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Math Help - sequence with convergent subsequence

  1. #1
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    sequence with convergent subsequence

    If (a_{n_k}) is a convergent subsequence of (a_n), show that lim inf _{n--> infinity} a_n <= lim_{k-->infinity} a_{n_k} <= lim sup _{n-->infinity} a_n.

    Im not sure what lim inf _{n--> infinity} a_n [tex] and lim sup _{n-->infinity} a_n mean, since inf a_n and sup a_n are a number
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  2. #2
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    Re: sequence with convergent subsequence

    Quote Originally Posted by wopashui View Post
    If (a_{n_k}) is a convergent subsequence of (a_n), show that lim inf _{n--> infinity} a_n <= lim_{k-->infinity} a_{n_k} <= lim sup _{n-->infinity} a_n.
    This basically an easy question.
    By definition of subsequence it must be the case that k\le n_k.
    Thus \{a_{n_k}:n_k\ge N\}\subseteq\{a_n:n\ge N\}.
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  3. #3
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    Re: sequence with convergent subsequence

    Quote Originally Posted by Plato View Post
    This basically an easy question.
    By definition of subsequence it must be the case that k\le n_k.
    Thus \{a_{n_k}:n_k\ge N\}\subseteq\{a_n:n\ge N\}.
    hmm, but we are talking about the limit here, not just K and n_k
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  4. #4
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    Re: sequence with convergent subsequence

    Quote Originally Posted by wopashui View Post
    hmm, but we are talking about the limit here, not just K and n_k
    Are you sure that you understand what ~\liminf~ means?
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    Re: sequence with convergent subsequence

    Quote Originally Posted by Plato View Post
    Are you sure that you understand what ~\liminf~ means?
    that is what im not sure about, I dunno what it means
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  6. #6
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    Re: sequence with convergent subsequence

    Quote Originally Posted by wopashui View Post
    that is what im not sure about, I dunno what it means
    Well then, no wonder you cannot answer questions about it.
    Does it make sense to you that if
    A\subseteq B then \inf(B)\le\inf(A)
    and \sup(A)\le\sup(B)~?

    Suppose that (a_n) is a sequence.
    \limsup(a_n)=\lim _{n \to \infty } \sup \left\{ {a_k :k \geqslant n} \right\}
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  7. #7
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    Re: sequence with convergent subsequence

    we need to first show that supa_n and infa_n exist, do we?
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