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**ThePerfectHacker** 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.