HallsofIvy,
how can I use compactness directly if f is not supposed to be continuous function? I got no idea...thanks!
fibfreak,
what does "f has a limit at each x in [a,b]" imply? Let e>0 be given, you can associate a neighborhood with radius=delta(e,x) to each x in [a,b] such that, any y!=x and y in this neighborhood implies |f(y)-A(x)|<e, where A(x) is the limit of f at x.
\cup_{x\in [a,b]} N_{\delta(\epsilon,x)}(x)} covers [a,b], by the compactness of [a,b], you can choose finitely many neighborhoods that covers [a,b], say N_1,N_2...N_s. in each neighborhoods, f has an upperbound A(x_i)+e and a lower bound A(x_i)-e. then max{A(x_1)+e,...A(x_s)+e} is the upper bound of f and min{A(x_1)-e,...A(x_s)-e} is the lower bound of f.
Also you can prove it by supposing f is not bounded. Find a sequence {x_n} such that f(x_n)>n. being an infinite subset of [a,b] (compact), {x_n} has a limit point in [a,b], say x. prove that f has no limit at this x and you get a contradiction.
However, there's a problem that puzzles me a bit.
What if f is defined (which means f(x) has some value for each x) on each x in [a,b]? can we say that f is bounded? since max{f(x), x in [a,b]} fails to work. I've no idea.
