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 $⋂_{n = 1}^{\infty} F_{n} \neq \emptyset$ .

The statement is false. Consider the nested sequence of closed sets: $R, R ∖ [0, 1], R ∖ [0, 2], R ∖ [0, 3], . . .$ . 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 $n \in N$ , assume we are given a closed interval

I_{n} = [a_{n}, b_{n}] = {x \in R : a_{n} \leq x \leq 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

⋂_{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 $R$ is closed, and other half-open sets like $[0, R)$ are also closed.

Source

Abbott, p18, p88 (1st edition)

27.03.2021