Proof. Let \(\, A_n = \big [a_n\, ,\, b_n\big ]\), for each \(n\in \mathbb {N}\) then \(\, A_{n + 1} \subset A_n \, \implies \, a_n \leq a_{n + 1} \leq b_{n + 1} \leq b_n\, \,\forall n\) and
\(\displaystyle {\lim \limits _{n\rightarrow \infty }\big (}\) length of \(A_n\big ) = \displaystyle {\lim \limits _{n\rightarrow \infty }\big (b_n - a_n\big ) = 0}\). Then \(\displaystyle {\lim _{n\rightarrow \infty } b_n = \lim _{n \rightarrow \infty } a_n = L}\,\) where
\(\, L =\sup \big \{ a_n : \, n \in \mathbb {N}\big \} = \inf \big \{ b_n: \, n \in \mathbb {N}\big \}\,\) thus, \(\, a_n\leq L \leq b_n \, \forall n \in \mathbb {N} \, \implies \, L \in \displaystyle {\bigcap ^{\infty }_{n = 1} \, A_n}\).
If \(L^* \neq L\), then for \(\varepsilon = \begin {vmatrix} L - L^*\\ \end {vmatrix} > 0\) since \(\big (b_n - a_n\big ) \longrightarrow 0\) as \(n\longrightarrow \infty ,\, \exists m \in \mathbb {N} \ni \begin {vmatrix} b_n - a_n\\ \end {vmatrix}< \begin {vmatrix} L - L^*\\ \end {vmatrix}\hspace {0.2cm}\forall n\geq m\). It implies that \(L^*\not \in A_n \hspace {0.2cm} \forall n \geq m\), for \(L \in A_n \hspace {0.2cm} \forall n \in \mathbb {N}\). Hence, \(\displaystyle {\bigcap ^{\infty }_{n = 1}\, A_n}\,\) consists of exactly on point.

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