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Bounded function on an interval

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data
    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

    3. The attempt at a solution
    I have no idea
  2. jcsd
  3. Nov 23, 2008 #2
    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.
  4. Nov 23, 2008 #3
    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?
  5. Nov 23, 2008 #4
    If you infintely 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).
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