Characteristic Function of Union/Variant 3

Theorem
Let $A, B \subseteq S$.

Let $\chi_{A \cup B}$ be the characteristic function of their union $A \cup B$.

Then:
 * $\chi_{A \cup B} = \max \set {\chi_A, \chi_B}$

where $\max$ denotes the max operation.

Proof
Suppose $\map {\chi_{A \cup B} } s = 0$.

Then $s \notin A \cup B$, so $s \notin A$ and $s \notin B$.

Hence:
 * $\map {\chi_A} s = \map {\chi_B} s = 0$

and by definition of max operation:


 * $\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$

Conversely, suppose:


 * $\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$

Then it follows that:
 * $\map {\chi_A} s = \map {\chi_B} s = 0$

because characteristic functions are $0$ or $1$.

Hence $s \notin A$ and $s \notin B$, so $s \notin A \cup B$.

That is:
 * $\map {\chi_{A \cup B} } s = 0$

Above considerations give:


 * $\map {\chi_{A \cup B} } s = 0 \iff \max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$

and applying Characteristic Function Determined by 0-Fiber, the result follows.