= Solution
The quotient map $q:S^2\to X$ is continuous and surjective by the <quotient topology>, so $X$ is compact as a continuous image of a compact space. To prove the Hausdorff property explicitly, let $[u]\ne[v]$ and set $d=\min(\|u-v\|,\|u+v\|)>0$. Choose $r<d/3$. The saturated open sets
$$
U=(B_r(u)\cup B_r(-u))\cap S^2,\qquad
V=(B_r(v)\cup B_r(-v))\cap S^2
$$
are disjoint by the <triangle inequality>. Each contains both antipodes of every one of its points, so $q^{-1}(qU)=U$ and similarly for $V$. Consequently $qU,qV$ are disjoint open quotient neighborhoods separating the classes. Thus \b[the quotient is compact and Hausdorff]. It is the <real projective plane>; equivalently the <metric on the antipodal sphere quotient> is $d_X([u],[v])=\min(\|u-v\|,\|u+v\|)$ and induces this topology.
Back to article page
