# Prove that subset $S$ of $[0, 1]$ of total length greater than $0.6$ contains two elements such that their difference is excatly $0.1$

Let $S$ be a subset of $[0, 1]$ consisting of a finite number of intervals. How to prove that if the total length of intervals from $S$ is greater than $0.6$ then $S$ contains two numbers such that their difference is exactly $0.1$?

If there are no two numbers in $S$ with difference exactly $0.1$, then the length of the intervals in $S$ has to be $\leq0.1$ (it can only equal $0.1$ if the interval is open) and also the difference between the upper bound of an interval with the lower bound of the consecutive interval to the right should be $\geq 0.1$. But since length of $S$ is $0.6$, there cannot be more than 5 intervals in $S$ with 4 gaps in between them, where the sum of the lengths of the gaps is $0.4$. But then length of $S$ cannot be more than $0.5$. Hence contradiction.
• Why should the distances between consecutive intervals be $\geq 1$? If the size of these intervals is small, having a distance shorter than $1$ is no problem, no? – Peter May 12 '17 at 12:36
• @Peter If we assume that there are no two numbers with difference exactly $0.1$ then the distance between consecutive intervals should be $\ge 1$ because otherwise some element $x$ from one of the intervals could be present as $x + 0.1$ in the next one. – user128409235 May 12 '17 at 13:15
• What if we have the two intervals $[0, 0.01]$ and $[0.02,0.03]$? Their distance is less than $0.1$, but $\forall x \in [0, 0.01] \forall y \in [0.02,0.03]: |y-x| \leq 0.03 < 0.1$. I do not see how the distance of less than $0.1$ is a problem here. – Peter May 12 '17 at 13:22