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Thread: Bolzano-Weierstrauss: A Complex Version

  1. #1
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    Bolzano-Weierstrauss: A Complex Version

    Let $\displaystyle (z_n)$ be a bounded sequence of complex numbers. Does there exist a subsequence $\displaystyle (z_{n_k})$ 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 $\displaystyle (s_n)$ be a bounded sequence. Then there exists a monotone subsequence (a theorem from before) $\displaystyle (s_{n_k})$, which is certainly bounded. Thus, $\displaystyle (s_{n_k})$ 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 $\displaystyle (z_n)$ be a bounded sequence of complex numbers. Does there exist a subsequence $\displaystyle (z_{n_k})$ 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 $\displaystyle (s_n)$ be a bounded sequence. Then there exists a monotone subsequence (a theorem from before) $\displaystyle (s_{n_k})$, which is certainly bounded. Thus, $\displaystyle (s_{n_k})$ 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 $\displaystyle (z_n)$ is a bounded sequence of complex numbers then $\displaystyle (Re(z_n))$ is a bounded sequence of real numbers so there
    is a subsequence $\displaystyle (z_{n_k})$ such that $\displaystyle (Re(z_{n_k}))$ converges.

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

    So $\displaystyle (z_{n_{k_j}})$ converges.

    RonL
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