# Is there a way to show the infinite Euclidean plan in a finite way?

I'm a huge fan of hyperbolic geometry. I find the Poincaré disk pretty impressive because it represents an infinite espace within a finite space.

Is there a way to show the infinite Euclidean 2D space (aka an Euclidean plan) in a finite way? Is it feasable or just unthinkable?

$\newcommand{\Reals}{\mathbf{R}}$If $\phi:\Reals^{2} \to D$ is a bijection to a bounded set, then $D$ can be viewed as a model of the Cartesian/Euclidean plane. With no further constraints, however, such a model is unlikely to satisfy: Lines map to (effectively) arbitrary curves, rigid motions look "far from rigid" in $D$, and so forth.
• Stereographic projection from $(0, 0, 1)$ (below, left) $$\phi(u, v) = (x, y, z) = \frac{(2u, 2v, u^{2} + v^{2} - 1)}{u^{2} + v^{2} + 1},\qquad \phi^{-1}(x, y, z) = (u, v) = \frac{(x, y)}{1 - z},$$ maps the plane to the complement of the north pole $N = (0, 0, 1)$ in the unit sphere $S^{2} \subset \Reals^{3}$.
These maps are conformal (angle-preserving), Euclidean lines map to circles through $N$, and Euclidean rigid motions look not unlike hyperbolic rigid motions acting on the Poincaré disk.
• Central/gnomonic projection from $(0, 0, 1)$ (below, right) $$\phi(u, v) = (x, y, z) = \frac{(u, v, -1 + \sqrt{u^{2} + v^{2} + 1})}{\sqrt{u^{2} + v^{2} + 1}},\qquad \phi^{-1}(x, y, z) = (u, v) = \frac{(x, y)}{1 - z}$$ maps the plane to the "southern hemisphere" $x^{2} + y^{2} + (z - 1)^{2} = 1$, $0 \leq z < 1$. (If desired, map the open hemisphere to the open unit disk, either by vertical projection, or by stereographic projection from $(0, 0, 2)$.)