Is there a way to show the infinite Euclidean plane in a finite way? I'm a huge fan of hyperbolic geometry. I find the Poincaré disk pretty impressive because it represents an infinite space within a finite space.
Is there a way to show the infinite Euclidean 2D space (aka an Euclidean plane) in a finite way? Is it feasible or just unthinkable?
 A: $\newcommand{\Reals}{\mathbf{R}}$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.
Here are a couple of "relatively nice" representations of the Euclidean plane in a bounded spatial or planar region:


*

*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 $(0, 0, 2)$.)
Every Euclidean line determines a unique plane through the center of the sphere, and therefore maps to half a great circle. Again, Euclidean rigid motions look not unlike hyperbolic rigid motions acting on a disk model of the hyperbolic plane.

A: [Updated about a week later]
I was interested in this same issue. Inspired by the Poincaré disk, I wanted to draw a disk on which an infinite plane was mapped. The plane is “decorated” with a procedural texture, so this is like the computer graphics operation of texture mapping.
Rendering the disk uses the inverse of the plane-to-disk map: given a position (of a pixel) on the disk, the inverse transform finds the corresponding point on the infinite “source” plane. If that “destination” point $d$ is on the disk, $\left\lVert d \right\rVert <1$, then the corresponding “source” point $s$ on the input texture is defined by this mapping, parameterized by scale $a$ and exponent $b$.
$$s= \frac{d} {a(1-\left\lVert d \right\rVert^b)}$$
Shown below is a portion of source texture plane. This texture is defined procedurally across all of ℝ², or at least the part spanned by two floating point numbers:

This shows the mapping for $a=1$ and $b=1$:

The Poincaré disk model itself, via Douglas Dunham, with $a=1/2$ and $b=2$:

Here for $b=5$ we see the map for $a=1/3$, $a=1$, and $a=3$:



