, c+\delta)$. But that means $f$ is bounded on $[a, c+ \delta]$, contrary to $c = \sup A$.
A: Here's the most concise proof I know: The continuous image of a compact set is itself compact. Now, $[a,b]$ is compact so $f([a,b])$ is compact. By the Heine-Borel theorem, an interval in $\mathbb{R}$ is compact if and only if it is closed and bounded. Therefore, $f([a,b])$ is bounded.
A: The distance (function) of points from the set $\{0\}$ is uniformly continuous, so it is bounded on any compact set, so the compact set can be enclosed in a ball with a radius of that distance. Consider the compact set $f([0,1])$.
