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Thread: nested interval theorem

  1. #1
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    nested interval theorem

    Prove that if $\displaystyle I_n = [a_{n}, b_{n}] $ is a nested collection of closed intervals such that $\displaystyle \lim_{n \to \infty} b_{n} - a_{n} = 0 $, then there is an $\displaystyle x \in \mathbb{R} $ such that $\displaystyle \bigcap_{n \in \mathbb{N}} I_N = \{x \} $.

    So $\displaystyle a_n $ is an increasing sequence and $\displaystyle b_n $ is a decreasing sequence. So choose an $\displaystyle N \in \mathbb{N} $ so that $\displaystyle |b_{N}- a_{N}| < \varepsilon $. Then $\displaystyle a_{N} \leq a_{n} < b_{n} \leq b_{N} $ for all $\displaystyle n \geq N $. So $\displaystyle a_{n} \to \alpha $ and $\displaystyle b_n \to \beta $ for $\displaystyle \alpha, \beta \in \mathbb{R} $. We need to show that $\displaystyle x = \alpha = \beta $.

    Then $\displaystyle a_{N} \leq \sup \{a_{n}: n \geq N \} $ and $\displaystyle b_{N} \geq \inf \{b_{n}: n \geq N \} $.

    Now what?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by particlejohn View Post
    Prove that if $\displaystyle I_n = [a_{n}, b_{n}] $ is a nested collection of closed intervals such that $\displaystyle \lim_{n \to \infty} b_{n} - a_{n} = 0 $, then there is an $\displaystyle x \in \mathbb{R} $ such that $\displaystyle \bigcap_{n \in \mathbb{N}} I_N = \{x \} $.

    So $\displaystyle a_n $ is an increasing sequence and $\displaystyle b_n $ is a decreasing sequence. So choose an $\displaystyle N \in \mathbb{N} $ so that $\displaystyle |b_{N}- a_{N}| < \varepsilon $. Then $\displaystyle a_{N} \leq a_{n} < b_{n} \leq b_{N} $ for all $\displaystyle n \geq N $. So $\displaystyle a_{n} \to \alpha $ and $\displaystyle b_n \to \beta $ for $\displaystyle \alpha, \beta \in \mathbb{R} $. We need to show that $\displaystyle x = \alpha = \beta $.

    Then $\displaystyle a_{N} \leq \sup \{a_{n}: n \geq N \} $ and $\displaystyle b_{N} \geq \inf \{b_{n}: n \geq N \} $.

    Now what?
    consider the set $\displaystyle \{a_{n}: n \geq N \}$. $\displaystyle a_1 \leq a_n \leq b_n \leq b_1$ for all n.. this means that the set is bounded above by $\displaystyle b_1$ and hence supremum exists and $\displaystyle a_n \leq \sup \{a_{n}: n \geq N \}$ for all $\displaystyle n$.

    show that $\displaystyle \sup \{a_{n}: n \geq N \} \leq b_n$ for all $\displaystyle n$.

    let $\displaystyle x=\sup \{a_{n}: n \geq N \}$ and fix $\displaystyle n \in \mathbb{N}$

    consider $\displaystyle A_k = \{a_k | k \in \mathbb{N}\}$

    If $\displaystyle n \leq k$, we have $\displaystyle a_k \leq b_k \leq b_n$ for all $\displaystyle k \geq n$

    If $\displaystyle n > k$, we get $\displaystyle a_k \leq a_n \leq b_n$ for all $\displaystyle k < n$

    This means that $\displaystyle a_k \leq b_n$ for all $\displaystyle k$ and for all $\displaystyle n$..

    Thus, $\displaystyle b_n$ is an upper bound for the set $\displaystyle \{a_{n}: n \geq N \}$

    Therefore $\displaystyle \sup \{a_{n}: n \geq N \} = x \leq b_n$ for all $\displaystyle n$ and hence $\displaystyle a_n \leq x \leq b_n$ for all $\displaystyle n$..
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  3. #3
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    Quote Originally Posted by kalagota View Post
    consider the set $\displaystyle \{a_{n}: n \geq N \}$. $\displaystyle a_1 \leq a_n \leq b_n \leq b_1$ for all n.. this means that the set is bounded above by $\displaystyle b_1$ and hence supremum exists and $\displaystyle a_n \leq \sup \{a_{n}: n \geq N \}$ for all $\displaystyle n$.

    show that $\displaystyle \sup \{a_{n}: n \geq N \} \leq b_n$ for all $\displaystyle n$.

    let $\displaystyle x=\sup \{a_{n}: n \geq N \}$ and fix $\displaystyle n \in \mathbb{N}$


    consider $\displaystyle A_k = \{a_k | k \in \mathbb{N}\}$

    If $\displaystyle n \leq k$, we have $\displaystyle a_k \leq b_k \leq b_n$ for all $\displaystyle k \geq n$

    If $\displaystyle n > k$, we get $\displaystyle a_k \leq a_n \leq b_n$ for all $\displaystyle k < n$

    This means that $\displaystyle a_k \leq b_n$ for all $\displaystyle k$ and for all $\displaystyle n$..

    Thus, $\displaystyle b_n$ is an upper bound for the set $\displaystyle \{a_{n}: n \geq N \}$

    Therefore $\displaystyle \sup \{a_{n}: n \geq N \} = x \leq b_n$ for all $\displaystyle n$ and hence $\displaystyle a_n \leq x \leq b_n$ for all $\displaystyle n$..
    So $\displaystyle A_1 = \{a_1 \}, \ A_2 = \{a_2 \} \ \ldots $? Each one is a singleton set? Does this only show the existence of $\displaystyle x $ in the interval (e.g. it does not show that the intersection of the intervals is $\displaystyle x $)?
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  4. #4
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    I find this very strange! You do not connect this with previous problems that we have helped you with.
    Let $\displaystyle A = \left\{ {a_n :n \in Z^ + } \right\}\,\& \,B = \left\{ {b_n :n \in Z^ + } \right\}$.
    Now any element of $\displaystyle B$ is an upper bound for $\displaystyle A$.
    Now apply one of the first problems we did.
    Do you see that $\displaystyle \sup \left\{ A \right\} = \inf \left\{ B \right\} = x$ is in the intersection?
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  5. #5
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    yes I see that (Rudin does this to prove greatest lower bound property). I was just thinking that we would have to prove uniqueness also.

    But now I see that if $\displaystyle x,y \in \bigcap_{n \in \mathbb{N}} I_N $, then $\displaystyle x =y $.


    Quote Originally Posted by Plato View Post
    I find this very strange! You do not connect this with previous problems that we have helped you with.
    Let $\displaystyle A = \left\{ {a_n :n \in Z^ + } \right\}\,\& \,B = \left\{ {b_n :n \in Z^ + } \right\}$.
    Now any element of $\displaystyle B$ is an upper bound for $\displaystyle A$.
    Now apply one of the first problems we did.
    Do you see that $\displaystyle \sup \left\{ A \right\} = \inf \left\{ B \right\} = x$ is in the intersection?
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  6. #6
    MHF Contributor kalagota's Avatar
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    the proof i have given shows that there will always an element in $\displaystyle [a_n,b_n]$ for all $\displaystyle n$. and what happens when $\displaystyle \lim b_b - a_n = 0$? it must be the element.
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