Answer to Question #263075 in Discrete Mathematics for Amana Ateba

Question

Prove that the map f : C → R

2

, x + iy 7→ (x, y) is a bijection, i.e., a cardinal equivalence

C ≈ R

Answer 1

Let us prove that the map $f : \mathbb C → \R^2,\ f(x + iy)= (x, y),$ is a bijection.

Let $f(x_1 + iy_1)=f(x_2 + iy_2).$ Then $(x_1, y_1)=(x_2, y_2),$ and hence $x_1=x_2$ and $y_1=y_2.$ We conclude that $x_1 + iy_1=x_2 + iy_2,$ and thus the map $f$ is injective.

Let $(a,b)\in\R^2$ be arbitrary. Then for $z=a+ib$ we get that $f(z)=f(a+ib)=(a,b),$ and consequently, $f$ is surjective.

We conclude that $f : \mathbb C → \R^2$ is a bijection.