Exact relationship between topologies of $\mathbb C^n$ and $\mathbb R^{2n}$ I want to prove the compactness of a subset of $\mathbb C^n$. I am very tempted to say "$\mathbb C^n$ can be thought of as $\mathbb R^{2n}$," so that I can use Heine-borel in $\mathbb R^{2n}$. My question is, how do make precise the statement "$\mathbb C^n$ can be thought of as $\mathbb R^{2n}$"? I think most people take this for granted, but I have never seen an explicit homeomorphism $\varphi$ between the two spaces to merit such a statement.
natural idea:
Of course the most natural candidate is $\varphi:\mathbb C^n \to\mathbb R^{2n}$ where $\varphi(z_1,\ldots,z_n)=(\mbox{Re}(z_1),\ldots,\mbox{Re}(z_n),\mbox{Im}(z_1),\ldots,\mbox{Im}(z_n))$ and $\varphi$ is an isometry if $\mathbb C^{n}$ and $\mathbb R^{2n}$ are equipped with the 2-norm. Is this what people have in mind when they say "$\mathbb C^n$ can be thought of as $\mathbb R^{2n}$"? What happens if the norm is a $p$-norm?
 A: Yes, that's it.
Note that it is obvious that $\mathbb{C}^n$ is a vector space over the field $\mathbb{R}$ simply because $\mathbb{R}$ is a subfield of $\mathbb{C}$ and that $\dim_\mathbb{R}(\mathbb{C}^n) = 2n$. There are many isomorphisms of real vector spaces $\varphi: \mathbb{C}^n \to \mathbb{R}^{2n}$, the most obvious probably being $\varphi(z_1,\ldots,z_n) = (\mbox{Re}(z_1),\mbox{Im}(z_1), \ldots, \mbox{Re}(z_n),\mbox{Im}(z_n))$. Clearly $\lVert (z_1,\ldots,z_n) \rVert_2^\mathbb{C} = \lVert \varphi(z_1,\ldots,z_n) \rVert_2^\mathbb{R}$.
$\varphi$ is not an isometry for a $p$-norm with $p \ne 2$. This can be most easily seen for $n = 1$. We have $\lVert z \rVert_p^\mathbb{C} = \lvert z \vert$, but $\lVert (\mbox{Re}(z), \mbox{Im}(z)) \rVert_p^\mathbb{R} \ne \lVert (\mbox{Re}(z), \mbox{Im}(z)) \rVert_2^\mathbb{R} = \lvert z \vert$ for a general $z$.
However, that it is irrelevant as long as we are only interested in the topology of $\mathbb{C}^n$. For example, the compactness of a subset of $\mathbb{C}^n$ only depends on its topology.
As Calvin Khor has pointed out, all norms on a finite-dimensional (real or complex) vector space are equivalent and therefore generate the same topology. 
