Results 1 to 6 of 6

Math Help - nested interval theorem

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    170

    nested interval theorem

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

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

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

    Now what?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by particlejohn View Post
    Prove that if  I_n = [a_{n}, b_{n}] is a nested collection of closed intervals such that  \lim_{n \to \infty} b_{n} - a_{n} = 0 , then there is an  x \in \mathbb{R} such that  \bigcap_{n \in \mathbb{N}} I_N = \{x \} .

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

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

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

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

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

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

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

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

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

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

    Therefore \sup \{a_{n}: n \geq N \} = x \leq b_n for all n and hence a_n \leq x \leq b_n for all n..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    170
    Quote Originally Posted by kalagota View Post
    consider the set \{a_{n}: n \geq N \}. a_1 \leq a_n \leq b_n \leq b_1 for all n.. this means that the set is bounded above by b_1 and hence supremum exists and a_n \leq \sup \{a_{n}: n \geq N \} for all n.

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

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


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

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

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

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

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

    Therefore \sup \{a_{n}: n \geq N \} = x \leq b_n for all n and hence a_n \leq x \leq b_n for all n..
    So  A_1 = \{a_1 \}, \ A_2 = \{a_2 \} \ \ldots ? Each one is a singleton set? Does this only show the existence of  x in the interval (e.g. it does not show that the intersection of the intervals is  x )?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,910
    Thanks
    1759
    Awards
    1
    I find this very strange! You do not connect this with previous problems that we have helped you with.
    Let A = \left\{ {a_n :n \in Z^ +  } \right\}\,\& \,B = \left\{ {b_n :n \in Z^ +  } \right\}.
    Now any element of B is an upper bound for A.
    Now apply one of the first problems we did.
    Do you see that \sup \left\{ A \right\} = \inf \left\{ B \right\} = x is in the intersection?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2008
    Posts
    170
    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  x,y \in \bigcap_{n \in \mathbb{N}}  I_N , then  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 A = \left\{ {a_n :n \in Z^ +  } \right\}\,\& \,B = \left\{ {b_n :n \in Z^ +  } \right\}.
    Now any element of B is an upper bound for A.
    Now apply one of the first problems we did.
    Do you see that \sup \left\{ A \right\} = \inf \left\{ B \right\} = x is in the intersection?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    the proof i have given shows that there will always an element in [a_n,b_n] for all n. and what happens when \lim b_b - a_n = 0? it must be the element.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with nested interval proof
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 6th 2010, 02:04 PM
  2. Replies: 4
    Last Post: February 27th 2010, 03:44 PM
  3. Replies: 6
    Last Post: May 5th 2009, 07:49 AM
  4. nested interval thm in R^n
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 24th 2008, 10:00 AM
  5. Nested Interval
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 28th 2006, 11:46 AM

Search Tags


/mathhelpforum @mathhelpforum