Solution
= Solution
The earliest end is $1$, so strategy 2 first meets at \b[$t_1=1$], removing $I_1$ and $I_2$. Among the remaining intervals the earliest end is $4$, so it next meets at \b[$t_2=4$], removing $I_3$ and $I_4$. This is the two-meeting optimum proved in part (b). The closed endpoints are essential: attendance at exactly an interval's endpoint is permitted.