91Ó°ÊÓ

If \(\Psi\) is a universe and \(A \subseteq ?\), we define the characteristic function of \(A\) by \(\chi_{A}: \mathscr{U} \rightarrow\\{0,1\\}\), where $$ \chi_{A}(x)= \begin{cases}1, & x \in A \\ 0, & x \notin A\end{cases} $$ For sets \(A, B \subseteq \mathscr{\text { , prove each of the following: }}\) a) \(\chi_{A \cap B}=\chi_{A} \cdot \chi_{B}\) where \(\left(\chi_{A} \cdot \chi_{B}\right)(x)=\chi_{A}(x) \cdot \chi_{B}(x)\) b) \(\chi_{A \cup B}=\chi_{A}+\chi_{B}-\chi_{A \cap B}\) c) \(\chi_{A}=1-\chi_{A}\), where \(\left(1-\chi_{A}\right)(x)=1(x)-\chi_{A}(x)=\) \(1-\chi_{A}(x)\) (For \(q\) finite, placing the elements of \(q\) in a fixed order results in a one-to-one correspondence between subsets \(A\) of \(q\) and the arrays of 0 's and 1 's obtained as the images of \(\mathscr{L}\) under \(\chi_{A}\). These arrays can then be used for the computer storage and manipulation of certain subsets of \(q\).)

Step by step solution

01

Understand the characteristic function

The characteristic function of a set assigns a value 1 for every element in the set and 0 for all other elements.

02

Prove \(\chi_{A \cap B}=\chi_{A} \cdot \chi_{B}\)

Observe that the only time \(\chi_{A \cap B}(x)\) is equal to 1 is when \(x\) is in both \(A\) and \(B\). This is exactly the same condition when \(\chi_{A}(x) \cdot \chi_{B}(x)\) equals 1, because both of \(\chi_{A}(x)\) and \(\chi_{B}(x)\) should be 1 (which means \(x\) should be in both \(A\) and \(B\)). Hence the equality holds.

03

Prove \(\chi_{A \cup B}=\chi_{A}+\chi_{B}-\chi_{A \cap B}\)

The characteristic function of \(A \cup B\) equals 1 if \(x\) is in either \(A\), \(B\) or both. So, \(\chi_{A \cup B}(x)\) equals \(\chi_{A}(x) + \chi_{B}(x)\) minus the case when \(x\) is in both \(A\) and \(B\) (because it is counted twice), which is \(\chi_{A \cap B}(x)\). Therefore, the equality holds true.

04

Prove \(\chi_{A}=1-\chi_{A}\)

This case represents the complement of set \(A\). When \(x\) is in \(A\), \(\chi_{A}(x) = 1\) and \(1 - \chi_{A}(x) = 0\), and when \(x\) is not in \(A\), \(\chi_{A}(x) = 0\) and \(1 - \chi_{A}(x) = 1\). Therefore \(\chi_{A}=1-\chi_{A}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Set Intersection

In set theory, the intersection of two sets, denoted as \(A \cap B\), is a new set that contains all the elements that are present in both set \(A\) and set \(B\). Understanding this concept through the lens of the characteristic function can be particularly insightful.
When we talk about the characteristic function of the intersection \(\chi_{A \cap B}\), it assigns 1 only if an element \(x\) belongs to both \(A\) and \(B\). In terms of multiplication, \(\chi_{A}(x) \cdot \chi_{B}(x)\) results in 1 only when both \(\chi_{A}(x)\) and \(\chi_{B}(x)\) are 1, which happens precisely when \(x\) is in the intersection.

• This makes \(\chi_{A \cap B} = \chi_{A} \cdot \chi_{B}\), as the product of the two functions highlights the shared elements between the sets.

This logical operation not only aligns with our intuition about intersection but also reveals the elegance of applying characteristic functions.

Set Union

Set Union, represented as \(A \cup B\), consists of all elements that belong to either set \(A\), set \(B\), or both. The characteristic function \(\chi_{A \cup B}\) must reflect this by returning 1 if an element \(x\) is in at least one of the sets.
A simple way to achieve this is by using addition: \(\chi_{A}(x) + \chi_{B}(x)\). This configuration correctly counts elements that are in either set. However, when \(x\) belongs to both \(A\) and \(B\), it is double-counted.
To adjust for this, we subtract the intersection's characteristic:

• \(\chi_{A \cup B} = \chi_{A} + \chi_{B} - \chi_{A \cap B}\).

This subtraction removes the extra count of those shared elements, accurately reflecting the union's full complement of elements.

Complement of a Set

The complement of a set \(A\), symbolized as \(A^c\) or sometimes \(\overline{A}\), contains all elements not in \(A\) but within the universal set \(\Psi\). The role of the characteristic function here is to toggle between two states: in set \(A\) and not in set \(A\).
Using the formula \(1 - \chi_{A}\), the characteristic function for the complement \(\chi_{A^c}\) is constructed.
Here's how it functions:

• If an element \(x\) is in \(A\), then \(\chi_{A}(x) = 1\), resulting in \(1 - \chi_{A}(x) = 0\), which correctly indicates that \(x\) is not in \(A^c\).
• If \(x\) is not in \(A\), \(\chi_{A}(x) = 0\), making \(1 - \chi_{A}(x) = 1\), indicating inclusion in \(A^c\).

This simple arithmetic adjustment successfully captures the essence of a complement: being everything outside the original set within a universal perspective.