定理：单调有界数列必有极限
证明：仅证明单调递增有界数列必有极限，单调递减数列类似。
设{(a_{n})}为单调递增数列，且有上界。
把该数列各项用十进制无限小数形式表示如下：
(quadquadquadquadquadquad)(a_{1}=A_{1}.b_{11}b_{12}b_{13})......
(quadquadquadquadquadquad)(a_{2}=A_{2}.b_{21}b_{22}b_{23})......
(quadquadquadquadquadquad)(a_{3}=A_{3}.b_{31}b_{32}b_{33})......
(quadquadquadquadquadquad)(quadquadquad)..........
(quadquadquadquadquadquad)(quadquadquad)..........
其中(A_{1},A_{2},A_{3}...是整数部分，b_{ij}是小数部分的数字)，为0-9中的数字。(\)
(因为数列{a_{n}}是有界数列，所以整数部分不会无限增大，又因为{a_{n}}是有界数列，)(\)
(所以整数数列A_{n}在达到最大值之后将固定不变，记该最大整数为A，并设A在第N_{0}行)(\)
(上出现。考察第二列，即各行数字的小数部分的第一列b_{11}, b_{21}, b_{31}...)(\)
(考察第A_{N_{0}}行和以下各行。)(\)
(设x_{1}是出现在该列的最大数字，设其第一次出现在第N_{1}行上，其中N_{1}geqslant N_{0}).(\)
(则x_{1}一旦出现，不会改变，因为{a_{n}}是递增数列，且该行数字的整数部分，已经是最大值A)。(\)
(继续考察各行数字的小数部分的第二列b_{21}, b_{31}, b_{41}...)
(同理可知，该列将出现一个最大固定值x_{2}，假设出现在第N_{2}行，这里N_{2}geqslant N_{1})(\)
(quadquadquadquadquadquad)(a_{1}=A_{1}.b_{11}b_{12}b_{13})......
(quadquadquadquadquadquad)(a_{2}=A_{2}.b_{21}b_{22}b_{23})......
(quadquadquadquadquadquad)(a_{3}=A_{3}.b_{31}b_{32}b_{33})......
(quadquadquadquadquadquad)(quadquadquad)...............
(quadquadquadquadquadquad)(quadquadquad)...............
(第N_{0}行:quadquad)(a_{N_{0}}=A.b_{N_{0}1}b_{N_{0}2}b_{N_{0}3})......
(quadquadquadquadquadquad)(quadquadquad)...............
(quadquadquadquadquadquad)(quadquadquad)...............
(第N_{1}行:quadquad)(a_{N_{1}}=A.x_{1}b_{N_{1}2}b_{N_{1}3})......
(quadquadquadquadquadquad)(quadquadquad)...............
(quadquadquadquadquadquad)(quadquadquad)...............
(第N_{2}行:quadquad)(a_{N_{2}}=A.x_{1}x_{2}b_{N_{2}3})......
(quadquadquadquadquadquad)(quadquadquad)...............
(quadquadquadquadquadquad)(quadquadquad)...............
重复以上过程，得到(quadquad)a=(A.x_{1}x_{2}x_{3}x_{4}x_{5}x_{6})......
(下面证明，上式的a就是数列{a_{n}}的极限)
(forallvarepsilon)>0，设m(in)(N^{+}),使得(10^{-m}<varepsilon)
则有(a_{N_{m}})=A.(x_{1}x_{2}x_{3})......(x_{m}b_{N_{m},m+1})......
注：(a_{N_{m}})是指整数部分为A，小数部分的前m-1位，均为各自列的最大值(x_{1})-(x_{m-1})，(\)
(quadquad)(b_{n,m})这一列最大数字(x_{m})第一次出现的那个数字(\)
那么，(forall)n>(N_{m}), (A_{n})=A.(x_{1}x_{2}x_{3})......(x_{m}b_{n,m+1})......
a与(a_{n}) 的整数部分和前m位小数完全一样
|a-(a_{n})|=|A.(x_{1}x_{2}x_{3})......(x_{m}x_{m+1})
(quadquad)-A.(x_{1}x_{2}x_{3})......(x_{m}b_{n,m+1})|<(frac{1}{10^{m}})<(varepsilon)
所以 (lim_{n oinfty}a_{n}) = (A.x_{1}x_{2}x_{3}x_{4}x_{5}x_{6})......
即 (lim_{n oinfty}a_{n} = a)
证毕.