Countable or uncountable set 8 signs Let S be a set of pairwise disjoint 8-like symbols on the plane. (The 8s may be inside each other as well) Prove that S is at most countable.
Now I know you can "map" a set of disjoint intervals in R to a countable set (e.g. Q :rational numbers) and solve similar problems like this, but the fact that the 8s can go inside each other is hindering my progress with my conventional approach...
 A: Let $\mathcal{E}$ denote the set of all your figure eights. Then, define a map $f:\mathcal{E}\to\mathbb{Q}^2\times\mathbb{Q}^2$ by taking $E\in\mathcal{E}$ to a chosen pair of rational ordered pairs, one sitting inside each loop. Show that if two such figure eights were to have the same chosen ordered pair, they must interesect, which is impossible. Thus, $f$ is an injection and so $\mathcal{E}$ is countable.
A: Let us associate each 8 with its center. If two 8's have exactly the same size and are too close together, then they will intersect. The same holds even if they only have almost the same size. That is, there are some $\epsilon,\delta$ such that two nonintersecting 8's of size between $r$ and $(1+\epsilon)r$ cannot be $\delta r$-close (that is, their centers need to be at distance at least $\delta r$.
Let us denote by $S_r$ the set of 8's of size between $r$ and $(1+\epsilon)r$. The circles at radius $\delta r/3$ around the centers are disjoint, and so if we choose for each 8 in $S_r$ a rational point inside the circle at radius $\delta r/3$ around its center, then these rational points will be distinct. This shows that $S_r$ is countable.
The set of all 8's is the union of countably many sets of the form $S_r$ (for example, we can take $r = (1+\epsilon)^z$ for all integers $z$). Since the countable union of countable sets is countable, we conclude that the number of 8's is countable.
