Bounded function on an interval

[kbfrob]

Homework Statement

Let I=[a,b] and let f:I->R be a function (not necessarily continuous) with the property that for every x in I, f is bounded in a neighborhood of x.
Prove that f is bounded on I

The Attempt at a Solution

I have no idea

 

[Unco]

Hi kbfrob,

If you're familiar with the notion of compactness in terms of open covers, let x be a point in I and pick an open neighbourhood of x; then f is bounded on that open neighbourhood by your hypothesis. The collection of such open neighbourhoods (each corresponding to a point in I) provide an open cover for I, which is compact, so this collection has a finite open subcover of I.

I'll let you try and use these ideas to define an upper bound for f. Show us your work if you get stuck.

 

[kbfrob]

that is exactly what i tried to do, but I'm not familiar with the idea of compactness. Essentially what i was thinking about doing was starting at a, which is bounded in a neighborhood of (d1). then go to a+(d1), which is bounded in a neighborhood of d2. then go to a+(d1)+(d2)...etc. what i can't figure out is how to show that there will be a finite number of di's that cover [a,b]. Is there a way to show this without using compactness?

 

[Pere Callahan]

that is exactly what i tried to do, but I'm not familiar with the idea of compactness. Essentially what i was thinking about doing was starting at a, which is bounded in a neighborhood of (d1). then go to a+(d1), which is bounded in a neighborhood of d2. then go to a+(d1)+(d2)...etc. what i can't figure out is how to show that there will be a finite number of di's that cover [a,b]. Is there a way to show this without using compactness?

If you infinitely many d_i's, they would form a bounded sequences (bounded because they are in [a,b]) and by Bolzano-Weierstrass such a sequence must have a accumulation point d. (Think about what happens if it has more than one, or maybe infinitely many accumulation points!). You then know that f is bounded in a neighbourhood of d and this neighbourhood covers all but finitely many of your d_i's (if d is the only accumulation point).