How do I show: $$χ_{A∩B}(s) = \min(χ_A(s), χ_B(s)) $$ I know $χ_{A∩B}(s) = χ_A(s)χ_B(s)$
For $χ_{A∩B}(s)$ it is either $1$ or $0$ so if either $χ_A(s)$ or $χ_B(s)$ is $0$ then the statement holds true, as $χ_{A∩B}(s) = 0$. Similarly if both $χ_A(s)$ and $χ_B(s)$ are $1$ then $χ_{A∩B}(s) = 1$ and the statement still holds true.
However I do not know how to show: $$χ_{A∪B}(s) = \max(χ_A(s), χ_B(s)) $$
Where I know $χ_{A∪B}(s) = χ_A(s) + χ_B(s) − χ_{A∩B}(s)$
If $x \in A\cup B$, then $\chi_{A \cup B}(x) = 1$, and $x \in A$ or (including) $x \in B$, and $\max\{1,0\} = \max\{0,1\} = \max\{1,1\}= 1$, so we're good. On the other hand, if $x \not\in A\cup B$, then $\chi_{A \cup B}(x) = 0$, $x$ is in neither $A$ nor $B$, and $\max\{0,0\} = 0$, so we're also good.