Does the Plane Separation Axiom hold for the projective sphere?

Some background:

We'll denote the projective sphere as $\mathbb{S}=\{((x,y,z),(-x,-y,-z)):x^2+y^2+z^2=1\}$. That is, we'll consider the points of the projective sphere as the pair of antipodal points of the usual unit sphere.

A line in $\mathbb{S}$ is sort of like a great circle on the unit sphere. Simply, we will let a line be defined by three parameters: $a,b,c$. $$L_{a,b,c}=\{((x,y,z),(-x,-y,-z)):ax+by+cz=0 \text{ and }x^2+y^2+z^2=1\}$$

The Plane Separation Axiom says that for a given plane $P$, a line $l$ should divide $P$ into two non-empty, convex half-planes $H_1$ and $H_2$. Furthermore, if $x \in H_1$ and $y \in H_2$, then the line segment $\overline{xy}$ intersects $l$ in a non-empty fashion.

I haven't even been able to determine whether a line would divide $\mathbb{S}$ into two half-planes. For the unit sphere, I think we can take all the points $\{(x,y,z):ax+by+cz >0\}$ as one half-plane and $\{(x,y,z):ax+by+cz < 0\}$ as the other. However, this doesn't seem to work for $\mathbb{S}$.

If we do go ahead and assume there would be two non-empty, convex half-planes, I was thinking that we could find the line that the line segment $\overline{xy}$ is a part of and somehow see if it intersects $l$.

Any hints or ideas are much appreciated. Thank you in advance.

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    $\begingroup$ think of the projective plane and wether the line at infinity separates it. Think of two "affine" points each in one half plane ad try to joins them through the line at infinity without crossing the "affine" line defining the half planes $\endgroup$ – marwalix Jan 10 '15 at 23:14

What you call 'projective sphere' is just a realization of the projective plane. (I guess you rather wanted $\Bbb S=\{\{(x,y,z),\,(-x,-y,-z)\}\mid x^2+y^2+z^2=1\}$.)

As you correctly note, a line on $\Bbb S$ is 'sort of' a great circle. Correctly, a line is the image of a great circle under the canonical map $S^2\to\Bbb S$ where $S^2$ denotes the unit sphere.

And yes, a line on the sphere divides it into two parts, but these two parts already coincide in $\Bbb S$, so $\Bbb S$ is not divided into two parts.

The same goes if you consider projective plane as the Euclidean plane equipped with points in the infinity: starting from one half plane, going through a point in the infinity we can arrive to the other half plane. So, in the projective plane these are connected to each other, so is one part only.


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