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For every collection of events\({A_i}\left( {i \in I} \right)\), show that

\({\left( {\bigcup\limits_{i \in I} {{A_i}} } \right)^c} = \bigcap\limits_{i \in I} {{A_i}^c} \)and\({\left( {\bigcap\limits_{i \in I} {{A_i}} } \right)^c} = \bigcup\limits_{i \in I} {{A_i}^c} \).

Short Answer

Expert verified

We have to prove that \({\left( {\bigcup\limits_{i \in I} {{A_i}} } \right)^c} = \bigcap\limits_{i \in I} {{A_i}^c} \)and \({\left( {\bigcap\limits_{i \in I} {{A_i}} } \right)^c} = \bigcup\limits_{i \in I} {{A_i}^c} \).

Step by step solution

01

Given information

\({A_i}\)is a collection of events, where\(i \in I\).

02

Prove\({\left( {\bigcup\limits_{i \in I} {{A_i}} } \right)^c} = \bigcap\limits_{i \in I} {{A_i}^c} \)

Let x be an arbitrary event, such that,

\(\begin{aligned}{l}x \in {\left( {\bigcup\limits_{i \in I} {{A_i}} } \right)^c}\\ \Rightarrow x \notin {A_i}\left( {for\;any\;i \in I} \right)\\ \Rightarrow x \in A_i^c\;\left( {\forall i \in I} \right)\\ \Rightarrow x \in A_1^c\;{\rm{and}}\;x \in A_2^c\;{\rm{and}}\;x \in A_3^c\;{\rm{and}}\;{\rm{so}}\;{\rm{on}}\\ \Rightarrow {\rm{x}} \in \bigcap\limits_{i \in I} {A_i^c} \end{aligned}\)

Hence, \({\left( {\bigcup\limits_{i \in I} {{A_i}} } \right)^c} = \bigcap\limits_{i \in I} {{A_i}^c} \).

03

Prove\({\left( {\bigcap\limits_{i \in I} {{A_i}} } \right)^c} = \bigcup\limits_{i \in I} {{A_i}^c} \) 

Let y be an arbitrary event, such that,

\(\begin{aligned}{}y \in {\left( {\bigcap\limits_{i \in I} {{A_i}} } \right)^c}\\ \Rightarrow y \notin \bigcap\limits_{i \in I} {{A_i}} \\ \Rightarrow y \in {A_i}^c\;\left( {{\rm{for some }}i \in I} \right)\\ \Rightarrow y \in {A_1}^c{\rm{ or }}y \in {A_2}^c{\rm{ or }}y \in {A_3}^c{\rm{ and so on}}\\ \Rightarrow y \in \bigcup\limits_{i \in I} {{A_i}^c} \end{aligned}\)

Hence, \({\left( {\bigcap\limits_{i \in I} {{A_i}} } \right)^c} = \bigcup\limits_{i \in I} {{A_i}^c} \).

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