Intuition for the compactness of real projective space $\mathbb{R}\mathbb{P}^n$.

Question

I want to have an intuition for why the $n$-dimensional real projective space defined as $$\mathbb{R}\mathbb{P}^n:=\mbox{set of 1-dimensional subspaces of }\mathbb{R}^{n+1}$$ is compact. I don't see how it is bounded since the subspaces can "extend" to as much as I want.

But I know the proof:

Define a relation on $\mathbb{R}^{n+1}\setminus\{0\}$ via $x\sim y \Leftrightarrow \ x=\lambda y, \ \lambda\in \mathbb{R}\setminus\{0\}.$ Then, w.r.t. this equivalence relation, $$\mathbb{R}^{n+1}\setminus\{0\}/\sim \ \equiv \mathbb{R}\mathbb{P}^n.$$ There is a homeomorphism from $\mathbb{R}\mathbb{P}^n \mbox{ to } \mathbb{S}^n/\sim$ where $x\sim -x, \forall x\in \mathbb{S}^n$. It is $$f:\mathbb{R}\mathbb{P}^n \to \mathbb{S}^n/\sim, \ [x]\mapsto \bigg[\frac{x}{||x||}\bigg].$$ The compactness of $\mathbb{R}\mathbb{P}^n$ follows from $\mathbb{S}^n/\sim$ being compact.

To repeat: My question is about having an intuition. I cannot see how $\mathbb{R}\mathbb{P}^n$ is bounded. Perhaps, you should give me the precise set that will contain $\mathbb{R}\mathbb{P}^n.$

Answer 1

An open set in $\mathbb{RP}^n$ is basically an open cone in $\mathbb{R}^{n+1}$, or it can also be seen as just an open set on $S^{n}$ that is symmetric abot the origin, so $U \subset S^n$ such that $U = -U$. Thus any open cover of $S^n$ gives an open cover of the sphere. More generally the point is that every ray corresponds (uniuqe up to sign) to a point on the sphere, so though the rays look unbounded the point is. For yet another perspective, by quotienting the rays down to points, you are basically killing the unbounded part of $\mathbb{R}^{n+1}-\{0\}$.

Answer 2

"Bounded" isn't really the right intuitive notion to be thinking about, because the topology of $\mathbb R \mathbb P^n$ is not defined by a metric. Instead, I think the clearest intuition comes from thinking about sequential compactness: Every sequence in $\mathbb R\mathbb P^n$ has a convergent subsequence. (For manifolds, compactness and sequential compactness are equivalent.)

Suppose $\{\xi_i\}$ is any sequence of points in $\mathbb R \mathbb P^n$ (i.e., any sequence of $1$-dimensional subspaces of $\mathbb R^{n+1}$). Each $\xi_i$ intersects the unit sphere in two points, say $\{x_i,-x_i\}$. Because $\mathbb S^n$ is compact, the sequence $\{x_i\}$ has a convergent subsequence $\{x_{i_j}\}$. And because the quotient map $q\colon \mathbb R^{n+1}\smallsetminus \{0\}\to \mathbb P^n$ is continuous, it follows that $\xi_{i_j} = q(x_{i_j})$ converges.

Tl;dr: $1$-dimensional subspaces are determined by where they meet the unit sphere, and compactness of the sphere guarantees that subspaces can't wander too far away.