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Math and science::Analysis

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

\[ I_n = [a_n, b_n] = \{x \in \mathbb{R} : a_n \le x \le b_n \} \]

Assume also that each \( I_n \) contains \( I_{n+1} \). Then, the resulting nested sequence of closed intervals:

\[ I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 ... \]
has a nonempty intersection; that is \( \bigcap_{n=1}^{\infty} I_n \neq \emptyset \).

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.

Source

Abbott, p18, p88 (1st edition)