Characteristic Function of Intersection/Variant 1
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Theorem
Let $A, B \subseteq S$.
Let $\chi_{A \mathop \cap B}$ be the characteristic function of their intersection $A \cap B$.
Then:
- $\chi_{A \mathop \cap B} = \chi_A \chi_B$
Proof
By Characteristic Function Determined by 1-Fiber, it suffices to show that:
- $\map {\chi_A} s \map {\chi_B} s = 1 \iff s \in A \cap B$
Now, both $\chi_A$ and $\chi_B$ are characteristic functions.
It follows that, for any $s \in S$:
- $\map {\chi_A} s \map {\chi_B} s = 1 \iff \map {\chi_A} s = \map {\chi_B} s = 1$
By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$.
That is, $s \in A \cap B$, by definition of set intersection.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $7 \ \text{(a)}$
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $5 \ \text{(i)}$