# 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!).

• Does this mean (1) Choose any family or disjoint 8-signs on the plane and prove that it's countable; or (2) Prove that for any family of disjoint 8-signs on the plane, no matter which one it is, it must be countable. Presumably the latter is meant, but standard English usage bears either interpretation. I'd disambiguate it simply by saying "every" instead of "any". Aug 22 '12 at 18:38

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.

• So how do we see that characters like M can be inserted uncountably many times? May 23 '21 at 19:28
• @MateInTwo: Nest them. May 23 '21 at 20:04

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.

• There is a rational point in the interior of each O too, but it doesn't stop you embedding uncountably many into the plane... Aug 22 '12 at 15:46
• @mt_: yes, but the same one is inside many O's. For any pair of 8's, there is one inside one and outside the other. Aug 22 '12 at 15:51
• Did You mean that I can build a bijective function between any $(a,b)\in \mathbb{Q}X\mathbb{Q}$ $to 8-sign$? but I do not understand the meaning of 8-sign and why is conected to a circle Aug 22 '12 at 15:54
• @Hernan: The problem with 8's (and T's and E's) is you can't pack them next to each other on a line segment, like you can concentric 0's. You don't have a bijection from any $(a,b)$, but for each pair of 8's you have an $(a,b)$. You can make all of these distinct, and there are only countably many available (you won't use them all) Aug 22 '12 at 15:58

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.

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 $$0. We can make it dimensionless saying it is the ratio of the minimal area of the interior of some 8-sign to the area of the "mother" loop. One can pack a finite number $$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.