Nested intervals

In mathematics, a sequence of nested intervals is understood as a collection of sets of real numbers

I_n

4 members of a sequence of nested intervals

such that each set I_n is an interval of the real line, for n = 1, 2, 3, ..., and that further

I_{n + 1} is a subset of I_n

for all n. In other words, the intervals diminish, with the left-hand end moving only towards the right, and the right-hand end only to the left.

The main question to be posed is the nature of the intersection of all the I_n. Without any further information, all that can be said is that the intersection J of all the I_n, i.e. the set of all points common to the intervals, is either the empty set, a point, or some interval.

The possibility of an empty intersection can be illustrated by the intersection when I_n is the open interval

(0, 2⁻ⁿ).

Here the intersection is empty, because no number x is both greater than 0 and less than every fraction 2⁻ⁿ.

The situation is different for closed intervals. The nested intervals theorem states that if each I_n is a closed and bounded interval, say

I_n = [a_n, b_n]

with

a_n ≤ b_n

then under the assumption of nesting, the intersection of the I_n is not empty. It may be a singleton set {c}, or another closed interval [a, b]. More explicitly, the requirement of nesting means that

a_n ≤ a_{n + 1}

and

b_n ≥ b_{n + 1}.

Moreover, if the length of the intervals converges to 0, then the intersection of the I_n is a singleton.

One can consider the complement of each interval, written as ${\displaystyle (-\infty ,a_{n})\cup (b_{n},\infty )}$. By De Morgan's laws, the complement of the intersection is a union of two disjoint open sets. By the connectedness of the real line there must be something between them. This shows that the intersection of (even an uncountable number of) nested, closed, and bounded intervals is nonempty.