Figure 8's In A Plane

Clearly the number is at least countably infinite, just by tiling translated copies of a single figure 8. Conversely, suppose we have a set of figure 8s drawn in the plane with no crossings. Each such figure 8 consists of two loops with disjoint interiors: we associate to each figure an arbitrary fixed pair of points, one point in the (open) interior of each loop, in some arbitrary order. By the density of rationals, we may furthermore choose the points in such a way that they have rational Cartesian coordinates.

Now suppose two figures A and B share the same first point x: then A and B each have exactly one loop enclosing x. Since these loops don't cross, one must be contained in the interior of the other; WLOG say B's x-loop is contained within A's x-loop. Then figure B must be entirely contained within A's x-loop to avoid crossing, so that the second point of B is inside A's x-loop; the second point of A is outside of A's x-loop, so it cannot be equal to that of B.

We conclude that no two figure 8s can share both points, so they are less numerous than the set of point pairs of rational coordinates, which is $\mathbb Q^4$ and thus countable.

It seems to me that the problem should state the figure 8s are all the same size? If they are of different sizes an infinite number of figure 8s can fit in any of the holes in the above diagram, and in an infinite number of arrangements in each hole, and they are clearly uncountable then.

Consider any tiling of 8s, since $\mathbb{Q}^2$ is dense in $\mathbb{R}^2$ we can pick a $q$ and $r$ in each of the circles that make up the 8. Two different 8s will be equal if and only if they are made up from the same circles, so different circles will have different a $(q,r)$ assigned. Consider the set $A=\{(q,r)\in \mathbb{Q}^2\times\mathbb{Q}^2, \; (q,r) \;\text{is assigned to an 8 of the tiling}\}\subset \mathbb{Q}^2\times\mathbb{Q}^2$ There´s a bijection between the set $A$ and the set of 8´s in the tiling. Since $\mathbb{Q}^2\times\mathbb{Q}^2$ is countable infinite we have that $A$ is at most countable infinite. Finding a countably infinite tiling is rather easy so the proof concludes. (Someone else already gave this answer but this one has more symbols :) )

Uncountably infinite. You can make smaller 8's inside the both loops of each 8 and then even smaller 8's inside the loops of the smaller 8's and even even smaller ones inside the loops inside the loops inside the loops of the biggest 8 and so on

Given a non intersecting collection of figure 8s, associate each figure 8 with two rational points (points with rational x and y coordinates), one from each loop. Then this mapping is injective. Since QxQ is countable, so is the collection of figure 8s.

I visualized the answer to this question by imagining that the infinite plane somehow has X coordinates that run from one to infinity , and Y coordinates that run from one to infinity .

I then remembered how we demonstrated that this set of rational numbers (integer fractions) is countably infinite. We imagined each fraction being a point in an infinite "quadrant one" of our coordinate system, the X coordinates going from one to infinity, the Y coordinates going from one to infinity. When we moved through every possible rational number in a diagonal pattern , starting with one over one, continuing with the next diagonal one over two, two over one, continuing with the next diagonal one over three, two over two, three over one, etc., we are able to create a one to one match between the set of rational numbers and the set of positive integers.

It is then easy to imagine that each of these "figure rights" is one of these fractions, the numerator inside the left circle, the denominator inside the right circle . If the rational numbers are countably infinite, then this infinite pattern of figure eights is countably infinite .

The only flaw in this imagined proof is the fact that we do not account for the fact that the infinite plane is theoretically four times larger than the plane of only the positive X and Y coordinates as described above. However this is easy to circumvent. We can imagine that the numbers in our infinite plane start at one at the axis , with all the numbers starting at the axis and going to the right , and all the even numbers starting just to the left of the axis and moving to the left. In this way, we can demonstrate that the infinite plane is equal in size weather we start at the origin and only move toward infinity into directions , or if we go to infinity in all four directions.

This may not be as elegant away to prove the point as some of the other answers , but it works for me.

Ome can map each integer coordinates to a 8. These coordinates can be mapped to integers. Start in a corner, move 1 right (x+1,y) then along the diagonal (x-1,y+1). When you hit X=0 move one up (x,y+1) then along the diagonal (x+1,y-1). When you hit Y=0 repeat the process.

You will end up with a trail covering the whole plane! Each coordinate would have also a number.

Here are the first mappings:

$\begin{aligned} (0,0) & \mapsto & 0 \\ (1,0) & \mapsto & 1 \\ (0,1) & \mapsto & 2 \\ (0,2) & \mapsto & 3 \\ (1,1) & \mapsto & 4 \\ (2,0) & \mapsto & 5 \\ (3,0) & \mapsto & 6 \\ (2,1) & \mapsto & 7 \\ (1,2) & \mapsto & 8 \\ (0,3) & \mapsto & 9 \end{aligned}$

Etc !