Statement

Nested interval theorem: Let

In=[an,bn]⊆RI_n=[a_n,b_n]\subseteq\mathbb R

be a sequence of nonempty bounded closed such that In+1⊆InI_{n+1}\subseteq I_n for all nn. Then

⋂n∈NIn≠∅.\bigcap_{n\in\mathbb N}I_n\neq\varnothing.

More precisely,

⋂n∈NIn=[sup⁡nan,inf⁡nbn],\bigcap_{n\in\mathbb N}I_n =\left[\sup_n a_n,\inf_n b_n\right],

using and . If additionally bn−an→0b_n-a_n\to 0, then the intersection consists of a single point.

Why boundedness matters

Closedness and nesting alone do not suffice for unbounded intervals: the sequence In=[n,∞)I_n=[n,\infty) has empty intersection. Boundedness makes every InI_n a compact subset of R\mathbb R, so the nonempty intersection conclusion follows from compactness and the finite intersection property. The singleton conclusion when bn−an→0b_n-a_n\to0 is the interval case of the .