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In the complex set of numbers, what sequences have at least one convergent subsequence?

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What do you mean by "complex set"? Do you mean a set of complex numbers? Or is it actually the set rather than the numbers that you have in mind? – Michael Hardy Oct 16 '11 at 16:48
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As $\mathbb{C}$ is just $\mathbb{R}^2$ with additional structure (namely complex multiplication), the answer is provided by the Bolzanon-Weierstrass theorem.

That is, every bounded sequence has a convergent subsequence. The proof goes exactly as in the real case.

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By the Bolzano-Weierstrass theorem, every bounded complex sequence has at least one convergent subsequence.

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