Prove that any family of disjoint 8-signs on the plane is countable I try to answer  the following question from Basic Set Theory by Shen and Vereshchagin :
1.
(a).Prove that any family of disjoint 8-signs on the plane is countable.(By an 8-sign we mean a union of two tangent circles of any size; the interior part of the circles is not included).
(b)Prove a similar statement for letters T or E on the plane(but not M or O!).
 A: A good place to start is to ask yourself what the difference between the shapes of 8, T, and E on the one hand and M and O on the other. One thing that stands out is that each of 8, T, and E has a point that does not have a neighborhood that looks like a line segment: 8 has the point of tangency of the two circles, T has the intersection of the two straight lines, and E has the intersection of the vertical line and the middle horizontal line. (The other two horizontal lines of the E) can be straightened out so that it looks like $\vdash$. The M, on the other hand, can be straightened out into a line segment, and each point of the O looks like a point on C if you look at it up close, so it also looks like it’s on a line segment.
I would start by trying to show that a collection of pairwise disjoint T’s in the plane must be countable: they’re simpler than the 8, but every 8 contains a (slightly deformed) T, so if you can do it for T’s, you’ve basically done it for 8’s.
Suppose that $\mathscr{T}$ is an uncountable collection of pairwise disjoint T’s in the plane. For each $T\in\mathscr{T}$ there are rational numbers $p_T,q_T$, and $r_T$ with $r_T>0$ such that if $C_T$ is the circle of radius $r_T$ centred at $\langle p_T,q_T\rangle$, then the intersection point of $T$ is inside $C_T$, and the three endpoints of $T$ are outside $C_T$. There are only countably many triples of rational numbers, so there must be some $p,q,r\in\Bbb Q$ such that 
$$\mathscr{T}_0=\{T\in\mathscr{T}:p_T=p,q_T=q,\text{ and }r_T=r\}$$ is uncountable.
Now trim the ends of each $T\in\mathscr{T}_0$ so that the three segments terminate exactly on the circle $C_T$; you now have a circle $C_T$ divided into three regions by the $T$ inside it. (If you bend the top bar of $T$ at the intersection point, this circle-with-T will look something like the Mercedes-Benz symbol.) Each of the regions must contain a point whose coordinates are both rational; call these points $x_T,y_T$, and $z_T$. There must be some $x,y,z\in\Bbb Q^2$ such that
$$\mathscr{T}_1=\{T\in\mathscr{T}_0:x_T=x,y_T=y,\text{ and }z_T=z\}$$
is uncountable. Can you show that if $T,T'\in\mathscr{T}_1$, then $T\cap T'\ne\varnothing$ and so get a contradiction? A sketch may help; note that $C_T=C_{T'}$.
I’ve added a suitable sketch below; $C=C_T=C_{T'}$, $P$ is the intersection point of $T'$, and $x,y$, and $z$ are as above.

A: Hint for a:  there is an element of $\mathbb {Q \times Q}$ in the interior of each 8.  For b the idea is the same, prove you can't pack them arbitrarily closely, while for M and O you can fill a line segment with each point corresponding to a different letter.
A: The following argument deals with the original question (a) about figure eights.
Let $(E_\iota)_{\iota\in I}$ be the given family of eights. For each $\iota\in I$ denote  by $\bar E_\iota$ the "filled in" eight $E_\iota$, and by $r_\iota\geq s_\iota>0$ its two  radii. If $E_\iota$ is encircled by some other $E_{\iota'}$ then necessarily $r_\iota+s_\iota<r_{\iota'}$.
For any $k\in{\mathbb Z}$ and any $\ell\in{\mathbb N}_{\geq1}$ denote by $I_{k\ell}$ the set of $\iota\in I$ such that
$$2^k \leq s_\iota <2^{k+1}\ , \qquad \ell\, 2^k\leq r_\iota< (\ell+1) 2^k\ ;$$
the idea being that two eights from the same $I_{k\ell}$ should have about the same radii. This circumstance can now be exploited in the following way:
Fix a $k$ and an $\ell$, and consider two eights $E_\iota$, $E_{\iota'}$ with $\iota$, $\iota'\in I_{k\ell}$.
Then $r_\iota +s_\iota\geq (\ell+1)\,2^k> r_{\iota'}$. It follows that neither of these two eights can encircle the other; whence $\bar E_\iota$ and $\bar E_{\iota'}$ are disjoint sets of positive area. As a consequence there can be only countable many eights $E_\iota$ with $\iota\in I_{k\ell}$. That is to say, $I_{k\ell}$ is countable.
As ${\mathbb Z}$ and ${\mathbb N}_{\geq1}$ are countable the full index set $I=\bigcup_{k\in{\mathbb Z},\ \ell\in{\mathbb N}_{\geq1}} I_{k\ell}$ is countable as well.
A: I asked the same question today on MO: https://mathoverflow.net/questions/371669/fundamental-group-and-countability and someone there gave the link to this thread.
I had an idea involving $p$-adic integers:
First let's consider one of the two loops defing the 8-sign, say the upper one, and consider its area equals 1. We'll put smallest 8-signs, all included in this same upper loop and with disjoint interiors of respective area at least $A$ with $0<A<1$. We can make it dimensionless saying it is the ratio of the minimal area of the interior of some 8-sign to the area of the "mother" loop. One can pack a finite number $N(A)$ of 8-signs of disjoint interiors of area at least $A$ into it. Let $p(A)$ the smallest prime greater or equal to $N(A)$. The iteration of this process inside each upper loop at each level shows the number thereof is upper bounded by $\sum_{i=1}^{\infty}p(A)^i$ which is countably infinite: it boils down to counting the number of $1$ in $1+(1+1+...+1)+(1+1+...+1)+...$, the $k$-th pair of parentheses containing a number of $1$ equal to $p^{k}$. This is an upper bound for the number of upper loops: to get an upper bound for the number of $8$ signs at each step, just replace $N(A)$ by $2N(A)$ and $p(A)$ by the smallest prime greater or equal to $2N(A)$, that we'll denote by $q(A)$. The same reasoning applies mutatis mutandis, and we end up with countably many 8-signs in any region of the plane where you can draw an 8-sign.
