Do two closed subsets of $[0, 1]$ with measure $\frac{1}{2}$ intersect?

Question

Let $A$ and $B$ be two closed subsets of $[0,1]$, each with a length of $1/2$. Is it always true that $A\cap B\neq \emptyset$?

My intuition is yes, because:

Either they intersect in their interior;
Or, they are interior-disjoint (i.e. $\operatorname{int}(A) = [0,1]\setminus B$), but in this case they will intersect at their boundary.

What is a formal proof to this claim?

Also, I will be thankful for references that discuss possible generalizations of this claim to more than two subsets (possibly in $\mathbb{R}^n$).

I looked at your profile, and, maybe, when I’ll have more time, we’ll have a talk. From time to time I deal with partitioning problems. For instance, when I was a PhD student, I used a variation of this idea combined with Helly Theorem to solve Steinhaus cake division game. — Alex Ravsky, May 04 '15 at 02:41
Some my other fair partitioning results are the following. Every open subset of $\Bbb R^m$ can be partitioned onto $n$ homeomorphic parts provided $n\ge 2^{m+1}-1$. No non-empty compact convex subset of $\Bbb R^m$ can be partitioned onto two congruent parts. Also it seems that every perfect $G_\delta$ subset of the real line can be partitioned onto arbitrary finite number of homeomorphic parts. — Alex Ravsky, May 04 '15 at 02:42
@AlexRavsky Indeed, the motivation to this question came from a cake-cutting problem: http://mathoverflow.net/questions/203060/fair-cake-cutting-between-groups I will be happy to discuss it with you if you have time. — Erel Segal-Halevi, May 04 '15 at 08:22

Michael Albanese · Accepted Answer · 2015-05-17T04:09:30.813

Let $A$, $B$ be two closed subsets of $[0, 1]$, both with measure $\frac{1}{2}$, and suppose $A\cap B = \emptyset$. Then

$$m([0, 1]\setminus(A\cup B)) = m([0, 1]) - (m(A) + m(B)) = 1 - \left(\tfrac{1}{2} + \tfrac{1}{2}\right) = 0.$$

But $[0, 1]\setminus(A\cup B)$ is open and the only open set with measure zero is the empty set, so $[0, 1]\setminus(A\cup B) = \emptyset$; i.e. $A\cup B = [0, 1]$.

As $A$ and $B$ are two non-empty disjoint closed sets with union $[0, 1]$, $A^c$ and $B^c$ are two non-empty disjoint open sets with union $[0, 1]$. But this is a contradiction as $[0, 1]$ is connected. Therefore, $A\cap B \neq \emptyset$.

The above argument can easily be adapted to prove the following generalisation: for any two closed subsets $A$, $B$ of $[0, 1]$ with $m(A) + m(B) = 1$, $A\cap B \neq \emptyset$.

score 3 · Answer 2 · edited May 04 '15 at 08:19

3

Also, I will be thankful for references that discuss possible generalizations of this claim to more than two subsets and to $\mathbb{R}^n$.

It seems the following.

Proposition. Let $K\subset\Bbb R^n$ be a convex body (that is, the interior of the set $K$ is non-empty), $K_1,\dots, K_n$ be a finite family of closed subsets of the set $K$ such that $\sum\mu(K_i)\ge\mu(K)$. There there exist different $K_i$ and $K_j$ with non-empty intersection.

Proof. Assume the converse. Since the set $K$ is connected, the set $\Delta=K\setminus\bigcup K_i$ is non-empty. Since $\Delta$ is an open subset of the convex body, $\mu(\Delta)>0$, a contradiction.

A generalization for the countable number of closed subsets fails already for the unit segment.

edited May 04 '15 at 08:19

Erel Segal-Halevi

11,509

answered May 03 '15 at 10:09

Alex Ravsky

106,166

1

Don't you have to assume that the sets $K_1,...,K_n$ are closed? – Erel Segal-Halevi May 03 '15 at 17:30
@ErelSegal-Halevi Yes. I corrected this misprint. – Alex Ravsky May 03 '15 at 17:47

Do two closed subsets of $[0, 1]$ with measure $\frac{1}{2}$ intersect?

2 Answers2