# Characteristic Function of Intersection/Variant 2

## Theorem

Let $A, B \subseteq S$.

Let $\chi_{A \cap B}$ be the characteristic function of their intersection $A \cap B$.

Then:

$\chi_{A \cap B} = \min \left\{{\chi_A, \chi_B}\right\}$

## Proof

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

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

By definition of characteristic function, we have:

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

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

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.

$\blacksquare$