Bisection and the nested intervals theorem

The bisection principle

Cut an interval in two, keep one half, start again: the length is divided by $2$ at each step. Starting from $[a, b]$, the interval obtained after $n$ cuts has length

$$b_n - a_n = \frac{b - a}{2^n} \xrightarrow[n \to +\infty]{} 0$$

On $[0, 1]$, the decrease is spectacular:

The marked bar is the first whose length drops below $10^{-3}$: ten cuts are enough, since $2^{10} = 1024 > 1000$.

The nested intervals theorem

Let $\left([a_n, b_n]\right)_{n \in \mathbb{N}}$ be a sequence of nested closed intervals, that is, satisfying $[a_{n+1}, b_{n+1}] \subset [a_n, b_n]$ for all $n$, and such that $b_n - a_n \to 0$. Then there exists a unique real number $c$ belonging to all these intervals, and $a_n \to c$, $b_n \to c$.

Proof. Nesting translates into $a_n \leq a_{n+1} \leq b_{n+1} \leq b_n$: the sequence $(a_n)$ is increasing, the sequence $(b_n)$ is decreasing, and $a_n \leq b_n \leq b_0$ for all $n$.

The sequence $(a_n)$ is therefore increasing and bounded above by $b_0$: it converges to a real number $c$. Likewise $(b_n)$ is decreasing and bounded below by $a_0$, so it converges to a real number $c'$. Finally

$$c' - c = \lim_{n \to +\infty} \left(b_n - a_n\right) = 0$$

hence $c = c'$. Passing to the limit in $a_n \leq c \leq b_n$ — legitimate since $(a_n)$ increases to $c$ and $(b_n)$ decreases to $c$ — shows that $c$ belongs to each interval. And if $d$ also belonged to them, we would have $\lvert d - c \rvert \leq b_n - a_n \to 0$, so $d = c$.

In other words, the two sequences $(a_n)$ and $(b_n)$ are adjacent, and their common limit is the only point of the intersection:

$$\bigcap_{n \in \mathbb{N}} \left[a_n, b_n\right] = \{c\}$$

Both hypotheses are essential

Sequence of intervals	Nested	Length to $0$	Intersection
$\left[0, 1 + \tfrac{1}{n}\right]$	yes	no	$[0, 1]$, not a point
$\left(0, \tfrac{1}{n}\right)$, not closed	yes	yes	empty
$\left[\tfrac{-1}{n}, \tfrac{1}{n}\right]$	yes	yes	$\{0\}$

The second row deserves a closer look: these are open intervals, and the intersection is empty. The theorem holds for segments, that is, for closed and bounded intervals — the same compactness requirement as in Heine's theorem.

Classic pitfall

Writing directly "the nested intervals converge to a point" without checking that the length tends to $0$. Without this hypothesis, the intersection is still nonempty, but it is an interval, not a singleton. Bisection, on the other hand, provides the $\dfrac{b-a}{2^n}$ decrease for free: this is precisely why it is the construction tool for the Bolzano-Weierstrass theorem.