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Intermediate value Theorem?

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Let I:[a,b] and let f: I→ℝ be a function with the property where for all x in I, the function f is bounded on a neighborhood Vδx(x) of x. Prove that f is bounded on I.




    2. The attempt at a solution

    Let I:[a,b] and let f: I→ℝ be a function with the property where for all x in I, the function f is bounded on a neighborhood Vδx(x) of x. The interval I consists of real numbers a and b where a < b. According to the density theorem, there exists a c in I where a<c<b wher c is a cluster point of I. Also, following the density theorem there exists a x in I where a < x < c or c < x < b such that |a-x| = |b-x| = |c-x|. Now given an ε > 0, there exists a δ > 0 where |x - c | < δ ....

    Is this right up to this point?
    If it is, how do I proceed from here?
     
  2. jcsd
  3. Mar 20, 2012 #2

    Dick

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    I really have know idea where you are going with that. Shouldn't you use that the interval [a,b] is compact?
     
  4. Mar 20, 2012 #3
    I don't really know where i am going with it either. I do not recall what compact means, is that the same thing as a closed bounded interval.

    My text said to solve it like this limit theorem, If A is contained in R and f : A→ R has a limit at c in ℝ the f is bounded on some neighborhood of c.

    I thought to have a neighborhood of c, i needed to have a cluster point, so i tried proving that [a,b] had a cluster point and then work towards bounding. I guess i am approaching it wrong.
     
  5. Mar 20, 2012 #4

    Dick

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    Yes, in your problem compact means closed and bounded. The property I'm looking for is that if [a,b] is closed and bounded and O is a set of open sets that cover [a,b], then there is a finite subset of O that also covers [a,b]. Do you know that?
     
  6. Mar 20, 2012 #5
    I did but I don't see how to use that here.
     
  7. Mar 20, 2012 #6

    Dick

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    Ok, then every point in [a,b] has a neighborhood in which it is bounded. Call those neighborhoods your collection O. The theorem tells you that [a,b] is covered by a finite number of neighborhoods on which is it bounded. Can you see how to combine the bounds on each of the finite number of neighborhoods to get a bound on all of [a,b]?
     
  8. Mar 20, 2012 #7
    That one i know.. let me think about that for a second.
     
  9. Mar 20, 2012 #8

    Dick

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    You can use either that or the open set formulation. They are really the same thing. To do it that way, assume f is unbounded. Then pick c_n such that |f(c_n)|>n for all n. What happens at a limit point of that sequence?
     
    Last edited: Mar 20, 2012
  10. Mar 20, 2012 #9
    how can i assume it is unbounded wshen i am told that it is bounded?
     
  11. Mar 20, 2012 #10
    Does this work? Have i established the conclusion by the end?

    Let I := [a,b] and let f: I→ℝ be a function with the property where for all x in I the function f is bounded on a neighborhood Vδx(x) of x. Since I is a closed set, I is also bounded. Therefore any sequence (Xn) in I must also be bounded. Thus by the Bolzano-Weirstrauss theorem, there exist a subsequence (Xnk) of (xn) that converges to a number x. Since I is closed and the elements of the subsequence (Xnk) belong to I it follows that x is in I. Then f(x) is continuous at x. so that (f(Xnk) ) converges to f(x). Then (f(Xnk) ) must be bounded. Since (Xn) are arbitrary sequences, f(x) is bounded on I.
     
  12. Mar 20, 2012 #11
    We already know that for each x there's a little nbd of x where f is bounded. Taken all together, one nbd for each x, that's a lot of neighborhoods. But what if we knew something about the domain that let us dramatically reduce the number of neighborhoods we need to deal with?

    If you review the definition of compactness, this will be very helpful.
     
  13. Mar 20, 2012 #12
    We haven't learned compactness yet. Our text has that two chapters down and we are not permitted to use any information in the text on HW if we haven't covered that section yet
     
  14. Mar 20, 2012 #13
    http://www.infoocean.info/avatar1.jpg [Broken]I really have know idea where you are going with that.
     
    Last edited by a moderator: May 5, 2017
  15. Mar 20, 2012 #14
    I don't know how to solve this then. everybody keeps saying use compactness but i can't...
     
    Last edited by a moderator: May 5, 2017
  16. Mar 20, 2012 #15

    Dick

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    The idea here is to use a proof by contradiction. You assume f is unbounded and try to show that leads to a contradiction with the assumption that it has a bounded neighborhood of every point. That would prove it must be bounded. If you choose c_n such that |f(c_n)|>n and c_n has a limit point c, can c have a neighborhood where it is bounded?
     
  17. Mar 20, 2012 #16
    ohhhh.
     
  18. Mar 20, 2012 #17

    Dick

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    And using that every sequence has a limit point is, in fact, using 'compactness'. You might not learn the abstract definition for a few chapters, but you are using it. That's why everyone is saying that.
     
  19. Mar 21, 2012 #18
    is it because c can't be in the set I if it is a cluster point for F
     
  20. Mar 21, 2012 #19

    Dick

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    What are F and c and what do you mean by that? Can you explain in full?
     
  21. Mar 21, 2012 #20
    I meant to reference your quote. So are we saying that c is a cluster point of f(x) and so can't be bounded as that would imply values greater than c and thus beyond the bound?
     
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