极限和下极限"/>
集合序列的上极限和下极限
设{ A n , n ≥ 1 {A_n,n \ge 1} An,n≥1 },求 ⋂ n = 1 ∞ ⋃ k = n ∞ A k \bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty A_k n=1⋂∞k=n⋃∞Ak 及 ⋃ n = 1 ∞ ⋂ k = n ∞ A k \bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k n=1⋃∞k=n⋂∞Ak,
其中 A 1 = { 1 , a } A_1=\left\{1,a \right\} A1={1,a}, A 2 = { 0 , b } A_2=\left\{0,b \right\} A2={0,b}, A 3 = { 1 , b } A_3=\left\{1,b \right\} A3={1,b}, A 4 = { 0 , b } A_4=\left\{0,b \right\} A4={0,b}, A 5 = { 1 , b } , ⋯ A_5=\left\{1,b \right\} ,\cdots A5={1,b},⋯。
解:解法如下:
一、 ⋂ n = 1 ∞ ⋃ k = n ∞ A k \bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty A_k n=1⋂∞k=n⋃∞Ak 的解法是先并再交。
当 n = 1 n=1 n=1时,记 P 1 = ⋃ k = n = 1 ∞ A k = { 0 , 1 , a , b } P_1=\bigcup\limits_{k=n=1}^\infty A_k=\left\{ 0,1,a,b\right\} P1=k=n=1⋃∞Ak={0,1,a,b},
当 n = 2 n=2 n=2时,记 P 2 = ⋃ k = n = 2 ∞ A k = { 0 , 1 , b } P_2=\bigcup\limits_{k=n=2}^\infty A_k=\left\{ 0,1,b\right\} P2=k=n=2⋃∞Ak={0,1,b},
⋯ ⋯ \cdots\cdots ⋯⋯
所以, ⋂ n = 1 ∞ ⋃ k = n ∞ A k = P 1 ⋂ P 2 ⋂ P 3 ⋂ ⋯ = { 0 , 1 , b } \bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty A_k=P_1\bigcap P_2\bigcap P_3\bigcap\cdots=\left\{ 0,1,b\right\} n=1⋂∞k=n⋃∞Ak=P1⋂P2⋂P3⋂⋯={0,1,b}
若记 lim n → ∞ s u p A n = ⋂ n = 1 ∞ ⋃ k = n ∞ A k \lim\limits_{n\rightarrow\infty}supA_n=\bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty A_k n→∞limsupAn=n=1⋂∞k=n⋃∞Ak
则, lim n → ∞ s u p A n = { w ∣ w 属 于 无 穷 多 个 A n } \lim\limits_{n\rightarrow\infty}supA_n=\left\{w|w属于无穷多个A_n\right\} n→∞limsupAn={w∣w属于无穷多个An}
二、 ⋃ n = 1 ∞ ⋂ k = n ∞ A k \bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k n=1⋃∞k=n⋂∞Ak 的解法是先交再并。
当 n = 1 n=1 n=1时,记 Q 1 = ⋂ k = n = 1 ∞ A k = ∅ Q_1=\bigcap\limits_{k=n=1}^\infty A_k=\varnothing Q1=k=n=1⋂∞Ak=∅,
当 n = 2 n=2 n=2时,记 Q 2 = ⋂ k = n = 2 ∞ A k = { b } Q_2=\bigcap\limits_{k=n=2}^\infty A_k=\left\{ b\right\} Q2=k=n=2⋂∞Ak={b},
⋯ ⋯ \cdots\cdots ⋯⋯
所以, ⋃ n = 1 ∞ ⋂ k = n ∞ A k = Q 1 ⋂ Q 2 ⋂ Q 3 ⋂ ⋯ = { b } \bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k=Q_1\bigcap Q_2\bigcap Q_3\bigcap\cdots=\left\{b\right\} n=1⋃∞k=n⋂∞Ak=Q1⋂Q2⋂Q3⋂⋯={b}
若记 lim n → ∞ i n f A n = ⋃ n = 1 ∞ ⋂ k = n ∞ A k \lim\limits_{n\rightarrow\infty}infA_n=\bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k n→∞liminfAn=n=1⋃∞k=n⋂∞Ak
则, lim n → ∞ i n f A n = { w ∣ w 至 多 不 属 于 有 限 多 个 A n } \lim\limits_{n\rightarrow\infty}infA_n=\left\{w|w至多不属于有限多个A_n\right\} n→∞liminfAn={w∣w至多不属于有限多个An}
本文的LaTeX代码如下:
设{ ${A_n,n \ge 1}$ },求$\bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty A_k$ 及$\bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k$,
其中$A_1=\left\{1,a \right\}$,$A_2=\left\{0,b \right\}$,$A_3=\left\{1,b \right\}$,$A_4=\left\{0,b \right\}$,$A_5=\left\{1,b \right\} ,\cdots$。
解:解法如下:
一、$\bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty A_k$ 的解法是先并再交。
当$n=1$时,记$P_1=\bigcup\limits_{k=n=1}^\infty A_k=\left\{ 0,1,a,b\right\}$,
当$n=2$时,记$P_2=\bigcup\limits_{k=n=2}^\infty A_k=\left\{ 0,1,b\right\}$,
$\cdots\cdots$
所以,$\bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty A_k=P_1\bigcap P_2\bigcap P_3\bigcap\cdots=\left\{ 0,1,b\right\}$
若记$\lim\limits_{n\rightarrow\infty}supA_n=\bigcap\limits_{n=1}^\infty \bigcup\limits_{k=n}^\infty A_k$
则,$\lim\limits_{n\rightarrow\infty}supA_n=\left\{w|w属于无穷多个A_n\right\}$二、$\bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k$ 的解法是先交再并。
当$n=1$时,记$Q_1=\bigcap\limits_{k=n=1}^\infty A_k=\varnothing$,
当$n=2$时,记$Q_2=\bigcap\limits_{k=n=2}^\infty A_k=\left\{ b\right\}$,
$\cdots\cdots$
所以,$\bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k=Q_1\bigcap Q_2\bigcap Q_3\bigcap\cdots=\left\{b\right\}$
若记$\lim\limits_{n\rightarrow\infty}infA_n=\bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k$
则,$\lim\limits_{n\rightarrow\infty}infA_n=\left\{w|w至多不属于有限多个A_n\right\}$
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集合序列的上极限和下极限
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