Question

Let S be a subset of a universal set U. The characteristic function of S is the function from U to the set {0,1} such that fs(x) = 1 if x belongs to S and if x does not belong to S and fs(x) = 0. Let A and B be sets. Show that for all $x\in U$

$\begin{array}{r}\left(a\right)f_{A\cap B}\left(x\right)=f_A\left(x\right)\cdot f_B\left(x\right)\\ \left(b\right)f_{A\cup B}\left(x\right)=f_A\left(x\right)+f_B\left(x\right)-f_A\left(x\right)\cdot f_B\left(x\right)\\ \left(c\right)f_{\overline{A}}\left(x\right)=1-f_A\left(x\right)\\ \left(d\right)f_{A\oplus B}\left(x\right)=f_A\left(x\right)+f_B\left(x\right)-2f_A\left(x\right)\cdot f_B\left(x\right)\end{array}$

Short answer

$\begin{array}{r}\left(a\right)f_{A\cap B}\left(x\right)=f_A\left(x\right)\cdot f_B\left(x\right)\\ \left(b\right)f_{A\cup B}\left(x\right)=f_A\left(x\right)+f_B\left(x\right)-f_A\left(x\right)\cdot f_B\left(x\right)\\ \left(c\right)f_{\overline{A}}\left(x\right)=1-f_A\left(x\right)\\ \left(d\right)f_{A\oplus B}\left(x\right)=f_A\left(x\right)+f_B\left(x\right)-2f_A\left(x\right)\cdot f_B\left(x\right)\end{array}$

Step-by-step solution

Step: 1

a)

$\begin{array}{r}{f}_{A\cap B}\left(x\right)=1\\ \leftrightarrow x\in A\cap B\\ \leftrightarrow x\in \text{Aand}x\in B\\ \leftrightarrow f_A\left(x\right)=1\text{and}{f}_B\left(x\right)=1\\ \leftrightarrow f_A\left(x\right)\cdot f_B\left(x\right)=1\end{array}$

Step: 2

b)

$\begin{array}{r}{f}_{A\cup B}\left(x\right)=1\\ \leftrightarrow x\in \text{Aor}x\in B\\ \leftrightarrow f_A\left(x\right)={\mathrm{lorf}}_B\left(x\right)=1\\ \leftrightarrow f_A\left(x\right)+f_B\left(x\right)-f_A\left(x\right)\cdot f_B\left(x\right)=1\end{array}$

Step: 3

c)

$\begin{array}{r}{f}_{\overline{A}}\left(x\right)=1\\ \leftrightarrow x\in \overline{A}\\ \leftrightarrow x\notin A\\ \leftrightarrow f_A\left(x\right)=0\\ \leftrightarrow 1-f_A\left(x\right)=1\end{array}$

d)

$\begin{array}{r}{f}_{A\oplus B}\left(x\right)=1\\ \leftrightarrow x\in A\oplus B\\ \leftrightarrow (x\in \text{Aand}x\notin B)\text{or}(x\notin \text{Aand}x\in B)\\ \leftrightarrow f_A\left(x\right)+f_B\left(x\right)-2f_A\left(x\right)\cdot f_B\left(x\right)=1\end{array}$