How can $\Bbb RP^1 \cong S^1$ be true if they have different cardinalities?

Question

Prove that $\Bbb RP^1 \cong S^1$

$$\Bbb R P^1= \{[(x_0, x_1)]: (x_0, x_1) \in \Bbb R \times \Bbb R \text{ and } (x_0, x_1) \sim(y_0, y_1) \text{ iff } (x_0,x_1) = \lambda(y_0, y_1) \text{ for some $\lambda \in R-\{0\}$ }\}$$

is the set of lines through the origin, excluding the origin, and

$S^1 = \{(z_1,z_2) \in \Bbb R \times \Bbb R : z_1^2 + z_2^2=1\}$.

Superimposing the two on the plane you can see that each line through the origin intersects the circle in exactly two antipodal points.

But wouldn't this imply a map between the two spaces couldn't be bijective since one has the cardinality of points on a circle and the other would have the cardinality of points on the hemi-circle?

Answer 1

There is a homeomorphism between ${\mathbb R}P^1$ and $S^1$ allright, but it is slightly more involved than your map. You arrived at a $1:2$ correspondence. Now think about a map that makes out of two opposite points on $S^2$ one single point on $S^2$ in a nice way. A hint: Maybe you remember $z\mapsto z^2$ from complex analysis.

To expand on the hint: The map $\psi:\ z\mapsto z^2$ restricted to $S^1\subset{\mathbb C}$ maps two opposite points $\pm z\in S^1$ to one single point. Now consider the map $\psi\circ f$, where $f$ is the map you hinted at in your question.

Answer 2

$\mathbb{R}P^1$ and $S^1$ are both 1-dimensional compact connected topological manifolds, and there is only one such manifold up to homeomorphisms. Therefore they are homeormorphic and have the same cardinality.