# What is the set with characteristic function $\chi_A(x) + \chi_B(x)-\chi_A(x)\chi_B(x)$?

Suppose that $A$ and $B$ are subsets of $X$

Find the subset $C$ whose characteristic function is given by:

$\chi_C(x)=\chi_A(x) + \chi_B(x)-\chi_A(x)\chi_B(x)$

The answer given is $C=(A\cup B) - (A \cap B)$

My answer is $A \cup B$ because:

\begin{align} x\in A \cup B & \Rightarrow & x \in A \text{ or } x\in B \\ & \Rightarrow & \chi_A(x)=1 \text { or } \chi_B(x)=1 \\ & \Rightarrow & \chi_C(x)=1\\ \end{align}

\begin{align} x\notin A \cup B & \Rightarrow & x \in (A^c \cap B^c) \\ & \Rightarrow & x \notin A \text { and } x \notin B \\ & \Rightarrow & \chi_A(x)=0 \text { and } \chi_B(x)=0 \\ & \Rightarrow & \chi_C(x)=0 \end{align}

What is wrong with my answer and how to I get the correct answer?

• You are correct, the "answer given" is false. – Ewan Delanoy Aug 14 '14 at 9:00
• If $x\in A\cup B$, then $x\in A\cap B^c$ or $x\in A^c\cap B$ or $x\in A\cap B$. – Cm7F7Bb Aug 14 '14 at 9:04

Your reasoning is perfectly valid. The given answer is wrong and your answer is correct. One easily checks that if $x\in A\cap B$ then $\chi_C(x)=1+1-1=1$, so $A\cap B\subseteq C$ contrary to what the given answer suggests. Indeed, the characteristic function of $A\triangle B$ is
$$\begin{array}{ll} \chi_{A\triangle B}(x) & =\chi_A(x)\left(1-\chi_B(x)\right)+\chi_B(x)\left(1-\chi_A(x)\right) \\ & =\chi_A(x)+\chi_B(x)-\color{Red}{2}\chi_A(x)\chi_B(x). \end{array}$$
(The symmetric difference is defined by $A\triangle B=(A\cup B)-(A\cap B)$.)
For subsets A and B of a non empty set X: $$χ_ϕ=0\,\,\,\,\ \text{and}\,\,\,\,\ χ_X=1$$ $$A⊆B ⇔ χ_A≤χ_B$$ $$χ_{A^c }=1-χ_A$$ $$χ_{A∩B}=χ_A χ_B=\min(χ_A,χ_B )$$ $$χ_{A∪B}=χ_A+χ_B-χ_{A∩B}=\max(χ_A,χ_B )$$ $$χ_{A-B}=χ_A-χ_{A∩B}.$$ $$χ_{A△B}=χ_A+χ_B-2χ_{A∩B}$$ $$χ_{A×B}=χ_A χ_B$$