Can space be augmented with a plane at infinity so that parallel planes intersect at a line at infinity?

Question

The real plane can be augmented with a line at infinity such that two parallel lines intersect at a point at infinity, and the set of all such points forms a line.

In space (3 dimensional solid geometry), can a similar augmentation be made? Can we augment space, adding a plane at infinity, such that parallel planes intersect in a line at infinity?

I ask because augmenting space accordingly would seem to cause a contradiction: In general, if line $\ell$ is parallel to plane $P$, and $\ell$ lies within plane Q, then $\ell$ is also parallel to any line $m$ formed by the intersection of planes $P$ and $Q$. How would that apply to the line at infinity? That is, if $P$ and $Q$ are parallel, and $\ell_1, \ell_2$ are distinct intersecting lines which are both parallel to $P$, how can they both be parallel to the line at infinity?

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To clarify the geometric objection, we have in Euclidean geometry:

Let planes $P, Q$ intersect in line $m$. Let line $\ell$ be parallel to $P$ and lie within $Q$. Then $\ell \parallel m$.

Now, let's assume $P, Q$ are parallel and hence intersect at line $m$ at infinity. Then by the theorem above, any line $\ell$ parallel to $P$ must be parallel to $m$.

But consider two intersecting lines $\ell_1, \ell_2$, both of which are parallel to $P$ and lie within $Q$. By the above, they must both be parallel to $m$. Since parallelism is transitive, then $\ell_1 \parallel \ell_2$. But this is impossible, since they are assumed to intersect, giving us a contradiction.

Without the line at infinity, this contradiction cannot arise, because two intersecting lines, both lying in plane $Q$, cannot both be parallel to plane $P$ if $P$ intersects $Q$.

Answer 1

Yes, you can extend Euclidean 3-dimensional space to projective 3-space by adding a plane at infinity. A projective plane at infinity, to be precise.

The conflict you observed is due to an incorrect assumption of transitivity. Let me explain in more detail, and do so in the projective plane because that's enough to illustrate this point.

What does it mean for two distinct lines in the plane to be parallel? In projective terminology, it means that the unique point of intersection lies on the line at infinity. Now what happens if you intersect the line at infinity with a finite line? You get a point if intersection that lies on the line at infinity. So by the definition I just gave, you have to consider them parallel. The line at infinity is parallel to every finite line, in this sense.

That of course breaks transitivity. If you say finite line $a$ is parallel to the line at infinity, and the line at infinity is parallel to finite line $b$, that doesn't mean $a$ and $b$ share the same point of intersection with the line at infinity, so they don't have to be parallel to one another.

In a way this problem is very similar to some basic arithmetic. For positive integers $a,b,c$, if $ac=bc$ then you would conclude that $a=b$. If you slightly increase your domain of consideration to include zero, then suddenly that's no longer the case: you have the case $c=0$ to consider, which plays a special role for multiplication. It forces you to do case distinctions: either $a=b$ or $c=0$ (or maybe both).

The same happens when you go from Euclidean to projective geometry. If you have $a\Vert c$ and $b\Vert c$ then in Euclidean geometry you conclude that $a\Vert b$. But as you extend the scope to the projective plane, suddenly you have a case distinction, with $c$ being the line at infinity as the second alternative. And again that's because the newly added element plays a special role for the operation you consider.

Your 3d setup is just a more complicated setup where this lack of transitivity defies your initial intuition.