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索引集合族

了解集合的索引族。

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索引集合族

了解集合的索引族。

{\cal A} = {A_i \mid i ∈ I}

\begin{align*} {\cal A} &={A_1, A_2, A_3}\A_1 &= {11,13, 15, 17} \ A_2 &= {21, 23, 25, 27}\ A_3 &= {31, 33, 35, 37}\end{align*}

\begin{align*} {\cal A} &={A_a, A_b, A_c,A_d,A_e}\A_a &= {11} \ A_b &= {12, 22}\ A_c &= {13, 23, 33}\ A_d &= {14, 24, 34, 44}\ A_e &= {15,25,35,45,55}\end{align*}

\bigcup_{i \in I} A_i = {x\mid x\in A_i \text{ for some } i\in I}

\bigcap_{i \in I} A_i = {x\mid x\in A_i \text{ for every } i\in I}

\begin{align*}{\cal A} &= {A_i \mid i\in I} \ A_i &= {1,2,3,\ldots,i} \end{align*}

\begin{align*} \bigcup_{i \in I} A_i &= {1,2,3,\ldots } \ \bigcap_{i \in I} A_i &= {1} \end{align*}

\begin{align*}B \cap \left( \bigcup_{i \in I} A_i \right) &= \bigcup_{i \in I} \left(B\cap A_i \right) \ B \cup \left( \bigcap_{i \in I} A_i \right) &= \bigcap_{i \in I} \left( B\cup A_i\right) \end{align*}

\begin{align*}\overline{\left( \bigcup_{i \in I} A_i \right)} &= \bigcap_{i \in I} \overline{A_i} \ \overline{\left( \bigcap_{i \in I} A_i \right)} &= \bigcup_{i \in I} \overline{A_i} \end{align*}

{\cal A} = {A_i\mid i\in I}

A_i = {10\times (i-1)+1, 10\times (i-1)+2, 10\times (i-1)+3, \ldots, 10\times (i-1)+15 }

\begin{align*}A_1 &= {1,2,3,\ldots, 15}\ A_2 &= {11,12,13,\ldots, 25}\ A_3 &= {21,22,23,\ldots, 35}\ \vdots\space\space &= \qquad \vdots \A_9 &= {81,82,83,\ldots, 95}\end{align*}

\begin{equation}\bigcup_{i \in I} A_i = {1,2,3,\ldots,95 }\end{equation}

\begin{equation}\bigcap_{i \in I} A_i = {}\end{equation}

\begin{align*}\mathbb U &= {1,2,3,4,5,\ldots, 100}\ B &= {5,10,15,20,25,\ldots,100}\end{align*}

\begin{align*}B \cap \left( \bigcup_{i \in I} A_i \right) &= {5,10,15,\ldots,100}\cap {1,2,3,\dots,95} = {5,10,15,\dots,95} \ \bigcup_{i \in I} \left(B\cap A_i\right) &={5,10,15}\cup{15,20,25}\cup\cdots\cup{85,90,95} = {5,10,15,\dots,95} \end{align*}

\begin{align*}B \cup \left( \bigcap_{i \in I} A_i \right) &= B\cup \emptyset = B\ \bigcap_{i \in I} \left(B\cup A_i\right)&= B\end{align*}

\begin{align*}\overline{\left( \bigcup_{i \in I} A_i \right)} &= {96,97,98,99,100}\ \bigcap_{i \in I} \overline{A_i}&={16,17,18,\ldots,100}\cap\cdots\cap{1,2,3,\ldots,80,96,97,98,99,100}\&={96,97,98,99,100}\end{align*}

\begin{align*} \overline{\left( \bigcap_{i \in I} A_i \right)} &= \overline{\emptyset} = \mathbb U \ \bigcup_{i \in I} \overline{A_i} &={16,17,18,\ldots,100} \cup \cdots \cup {1,2,3,\ldots,80,96,97,98,99,100}\ &={1,2,3,\ldots,100} \ &= {\mathbb U}\end{align*}

Quiz

单选题

考虑以下集合的索引族: \begin{align*}{\cal A} &= \{A_i \mid i\in I\}. \\ A_i &= \{1,2,3,\ldots,i+1\}. \end{align*} 此处,I={1,2,3,…}I = \{1,2,3,\ldots\}

下列哪个选项等于下面给出的集合? ⋃i∈IAi\bigcup_{i \in I} A_i

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函数

序列和元组