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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...

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Intuitively the problem is the 8's can't pack against each other like line segments. There must be a small disk between them even if one is inside the other. In that disk must be an element of $\mathbb{Q \times Q}$. – Ross Millikan Nov 2 '11 at 0:21
The first thing you need is a rigorous definition of what qualifies as an "8-like symbol". – Henning Makholm Nov 2 '11 at 0:21
@Ross, some subtlety will be needed, because there are uncountably many pairwise disjoint "0-like" curves in the plane (such as for example all concentric circles). – Henning Makholm Nov 2 '11 at 0:24
@HenningMakholm: I agree. That is exactly why the gaps between are important and we need to define 8-like curves in a way that requires them. – Ross Millikan Nov 2 '11 at 0:29
@RossMillikan: The 8s can pack against each other (and even fill the plane) - e.g. 8s consisting of two rectangles. It's the "second o" that creates the "gaps" which prevent the concentric circle argument from working with 8s. – David Bevan Nov 2 '11 at 9:15
up vote 6 down vote accepted

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.

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I am confused, but I don't know why :( First question: what does Q^2 x Q^2 denote? Q x Q denotes a "2-D pair" in the rationals right, so what does this mean? – MathMathCookie Nov 2 '11 at 3:39
It means ordered pairs of ordered pairs, so a typical element would look like 4((p,q),(r,s)) – Alex Youcis Nov 2 '11 at 3:41
Oh, I see. I still don't understand this part: "they must intersect, which is impossible" Could you write out more? Thank you so much. – MathMathCookie Nov 3 '11 at 1:41
Think about it, if you had two figure eights, $E$ and $E'$ which share a point in each of their loops. How could they do this and not intersect? I can't think of a simple analytical proof, but it's pretty clear by inspection. – Alex Youcis Nov 3 '11 at 2:47
If figure eight $A$ is inside a loop of figure eight $B$, $A$'s index points cannot include a point from the other loop of $B$. If $A$ and $B$ are external to each other, then they can have no index points in common. – David Bevan Nov 3 '11 at 8:26

Hint: Suppose we have a lower bound on the size of the 8s. Show that no two of them can be very close, and conclude that there are only countably many of them.

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It does work for other figures :) – Yuval Filmus Nov 2 '11 at 10:11

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