Prove that f is bounded

1. Nov 2, 2008

fibfreak

1. The problem statement, all variables and given/known data

Let f:[a, b] -> R have a limit at each x in [a, b]. Prove that f is bounded.

2. Relevant equations

None

3. The attempt at a solution

No idea on how to start the proof. Completely lost.

Thank you

2. Nov 2, 2008

HallsofIvy

Staff Emeritus
The problem with not showing any work at all is that we have no idea what you know and can use. For example, I think there is a fairly simple proof using the fact that [a, b] is compact. Do you know anything about "compactness"?

3. Nov 2, 2008

boombaby

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.

4. Nov 2, 2008

HallsofIvy

Staff Emeritus
For example, suppose f is defined on [0, 1] by: f(x)= 0 if x is not of the form 1/n for n a positive integer, f(1/n)= n. That function is defined for all x in [0,1] but is NOT bounded. Of course, then the limit as x goes to 0 is not defined.

5. Nov 2, 2008

boombaby

AH, yes....I forgot this one...Thanks a lot!

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