Exercise: nested sequences of sets
The following statement is false
If \( F_1 \supseteq F_2 \supseteq F_3 \supseteq F_4 \supseteq ... \) is a nested sequence of nonempty closed sets, then the intersection \( \bigcap_{n=1}^{\infty} F_n \neq \emptyset \).
The statement is false. Consider the nested sequence of closed sets: \( \mathbb{R}, \mathbb{R} \setminus [0, 1], \mathbb{R} \setminus [0, 2], \mathbb{R} \setminus [0, 3], ... \). The infinite intersection is empty.
If this seems to contradict the nested interval property, recall its details to see why there is no contradiction:
Nested Interval Property
For each \( n \in \mathbb{N} \), assume we are given a closed interval
Assume also that each \( I_n \) contains \( I_{n+1} \). Then, the resulting nested sequence of closed intervals:
This property can be generalized by replacing "closed interval" with "compact set". But it can't be replaced with "closed set", as the above counter-example shows. The issue is that \( \mathbb{R} \) is closed, and other half-open sets like \( [0, \mathbb{R}) \) are also closed.