Characteristic Function of Intersection

Theorem
Let $A, B \subseteq S$.

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

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


 * $(1):\quad \chi_A \left({s}\right) \chi_B \left({s}\right) = 1 \iff s \in A \cap B$
 * $(2):\quad \min \left\{{\chi_A \left({s}\right), \chi_B \left({s}\right)}\right\} = 1 \iff s \in A \cap B$

Proof of $(1)$
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.

Proof of $(2)$
By definition of characteristic function, have


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

iff $\chi_A \left({s}\right) = \chi_B \left({s}\right) = 1$ because $\chi_A, \chi_B$ take values in $\left\{{0, 1}\right\}$.

But as before, these equalities are equivalent to $s \in A \cap B$.