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Are the points that lie on the outside edge of a Poincare Disc considered to be in the hyperbolic plane? In other words, are points C, D, J, Y, and Z inside the plane?

enter image description here

Community
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Dana Hill
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  • Hyperbolic geometry obeys the same axioms as Euclidean geometry, except that the parallel postulate is replaced by the hyperbolic parallel postulate. If you included the boundary points in the Poincaré disc model, it would also violate one of the betweenness axioms ("Given any two points $A$ and $B$, there is a point $C$ on the line joining them such that $B$ is between $A$ and $C$ along the line.") So it would no longer be a model of hyperbolic geometry. – Micah Aug 06 '17 at 04:47
  • Maybe best is to look at https://en.wikipedia.org/wiki/Ideal_point will add new arguments from the answer to this article – Willemien Aug 06 '17 at 08:05

2 Answers2

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No, the perimeter is not part of the model.

The lines that approach a mutual point in the perimeter are considered parallel.

In hyperbolic geometry, two distinct parallel lines are further classified as one of two types of parallel

  • limit parallel: if the two lines approach a common point on the perimeter of the circle
  • ultraparallel: if they are not limit parallel.

In your picture, the line through $AB$ and the line through $KL$ are ultraparallel, while $EF$ is limit parallel to $AB$.

rschwieb
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No, they are not elements of the hyperbolic plane. The boundary circle is sometimes called "ideal boundary" because of that. Its elements can be thought of as equivalence classes of geodesic rays: two such rays are equivalent (i.e., go to the same ideal boundary point) if they stay at bounded distance from each other when parametrized by arclength.

In intuitive terms, it's like points of the horizon vs points on the Earth's surface.

Using the aforementioned idea of geodesic rays, one can give the ideal boundary more structure: see Wikipedia.