定理：單調有界數列必有極限
證明：僅證明單調遞增有界數列必有極限，單調遞減數列類似。
設{\(a_{n}\)}為單調遞增數列，且有上界。
把該數列各項用十進制無限小數形式表示如下：
\(\quad\quad\quad\quad\quad\quad\)\(a_{1}=A_{1}.b_{11}b_{12}b_{13}\)......
\(\quad\quad\quad\quad\quad\quad\)\(a_{2}=A_{2}.b_{21}b_{22}b_{23}\)......
\(\quad\quad\quad\quad\quad\quad\)\(a_{3}=A_{3}.b_{31}b_{32}b_{33}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)..........
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)..........
其中\(A_{1},A_{2},A_{3}...是整數部分，b_{ij}是小數部分的數字\)，為0-9中的數字。\(\\\)
\(因為數列{a_{n}}是有界數列，所以整數部分不會無限增大，又因為{a_{n}}是有界數列，\)\(\\\)
\(所以整數數列A_{n}在達到最大值之后將固定不變，記該最大整數為A，並設A在第N_{0}行\)\(\\\)
\(上出現。考察第二列，即各行數字的小數部分的第一列b_{11}, b_{21}, b_{31}...\)\(\\\)
\(考察第A_{N_{0}}行和以下各行。\)\(\\\)
\(設x_{1}是出現在該列的最大數字，設其第一次出現在第N_{1}行上，其中N_{1}\geqslant N_{0}\).\(\\\)
\(則x_{1}一旦出現，不會改變，因為{a_{n}}是遞增數列，且該行數字的整數部分，已經是最大值A\)。\(\\\)
\(繼續考察各行數字的小數部分的第二列b_{21}, b_{31}, b_{41}...\)
\(同理可知，該列將出現一個最大固定值x_{2}，假設出現在第N_{2}行，這里N_{2}\geqslant N_{1}\)\(\\\)
\(\quad\quad\quad\quad\quad\quad\)\(a_{1}=A_{1}.b_{11}b_{12}b_{13}\)......
\(\quad\quad\quad\quad\quad\quad\)\(a_{2}=A_{2}.b_{21}b_{22}b_{23}\)......
\(\quad\quad\quad\quad\quad\quad\)\(a_{3}=A_{3}.b_{31}b_{32}b_{33}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(第N_{0}行:\quad\quad\)\(a_{N_{0}}=A.b_{N_{0}1}b_{N_{0}2}b_{N_{0}3}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(第N_{1}行:\quad\quad\)\(a_{N_{1}}=A.x_{1}b_{N_{1}2}b_{N_{1}3}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(第N_{2}行:\quad\quad\)\(a_{N_{2}}=A.x_{1}x_{2}b_{N_{2}3}\)......
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
\(\quad\quad\quad\quad\quad\quad\)\(\quad\quad\quad\)...............
重復以上過程，得到\(\quad\quad\)a=\(A.x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}\)......
\(下面證明，上式的a就是數列{a_{n}}的極限\)
\(\forall\varepsilon\)>0，設m\(\in\)\(N^{+}\),使得\(10^{-m}<\varepsilon\)
則有\(a_{N_{m}}\)=A.\(x_{1}x_{2}x_{3}\)......\(x_{m}b_{N_{m},m+1}\)......
注：\(a_{N_{m}}\)是指整數部分為A，小數部分的前m-1位，均為各自列的最大值\(x_{1}\)-\(x_{m-1}\)，\(\\\)
\(\quad\quad\)\(b_{n,m}\)這一列最大數字\(x_{m}\)第一次出現的那個數字\(\\\)
那么，\(\forall\)n>\(N_{m}\), \(A_{n}\)=A.\(x_{1}x_{2}x_{3}\)......\(x_{m}b_{n,m+1}\)......
a與\(a_{n}\) 的整數部分和前m位小數完全一樣
|a-\(a_{n}\)|=|A.\(x_{1}x_{2}x_{3}\)......\(x_{m}x_{m+1}\)
\(\quad\quad\)-A.\(x_{1}x_{2}x_{3}\)......\(x_{m}b_{n,m+1}\)|<\(\frac{1}{10^{m}}\)<\(\varepsilon\)
所以 \(\lim_{n\to\infty}a_{n}\) = \(A.x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}\)......
即 \(\lim_{n\to\infty}a_{n} = a\)
證畢.