1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Nested Interval Theorem

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Nested interval theorem:
    Suppose [an,bn] is a nested sequence of closed intervals, i.e. [an+1,bn+1] is contained in [an,bn] for all n≥1. Then the intersection of all these intervals is nonempty.

    an is an increasing sequence bounded above by b1 (or any bn), so sup an = a exists and a≤bm for all m.
    So "a" is a lower bound of bm
    Thus let b = inf bm, then we have a≤b (since any lower bound is ≤ its greatest lower bound).
    Then [a,b] is contained in [an,bn] for all n since an≤a and b≤bn for all n (because a is an upper bound of an and b is a lower bound of bn)
    Therefore, the intersection is nonempty.

    1) In this theorem, it says that "...the seqeunce of CLOSED intervals...". Is it enough to have closed intervals, or must the endpoints also be FINITE numbers? Can any of the endpoints be -∞ or +∞? (reacall that intervals like [a,∞) are classified as closed interval)

    2) In the proof, I am confused and I really don't understand why "a≤bm for all m". Can somebody please explain this?

    2. Relevant equations

    3. The attempt at a solution

    Thank you! :)
  2. jcsd
  3. Jan 14, 2010 #2
    [tex] \bigcap_{n=1}^\infty [n,\infty)=\emptyset[/tex]
    Last edited: Jan 14, 2010
  4. Jan 14, 2010 #3
    bm is an upper bound for the sequence (an). By definition of the supremum, sup an is less than or equal to bm
  5. Jan 14, 2010 #4
    But [n,∞) satisifies ALL assumptions as stated in the theorem (it is a nested sequence of closed intervals). So is the theorem wrong? What should be the correct statement of the theorem?
  6. Jan 14, 2010 #5
    I think it should be "... a sequence of closed, bounded, non-empty intervals in [tex]\mathbb{R}[/tex]..."
  7. Jan 14, 2010 #6
    I agree that for each m, bm is an upper bound for the sequence (an).
    But I still don't understand why "sup an≤bm for all m."
    By definition of upper bound, sup an≥an for all n, but this idea seems to be unconnected with the above and I don't see how sup an can possibly be related to bm. They are not even talking about the same sequence.

    Can someone please explain this in more detail?

  8. Jan 14, 2010 #7
    A number s is the supremum for a sequence if :
    1) s is an upper bound for the sequence
    2) if p is any upper bound for the sequence, s is less than or equal to p

    so we are talking about 2) here. bm is an upper bound for the sequence; thus by definition sup an<=bm. It doesn't matter whether or not bm is related to a sequence. All that matters is whether it is an upper bound.

    If this isn't good enough, then what about this: sup an is the least upper bound. suppose for contradiction bm<sup an. remember that bm is an upper bound for an. but then sup an wouldn't be the least upper bound because there is another upper bound, bm, that is less than it. so we must have bm>=sup an
  9. Jan 15, 2010 #8
    Thanks, so it just follows from the fact that the least upper bound of (a_n) is less than or equal to any upper bound of (a_n). Got it!:)

    One more question...

    How can we PROVE that bm is an upper bound for the sequence (an) for ALL m=1,2,3,...?

    When I draw a picture, I can see that this is the case, but I am not sure how to write it out and PROVE rigorously.
    We know that
    and b1≥b2≥b3≥...
    But from here on, I don't see why it follows that bm is an upper bound for the sequence (an) for ALL m=1,2,3,... .

  10. Jan 15, 2010 #9
    I think it comes from the hypothesis of nested intervals. If there exists an m such that bm is not an upper bound, then ak>bm for some k. So [ak,bk] would be disjoint from [am,bm], which contradicts the assumption of nested intervals. When you have a sequence of nested intervals, no two intervals can be disjoint.
  11. Jan 15, 2010 #10
    Makes sense. Thank you! :)
  12. Jan 15, 2010 #11
    Another proof:
    (an) is an increasing sequence bounded above (by b1), so
    lim an = sup an = a

    (bn) is a decreasing sequence bounded below (by a1), so
    lim bn = inf bn = b

    an≤bn for all n
    => a ≤ b

    Thus, ak ≤ a ≤ b ≤ bk for all k≥1. So the point a belongs to the intersection and the intersection is nonempty.

    Why is the red part true? How can we prove it?

  13. Jan 16, 2010 #12
    Any of the b_m is an upper bound for the set of all a_n, so the LEAST upper bound a must be less than or equal to b_m for any m. Now you figure out why b must in turn be greater than or equal to a.
  14. Jan 16, 2010 #13
    Yes, I already understood that a ≤ bm for all m. It just follows from the fact that the least upper bound of (a_n) is less than or equal to any upper bound of (a_n). (see post #8)

    And now my problem is about the second proof.

    How can we prove that "an ≤ bn for all n IMPLIES lim an ≤ lim bn"?

    It seems clear in my mind, but how can we prove rigorously using epsilon?

  15. Jan 16, 2010 #14


    User Avatar
    Gold Member

    If it helps, here's how I might tackle this problem. You already know that [itex]a_n \leq b_m[/itex] for all [itex]m,n \in \mathbb{N}[/itex]. From this we can conclude that the sequence [itex]\{a_n\}[/itex] has a least upper bound which we denote [itex]\alpha[/itex]. Similarly, the sequence [itex]\{b_n\}[/itex] has a greatest lower bound which we denote [itex]\beta[/itex]. Now, from this we have only three possible scenarios:
    • [tex]a_n \leq \alpha < \beta \leq b_n[/tex]
    • [tex]a_n \leq \alpha = \beta \leq b_n[/tex]
    • [tex]a_n \leq \beta < \alpha \leq b_n[/tex]
    If either of the first two cases hold, we know that the theorem we're trying to prove is true. If the third case is true however, we know that the intersection of all intervals [itex][a_n,b_n][/itex] is necessarily empty. Thus, if you can rule the third case out, then you'll have proved the theorem. So, suppose that the third case is true and let [itex]\alpha - \beta = 2\varepsilon > 0[/itex]. Can you derive a contradiction from this?
    Last edited: Jan 16, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook