Let A be a set. Define the characteristic function \(\chi_A(x)\) of A. Then, given two sets A and B, express the characteristic function of \(A \cup B\), \(A \cap B\), and \(A^c\)... Let A be a set. Define the characteristic function \(\chi_A(x)\) of A. Then, given two sets A and B, express the characteristic function of \(A \cup B\), \(A \cap B\), and \(A^c\) in terms of \(\chi_A(x)\) and \(\chi_B(x)\). Where complements are defined with respect to a universal set, U. Read more

Steps to Solve

1. Characteristic function of the union $A \cup B$

The characteristic function of $A \cup B$, denoted as $\chi_{A \cup B}(x)$, is 1 if $x$ belongs to either $A$ or $B$ (or both), and 0 if $x$ belongs to neither $A$ nor $B$. We can express this using the individual characteristic functions $\chi_A(x)$ and $\chi_B(x)$. If either $\chi_A(x) = 1$ or $\chi_B(x) = 1$, then $\chi_{A \cup B}(x) = 1$. If both $\chi_A(x) = 0$ and $\chi_B(x) = 0$, then $\chi_{A \cup B}(x) = 0$. We can express this as $\chi_{A \cup B}(x) = \chi_A(x) + \chi_B(x) - \chi_A(x)\chi_B(x)$. This ensures that if $x$ is in both $A$ and $B$, we don't count it twice.

2. Characteristic function of the intersection $A \cap B$

The characteristic function of $A \cap B$, denoted as $\chi_{A \cap B}(x)$, is 1 if $x$ belongs to both $A$ and $B$, and 0 otherwise. This means $\chi_{A \cap B}(x) = 1$ if and only if both $\chi_A(x) = 1$ and $\chi_B(x) = 1$. This can be expressed as the product of the individual characteristic functions: $\chi_{A \cap B}(x) = \chi_A(x) \chi_B(x)$.

3. Characteristic function of the complement $A^c$

The characteristic function of the complement of $A$, denoted as $\chi_{A^c}(x)$, is 1 if $x$ does not belong to $A$, and 0 if $x$ does belong to $A$. This means $\chi_{A^c}(x) = 1$ if $\chi_A(x) = 0$, and $\chi_{A^c}(x) = 0$ if $\chi_A(x) = 1$. We can express this as $\chi_{A^c}(x) = 1 - \chi_A(x)$.