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Math Help - Bolzano-Weierstrauss: A Complex Version

  1. #1
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    Bolzano-Weierstrauss: A Complex Version

    Let (z_n) be a bounded sequence of complex numbers. Does there exist a subsequence (z_{n_k}) which is convergent?

    I am asking this because my book on Complex Analysis was proving a theorem and it did it by contradiction. I realized that if the Bolzano-Weierstrauss theorem is actually true then we can avoid that method and do it directly. But I wondered if it is really true.

    The proof of the real version of the Bolzano-Weierstrauss theorem we did was a follows:

    Proof: Let (s_n) be a bounded sequence. Then there exists a monotone subsequence (a theorem from before) (s_{n_k}), which is certainly bounded. Thus, (s_{n_k}) is convergent.

    My professor said that this is a bad way to proof this theorem and my book on real analysis says "...our proof of this theorem is somewhat nonstandard...".

    I am begining to think this is what my professor meant, i.e. a bad proof.

    The main idea is to construct a monotone subsequence, but the complex numbers are not ordered. So it is difficult to use this approach.
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  2. #2
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    Quote Originally Posted by ThePerfectHacker View Post
    Let (z_n) be a bounded sequence of complex numbers. Does there exist a subsequence (z_{n_k}) which is convergent?

    I am asking this because my book on Complex Analysis was proving a theorem and it did it by contradiction. I realized that if the Bolzano-Weierstrauss theorem is actually true then we can avoid that method and do it directly. But I wondered if it is really true.

    The proof of the real version of the Bolzano-Weierstrauss theorem we did was a follows:

    Proof: Let (s_n) be a bounded sequence. Then there exists a monotone subsequence (a theorem from before) (s_{n_k}), which is certainly bounded. Thus, (s_{n_k}) is convergent.

    My professor said that this is a bad way to proof this theorem and my book on real analysis says "...our proof of this theorem is somewhat nonstandard...".

    I am begining to think this is what my professor meant, i.e. a bad proof.

    The main idea is to construct a monotone subsequence, but the complex numbers are not ordered. So it is difficult to use this approach.
    If (z_n) is a bounded sequence of complex numbers then (Re(z_n)) is a bounded sequence of real numbers so there
    is a subsequence (z_{n_k}) such that (Re(z_{n_k})) converges.

    Now (Im(z_{n_k})) is a bounded sequence and so (z_{n_k})has a subsequence (z_{n_{k_j}}) such that (Im(z_{n_{k_j}})) converges.

    So (z_{n_{k_j}}) converges.

    RonL
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