Prove that the real projective line cannot be embedded into Euclidean space

Question

Can the real projective line $RP^1$ be embedded into $\mathbb{R}^n$ for any $n$?

At first, I thought it could because $RP^2$ can be embedded in four-dimensional space, and $RP^1$ seemed like a simpler object to deal with than $RP^2$. But after drawing diagrams and attempting to find an embedding, I'm getting less and less sure that it can be embedded... unrigorously it always seems to require that 'point at infinity' that comes from say the one-point compactification of $\mathbb{R}$, so is there a way that I could prove this one way or the other?

Answer 1

Well, $\mathbb{R}P^1$ is a subspace of $\mathbb{R}P^2$ (namely, any projective line in the projective plane), so since $\mathbb{R}P^2$ embeds in $\mathbb{R}^4$, so does $\mathbb{R}P^1$. But actually, you can do better: $\mathbb{R}P^1$ is just a circle, so it embeds in $\mathbb{R}^2$. Indeed, if you remove a point from a circle, the resulting space is homeomorphic to $\mathbb{R}$, so a circle is the 1-point compactification of $\mathbb{R}$ and hence homeomorphic to $\mathbb{R}P^1$.

Answer 2

$\mathbb{R}P^1$ is just the circle $S^1$ and this cannot be embedded in the reals: an infinite connected subset of $\mathbb{R}$ (is an interval so) always has a cut-point (a point we can remove to leave a disconnected subset), and the circle remains connected if we remove any point.

Answer 3

In lieu of a thousand words, here's an embedding of the real projective line (the set of lines in the plane through the origin) in the Cartesian plane: