geometrynoneuclidean-geometryprojective-geometry
When talking to someone who knows basic mathematics but not really in-depth, how would you explain that parallel lines can touch?
I am referring to Non-Euclidian/ Projective Geometry.
Edit:
Why do parallel lines in the Euclidean plane (correspond with lines that) cross in the projective plane?
According to the Wikipedia article absolute geometry the first 28 propositions of Euclid's Elements do not require the parallel postulate and hence may be proven for neutral or "absolute" geometry.
Prop. 11 is the construction of a perpendicular to a line from a point on the line. Prop. 12 is the construction of a line perpendicular to a given line through a point not on the line. So, combined with OP's proof that all lines perpendicular to a given line are parallel, the result is established: The family of parallel lines perpendicular to a given line covers all points.
This answer relates to the first bullet point of the OP.
(The second bullet point has been answered in a note due to Arthur. )
The question can be interpreted in the case of the Euclidean and in the case of Hyperbolic geometry.
Euclid defines the parallel to a straight the following way: let $a$ and $P$ be a straight and a point, resp. in a plane. The straight $b$ through $P$ is parallel to $a$ if it does not have a common point with ot. (An equivalent formulation of Euclid's parallel postulate claims hat there is only one parallel through $P$ to $a$.
Bolyai and Lobachevsky define the parallel to $a$ through $P$ the following way: Construct a straight through $P$ in the common plane of $a$ and $P$. Let $b$ intersect $a$ then turn $b$ around $P$ so that the intersection point gets further and further away from $P$, there will be a situation when $b$ does not meet $a$ the first time. Then there will be a situation when $b$ intersect $a$ again the first time. (This latter definition allows that $a$ has two parallels through $P$. The hyperbolic parallel postulate claims that there are two parallels to through $P$ to $a$.)
In both cases the essence of parallelism is that parallel lines do not meet.
Now, consider the Klein model of hyperbolic geometry. This model embeds the hyperbolic plane (space) in the Euclidean plane (space) by taking am open disk (sphere) whose (internal) points are the point of the hyperbolic plan (space) and whose chords are the straights.
The following figure depicts the Klein model of the hyperbolic plane.
Here $P$ is a hyperbolic point, $a$, $b$, and $b'$ are parallel hyperbolic straight lines to $a$ in the Bolyay-Lobachevsky sense. Apparently the two parallels meet on the circle (in the infinity). There is no contradiction because the circle itself does not belong to the model of the hyperbolic plane. But it is so good to be mystic and say that the parallels meet in the infinity.
But what about the parallels of the Euclidean geometry that do not meet but in the infinity?
It is harder to intuit the Euclidean parallels as pairs of straight lines meeting in infinity. It helps if one can imagine the Euclidean plane modeled within itself the following way.
Take the inside of an Euclidean circle and remove its center. Now, consider the Euclidean circles that go through the center and are within the circle. These Euclidean circles will model the Euclidean straight lines. Every pair of straight lines meet in the center (not belonging to the model plane). Some of these model straight have another meeting point and some don't. In the figure above $a$ and $b$ are parallels having only one "common point in the infinity". Now, we know that the center of the big circle represents the point in the infinity (not belonging to the model), in which point parallels meet.
Note. Actually the inverse images of the Euclidean straights wrt the black circle are the images of the Euclidean straight.
Best Answer
Lines are parallel if they lie in the same plane and they don't intersect. In other geometries, there may be no parallel lines, lines may not have a common point but they may have a common limit point at infinity, or they may just not intersect.
You might be thinking about Projective Geometry where a "point at infinity" is added to every family of parallel lines and the set of all points at infinity is called the "line at infinity".
Addendum
Typically, in formal geometry, points, lines, and planes are not defined. But postulates define their "baseline" behavior. The postulate that every geometry seems to agree on is the one that states
L1: Given two distinct points in a plane, there is exactly one line in that plane that contains them.
The "dual" of that postulate is
DL1: Given two distinct lines in a plane, there is exactly one point in that plane that belongs to both lines.
Since Euclidean geometry contains parallel lines, DL2 is false. But Projective geometry accepts DL2 as a postulate. The big question is, "Does there exists a geometry that satisfies the postulates of Projective geometry?" Yes there does.
The creation of such a geometry is really quite clever. You start with a Euclidean plane and you add points to it as follows. Pick any line in the plane. To that line and all lines parallel to it, you add one extra point, a point at infinity. This is a set thing. We are treating a Euclidean line, $l$, as a set of points and we are adding a non Euclidean point $p$ to that set, $l' = l \cup \{p\}$.
Adding this point to those lines means that those lines are no longer parallel. Define the set of all points at infinity to be the line at infinity. The Projective plane is the Euclidian plane with all points at infinity and the line at infinity added to it. This particular Projective plane can be proved to satisfy DL2.
In the other direction, pick any line, $l$, in the projective plane, $\mathbb P^2$, and remove it. What you end up with is the Euclidean plane, $\mathbb E^2 \cong \mathbb P^2 - \{l\}$. Some lines will still intersect. Those that intersected at a point on the line that was removed will now be parallel.