Characteristic Function of Disjoint Union

Theorem
Let $X$ be a set.

Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint subsets of $X$.

Let:


 * $\ds D = \bigcup_{n \mathop = 1}^\infty D_n$

Then:


 * $\ds \chi_D = \sum_{n \mathop = 1}^\infty \chi_{D_n}$

where:


 * $\chi_D$ is the characteristic function of $D$
 * $\chi_{D_n}$ is the characteristic function of $D_n$.

Proof
We aim to show that:


 * $\ds \sum_{n \mathop = 1}^\infty \map {\chi_{D_n} } x = \begin{cases}1 & x \in D \\ 0 & x \in X \setminus D\end{cases}$

at which point we will have the demand from the definition of a characteristic function.

Let $x \in D$.

Then:


 * $\ds x \in \bigcup_{n \mathop = 1}^\infty D_n$

From the definition of set union, we have:


 * $x \in D_n$ for some $n \in \N$.

Since the sets $\sequence {D_n}_{n \mathop \in \N}$ are pairwise disjoint, we have:


 * $x \in D_n$ for precisely one $n \in \N$.

That is, form the definition of a characteristic function:


 * $\map {\chi_{D_n} } x = 1$ for precisely one $n \in \N$.

So that:


 * $\ds \sum_{n \mathop = 1}^\infty \map {\chi_{D_n} } x = 1$

as desired.

Let $x \in X \setminus D$.

Then:


 * $\ds x \in X \setminus \bigcup_{n \mathop = 1}^\infty D_n$

so from the set difference:


 * $x \in X$ and $\ds x \not \in \bigcup_{n \mathop = 1}^\infty D_n$.

So, from the definition of set union:


 * $x \not \in D_n$ for all $n \in \N$.

That is, form the definition of a characteristic function:


 * $\map {\chi_{D_n} } x = 0$ for all $n \in \N$.

So that:


 * $\ds \sum_{n \mathop = 1}^\infty \map {\chi_{D_n} } x = 0$

as desired.