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Math Help - Bolzano-Weierstrass Theorem

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    Bolzano-Weierstrass Theorem

    Prove that a set which is both bounded and infinite has a limit point. For  I = [a,b] , let  I^{L} = [(a+b)/2, b] and  I^{R} = [a, (a+b)/2] . Since  S is bounded, then there is an interval  I_{1} = [-B, B] containing  S . Now if  S is infinite, then does this imply that both  I_{1}^{R} \cap S and  I_{1}^{L} \cap S are both infinite?

    In other words, are  [0,B] \cap S and  [-B,0] \cap S both infinite?
    Last edited by particlejohn; July 8th 2008 at 02:12 PM.
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    Eventually we know that the length of  I_n is  B/2^{n-2} \to 0 . Then let  \varepsilon > 0 , and choose an  n \in \bold{N} such that  B/2^{n-2} < \varepsilon . So  x_0 \in I_n \subset (x_{0}- \varepsilon, x_{0} + \varepsilon) . And  S \cap (x_{0}- \varepsilon, x_{0}+ \varepsilon) \backslash \{x_0 \} \neq \emptyset which shows that  x_0 is a limit point of  S .

    But why can't we say that both  [-B,0] \cap S and  [0,B] \cap S are both infinite? Why at least?
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    First you cannot just I = \left[ {a,b} \right]. You can say that I is a bounded infinite set.
    There is a point x_0  \in I such there are infinitely many points in I either greater than or less that x_0.
    We will say greater than x_0.
    Define a set T as follows: x \in T\quad iff there are infinitely many terms of I greater than x.
    1. T is not empty. (WHY?)
    2. T has an upper bound. (WHY?)
    3. T has a least upper bound, \gamma. (WHY?)
    4. For each \delta > 0 there are infinitely many terms of I that are greater that \gamma - \delta. (WHY?)
    5. That means that \gamma is a limit point of I. (WHY)?
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    Why can't you say  I = [a,b] ? It is compact and infinite right? You could have an infinite bounded set that is just constant (e.g.  \{1,1,1, \ldots \} ). Actually, that was purely notational. I did suppose that  S is a bounded infinite set such that  [-B,B] contains  S . In this case,  a = -B and  b = B . And I split up the interval into 2 sub intervals:  [-B, 0] and  [0, B ] and continued in this way, getting a series of nested intervals.

    So  T = \{x : \ \text{there are infinitely many terms of} \ I \ \text{greater than} \ x \}

    1.  T is not empty, because for  x = a ,  y = \frac{a+b}{2} > x by Archmidean property.

    2.  T has an upper bound because for  \iota > b , then  \iota > x for all  x \in [a,b] .

    3. By least upper bound property,  T \subset \bold{R} (an ordered set),  T is not empty, and  T is bounded above. Thus  \sup T exists in  \bold{R} .

    4.  (\gamma - \delta, \gamma) \cap T is an infinite set.


    Quote Originally Posted by Plato View Post
    First you cannot just I = \left[ {a,b} \right]. You can say that I is a bounded infinite set.
    There is a point x_0  \in I such there are infinitely many points in I either greater than or less that x_0.
    We will say greater than x_0.
    Define a set T as follows: x \in T\quad iff there are infinitely many terms of I greater than x.
    1. T is not empty. (WHY?)
    2. T has an upper bound. (WHY?)
    3. T has a least upper bound, \gamma. (WHY?)
    4. For each \delta > 0 there are infinitely many terms of I that are greater that \gamma - \delta. (WHY?)
    5. That means that \gamma is a limit point of I. (WHY)?
    Last edited by particlejohn; July 8th 2008 at 02:12 PM.
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    What makes you think that I is compact?
    It does not say that I is closed, only inifinite and bounded.
    We could have I = \left\{ {1 - \frac{1}{n}:n \in Z^ +  } \right\}
    That is infinite and bounded but not compact.
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    yeah that was just a notational thing. I had assumed a bounded infinite set contained in  [-B,B] .
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    The key question if the following: if  [-L,0] and  [0,L ] are they both infinite?
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    MHF Contributor kalagota's Avatar
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    no, just at least one..

    Let us consider the set I. Since it is bounded, there is an L such that I = [-L,L].

    Bisect I into {I_1}' = [-L,0] and {I_1}'' = [0,L].
    At least one of these has the properties:
    1. intersection with I is not empty; and (WHY?)
    2. Infinite (WHY?)

    Name that set I_1. Again, bisect I_1 into {I_2}' and {I_2}''. again, at least one of these has the properties:
    1. intersection with I_1 is not empty; and (WHY?)
    2. Infinite (WHY?)

    Name that set I_2..

    Do this continuously and form the sequence of nested intervals \{I_n\}.

    Use the Property of Nested interval, \exists \ x such that x \in I_n for all n.

    ... now, can you do the rest? you only need to show that every neighborhood of x contains infinitely many points from I_n and hence from I. (maybe, two steps or three more.. Ü)
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    1. Consider  L = 1 . Then we have  I = [-1,1] . Then  I_{1}' = [-1,0] and  I_{1}'' = [0,1] . Both of these have a nonempty intersection with  I . I don't see a case where one interval has an empty intersection with  I and another interval has a nonempty intersection with  I .

    2. Closed intervals are infinite.

    3. Choose  I_1 = I_{1}' . Repeat step 1. Do this continuously. As we continue, the length of the intervals,  L/2^{n-2} \to 0 .


    Quote Originally Posted by kalagota View Post
    no, just at least one..

    Let us consider the set I. Since it is bounded, there is an L such that I = [-L,L].

    Bisect I into {I_1}' = [-L,0] and {I_1}'' = [0,L].
    At least one of these has the properties:
    1. intersection with I is not empty; and (WHY?)
    2. Infinite (WHY?)

    Name that set I_1. Again, bisect I_1 into {I_2}' and {I_2}''. again, at least one of these has the properties:
    1. intersection with I_1 is not empty; and (WHY?)
    2. Infinite (WHY?)

    Name that set I_2..

    Do this continuously and form the sequence of nested intervals \{I_n\}.

    Use the Property of Nested interval, \exists \ x such that x \in I_n for all n.

    ... now, can you do the rest? you only need to show that every neighborhood of x contains infinitely many points from I_n and hence from I. (maybe, two steps or three more.. Ü)
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  10. #10
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by particlejohn View Post
    1. Consider  L = 1 . Then we have  I = [-1,1] . Then  I_{1}' = [-1,0] and  I_{1}'' = [0,1] . Both of these have a nonempty intersection with  I . I don't see a case where one interval has an empty intersection with  I and another interval has a nonempty intersection with  I .

    2. Closed intervals are infinite.

    3. Choose  I_1 = I_{1}' . Repeat step 1. Do this continuously. As we continue, the length of the intervals,  L/2^{n-2} \to 0 .
    no! what if your I=\left\{2 + \frac{1}{n} : n\in \mathbb{N}\right\}

    the questions states that for a given bounded and infinite interval.. Show that it has a limit point..

    what you did is to take the L=1 without the interval..
    so how would you know if this L is sufficient to bound your arbitrary interval I?
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  11. #11
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by particlejohn View Post
    1. Consider  L = 1 . Then we have  I = [-1,1] . Then  I_{1}' = [-1,0] and  I_{1}'' = [0,1] . Both of these have a nonempty intersection with  I . I don't see a case where one interval has an empty intersection with  I and another interval has a nonempty intersection with  I .
    it really doesn't matter whether both intervals satisfy the conditions since you can do it simultaneously.. your job is to show the existence of a limit point.
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    I was just illustrating the fact that why do we have at least one interval whose intersection with  S is nonempty? This implies that the other intervals intersection with  S is empty. But I cant find a case for that.



    Quote Originally Posted by kalagota View Post
    no! what if your I=\left\{2 + \frac{1}{n} : n\in \mathbb{N}\right\}

    the questions states that for a given bounded and infinite interval.. Show that it has a limit point..

    what you did is to take the L=1 without the interval..
    so how would you know if this L is sufficient to bound your arbitrary interval I?
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    Quote Originally Posted by particlejohn View Post
    Eventually we know that the length of  I_n is  B/2^{n-2} \to 0 . Then let  \varepsilon > 0 , and choose an  n \in \bold{N} such that  B/2^{n-2} < \varepsilon . So  x_0 \in I_n \subset (x_{0}- \varepsilon, x_{0} + \varepsilon) . And  S \cap (x_{0}- \varepsilon, x_{0}+ \varepsilon) \backslash \{x_0 \} \neq \emptyset which shows that  x_0 is a limit point of  S .
    I have already done that.
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  14. #14
    MHF Contributor kalagota's Avatar
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    I see, your question is why at least, right?

    because the union of a finite and an infinite sets is infinite..
    however, union of finite sets is still finite..
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    can you give me an example of where one's intersection is empty and another intersection is not?

    Quote Originally Posted by kalagota View Post
    I see, your question is why at least, right?

    because the union of a finite and an infinite sets is infinite..
    however, union of finite sets is still finite..
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