- #1
SpatialVacancy
- 24
- 0
Constructing Proofs help!
Here is the problem:
Given a set [tex]S[/tex] and subset [tex]A[/tex], the characteristic function of A, denoted [tex]\chi_A[/tex], is the function defined from [tex]S[/tex] to [tex]\mathbb{Z}[/tex] with the property that for all [tex]u \ \epsilon \ S[/tex]:
[tex]
\chi_A(u)=
\begin{cases}
1 & \text{if u $ \epsilon \ A$} \\
0 & \text{if u $ is not \ \epsilon \ A$}
\end{cases}
[/tex]
Show that each of the following holds for all subsets [tex]A[/tex] and [tex]B[/tex] of [tex]S[/tex] and all [tex]u \ \epsilon \ S[/tex].
a. [tex]\chi_{A \cap B}(u)= \chi_A (u) \cdot \chi_B (u)[/tex]
b. [tex]\chi_{A \cup B}(u)= \chi_A (u) + \chi_B (u) - \chi_A (u) \cdot \chi_B (u)[/tex]
I have NO IDEA what this problem is asking...can someone please help!
Here is the problem:
Given a set [tex]S[/tex] and subset [tex]A[/tex], the characteristic function of A, denoted [tex]\chi_A[/tex], is the function defined from [tex]S[/tex] to [tex]\mathbb{Z}[/tex] with the property that for all [tex]u \ \epsilon \ S[/tex]:
[tex]
\chi_A(u)=
\begin{cases}
1 & \text{if u $ \epsilon \ A$} \\
0 & \text{if u $ is not \ \epsilon \ A$}
\end{cases}
[/tex]
Show that each of the following holds for all subsets [tex]A[/tex] and [tex]B[/tex] of [tex]S[/tex] and all [tex]u \ \epsilon \ S[/tex].
a. [tex]\chi_{A \cap B}(u)= \chi_A (u) \cdot \chi_B (u)[/tex]
b. [tex]\chi_{A \cup B}(u)= \chi_A (u) + \chi_B (u) - \chi_A (u) \cdot \chi_B (u)[/tex]
I have NO IDEA what this problem is asking...can someone please help!