Constructing Proofs: Solving Set Functions with Characteristic Functions

[SpatialVacancy]

Constructing Proofs help!

Here is the problem:

Given a set $$S$$ and subset $$A$$, the characteristic function of A, denoted $$\chi_A$$, is the function defined from $$S$$ to $$\mathbb{Z}$$ with the property that for all $$u \ \epsilon \ S$$:

$$\chi_A(u)=<br /> \begin{cases}<br /> 1 & \text{if u \epsilon \ A} \\<br /> 0 & \text{if u is not \ \epsilon \ A}<br /> \end{cases}$$
​

Show that each of the following holds for all subsets $$A$$ and $$B$$ of $$S$$ and all $$u \ \epsilon \ S$$.

a. $$\chi_{A \cap B}(u)= \chi_A (u) \cdot \chi_B (u)$$
b. $$\chi_{A \cup B}(u)= \chi_A (u) + \chi_B (u) - \chi_A (u) \cdot \chi_B (u)$$



I have NO IDEA what this problem is asking...can someone please help!

 

[Galileo]

The brute straightforward way is:
Compute both sides of the equations for all cases:
(I) u is in A and B
(II) u is in A or B, but not both.
(III) u is not in A and not in B.

 

[xanthym]

.
You might present the proof in "Truth Table" format and show equivalence via Table equality. For example:

.
.-----------------------> [B]Χ(A ∩ B)[/B]

                           Member
                             B
                      YES         NO

             YES       [B]1          0[/B]
     Member   
       A
             NO        [B]0          0[/B]




.----------------------> [B]Χ(A)*Χ(B)[/B]

                           Member
                             B
                      YES         NO

             YES    (1)*(1)    (1)*(0)
     Member            [B]1          0 [/B]
       A
             NO     (0)*(1)    (0)*(0)
                       [B]0          0[/B]