Nested intervals property for reals
Nested interval property
A sequence of nested closed intervals of reals has a nonempty intersection.
In other words, if one considers a sequence of nested closed intervals \( (I_n)_{n=0}^{\infty} \) such that for each \( n \in \mathbb{N} \), \( I_n = [a_n, b_n] \) for some \( a_n, b_n \in \mathbb{R} \) and \( I_{n+1} \subseteq I_n \), then It holds that \( \cap_{n=1}^{\infty} I_n \neq \emptyset \).
Proof on the reverse.
Proof
The proof uses the property that sets of reals have a supremum; this is used to find an \( x \in \mathbb{R} \) such that \( x \in \cap_{n=0}^{\infty} I_n \).
Proof. Let \( A = \{a_n : n \in \mathbb{N} \} \) be the set of left-endpoints for the intervals. Note that for every \( i \in \mathbb{N} \), \( b_i \) is an upper bound for \( A \). Let \( x = \sup A \).
- \( \forall i \in \mathbb{N}, x \le b_i \), as \( x \) is the least upper bound for \( A \), and every \( b_i \) is an upper bound.
- \( \forall i \in \mathbb{N}, x \ge a_i \), as \( x \) is an upper bound for \( A \) and all \( a_i \in A \).
So, we have \( \forall i \in \mathbb{N}, a_n \le x \le b_n \), which means that \( \forall i \in \mathbb{N}, x \in I_n \), and furthermore, \( x \in \cap_{n=0}^{\infty} I_n \) meaning that this intersection is not empty.
Visualization of the interval endpoints
