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Is the set of complex numbers homeomorphic to $\mathbb{R}^2$?

They are isomorphic. Are they homeomorphic?

hjhjhj57
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Junior Soares
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  • What is required to prove the two are isomorphic? What is required to show the two are homeomorphic? – graydad May 08 '15 at 18:18
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    As topological spaces, they are the same space, so they are obviously homeomorphic. – Crostul May 08 '15 at 18:19
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    @Crostul: That depends on the exact construction of $\mathbb C$ you're using. It's quite common to define it as $\mathbb R^2$ with additional field operations, but other options include the ring $\mathbb R[X]/\langle X^2+1\rangle$ or the ring of real matrices of the form $\begin{pmatrix}a&b\-b&a\end{pmatrix}$. – hmakholm left over Monica May 08 '15 at 18:35
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    @Crostul I find that a very vague and handwavy comment that does not help the asker of the question – Perturbative May 05 '18 at 22:55

1 Answers1

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Yes. The topologies of both spaces are induced by the Pythagorean norm, so the bijection $$ \left\{ \begin{matrix} \Bbb{R}^{2} & \to & \Bbb{C} \\ (x,y) & \mapsto & x + i y \end{matrix} \right\} $$ maps open balls of $ \Bbb{R}^{2} $ to open balls of $ \Bbb{C} $ and vice-versa.

Berrick Caleb Fillmore
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    Or in other words, an isometry (i.e. an isomorphism of metric spaces) is always a homeomorphism (i.e. an isomorphism of topological spaces). – hmakholm left over Monica May 08 '15 at 18:32
  • @Henning: Exactly, Mr Makholm! :) – Berrick Caleb Fillmore May 08 '15 at 18:35
  • What about the fact that the complex plane can be mapped onto the surface of a sphere, so that in particular "infinity" behaves as a single point? Whereas in the $2D$ real space there are in general differences between "+ infinity" and "- infinity" on e.g. the real axis? – M. Wind May 08 '15 at 18:41
  • @M.Wind: The set of points in $ \mathbb{C} $ that are sufficiently far away from a given compact subset of $ \mathbb{C} $ has only one connected component, whereas the set of points in $ \mathbb{R} $ that are sufficiently far away from a given compact subset of $ \mathbb{R} $ has two (and no more) connected components. This is the reason why $ \mathbb{C} $ can have only one point at infinity, whereas $ \mathbb{R} $ can have up to two points at infinity. – Berrick Caleb Fillmore Nov 01 '16 at 08:47