Characteristic Function of Union

Theorem
Let $A, B \subseteq S$.

Then:


 * $(1):\quad \chi_{A \cup B} = \min \left\{{\chi_A + \chi_B, 1}\right\}$
 * $(2):\quad \chi_{A \cup B} = \chi_A + \chi_B - \chi_{A \cap B}$
 * $(3):\quad \chi_{A \cup B} = \max \left\{{\chi_A, \chi_B}\right\}$

where $\chi$ denotes characteristic function.

Proof of $(1)$
By Characteristic Function Determined by 1-Fiber, it suffices to show:


 * $\min \left\{{\chi_A \left({s}\right) + \chi_B \left({s}\right), 1}\right\} = 1 \iff s \in A \cup B$

By the nature of the minimum, this amounts to showing that:


 * $\chi_A \left({s}\right) + \chi_B \left({s}\right) \ge 1 \iff s \in A \cup B$

As $\chi_A, \chi_B$ are characteristic functions, the left-hand side amounts to:


 * $s \in A \lor s \in B$

which is precisely the definition of $s \in A \cup B$.

Proof of $(2)$
From Subset of Union:
 * $A, B \subseteq A \cup B$

From Intersection is Subset of Union:
 * $A \cap B \subseteq A \cup B$

Thus from Characteristic Function of Subset:


 * $\chi_{A \cup B} \left({s}\right) = 0 \implies \chi_A \left({s}\right) = \chi_B \left({s}\right) = \chi_{A \cap B} \left({s}\right) = 0$

Now suppose that $\chi_A \left({s}\right) + \chi_B \left({s}\right) - \chi_{A \cap B} \left({s}\right) = 0$.

That is, $\chi_A \left({s}\right) + \chi_B \left({s}\right) = \chi_{A \cap B} \left({s}\right)$.

Suppose the latter equals $1$, i.e. $s \in A \cap B$.

Then both $s \in A$ and $s \in B$, so by definition of characteristic function:


 * $\chi_A \left({s}\right) + \chi_B \left({s}\right) = 1 + 1 = 2$

Since $2 \ne 1$, it follows that $\chi_{A \cap B} \left({s}\right) \ne 1$, i.e. it equals $0$.

Thence, it follows that $\chi_A \left({s}\right) + \chi_B \left({s}\right) = 0$.

This only happens when $\chi_A \left({s}\right) = \chi_B \left({s}\right) = 0$.

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

It finally follows that $\chi_{A \cup B} \left({s}\right) = 0$.

It is now established that:


 * $\chi_{A \cup B} \left({s}\right) = 0 \iff \chi_A \left({s}\right) + \chi_B \left({s}\right) - \chi_{A \cap B} \left({s}\right) = 0$

and from Characteristic Function Determined by 0-Fiber, $(2)$ follows.

Proof of $(3)$
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, $(3)$ follows.