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Math Help - prove limsup-liminf<2

  1. #1
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    prove limsup-liminf<2

    Suppose that there exists N \in N, such that |a_n-a_m|<1 for all n,m>=N (this will be true, in particular, if (a_n) is cauchy). Prove that (limsup_{n-->\infty}a_n)-(liminf_{n-->\infty}a_n) <=2.

    it seems like an easy proof, but i can't find the connection between limsup and cauchy sequence, need some help.
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  2. #2
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    Re: prove limsup-liminf<2

    Quote Originally Posted by wopashui View Post
    Suppose that there exists N \in N, such that |a_n-a_m|<1 for all n,m>=N (this will be true, in particular, if (a_n) is cauchy). Prove that (limsup_{n-->\infty}a_n)-(liminf_{n-->\infty}a_n) <=2.

    it seems like an easy proof, but i can't find the connection between limsup and cauchy sequence, need some help.
    I will leave the epsilons to you but here is an idea. First there exists a subsequence
    a_{n}_k
    that converges to the limit supreemum and another subsequence
    a_{n}_j
    that converges to the lim inf

    Now

    |L_{\text{sup}}-L_{\text{inf}}| \le  |L_{\text{sup}}-a_n+a_n-L_{\text{inf}}| \le |L_{\text{sup}}-a_n|+|a_n-L_{\text{inf}}| = |L_{\text{sup}}-a_{n}_k+a_{n}_k-a_n|+|a_n-a_{n}_j+a_{n}_j-L_{\text{inf}}|

    Now just use the triangle inequality again and find the n,j, and k's to show what you want
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  3. #3
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    Re: prove limsup-liminf<2

    Quote Originally Posted by wopashui View Post
    Suppose that there exists N \in N, such that |a_n-a_m|<1 for all n,m>=N (this will be true, in particular, if (a_n) is cauchy). Prove that (limsup_{n-->\infty}a_n)-(liminf_{n-->\infty}a_n) <=2.
    it seems like an easy proof, but i can't find the connection between limsup and cauchy sequence, need some help.
    I think that you have the wrong idea here. The given does not prove that the sequence is Cauchy. Rather it asks you to note that the conclusion would apply to Cauchy sequences.

    First, you must prove the the \liminf~\&~\limsup both exist.
    To do that notice that from the given \left( {\forall n \geqslant N} \right)\left[ {\left| {a_n  - a_N } \right| \leqslant 1\, \Rightarrow \, - 1 + a_N  \leqslant a_n  \leqslant 1 + a_N } \right]
    Thus they exist and if \lambda=\liminf{a_n}~\&~\sigma= \limsup{a_n}  then -1+a_N\le\lambda~\&~\sigma\le 1+a_N.

    Now you finish
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    Re: prove limsup-liminf<2

    thanks, so is TheEmptySet's approach correct?
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    Re: prove limsup-liminf<2

    Quote Originally Posted by Plato View Post
    I think that you have the wrong idea here. The given does not prove that the sequence is Cauchy. Rather it asks you to note that the conclusion would apply to Cauchy sequences.

    First, you must prove the the \liminf~\&~\limsup both exist.
    To do that notice that from the given \left( {\forall n \geqslant N} \right)\left[ {\left| {a_n - a_N } \right| \leqslant 1\, \Rightarrow \, - 1 + a_N \leqslant a_n \leqslant 1 + a_N } \right]
    Thus they exist and if \lambda=\liminf{a_n}~\&~\sigma= \limsup{a_n} then -1+a_N\le\lambda~\&~\sigma\le 1+a_N.

    Now you finish
    why are you using N not m here? is it because if a_m is not bounded, then a_n will not be bounded, but a_N is fixed, so a_n is bounded. but why do we have <= not strictly < here?
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    Re: prove limsup-liminf<2

    Quote Originally Posted by wopashui View Post
    why are you using N not m here? is it because if a_m is not bounded, then a_n will not be bounded, but a_N is fixed, so a_n is bounded. but why do we have <= not strictly < here?
    Do you understand any of this?
    It was convenient to use a fixed a_N~,~N=m.

    Surely you understand that if |a-b|<1 then it is also true that |a-b|\le 1.
    Last edited by Plato; December 6th 2011 at 03:40 PM.
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