Characteristic Function of Union/Variant 2
Theorem
Let $A, B \subseteq S$.
Let $\chi_{A \mathop \cup B}$ be the characteristic function of their union $A \cup B$.
Then:
- $\chi_{A \mathop \cup B} = \chi_A + \chi_B - \chi_{A \mathop \cap B}$
Proof
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:
- $\map {\chi_{A \mathop \cup B} } s = 0 \implies \map {\chi_A} s = \map {\chi_B} s = \map {\chi_{A \mathop \cap B} } s = 0$
Now suppose that $\map {\chi_A} s + \map {\chi_B} s - \map {\chi_{A \mathop \cap B} } s = 0$.
That is:
- $\map {\chi_A} s + \map {\chi_B} s = \map {\chi_{A \mathop \cap B} } s$
Suppose the latter equals $1$, that is $s \in A \cap B$.
Then both $s \in A$ and $s \in B$, so by definition of characteristic function:
- $\map {\chi_A} s + \map {\chi_B} s = 1 + 1 = 2$
Since $2 \ne 1$, it follows that $\map {\chi_{A \mathop \cap B} } s \ne 1$, that is it equals $0$.
Thence, it follows that:
- $\map {\chi_A} s + \map {\chi_B} s = 0$.
This only happens when $\map {\chi_A} s = \map {\chi_B} s = 0$.
Thus, $s \notin A$ and $s \notin B$, so $s \notin A \cup B$.
It finally follows that $\map {\chi_{A \mathop \cup B} } s = 0$.
It is now established that:
- $\map {\chi_{A \mathop \cup B} } s = 0 \iff \map {\chi_A} s + \map {\chi_B} s - \map {\chi_{A \mathop \cap B} } s = 0$
and from Characteristic Function Determined by 0-Fiber, the result follows.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $7 \ \text{(b)}$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $5 \ \text{(iv)}$