327: Projective Hyperplane Is Hausdorff

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that projective hyperplane is Hausdorff

---

Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of projective hyperplane.
• The reader knows a definition of Hausdorff topological space.

---

Target Context

• The reader will have a description and a proof of the proposition that any projective hyperplane is Hausdorff.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

Any projective hyperplane, $S^d/$, is Hausdorff.

---

2: Proof

Let $f$ be the quotient map, $f:S^d\to S^d/$, that maps any pair of antipodal points to the equivalent class. For any points, $p_1,p_2\in S^d/$ such that $p_1\ne p_2$, there are $p_1^{\prime },p_2^{\prime }\in S^d$ such that $f(p_i^{\prime })=p_i$ and $p_1^{\prime }\ne p_2^{\prime }$ and a hemisphere excluding the border ("hemisphere" hereafter always means excluding the border) on which (the hemisphere) $p_1^{\prime }$ and $p_2^{\prime }$ are on, because there is the geodesic that passes $p_1^{\prime }$ and $p_2^{\prime }$ and we can take the middle point of the geodesic as the pole of the hemisphere ($p_i^{\prime }$ is not on the border as $p_1^{\prime }$ and $p_2^{\prime }$ are not antipodal).

There are some disjoint open balls, $B_{p_i^{\prime }-ϵ}^{\prime }$, around $p_i^{\prime }$s contained in the hemisphere (which we do not prove intricately here, but will be obvious intuitively). $U_{p_i}:=f(B_{p_i^{\prime }-ϵ}^{\prime })$. $p_i\in U_{p_i}$. $f^{-1}(U_{p_i})=B_{p_i^{\prime }-ϵ}^{\prime }\cup B_{p_i^{\prime }-ϵ}^{″}$ where $B_{p_i^{\prime }-ϵ}^{″}$ is the antipodal image of $B_{p_i^{\prime }-ϵ}^{\prime }$. $B_{p_i^{\prime }-ϵ}^{″}$ is open on $S^d$, because it is an open ball around the antipodal point of $p_i^{\prime }$, so, $f^{-1}(U_{p_i})$ is open on $S^d$, and so, $U_{p_i}$ is open on $S^d/$ by the definition of quotient topology. $U_{p_1}\cap U_{p_2}=\mathrm{\varnothing }$, obviously.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>