Let
be a bounded sequence of complex numbers. Does there exist a subsequence
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
be a bounded sequence. Then there exists a monotone subsequence (a theorem from before)
, which is certainly bounded. Thus,
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.