In Theorem 3.11 in Rudin, it says
In $\mathbb{R}^k$, every Cauchy sequence converges.
But I'm wondering how about in $\mathbb{C}$ or $\mathbb{C}^k$, does every Cauchy sequence still converge? I only learned basic concepts about convergence so haven't been equipped with more proof skills yet. So thank you for explaining!
The space $\mathbb C^k$ with the usual distance metric is actually equivalent to $\mathbb R^{2k}$ with the usual metric. If $z_j=a_j+ib_j$ for $j=1,\dots,k$ then we can define:
$$f:\mathbb C^k \to\mathbb R^{2k}\; ;\,(z_j)\mapsto (a_1,b_1,a_2,b_2,\dots,a_k,b_k)$$
This has an obvious inverse function $g:\mathbb R^{2k}\to \mathbb C^k$.
This function has the nice property that $d(\mathbf z,\mathbf w)=d(f(\mathbf z),f(\mathbf w))$. So any Cauchy sequence in $\mathbf C^k$ goes to a Cauchy seuence in $\mathbf R^{2k}$, which converges to some vector $\mathbf v\in\mathbb R^{2k}$. Then the original Cauchy sequence converges to $g(\mathbf v)$.
It's just brute force to prove this, but it hinges on the result that:
$$|a+bi-(c+di)|^2 = (a-c)^2 + (b-d)^2.$$
From the point of view of distance, these two spaces are exactly the same.
They are different when you start exploring differentiability in them.