In the complex set of numbers, what sequences have at least one convergent subsequence?
When is a complex subsequence convergent?
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$\begingroup$
sequences-and-series
complex-numbers
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0What 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? – 2011-10-16
2 Answers
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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.