Let $A$ be a set and $E\subset A$. The function $\chi_E:A\to \{0,1\}$ is defined by:
$\chi_E(x) = \begin{cases} 1 & \text{for } x \in E \cr 0 & \text{for } x \notin E\end{cases}$
Prove that for $F \subset A$,
- $\chi_{E \cap F} =\chi_E\cdot \chi_F$
- $\chi_{E \cup F} =\chi_E+ \chi_F-\chi_{E \cap F}$
- Find a similar expression for $\chi_{E \cup F \cup G}$
For the first:
$E\cap F =\{x : x \in E \wedge x \in F\}$
So
$\chi_{E\cap F}(x) = \begin{cases} 1 & \text{for } x \in E \wedge x \in F\cr 0 & \text{for } x \notin E\vee x \notin F\end{cases}$
$\chi_{E\cap F}(x) = \begin{cases} 1 & \text{for } x \in E \wedge x \in F\cr 0 & \text{for } x \notin E\wedge x \in F \cr 0& \text{for } x \in E\wedge x \notin F \cr 0& \text{for } x \notin E\wedge x \notin F \end{cases}=\chi_E\cdot \chi_F(x)$
Should the same approach (if correct) be applied to 2.?