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I'm trying to prove Bolzano-Weierstrass Theorem to the complex case, i.e., if $(z_n)$ a complex sequence is bounded, then there is a subsequence of $z_n$ which converges.

I'm trying to use the real case of the Bolzano-Weierstrass theorem to prove the complex case without success. I need help.

Thanks

user42912
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  • Perhaps first prove the theorem for $\mathbb R^2$. – GEdgar Apr 24 '15 at 20:32
  • The theorem states that: any sequence on a compact space has a convergent subsequence, compactness can be found here http://www.maths.kisogo.com/index.php?title=Compactness - so there is no "complex case", closed and bounded => compact in a real space (and you can identify $\mathbb{C}$ naturally with $\mathbb{R^2}$). – Alec Teal Apr 24 '15 at 20:36
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    @AlecTeal: we can probably assume that the op has only seen a proof in the case of $\Bbb R$. – Alexandre Halm Apr 24 '15 at 20:40

3 Answers3

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If the sequence $(z_n)$ is bounded, then so is $(\Re(z_n))$. Thus there exists a subsequence of $(z_n)$ for which the real part converges. Let's call this sequence $(w_n)$. As $(w_n)$ is a subsequence of $(z_n)$, it is also bounded, and thus $(\Im(w_n))$ is bounded, so we can find a subsequence of $(w_n)$ for which the imaginary part converges. Thus we have found a subsequence of $(z_n)$ for which both the real and the imaginary parts converge.

Ben Frankel
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HINT: Whether you write $z_n=x_n+iy_n$ or $z_n=r_ne^{i\theta_n}$, your problem boils down to showing that if $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$ are bounded sequences of real numbers, then there is a strictly increasing sequence $\langle n_k:k\in\Bbb N\rangle$ in $\Bbb N$ such that $\langle a_{n_k}:k\in\Bbb N\rangle$ and $\langle b_{n_k}:k\in\Bbb N\rangle$ are both convergent. Do it one sequence at a time. First get a convergent subsequence of $\langle a_n:n\in\Bbb N\rangle$, then get a convergent subsequence of the corresponding subsequence of $\langle b_n:n\in\Bbb N\rangle$.

Brian M. Scott
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You could use BW a first time to extract a converging sub sequence of $\Re(z_n)$ and then a second time on $\Im(z_n)$ (you need to figure out how to make both converge together)

Alexandre Halm
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