Clarification about $\mathbb{R^2} \cong \mathbb{C}$

I often hear of identification $\mathbb{R^2} \cong \mathbb{C}$. Exactly what kind of isomorphism is there? Are we considering groups? Fields? Topological spaces? Or is it even a strict equality?

For example, can we say that the Heine-Borel theorem about compact sets in $\mathbb{R^n}$ holds for $\mathbb{C}$, meaning that the identification mentionned above is a homeomorphism of topological space?

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Topological spaces. – Brian M. Scott Nov 9 '12 at 17:10
Additive groups too. – Andreas Blass Nov 9 '12 at 17:13
Even Lie groups ) – Dan Shved Nov 9 '12 at 17:17

1 Answer

Consider the bijection $\Bbb R^2\to\Bbb C$ given by $\langle a,b\rangle\mapsto a+bi$. This is a topological homeomorphism, and an isomorphism of normed vector spaces over $\Bbb R$. If we impose a multiplicative structure on $\Bbb R^2$ given by $$\langle a,b\rangle\cdot\langle c,d\rangle:=\langle ac-bd,ad+bc\rangle,$$ then we have given $\Bbb R^2$ a field structure and the above map is a field isomorphism. In fact, we may define $\Bbb C$ that way, as $\Bbb R^2$ with componentwise addition, multiplication defined as above, and $i:=\langle 0,1\rangle$.

Since they're isomorphic as normed vector spaces, then they're isomorphic as metric spaces with the absolute difference metric $d(z_1,z_2)=|z_1-z_2|$. Thus, we may apply Heine Borel to $\Bbb C$.

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Additionally, we may say that $\mathbb{R}^2$ and $\mathbb{C}$ are isomorphic as normed vector spaces. In particular, they are isomorphic as metric spaces, so, as per the OP's query, we may apply Heine-Borel to $\mathbb{C}$. – Eric M. Schmidt Nov 9 '12 at 17:38
Fair point, Eric (forgot to answer that part). – Cameron Buie Nov 9 '12 at 18:08