Solution to Prove that the map f : C → R 2 , x + iy 7→ … - Sikademy
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Archangel Macsika

Prove that the map f : C → R 2 , x + iy 7→ (x, y) is a bijection, i.e., a cardinal equivalence C ≈ R

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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.

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