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Bolzano's Theorem

  1. Jun 18, 2010 #1
    Hi,

    I actually have trouble understanding Bolzano's Theorem. Could someone please explain it to me?
     
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  3. Jun 18, 2010 #2

    Gib Z

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    The Bolzano Weierstrass Theorem?

    In short, its basically an extension of a simpler fact. We know that if we have a bounded sequence of real numbers ( a sequence is bounded if there exists a real number M that is larger in magnitude than all the sequence terms), then that sequence has a convergent subsequence. A proof of this should be in your textbook, the general idea is to show that limsup a_n or liminf a_n exist (because a_n is bounded), and then to find a sequence that at each step gets closer and closer to one of those values.

    Bolzano Weierstass basically says that if you have a bounded sequence in [tex]R^n[/tex] or [tex]C^n[/tex], not just in R, then the same result applies - bounded sequences of elements from those sets have a convergent subsequence.
     
  4. Jun 18, 2010 #3
  5. Jun 18, 2010 #4
    Yes, I am talking abt the special case of the Intermediate Value Theorem when f(intermediate value) is zero.
     
  6. Jun 19, 2010 #5

    Gib Z

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    Sorry, I never knew that special case had it's own name. Anyway, could you elaborate what about the theorem you don't understand? What it states, its proof, or something else?
     
  7. Jun 19, 2010 #6
    The Bolzano theorem, which is a special case of intermediate value theorem, states that if you have a continuous function on an interval [a,b], such that f(a) is positive and f(b) is negative, then there must exist a point "c" belonging to the interval (a,b) where f(c)=0.

    Continuity is essential.
     
  8. Jun 19, 2010 #7
    I understand the theorem statement, but I don't get the last part of the proof. I understand that the proof ends if the function value at the mean of any interval is zero. But if the function value at the mean of any interval is not zero for any size of the interval, then we have to consider the bounded sequence of end-points values at both sides of the curve. I don't understand how that consideration leads to the completeion of the proof.
     
  9. Jun 19, 2010 #8

    HallsofIvy

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    The proof relies on this much deeper theorem: a continuous function maps a connected set into a connected set. The only connected sets in the real numbers are the intervals.
     
  10. Jun 19, 2010 #9
    I'd be grateful to you if you could explain that theorem in more detail or point to a good source which expalins it.
     
  11. Jun 19, 2010 #10

    HallsofIvy

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  12. Jun 19, 2010 #11
    So the proof involves bisection of the intervals. We start with interval [a,b], where f(a)>0 and f(b)<0. Now we bisect the inteval and consider f((a+b)/2). If the function value at the midpoint is +ve then we select the interval [(a+b/2),b], else the interval [a,(a+b)/2]. So in this way we get a subinterval of [a,b], and the length is half of the previous one. Call this new subinterval [tex][a_{1},b_{1}][/tex]We continue the bisection on the interval in the above manner. So we get a sequence of nested subintervals.

    [tex][a,b]\supset[a_{1},b_{1}]\supset[a_{2},b_{2}]....[/tex].

    We get two sequences [tex]a_{n}[/tex] and [tex]b_{n}[/tex]. Now both sequence converge to the same point say "x". Then we assert that f(x)=0. If not zero, then there exist a neighborhood of x, where it is not zero, say [tex](x-\epsilon, x+\epsilon)[/tex] and function values are entirely +ve or -ve in the interval. But given this epslion i can choose a interval from the sequence of intervals which lies completely in [tex](x-\epsilon, x+\epsilon)[/tex]. But then the endpoints have +ve and -ve value, which is a contradiction.

    Hence f(x)=0. Proved.
     
  13. Jun 19, 2010 #12
    Thanks! Actually, the last three lines of the 2nd paragraph were the lines I could not understand before! Thank you so much! This has really helped.
     
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