Characteristic Function of Intersection

Theorem
Let $A, B \subseteq S$.

Then $\chi_{A \cap B} = \chi_A \chi_B$, where $\chi$ denotes characteristic function.

Proof
By Characteristic Function Determined by 1-Fiber, it suffices to show that:


 * $\chi_A \left({s}\right) \chi_B \left({s}\right) = 1 \iff s \in A \cap B$

Now, both $\chi_A$ and $\chi_B$ are characteristic functions.

It follows that, for any $s \in S$:


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

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.