Let $X$ be a set and $\mathcal{B}$ a subset of the set of maps from $X$ to $\mathbb{Z}_2$.
For any subset $A\subset X$ we define the characteristic mapping of A as the mapping $\chi_A:X\rightarrow \mathbb{Z}_2$ given by $\chi_A(x)=1$ if $x\in A$ and $\chi_A(x)=0$ if $x\notin A$.
Give $\mathcal{B}$ the operations:
$\odot: \mathcal{B} \times \mathcal{B} \rightarrow \mathcal{B}$ given by $(\chi_A \odot \chi_B)(x)=\chi_A(x) \cdot \chi_B(x)$ and
$\oplus : \mathcal{B} \times \mathcal{B} \rightarrow \mathcal{B}$ given by $(\chi_A \oplus \chi_B)(x)=\chi_A(x) + \chi_B(x) – \chi_A(x) \cdot \chi_B(x)$.
Show that $\mathcal{B}$ whith these operations is a ring.
As a part of this exercise, we have to see that $(\mathcal{B},\oplus)$ is a group.
I have considered the map $\chi_\emptyset$ as the neutral element for the addition $\oplus$ since $\chi_\emptyset (x)=0$ $\forall x \in X$.
But I have a problem when I try to find an opposite element because $\chi_A(x) + \chi_B(x) – \chi_A(x) \cdot \chi_B(x)$ has to be $0$ but it only happens if $x\in A^c \cap B^c$. So, what is our candidate for be an opposite of $\chi_A$ and how can I denote it? Is it correct to consider the map $\chi_\emptyset$ as the neutral element?
Best Answer
As you correctly observed, this definition doesn't produce additive inverses, for example $\chi_X$ doesn't have inverse, unless we interpret $\cdot, +, -$ as Boolean operations 'meet', 'join' and $x-y=x\cdot(-y)$ where $-0=1,\ -1=0$.
In terms of your original interpretation, the correct definition for $\oplus$ is any of the following:
In each of the above definitions, for a particular $x\in X$, the cases $(0,0)$ and $(1,1)$ are mapped to $0$ and $(1,0)$ and $(0,1)$ are mapped to $1$.
Using the first alternative, it is immediate that $\mathcal B$ is a ring, as it is ${\cong\Bbb Z_2}^{|X|}$.