A description/proof of that projective hyperplane is Hausdorff
Topics
About: topological space
The table of contents of this article
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, \(f: S^d \rightarrow S^d/~\), that maps any pair of antipodal points to the equivalent class. For any points, \(p_1, p_2 \in S^d/~\) such that \(p_1 \neq p_2\), there are \(p'_1, p'_2 \in S^d\) such that \(f (p'_i) = p_i\) and \(p'_1 \neq p'_2\) and a hemisphere excluding the border ("hemisphere" hereafter always means excluding the border) on which (the hemisphere) \(p'_1\) and \(p'_2\) are on, because there is the geodesic that passes \(p'_1\) and \(p'_2\) and we can take the middle point of the geodesic as the pole of the hemisphere (\(p'_i\) is not on the border as \(p'_1\) and \(p'_2\) are not antipodal).
There are some disjoint open balls, \(B'_{p'_i-\epsilon}\), around \(p'_i\)s contained in the hemisphere (which we do not prove intricately here, but will be obvious intuitively). \(U_{p_i} := f (B'_{p'_i-\epsilon})\). \(p_i \in U_{p_i}\). \(f^{-1} (U_{p_i}) = B'_{p'_i-\epsilon} \cup B''_{p'_i-\epsilon}\) where \(B''_{p'_i-\epsilon}\) is the antipodal image of \(B'_{p'_i-\epsilon}\). \(B''_{p'_i-\epsilon}\) is open on \(S^d\), because it is an open ball around the antipodal point of \(p'_i\), 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. \(U_{p_1} \cap U_{p_2} = \emptyset\), obviously.
