Prove that $\mathbb{C}$ with the absolute value norm is a Banach space.

Question

Prove that $\mathbb{C}$ with the absolute value norm is a Banach space.

In this link, we can read a proof for the real case. Is the proof for $\mathbb{C}$ the same?

Answer 1

If $(z_n)_{n=1}^\infty = (z_n)$ is Cauchy in the complex numbers, set $z_n = x_n + \sqrt{-1} y_n$, where $x_n$ is the real part of $z_n$ and $y_n$ the imaginary part. As $(z_n)$ is Cauchy, for $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for each $m,n \geq N$,

$|z_n -z_m| < \epsilon$,

from this, and the fact that $|x_n - x_m | \leq |z_n - z_m|$, deduce that $(x_n)$ and $(y_n)$ are Cauchy in the reals and thus convergent. To which complex number thoes $(z_n)$ then converges?

Answer 2

In the link you give, the proof of completeness of $\mathbb R$ consists in 3 steps:

1. Every Cauchy sequence in $\mathbb R$ is bounded.

2. (Bolzano-Weierstrass theorem) In $\mathbb R$, every bounded sequence has a convergent subsequence.

3. In $\mathbb R$, every Cauchy sequence which has a convergent subsequence is itself convergent.

Let us analyze the next sentence in your link: "An analogous proof shows that the space of complex numbers is also a Banach space."

Steps 1 and 3 hold in any metric space (the proof is the same), in particular in $\mathbb C$.

Step 2 is more specific but holds in $\mathbb C$ (and similarly, in any finite dimensional normed real vector space).

So the answer to your question is: Yes, "the" proof for $\mathbb C$ is the same (but of course, there are other proofs, like the one suggested by Victor's answer).

Answer 3

A Banach space is in particular a linear space, so you should mention what linear structure you impose on $\Bbb C$. It won’t matter though, since any $|\cdot|$-Cauchy sequence converges in $\Bbb C$ as $\Bbb C$ is a complete $|\cdot|$-metric space, so $\Bbb C$ - with any natural linear structure - is a Banach space.