, 2012

We shall construct two sets of intervals $A$ and $B$ such that the intervals in $A$ will be pairwise disjoint, the intervals in $B$ will be pairwise disjoint, and the union of all the intervals in $A$ and $B$ will be equal to $[0, 1]$. This construction will prove the claim of the problem because then either the sum of the lengths of all the intervals in $A$ or the sum of the lengths of all the intervals in $B$ will be greater than $\frac{1}{2}$. If not, the sum of the lengths of all the intervals in both sets would be smaller than $1$, which would contradict the fact that the union of all the intervals in both sets is equal to $[0, 1]$. The elements of one of the sets $A$ and $B$ will solve the problem.

Let $A$ and $B$ be empty sets. Among the intervals $I_1 = [a_1, b_1], I_2 = [a_2, b_2], \dots, I_{10} = [a_{10}, b_{10}]$ choose the one that contains $0$ and has the greatest right endpoint, $b_{A_1}$. Add this interval $I_{A_1}$ in the set $A$. Among the other intervals (different than $I_{A_1}$) choose the one that has a non-empty intersection with $I_{A_1}$ and the greatest right endpoint, $b_{B_1}$. Add this interval $I_{B_1}$ in the set $B$. Among the rest (intervals different than $I_{A_1}$ and $I_{B_1}$) choose the one that has a non-empty intersection with $I_{B_1}$ and the greatest right endpoint, $b_{A_2}$. Add this interval $I_{A_2}$ in the set $A$. Repeat the process as long as the right endpoints of the intervals are all different than $1$. Suppose, at the end of the process, the sets of intervals $A = \{I_{A_1}, I_{A_2}, \dots, I_{A_n}\}$ and $B = \{I_{B_1}, I_{B_2}, \dots, I_{B_m}\}$, where $n - 1 \le m \le n$ and $n + m \le 10$, are equal.

Let us prove that such sets have the desired properties. From the above construction we see that the right endpoints of the successively constructed intervals increase, i.e., for $i \le n - 1$ we have $b_{A_i} < b_{B_i}$ and $b_{B_i} < b_{A_{i+1}}$. Clearly, the union of all the intervals in both sets is equal to $[0, 1]$. Let us also prove that the elements in any of the sets are pairwise disjoint. Let $I_{A_j}$ and $I_{A_{j+1}}$ be two successively constructed intervals in $A$, and suppose that their intersection is empty. According to the definition, $I_{B_j}$ has the greatest right endpoint among all the intervals (different than $I_{B_i}$ and $I_{A_i}$ for $i < j$) that intersect $I_{A_j}$. The right endpoint of the interval $I_{B_j}$ thus needs to be greater than the right endpoint of the interval $I_{A_{j+1}}$. It needs to hold $b_{B_i} > b_{A_{i+1}}$, which is a contradiction. The elements of $A$ must be pairwise disjoint. We may show similarly that the elements of $B$ are pairwise disjoint. This proves the claim.