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Math Help - nested interval theorem with unbounded interval

  1. #1
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    nested interval theorem with unbounded interval

    I'm trying to show that the Nested Interval Theorem does NOT hold if the "first" interval in the sequence is unbounded.


    So, \big\{I_n\big\}_{n=1} ^\infty where I_{n} = [a_{n}, b_{n}],  a,b \in R


    with   I_{1} = [a_{1}, \infty), for example, implies that \bigcap_{n}^{\infty} A_{n} = \varnothing


    The only thing that makes sense to me that I can think of is that a_{1}, a_{2}, a_{3} ... a_{n} is an increasing sequence with no upper bound so eventually there is a possibility that for some large N [a_{N}, b_{N}] becomes an empty set. And the intersection of any set with the empty set is the empty set.

    Any suggestion or tips? Thanks
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  2. #2
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    Re: nested interval theorem with unbounded interval

    Quote Originally Posted by director View Post
    I'm trying to show that the Nested Interval Theorem does NOT hold if the "first" interval in the sequence is unbounded. So, \big\{I_n\big\}_{n=1} ^\infty where I_{n} = [a_{n}, b_{n}],  a,b \in R
    with   I_{1} = [a_{1}, \infty), for example, implies that \bigcap_{n}^{\infty} A_{n} = \varnothing
    I think that you need to tell us what you mean by " Nested Interval Theorem ", what is its statement?

    The examples you gave are confusing. Look at this:
    if n\in\mathbb{Z}^+ define A_n=[n,\infty) then \bigcap\limits_n {{A_n}}  = \emptyset .

    Please post what you are confused about.
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  3. #3
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    Re: nested interval theorem with unbounded interval

    Sorry for the confusion.

    Quote Originally Posted by Plato View Post
    if n\in\mathbb{Z}^+ define A_n=[n,\infty) then \bigcap\limits_n {{A_n}}  = \emptyset .
    Plato, I found your answer to a problem similar to the one quoted above and it makes sense to me. I see why the intersection is empty. Thanks
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