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:
 * $\chi_{A \cup B} = \max \left\{{\chi_A, \chi_B}\right\}$

Proof
Suppose $\chi_{A \cup B} \left({s}\right) = 0$.

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

Hence $\chi_A \left({s}\right) = \chi_B \left({s}\right) = 0$, and by definition of maximum:


 * $\max \left\{{\chi_A \left({s}\right), \chi_B \left({s}\right)}\right\} = 0$

Conversely, suppose:


 * $\max \left\{{\chi_A \left({s}\right), \chi_B \left({s}\right)}\right\} = 0$

Then it follows that $\chi_A \left({s}\right) = \chi_B \left({s}\right) = 0$ because characteristic functions are $0$ or $1$.

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

That is, $\chi_{A \cup B} \left({s}\right) = 0$.

Above considerations give:


 * $\chi_{A \cup B} \left({s}\right) = 0 \iff \max \left\{{\chi_A \left({s}\right), \chi_B \left({s}\right)}\right\} = 0$

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