Question

Let \(U\) be the universal set and \(A \cup B \cup C=U\). Then \([(A-B) \cup(B-C) \cup(C-A)]\) equals (a) \(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}\) (b) \(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}\) (c) \(\mathrm{A} \cup(\mathrm{B} \cap \mathrm{C})\) (d) \(\mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})\)

Short answer

The simplified expression (A ∩ B') ∪ (B ∩ C') does not match any of the given options. Therefore, none of the given options is the correct answer.

Step-by-step solution

Apply the formula of difference of sets

First, let's find the difference of sets in the expression: (A-B) = A ∩ B' (B-C) = B ∩ C' (C-A) = C ∩ A' Here, X' represents the complement of set X.

Apply the formula of union of sets

Now, apply the union operation to the resulting sets: [(A-B) ∪ (B-C) ∪ (C-A)] = [(A ∩ B') ∪ (B ∩ C') ∪ (C ∩ A')]

Distribute the intersection operation

Use the distributive property to simplify the expression. We distribute the intersection over the union: [(A ∩ B') ∪ (B ∩ C') ∪ (C ∩ A')] = (A ∩ B' ∩ (B ∪ C ∪ A')) ∪ (B ∩ C' ∩ (A ∪ B ∪ C'))

Apply given condition that A ∪ B ∪ C = U and simplify

We know that A ∪ B ∪ C = U, so we can rewrite the expression as: (A ∩ B' ∩ U') ∪ (B ∩ C' ∩ U') For any set X, X ∩ U = X. So, (A ∩ B' ∩ U') = A ∩ B' and (B ∩ C' ∩ U') = B ∩ C'. Therefore, the expression simplifies to: (A ∩ B') ∪ (B ∩ C')

Compare with given options

Now, compare the simplified expression with the given options: (a) A ∪ B ∪ C (b) A ∩ B ∩ C (c) A ∪ (B ∩ C) (d) A ∩ (B ∪ C) The simplified expression (A ∩ B') ∪ (B ∩ C') does not match any of the given options. Therefore, none of the given options is the correct answer.

Key concepts

Difference of Sets

Understanding the difference between sets is essential when working with various sets within mathematics. The difference of sets, usually denoted as \( A - B \), is found by taking all the elements that are in set \( A \) but not in set \( B \). Another way to express this is \( A \cap B' \), where \( B' \) represents the set complement of \( B \); that is, all the elements not in \( B \).

When we subtract one set from another, we're essentially filtering out any shared elements that the second set has. In the context of our exercise, the difference \( (A - B) \) means taking all the members of set \( A \) that are not also members of set \( B \). This concept helps to isolate the unique characteristics of set \( A \) relative to set \( B \).

Union of Sets

The union of sets is another fundamental concept in set theory and is represented by the symbol \( \cup \). The union of two or more sets consists of all elements that are present in any of the sets being united. For two sets, \( A \) and \( B \), the union is denoted as \( A \cup B \), and includes every element that is in \( A \), \( B \), or both.

Formally, for any element \( x \), \( x \) belongs to \( A \cup B \) if \( x \) is in either \( A \), \( B \), or in both. In the exercise provided, the union operation helps to merge the differences obtained from the sets, creating a set pooled from the unique elements of each initial set after the respective differences are accounted for.

Set Complement

The set complement is essential for various operations in set theory. If you have a universal set \( U \), which contains all possible elements under consideration, the complement of set \( A \), denoted as \( A' \), includes all elements that are in the universal set \( U \) but not in set \( A \).

In essence, the complement of a set provides us with the 'opposite' of that set within the bounds of the universal set. It's important to note that the idea of a complement depends on the universal set one is working within: since the universal set contains all possible elements, the complement of \( A \) is everything outside of \( A \). The complement plays a crucial role in set differences and in operations like De Morgan's laws that connect the concepts of union, intersection, and complements in set theory.

Distributive Property in Sets

Among the critical properties of set operations is the distributive property, which allows us to simplify expressions involving unions and intersections. The distributive property in set theory states that for any sets \( A \), \( B \), and \( C \), the following is true:

\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)
and
\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \).

This property allows us to 'distribute' one set over the union (or intersection) of two other sets, facilitating the simplification of set expressions. It's akin to the distributive property in arithmetic where \( a(b + c) = ab + ac \). During the exercise, we used the distributive property to simplify the union of intersections, which helped to bring the expression to a form that we could analyze and compare with the given options.