7
$\begingroup$

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 $\mathbb{R}^n$).

$\endgroup$
31
$\begingroup$

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 $m(A) + m(B) = 1$, $A\cap B \neq \emptyset$.

$\endgroup$
  • 3
    $\begingroup$ I like your proof! +1 $\endgroup$ – Nicolas May 3 '15 at 9:30
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Don't you have to assume that the sets $K_1,...,K_n$ are closed? $\endgroup$ – Erel Segal-Halevi May 3 '15 at 17:30
  • $\begingroup$ @ErelSegal-Halevi Yes. I corrected this misprint. $\endgroup$ – Alex Ravsky May 3 '15 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.