Nested Interval Property

Nested closed bounded intervals trap a single real number.

The idea

Nested Interval Property. Let $I_{1} \supseteq I_{2} \supseteq I_{3} \supseteq \cdots$ be closed bounded intervals, each containing the next. Then the intersection $\bigcap_{n} I_{n}$ is nonempty, and if the lengths of the intervals tend to $0$, the intersection is a single point.

The property pins down a number we cannot write directly: construct a shrinking chain of intervals that must contain it, and the property guarantees a point common to them all.

Write $I_{n} = [x_{n}, y_{n}]$. Nesting moves each left endpoint only rightward and each right endpoint only leftward, so $x_{1} \le x_{2} \le x_{3} \le \cdots$, and every $y_{n}$ is an upper bound for all the left endpoints. The left endpoints therefore form an increasing bounded sequence, and completeness of $\mathbb{R}$ supplies $x = \sup_{n} x_{n}$. This $x$ is at least every $x_{n}$ and at most every $y_{n}$, so it lies in every interval. When the lengths tend to $0$, no second common point can exist: two distinct common points would force every interval to be at least as long as the distance between them.

Both hypotheses are necessary. Bounded is what lets completeness apply. Closed is what makes the trapped point a member of each interval: the point a chain squeezes toward is typically an endpoint, and an open interval does not contain its endpoints.

Ways to work on it

• Walkthrough. Why nested closed bounded intervals share a point, and what closedness buys.
• Practice. Identify the point a nested decreasing chain shrinks to.
• Hardest. Construct √2 as the one point shared by a chain of nested intervals.

Not sure where to start? Take the ten-question placement test.