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Call a tack the one point union of three open intervals. Can you fit an uncountable number of them on the plane? Or is only a countable number?

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Why three? With two open intervals you get a figure eight. Is it known whether or not the plane can contain an uncountable number of figure eights? –  Harald Hanche-Olsen Jun 6 '10 at 15:25
    
I think that nonlinearity meant a one-point union of three closed intervals. –  Greg Kuperberg Jun 6 '10 at 15:31
    
All the comments are correct that I meant union, thanks. –  nonlinearality Jun 6 '10 at 17:03

2 Answers 2

up vote 26 down vote accepted

First of all, the one-point compactification of three open intervals is not a "tack", it's a three-leaf clover. I think that you mean a one-point union of three closed intervals; of course it doesn't matter if the other three endpoints are there or not. This topological type can be called a "Y" or a "T" or a "simple triod". R.L. Moore published a solution to your question in 1928. The answer is no. It was generalized in 1944 by his student Gail Young: You can only have countably many $(n-1)$-dimensional tacks in $\mathbb{R}^n$ for any $n \ge 2$. For her theorem, the name "tack" makes rather more sense, but she calls it a "$T_n$-set".

Actually Moore's theorem applies to a more general kind of triod, in which three tips of the "Y" are connected to the center by "irreducible continua", rather than necessarily intervals.


I don't know whether I might be spoiling a good question, but here in any case is a solution to the original question (see as both Moore and Young did something more general that takes more discussion). Following domotorp's hint, there is a principle of accumulation onto a countable set of outcomes pigeonhole principle for uncountable sets. If $f:A \to B$ is a function from an uncountable set $A$ to a countable set $B$, then there is an uncountable inverse image $A' = f^{-1}(b)$. If you want to show that $A$ does not exist, then you might as well replace it with $A'$. Unlike the finite pigeonhole principle, which becomes more limited with each such replacement, $A'$ has the same cardinality as $A$, so you haven't lost anything. You are even free to apply the uncountable pigeonhole principle again.

Suppose that you have uncountably many simple triods in the plane. Given a simple triod, we can choose a circle $C$ with rational radius and rational center with the branch point of the triod on the inside and the three tips on the outside. Since there are only countably many such circles, there are uncountably many triods with the same circle $C$. We can trim the segments of each such triod so that they stop when they first touch $C$, to make a pie with three slices (a Mercedes-Benz symbol). Then, given such a triod, we can pick a rational point in each of three slices of the pie. Since there are only countably many such triples of points, there must be uncountably many triods with the same three points $p$, $q$, and $r$. In particular there are two such triods, and a suitable version of the Jordan curve theorem implies that they intersect.

The argument can be simplified to just pick a rational triangle that functions as the circle, and whose corners function as the three separated points. But I think that there is something to learn from the variations together, namely that the infinite pigeonhole principle gives you a lot of control. For instance, with hardly any creativity, you can assume that the triods are all large.

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The "principle of accumulation onto a coutable set of outcomes" is also commonly known as the pigeon-hole principle. –  Joel David Hamkins Jun 6 '10 at 18:12
    
Sure, that's a good name; mine is a little silly. –  Greg Kuperberg Jun 6 '10 at 19:12
    
I learned this problem and its variations, such as "Is there a disjoint union of uncountably many Möbius bands in $\mathbb{R}^3$?" from Max Forrester's undergraduate talk "Möbius bands in space". There is a common general principle behind them, but I won't spoil a good problem of finding it. –  Victor Protsak Jun 6 '10 at 23:20

This is a well-known puzzle/problem, the trick is to make an injective mapping from any set of disjoint tacks to triples of $\mathbb Q^2$.

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