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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 $y$.

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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 $\Lambda_y:=\{\lambda u \,:\,u\in U_y, \lambda\in\Bbb R\}$.

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  • $\begingroup$ 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

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