Is f Bounded if It Has a Limit at Every Point on a Closed Interval?

[fibfreak]

Homework Statement

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

Homework Equations

None

The Attempt at a Solution

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

Thank you

 

[HallsofIvy]

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"?

 

[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.

 

[HallsofIvy]

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.

 

[boombaby]

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