# Thread: Bolzano-Weierstrauss: A Complex Version

1. ## Bolzano-Weierstrauss: A Complex Version

Let $(z_n)$ be a bounded sequence of complex numbers. Does there exist a subsequence $(z_{n_k})$ 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 $(s_n)$ be a bounded sequence. Then there exists a monotone subsequence (a theorem from before) $(s_{n_k})$, which is certainly bounded. Thus, $(s_{n_k})$ 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.

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

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

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

RonL