Nested intervals theorem

Remarks

The left endpoints $a_n$ increase and are bounded above by $b_1$. By the completeness axiom, they have a supremum $s$. Each $b_m$ is an upper bound of all $a_n$, so $a_n \le s \le b_m$ for all $n$ and $m$, and $s$ lies in every $I_n$. If two points $s < t$ lie in every $I_n$, then $b_n - a_n \ge t - s > 0$ for all $n$, so the lengths cannot tend to $0$.

Each hypothesis is needed. The open intervals $(0, 1/n)$ are nested and nonempty, but no real number lies in all of them. The closed unbounded sets $[n, \infty)$ are nested, but their intersection is empty. Over $\Q$, the intervals $\{q \in \Q : a_n \le q \le b_n\}$ with $a_n, b_n$ the decimal approximations of $\sqrt 2$ from below and above have empty intersection.

The theorem is the case $X = \R$ of Cantor's intersection theorem.

Going deeper

For a student

The theorem is one form of the completeness of the real numbers, and in an Archimedean ordered field it is equivalent to the least-upper-bound property. It drives bisection proofs: halve an interval, keep a half with a desired property, and repeat. This is how one proves the Bolzano–Weierstrass theorem and the intermediate value theorem, and how Cantor showed that $\R$ is uncountable.