Soient A A A et B B B tels que A ⊂ B A \subset B A ⊂ B . On définit 1 A : B ⟶ 0 , 1 \mathbb{1}_{A} : B \longrightarrow {0,1} 1 A : B ⟶ 0 , 1 telle que
∀ x ∈ B , 1 A ( x ) = { 1 si x ∈ A , 0 si x ∉ A . \forall x \in B,\quad
\mathbb{1}_A(x)=
\begin{cases}
1 & \text{si } x\in A,\\
0 & \text{si } x\notin A.
\end{cases} ∀ x ∈ B , 1 A ( x ) = { 1 0 si x ∈ A , si x ∈ / A .
Soit E E E un ensemble de cardinal fini et soient A 1 , . . . , A n A_{1}, ..., A_{n} A 1 , ... , A n tels que A 1 ∪ . . . ∪ A n ⊂ E A_1 \cup ... \cup A_{n} \subset E A 1 ∪ ... ∪ A n ⊂ E . On prend comme ensemble de définition de 1 A i 1_{A_{i}} 1 A i l’ensemble E E E .
Montrer les propriétés suivantes :
i) ∀ i ∈ ⟦ 1 , n ⟧ \forall i \in \llbracket 1,n\rrbracket ∀ i ∈ [ [ 1 , n ] ] , 1 A i = 1 − 1 A i ‾ 1_{A_{i}} = 1 - 1_{\overline{A_{i}}} 1 A i = 1 − 1 A i
ii) ∀ i , j ∈ ⟦ 1 , n ⟧ \forall i, j \in \llbracket 1,n\rrbracket ∀ i , j ∈ [ [ 1 , n ] ] , 1 A i ∩ A j = 1 A i × 1 A j 1_{A_{i}\cap A_{j}} = 1_{A_{i}} \times 1_{A_{j}} 1 A i ∩ A j = 1 A i × 1 A j
iii) ∀ k ∈ ⟦ 1 , n ⟧ \forall k \in \llbracket 1,n\rrbracket ∀ k ∈ [ [ 1 , n ] ] , 1 A 1 ‾ ∩ . . . ∩ A k ‾ = ∏ i = 1 k ( 1 − 1 A i ) 1_{\overline{A_1}\cap ... \cap \overline{A_k}} = \displaystyle \prod_{i=1}^{k}(1-1_{A_{i}}) 1 A 1 ∩ ... ∩ A k = i = 1 ∏ k ( 1 − 1 A i )
iv) ∀ i ∈ ⟦ 1 , n ⟧ , ∑ x ∈ E 1 A i ( x ) = c a r d ( A i ) \forall i \in \llbracket 1,n\rrbracket, \displaystyle \sum_ {x \in E} 1_{A_{i}}(x) = card (A_i) ∀ i ∈ [ [ 1 , n ] ] , x ∈ E ∑ 1 A i ( x ) = c a r d ( A i )
À l’aide des propriétés de la question 1, démontrer la formule :
∣ ⋃ i = 1 n A i ∣ = ∑ k = 1 n ( − 1 ) k + 1 ∑ I ⊆ ⟦ 1 , n ⟧ , ∣ I ∣ = k ∣ ⋂ i ∈ I A i ∣ \left|\bigcup_{i=1}^{n}A_i\right|=\sum_{k=1}^{n}(-1)^{k+1}\displaystyle\sum_{\substack{I\subseteq\llbracket 1,n\rrbracket,|I|=k}}\left|\bigcap_{i\in I}A_i\right| ∣ ⋃ i = 1 n A i ∣ = ∑ k = 1 n ( − 1 ) k + 1 I ⊆ [ [ 1 , n ] ] , ∣ I ∣ = k ∑ i ∈ I ⋂ A i .