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Math Help - a new definition to bolzano weirshtrass law regarding liminf that i cant understand..

  1. #1
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    a new definition to bolzano weirshtrass law regarding liminf that i cant understand..

    if a sequence Y_n is bounded then there is sub sequence Y_r_n which satisfies lim inf Y_n<=lim Y_r_n as n->+infinity i didnt here of that definition before the only definition i know about liminf is that it the supremum of all the infimums of the sequence using the definition i know this one
    Code:
     lim inf Y_n<=lim Y_r_n as n->+infinity
    doesnt make any sense ??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    if a sequence Y_n is bounded then there is sub sequence Y_r_n which satisfies lim inf Y_n<=lim Y_r_n as n->+infinity i didnt here of that definition before the only definition i know about liminf is that it the supremum of all the infimums of the sequence using the definition i know this one
    Code:
     lim inf Y_n<=lim Y_r_n as n->+infinity
    doesnt make any sense ??
    Let x_n = \inf\{ y_k | k\geq n\} and z_n = \sup\{ y_k | k\geq n\}.
    We know that \lim x_n \leq \lim z_n.
    Let \{y_{r_n}\} be a monotome subsequence.
    Then \{ y_{r_n} | n\geq k\} \subseteq  \{ y_n | n\geq k\} for all k\geq 1.
    Therefore, \sup \{ y_{r_n} | n\geq k\} \geq \sup \{ y_n | n\geq k \} and so z_{r_n} \geq z_n.
    This means, \lim z_n \leq \lim z_{r_n}.
    Thus, \lim x_n \leq \lim z_n \leq \lim z_{r_n}.
    By definition this means, \liminf y_n \leq \limsup y_{r_n}.
    However, \{y_{r_n}\} is monotone and bounded so \lim y_{r_n} exists.
    Thus, \lim y_{r_n} = \limsup y_{r_n}.
    Putting all of this together we get, \liminf y_n \leq \lim y_{r_n}.
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