# Playfair's Axiom for parallel planes instead of lines

In three-dimensional space, can Playfair's Axiom:

Given a line $a$ and a point $P$ not in $a$, there is at most one line in $P$ parallel to $a$.

be “replaced by“ the following axiom?

Given a plane $\alpha$ and a point $P$ not in $\alpha$, there is one and only one plane in $P$ parallel to $\alpha$.

To make the question precise, can Plaifair's Axiom be proved from the following set of axioms?

1. If a point and a plane have a line in common, the point must lie in the plane.
2. Any two distinct points have one and only one line in common.
3. If two distinct planes have a point in common, they have one and only one line in common.
4. Any line and any point not in this line have one and only one plane in common.
5. Any plane and any line not in the plane have at most one point in common.
6. Given a plane and a point not in the plane, there is one and only one plane in the point having no point in common with the given plane.
7. In every line lie at least two points.
8. There exist two lines having no plane in common.

where to lie in is a symmetric binary relation on a set partitioned by the three sorts Point, Line, Plane, “$A$ and $B$ have $C$ in common“ is short for “$A$ lies in $C$ and $B$ lies in $C$“, and lines $a$ and $b$ are parallel iff they have a plane but no point in common.

My first guess was yes, but now I'm starting to think Axiom 6 is weaker than Playfair's Axiom, since I can't think of a way to prove it. In case anyone is wondering, this question is not from a book or anything but arose quite naturally while trying to find nice axioms characterizing three-dimensional affine space.

• For the distinction between at most one and exactly one parallel line, see this previous Question. – hardmath Feb 27 '18 at 17:19
• The point here is that I am talking about parallel planes instead of lines! – Larry Feb 28 '18 at 6:36
• I can see that, but you also ask explicitly about whether "at most" can be replaced by "one and only one". As you point out we could define a pair of parallel lines in space. – hardmath Feb 28 '18 at 6:42
• It sounds to me like Given a plane α and a point P not in α, there is one and only one plane *through P parallel to α.* – Narasimham Mar 9 '18 at 7:42
• I just didn't want to introduce multiple words for the single relation of Incidence between points, lines and planes, but of course intuitively one can think of a plane in a point as going “through“ it. – Larry Mar 9 '18 at 8:20

Given a plane α and a point P not in α, there is one and only one plane through P parallel to α.

When parallel planes set is defined and a plane and a point on one of the other planes is given,then we have this pair as a unique combination because other planes and other points in the given plane fall off in the selection.

The above is strictly true in Euclidean geometry. There is a set of Euclidean planes parallel to α. The situation is the same in non-euclidean geometries as well.

However for non-euclidean geometries ( born out of negation of Playfair axiom, Riemann's categorization into the foundations of three geometries etc.) imho we need to at first look to how non-euclidean parallel planes are defined.

Elliptic geometry

Consider a series of concentric sphere manifolds M each each normal to a radial plane normal to a straight line axis of symmetry inhabited by co-tangential great circles of constant but different minimum Clairaut constant radii.

Then for any point $P$ i.e., on one spherical shell/manifold in which $P$ inhabits there is an infinite set of parallel elliptic planes/spheres in $\mathbb R^3$ not containing $P$.

Hyperbolic geometry

Consider a series of pseudospheres (hyper/hypo varieties) sharing a common radial plane of symmetry and perpendicular to single straight line axis of symmetry. Each pseudosphere is inhabited by unique Chebychev Net lines or asymptotic lines of zero normal curvature and constant geodesic torsion obeying Sine-Gordon equation. Their uniqueness is due to two asymptotic parallel lines to line of symmetry.

Then to any point $P$ there is an infinite set of parallel planes/pseudospheres in $\mathbb R^3$ not containing $P$ but containing the Chebychev Nets.

If it makes sense I may upload two images after bounty deadline expires.

• I am not quite sure how your answer relates to my question. Could you please elaborate? – Larry Mar 9 '18 at 9:21
• When a parallel planes set is defined and a plane and a point on one of the other planes is given,then we have this pair as a unique combination choice because other planes and other points in the given plane fall off in the selection. – Narasimham Mar 9 '18 at 12:30
• @ Larry: A unique plane exists throgh a given point $P$ parallel to a given plane $\alpha$... this conclusion has to be axiomatically arrived at.. right? – Narasimham Mar 9 '18 at 13:12
• No, that is one of my Axioms (Axiom 6). I want to prove Playfair's Axiom from this and the other 7 Axioms. I hope my use of the word “replace“ was not too ambiguous here. – Larry Mar 9 '18 at 15:29
• Considereing context of axioms 1-5 you stated, are Axiom 6 "there one and only one plane through the point having no point in common with the given plane" and "there is one and only one plane in P parallel to α" are equivalent statements? Is the important parallelism aspect/requirement fully brought out? – Narasimham Mar 9 '18 at 18:25