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Intersection nested, closed sequence of intervals

  1. Nov 21, 2013 #1
    The problem statement, all variables and given/known data.
    Let ##\{I_n\}_{n \in \mathbb N}## be a sequence of closed nested intervals and for each ##n \in \mathbb N## let ##\alpha_n## be the lenght of ##I_n##.
    Prove that ##lim_{n \to \infty}\alpha_n## exists and prove that if ##L=lim_{n \to \infty}\alpha_n>0##, then ##\bigcap_{n \in \mathbb N} I_n## is a closed interval of lenght ##L##.

    The attempt at a solution.

    I didn't have problems to prove the existence of the limit: if ##I_n=[a_n,b_n]##, then ##\alpha_n=b_n-a_n## and ##\{a_n\}_n, \{b_n\}_n## are increasing and decreasing bounded sequences respectively, so both are convergent ##\implies \{\alpha_n\}_n## is convergent and ##lim_{n \to \infty} \alpha_n=lim_{n \to \infty}b_n-a_n=lim_{n \to \infty}b_n-lim_{n \to \infty}a_n=b-a##.

    Now, I would like to say that if ##b-a>0 \implies \bigcap_{n \in \mathbb N} I_n=[a,b]##, but I don't know hot to prove this part.
     
  2. jcsd
  3. Nov 21, 2013 #2

    LCKurtz

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    Try showing if x isn't in [a,b] then x isn't in the intersection and if x is in [a,b] then x is in the intersection.
     
    Last edited: Nov 21, 2013
  4. Nov 21, 2013 #3

    Dick

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    I guess I'm not seeing anything difficult about this. Can you prove that if x<a then x is not in ##\bigcap_{n \in \mathbb N} I_n##? Same if x>b. And if a<=x<=b then it is? It's just reasoning about what 'intersection' means.
     
  5. Nov 22, 2013 #4
    Yes, it was just thinking the definition of intersection and apply it, thanks!
     
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