Problem 47 - Discrete Mathematics with Applications

Question

Prove each, where \(A, B,\) and \(C\) are any sets. $$\left(A^{\prime}\right)^{\prime}=A$$

Step-by-Step Solution

Answer

To prove that the double complement of a set A is equal to the set A itself, i.e., \((A^{\prime})^{\prime} = A\), we showed that every element in the double complement of A is in A and vice versa. We demonstrated that \((A^{\prime})^{\prime} \subseteq A\) and \(A \subseteq (A^{\prime})^{\prime}\) using the definition of set complement. This implies that \((A^{\prime})^{\prime} = A\), proving the equality for any sets A, B, and C.

Step 1: Definition of Set Complement

Recall the definition of a set complement: the complement of a set A, denoted by \(A^{\prime}\), is the set of all elements that are not in A. In other words, if x is an element of the universe \(U\), \(x \in A^{\prime}\) if and only if \(x \notin A\).

Step 2: Prove \((A^{\prime})^{\prime} \subseteq A \)

We want to show that every element in the double complement of A is an element of A. Let's take an element x in the double complement of A, i.e., \(x \in (A^{\prime})^{\prime}\). By the definition of complement, since x is in the double complement of A, it is not in the complement of A, i.e., \(x \notin A^{\prime}\). Now, by the definition of complement again, since \(x \notin A^{\prime}\), it means that \(x \in A\). Therefore, every element in the double complement of A is in A, i.e., \((A^{\prime})^{\prime} \subseteq A\).

Step 3: Prove \(A \subseteq (A^{\prime})^{\prime}\)

Now, we want to show that every element in A is an element in the double complement of A. Let's take an element x in A, i.e., \(x \in A\). By the definition of complement, since x is in A, it is not in the complement of A, i.e., \(x \notin A^{\prime}\). Now, by the definition of complement again, since \(x \notin A^{\prime}\), it means that \(x \in (A^{\prime})^{\prime}\). Therefore, every element in A is in the double complement of A, i.e., \(A \subseteq (A^{\prime})^{\prime}\).

Step 4: Conclusion

From Steps 2 and 3, we have demonstrated that \((A^{\prime})^{\prime} \subseteq A\) and \(A \subseteq (A^{\prime})^{\prime}\). This implies that \((A^{\prime})^{\prime} = A\). Thus, we have proven the equality for any sets A, B, and C.

Key Concepts

Set Complement

In set theory, the complement of a set is a fundamental concept. If you have a universe of elements, the complement of a particular set within that universe includes all elements that are not in the original set. For example, if the universe is the set of all natural numbers and your set is the even numbers, then the complement of that set is all the odd numbers.

• The notation for the complement of set A is denoted as \(A'\).
• If an element \(x\) is in the complement of \(A\), \(x\) is not in \(A\).
• The universe is often assumed to be a specific set from which all subsets are derived; in many problems, it is implicitly defined.

Understanding the complement helps broaden the analysis scope of set interactions, especially when dealing with operations like unions and intersections.

Subset

A subset is a set in which all elements are also contained within another set. It's important in set theory because it helps describe the relationships between different sets. If you have two sets, A and B, A is considered a subset of B if every element of A is also an element of B.

• The symbol to denote that A is a subset of B is \(A \subseteq B\).
• If A has every element that B does, and potentially more, B would not be a subset of A.
• An important thing to remember is that every set is a subset of itself, meaning \(A \subseteq A\).

Subsets are crucial when establishing basic results or proofs in set theory, like the relationships in complements and intersections.

Double Complement

The concept of a double complement refers to finding the complement of the complement of a set. It might sound a bit repetitive, but it's an interesting property in set theory.

• If \(A'\) is the complement of set A, then \((A')'\) is the double complement, meaning elements not in \(A'\), which are exactly the elements in A.
• The double complement essentially undoes the complement operation, leading back to the original set, \((A')' = A\).
• This feature underpins the logical consistency of set operations and is essential for many proofs and set equality assertions.

The concept demonstrates that complements are 'reversible' operations, restoring sets through inverse operations.

Equality of Sets

In set theory, two sets are said to be equal if they contain exactly the same elements. This equality implies that each element of one set belongs to the other and vice versa. To check the equality of sets, two main set inclusions need to be verified.

• For two sets A and B, the condition \(A = B\) holds if \(A \subseteq B\) and \(B \subseteq A\).
• In the context of double complements, proving both \((A')' \subseteq A\) and \(A \subseteq (A')'\) ensures \((A')' = A\).
• Such equalities confirm that operations like taking double complements do not alter the intrinsic structure of a set.

Recognizing equality between sets helps validate various properties in problems like proofs by contradiction or demonstrating intersections and unions accurately.