Characteristic Function Determined by 0-Fiber

Theorem
Let $A \subseteq S$.

Let $f: S \to \left\{{0, 1}\right\}$ be a mapping.

Denote by $\chi_A$ the characteristic function on $A$.

Then the following are equivalent:


 * $(1):\quad f = \chi_A$
 * $(2):\quad \forall s \in S: f \left({s}\right) = 0 \iff s \notin A$

Using the notion of a fiber, $(2)$ may also be expressed as:


 * $(2'):\quad f^{-1} \left({0}\right) = S \setminus A$

$(1)$ implies $(2)$
Follows directly from the definition of characteristic function.

$(2)$ implies $(1)$
Let $s \in S$.

Suppose that $s \notin A$.

Then by assumption, $f \left({s}\right) = 0$.

Also, by definition of characteristic function, $\chi_A \left({s}\right) = 0$.

Next, suppose that $s \in A$.

Then $f \left({s}\right) \ne 0$ by assumption.

As $f \left({s}\right) \in \left\{{0, 1}\right\}$, it follows that $f \left({s}\right) = 1$.

Again, by definition of characteristic function, also have $\chi_A \left({s}\right) = 1$.

Hence, for all $s \in S$, have $f \left({s}\right) = \chi_A \left({s}\right)$.

By Equality of Mappings, it follows that $f = \chi_A$.