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Thread: Bolzano-Weierstrass Theorem

  1. #1
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    Bolzano-Weierstrass Theorem

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

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

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

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

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

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

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

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


    Quote Originally Posted by Plato View Post
    First you cannot just $\displaystyle I = \left[ {a,b} \right]$. You can say that $\displaystyle I $ is a bounded infinite set.
    There is a point $\displaystyle x_0 \in I$ such there are infinitely many points in $\displaystyle I$ either greater than or less that $\displaystyle x_0$.
    We will say greater than $\displaystyle x_0$.
    Define a set $\displaystyle T$ as follows: $\displaystyle x \in T\quad iff $ there are infinitely many terms of $\displaystyle I$ greater than $\displaystyle x$.
    1. $\displaystyle T$ is not empty. (WHY?)
    2. T has an upper bound. (WHY?)
    3. T has a least upper bound, $\displaystyle \gamma$. (WHY?)
    4. For each $\displaystyle \delta > 0$ there are infinitely many terms of $\displaystyle I$ that are greater that $\displaystyle \gamma - \delta$. (WHY?)
    5. That means that $\displaystyle \gamma$ is a limit point of $\displaystyle I$. (WHY)?
    Last edited by particlejohn; Jul 8th 2008 at 02:12 PM.
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  5. #5
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    What makes you think that $\displaystyle I$ is compact?
    It does not say that $\displaystyle I$ is closed, only inifinite and bounded.
    We could have $\displaystyle I = \left\{ {1 - \frac{1}{n}:n \in Z^ + } \right\}$
    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 $\displaystyle [-B,B] $.
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    The key question if the following: if $\displaystyle [-L,0] $ and $\displaystyle [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 $\displaystyle I$. Since it is bounded, there is an $\displaystyle L$ such that $\displaystyle I = [-L,L]$.

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

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

    Name that set $\displaystyle I_2$..

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

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

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

    2. Closed intervals are infinite.

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


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

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

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

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

    Name that set $\displaystyle I_2$..

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

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

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

    2. Closed intervals are infinite.

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

    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 $\displaystyle L=1$ without the interval..
    so how would you know if this $\displaystyle L$ is sufficient to bound your arbitrary interval $\displaystyle I$?
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  11. #11
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by particlejohn View Post
    1. Consider $\displaystyle L = 1 $. Then we have $\displaystyle I = [-1,1] $. Then $\displaystyle I_{1}' = [-1,0] $ and $\displaystyle I_{1}'' = [0,1] $. Both of these have a nonempty intersection with $\displaystyle I $. I don't see a case where one interval has an empty intersection with $\displaystyle I $ and another interval has a nonempty intersection with $\displaystyle 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 $\displaystyle S $ is nonempty? This implies that the other intervals intersection with $\displaystyle S $ is empty. But I cant find a case for that.



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

    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 $\displaystyle L=1$ without the interval..
    so how would you know if this $\displaystyle L$ is sufficient to bound your arbitrary interval $\displaystyle I$?
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    Quote Originally Posted by particlejohn View Post
    Eventually we know that the length of $\displaystyle I_n $ is $\displaystyle B/2^{n-2} \to 0 $. Then let $\displaystyle \varepsilon > 0 $, and choose an $\displaystyle n \in \bold{N} $ such that $\displaystyle B/2^{n-2} < \varepsilon $. So $\displaystyle x_0 \in I_n \subset (x_{0}- \varepsilon, x_{0} + \varepsilon) $. And $\displaystyle S \cap (x_{0}- \varepsilon, x_{0}+ \varepsilon) \backslash \{x_0 \} \neq \emptyset $ which shows that $\displaystyle x_0 $ is a limit point of $\displaystyle 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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