In University I have been told that every norm on $\mathbb C^n$, for any $n\in\mathbb{N}$, is equivalent to every other such norm. I have a proof for this on any vector space on $\mathbb R$. Trouble is, the proof uses the Heine-Borel theorem, proving that it is extended to any vector space on $\mathbb{R}$ with the pseudo-euclidean norm where, given $B=\left\lbrace v_1,\dots,v_n\right\rbrace$ a base of the vector space, if a vector $v$ of the space can be written as $v=\sum_{i=1}^nx_iv_i$, then the pseudo-euclidean norm is $\left\|v\right\|=\sqrt{\sum_{i=1}^nx_i^2}$, and therefore stating that, being $N$ another norm, the set of vectors $v$ in the space such that $N(v)=1$, being closed ($N$ is continuous and that set is $N^{-1}(\lbrace1\rbrace)$ and $\lbrace1\rbrace$ is closed in $\mathbb{R}$) and bounded (since $\left\vert x_i\right\vert<1$ for all $i$). Now unless I'm much mistaken, the Heine-Borel theorem (i.e. every closed and bounded set is compact) holds only on $\mathbb{R}^n$, and not on $\mathbb{C}^n$. So the questions are:
1) Is it true that the topological equivalence above stated holds for any norm on any $\mathbb{C}^n$?
2) Does the Heine-Borel theorem hold also on $\mathbb{C}^n$?
3) If not, and if the answer to 1) is yes, how do I extend that proof to complex-number spaces?
Posting this to take this off the unanswered list.
Heine-Borel works in $\mathbb{R}^n$ as we have proved, but then $\mathbb{C}^n$ is canonically identified with $\mathbb{R}^{2n}$, and that isomorphism is clearly also a homeomorphism since the norms are left basically unchanged. So Heine-Borel can easily be transferred to $\mathbb{C}^n$. And this is the content of Cameron's comment.
Frank's comment adds that the topological equivalence of all norms, extended to $\mathbb{C}^n$ by the same proof using Heine-Borel on $\mathbb{C}^n$, generalizes to any finite dimensional space. And since any such space over $\mathbb{R}$/$\mathbb{C}$ is isomorphic to $\mathbb{R}^n$/$\mathbb{C}^n$, I guess this means any finite dimensional space on any field has only one norm up to topological equivalence.