I want to prove the Hausdorff property of the projective space with this definition: Define $\mathbb{P}^n$, the real projective space of dimension n to be the set of 1-dimensional linear subspaces (lines through the origin) in $\mathbb{R}^{n+1}$. There is a natural map $\pi:\mathbb{R}^{n+1}\to \mathbb{P}^n$ defined by sending a point $x$ to its span. We topologize $\mathbb{P}^n$ by giving it the quotient topology with respect to this map.

I know how to prove that $\mathbb{P}^n$ is Hausdorff if it is define by the sphere $\mathbb{S}^n$ with the antipodal points identified or using gruop Actions. It is Here.

I cannot created two disjoint open cones.

Let $x$ and $y$ be distinct points in $\mathbb{P}^n$. Let $l_x$ and $l_y$ be the corresponding lines in $\mathbb{R}^{n+1}$. The Hausdorff property of projective space follows from the fact that we can fit the lines into two open cones in $\mathbb{R}^{n+1}$ that only have $0$ in common, whose projections to projective space give disjoint open sets that contain $x$ and $y$.


Just project the open surfaces on $\Bbb S^n$ from the origin to get the disjoint cones.

You are given two lines in ℝn+1. Those intersect $\Bbb S^n$ in 2-2 points: $x,−x$ and $y,−y$, say. You wrote you know how to prove there are open neighborhoods, so assume $U_x$ and $U_y$ are disjoint open neighborhoods of $x$ and $y$, respectively, on the surface of $\Bbb S^n$. Then consider the cones $\Lambda_x:=\{\lambda u \,:\,u\in U_x, \lambda\in\Bbb R\}$ and $\Lambda_y:=\{\lambda u \,:\,u\in U_y, \lambda\in\Bbb R\}$.

  • $\begingroup$ I want to create two open cones. $\endgroup$ – Renato Targino Mar 9 '14 at 22:59
  • $\begingroup$ @Berci Maybe you can add the comment to your answer. $\endgroup$ – egreg Mar 9 '14 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.