令 $\{A_n \}$ 為 一組 sequence of sets，我們說 $\{A_n \}$ 為 montone non-decreasing 若 $A_1 \subset A_2 \subset ... $，我們用 $A_n \uparrow$ 表示 $\{A_n \}$ 為 monotone non-decreasing sets。
另一方面，我們說 $\{A_n \}$ 為 montone non-increasing 若 $A_1 \supset A_2 \supset ... $。我們用 $A_n \downarrow$ 表示 $\{A_n \}$ 為 monotone non-increasing sets。
FACT: 對 Monotone Sequence of Sets 其極限存在。
現在我們看個結果
Theorem:
令 $\{A_n \}$ 為 monotone sequence of sets，我們有
- 若 $A_n \uparrow $ ( 亦即 $\{A_n\} $ 為 monotone non-decreasing) 則 \[\lim_{n \rightarrow \infty}A_n = \bigcup_{n=1}^\infty A_n\]
- 若 $A_n \downarrow $ (亦即 $\{A_n\} $ 為 monotone non-increasing) 則 \[\lim_{n \rightarrow \infty}A_n = \bigcap_{n=1}^\infty A_n\]
Proof
先證 (1)：令 $\{A_n \}$ 為 monotone sequence of sets 且 $A_n \uparrow $ ，我們要證\[\mathop {\lim }\limits_{n \to \infty } {A_n} = \bigcup\limits_{n = 1}^\infty {{A_n}} \]注意到若 集合數列的極限存在等價為
\[\mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n} = \mathop {\lim }\limits_{n \to \infty } {A_n}
\]現在由 $A_n \uparrow $ 我們知道 $A_j \subset A_{j+1}$ 故
\[\bigcap\limits_{k \ge n}^{} {{A_k}} = {A_n}
\]現在觀察 $\lim \inf$
\[\mathop {\lim \inf }\limits_{n \to \infty } {A_n}: = \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{A_k}} } \right)} = \bigcup\limits_{n = 1}^\infty {{A_n}}
\]接著我們在觀察 $\lim \sup$ 可得
\[\mathop {\lim \sup }\limits_{n \to \infty } {A_n}: = \bigcap\limits_{n = 1}^\infty {\left( {\bigcup\limits_{k \ge n}^{} {{A_k}} } \right)} \subset \bigcup\limits_{k \ge 1}^{} {{A_k}} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n}\]又我們知道
\[\mathop {\lim \inf }\limits_{n \to \infty } {A_n} \subset \mathop {\lim \sup }\limits_{n \to \infty } {A_n}
\]故總合以上可推知
\[\left\{ \begin{array}{l}
\mathop {\lim \sup }\limits_{n \to \infty } {A_n} \subset \mathop {\lim \inf }\limits_{n \to \infty } {A_n}\\
\mathop {\lim \inf }\limits_{n \to \infty } {A_n} \subset \mathop {\lim \sup }\limits_{n \to \infty } {A_n}
\end{array} \right. \Rightarrow \mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n}\]且
\[\mathop {\lim \sup }\limits_{n \to \infty } {A_n} \subset \bigcup\limits_{n = 1}^\infty {{A_n}} \subset \mathop {\lim \sup }\limits_{n \to \infty } {A_n} \Rightarrow \bigcup\limits_{n = 1}^\infty {{A_n}} = \mathop {\lim \sup }\limits_{n \to \infty } {A_n}\]
接著我們證 (2)：令 $\{A_n \}$ 為 monotone sequence of sets 且 $A_n \downarrow $ ，我們要證\[\mathop {\lim }\limits_{n \to \infty } {A_n} = \bigcap\limits_{n = 1}^\infty {{A_n}} \]
同之前證明(1)的手法，若 集合數列的極限存在等價為
\[\mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n} = \mathop {\lim }\limits_{n \to \infty } {A_n}
\] 故現在由定義 $\lim \sup$ 與 $\lim \inf$可知
\[\left\{ \begin{array}{l}
\mathop {\lim \inf }\limits_{n \to \infty } {A_n}: = \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{A_k}} } \right)} \\
\mathop {\lim \sup }\limits_{n \to \infty } {A_n}: = \bigcap\limits_{n = 1}^\infty {\left( {\bigcup\limits_{k \ge n}^{} {{A_k}} } \right)}
\end{array} \right.\]由於 $\{A_n \} \downarrow$，亦即 $A_{n+1} \subset A_n$ 故上述 $\lim \sup$ 與 $\lim \inf$ 有如下關係
\[\begin{array}{l}
\left\{ \begin{array}{l}
\mathop {\lim \inf }\limits_{n \to \infty } {A_n}: = \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{A_k}} } \right)} \\
\mathop {\lim \sup }\limits_{n \to \infty } {A_n}: = \bigcap\limits_{n = 1}^\infty {\left( {\bigcup\limits_{k \ge n}^{} {{A_k}} } \right) = \bigcap\limits_{n = 1}^\infty {{A_n}} }
\end{array} \right.\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} \Rightarrow \left\{ \begin{array}{l}
\mathop {\lim \inf }\limits_{n \to \infty } {A_n} \subset \mathop {\lim \sup }\limits_{n \to \infty } {A_n}\\
\mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \bigcap\limits_{n = 1}^\infty {{A_n}} \subset \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{A_k}} } \right) = \mathop {\lim \inf }\limits_{n \to \infty } {A_n}}
\end{array} \right.\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} \Rightarrow \mathop {\lim }\limits_{n \to \infty } {A_n} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n} = \mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \bigcap\limits_{n = 1}^\infty {{A_n}} \ \ \ \ \ \square
\end{array}\]
\[\mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n} = \mathop {\lim }\limits_{n \to \infty } {A_n}
\]現在由 $A_n \uparrow $ 我們知道 $A_j \subset A_{j+1}$ 故
\[\bigcap\limits_{k \ge n}^{} {{A_k}} = {A_n}
\]現在觀察 $\lim \inf$
\[\mathop {\lim \inf }\limits_{n \to \infty } {A_n}: = \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{A_k}} } \right)} = \bigcup\limits_{n = 1}^\infty {{A_n}}
\]接著我們在觀察 $\lim \sup$ 可得
\[\mathop {\lim \sup }\limits_{n \to \infty } {A_n}: = \bigcap\limits_{n = 1}^\infty {\left( {\bigcup\limits_{k \ge n}^{} {{A_k}} } \right)} \subset \bigcup\limits_{k \ge 1}^{} {{A_k}} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n}\]又我們知道
\[\mathop {\lim \inf }\limits_{n \to \infty } {A_n} \subset \mathop {\lim \sup }\limits_{n \to \infty } {A_n}
\]故總合以上可推知
\[\left\{ \begin{array}{l}
\mathop {\lim \sup }\limits_{n \to \infty } {A_n} \subset \mathop {\lim \inf }\limits_{n \to \infty } {A_n}\\
\mathop {\lim \inf }\limits_{n \to \infty } {A_n} \subset \mathop {\lim \sup }\limits_{n \to \infty } {A_n}
\end{array} \right. \Rightarrow \mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n}\]且
\[\mathop {\lim \sup }\limits_{n \to \infty } {A_n} \subset \bigcup\limits_{n = 1}^\infty {{A_n}} \subset \mathop {\lim \sup }\limits_{n \to \infty } {A_n} \Rightarrow \bigcup\limits_{n = 1}^\infty {{A_n}} = \mathop {\lim \sup }\limits_{n \to \infty } {A_n}\]
接著我們證 (2)：令 $\{A_n \}$ 為 monotone sequence of sets 且 $A_n \downarrow $ ，我們要證\[\mathop {\lim }\limits_{n \to \infty } {A_n} = \bigcap\limits_{n = 1}^\infty {{A_n}} \]
同之前證明(1)的手法，若 集合數列的極限存在等價為
\[\mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n} = \mathop {\lim }\limits_{n \to \infty } {A_n}
\] 故現在由定義 $\lim \sup$ 與 $\lim \inf$可知
\[\left\{ \begin{array}{l}
\mathop {\lim \inf }\limits_{n \to \infty } {A_n}: = \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{A_k}} } \right)} \\
\mathop {\lim \sup }\limits_{n \to \infty } {A_n}: = \bigcap\limits_{n = 1}^\infty {\left( {\bigcup\limits_{k \ge n}^{} {{A_k}} } \right)}
\end{array} \right.\]由於 $\{A_n \} \downarrow$，亦即 $A_{n+1} \subset A_n$ 故上述 $\lim \sup$ 與 $\lim \inf$ 有如下關係
\[\begin{array}{l}
\left\{ \begin{array}{l}
\mathop {\lim \inf }\limits_{n \to \infty } {A_n}: = \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{A_k}} } \right)} \\
\mathop {\lim \sup }\limits_{n \to \infty } {A_n}: = \bigcap\limits_{n = 1}^\infty {\left( {\bigcup\limits_{k \ge n}^{} {{A_k}} } \right) = \bigcap\limits_{n = 1}^\infty {{A_n}} }
\end{array} \right.\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} \Rightarrow \left\{ \begin{array}{l}
\mathop {\lim \inf }\limits_{n \to \infty } {A_n} \subset \mathop {\lim \sup }\limits_{n \to \infty } {A_n}\\
\mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \bigcap\limits_{n = 1}^\infty {{A_n}} \subset \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{A_k}} } \right) = \mathop {\lim \inf }\limits_{n \to \infty } {A_n}}
\end{array} \right.\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} \Rightarrow \mathop {\lim }\limits_{n \to \infty } {A_n} = \mathop {\lim \inf }\limits_{n \to \infty } {A_n} = \mathop {\lim \sup }\limits_{n \to \infty } {A_n} = \bigcap\limits_{n = 1}^\infty {{A_n}} \ \ \ \ \ \square
\end{array}\]
上述定理有個重要的結果值得紀錄。
若現在我們取 $B_n$ 為任意 sequence of sets，則
\[\left\{ \begin{array}{l}
\mathop {\inf }\limits_{k \ge n} {B_n}: = \bigcap\limits_{k \ge n}^{} {{B_k}} \uparrow \\
\mathop {\sup }\limits_{k \ge n} {B_n}: = \bigcup\limits_{k \ge n}^{} {{B_k}} \downarrow
\end{array} \right.\]利用上述定理可得
\[\left\{ \begin{array}{l}
\mathop {\inf }\limits_{k \ge n} {B_n}: = \bigcap\limits_{k \ge n}^{} {{B_k}} \uparrow \begin{array}{*{20}{c}}
{}
\end{array} \Rightarrow \begin{array}{*{20}{c}}
{}
\end{array}\mathop {\lim }\limits_{n \to \infty } \left( {\bigcap\limits_{k \ge n}^{} {{B_k}} } \right) = \bigcup\limits_{n = 1}^\infty {\left( {\bigcap\limits_{k \ge n}^{} {{B_k}} } \right) = \mathop {\lim \inf }\limits_{n \to \infty } {B_n}} \\
\mathop {\sup }\limits_{k \ge n} {B_n}: = \bigcup\limits_{k \ge n}^{} {{B_k}} \downarrow \begin{array}{*{20}{c}}
{}
\end{array} \Rightarrow \begin{array}{*{20}{c}}
{}
\end{array}\mathop {\lim }\limits_{n \to \infty } \left( {\bigcup\limits_{k \ge n}^{} {{B_k}} } \right) = \bigcap\limits_{n = 1}^\infty {\left( {\bigcup\limits_{k \ge n}^{} {{B_k}} } \right) = \mathop {\lim \sup }\limits_{n \to \infty } {B_n}}
\end{array} \right.\]
讀者可嘗試以下幾個例子看看是否能正確使用上述定理
Exercise：
\[\begin{array}{l}
\mathop {\lim }\limits_{n \to \infty } \left[ {0,1 - \frac{1}{n}} \right] = [0,1)\\
\mathop {\lim }\limits_{n \to \infty } \left[ {0,1 - \frac{1}{n}} \right) = [0,1)\\
\mathop {\lim }\limits_{n \to \infty } \left[ {0,1 + \frac{1}{n}} \right] = [0,1]\\
\mathop {\lim }\limits_{n \to \infty } \left[ {0,1 + \frac{1}{n}} \right) = [0,1]
\end{array}\]
