Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. Suppose \(A_{1}, A_{2}, \ldots \underline{A_{n}}\) are sets in some universal set \(U,\) and \(n \geq 2 .\) Prove that \(\overline{A_{1} \cup A_{2} \cup \cdots \cup A_{n}}=\overline{A_{1}} \cap \overline{A_{2}} \cap \cdots \cap \overline{A_{n}}\)

Short answer

The given identity \(\overline{A_{1} \cup A_{2} \cup \cdots \cup A_{n}} = \overline{A_{1}} \cap \overline{A_{2}} \cap \cdots \cap \overline{A_{n}}\) can be proved by induction. The proof involves proving the base case (for \(n=2\)), then assuming the statement is true for some positive integer k, and showing that it holds for \(k+1\). Since the base case is true and the statement is also true for \(k+1\) if it's true for k, we can infer that the identity holds for all positive integers \(n \geq 2\).

Step-by-step solution

Understand the Base Case

We start by proving the base case. As the problem explicitly states \(n \geq 2\), it means our base case is when n=2. Therefore, we need to prove that \(\overline{A_1 \cup A_2} = \overline{A_1} \cap \overline{A_2}\)

Prove the Base Case

We know that an element is in \(\overline{A_1 \cup A_2}\) if and only if it is not in \(A_1 \cup A_2\), meaning it is not in \(A_1\) and it is not in \(A_2\). Therefore, it is in the intersection of \(\overline{A_1} \) and \( \overline{A_2} \). Similarly, if an element is in \(\overline{A_1} \cap \overline{A_2}\), it is not in \(A_1\) and it is not in \(A_2\), meaning it is not in their union \(A_1 \cup A_2\), i.e., it is in \(\overline{A_1 \cup A_2}\). Hence, we have proved \(\overline{A_1 \cup A_2} = \overline{A_1} \cap \overline{A_2}\)

Understand the Inductive Step

Next, we assume that the statement is true for n=k : \(\overline{A_1 \cup A_2 \cup ... \cup A_k} = \overline{A_1} \cap \overline{A_2} \cap ... \cap \overline{A_k}\). We now need to prove the validity of the statement for n=k+1: \(\overline{A_1 \cup A_2 \cup ... \cup A_k \cup A_{k+1}} = \overline{A_1} \cap \overline{A_2} \cap ... \cap \overline{A_k} \cap \overline{A_{k+1}}\)

Prove the Inductive Step

We begin the proof by rewriting the left side of the equation as follow: \(\overline{A_1 \cup A_2 \cup ... \cup A_k \cup A_{k+1}} = \overline{(A_1 \cup A_2 \cup ... \cup A_k) \cup A_{k+1}}\). We know from our base case that this equals to \(\overline{A_1 \cup A_2 \cup ... \cup A_k} \cap \overline{A_{k+1}}\). And from our inductive assumption, we replace the left term with \(\overline{A_1} \cap \overline{A_2} \cap ... \cap \overline{A_k} \cap \overline{A_{k+1}}\). Hence, the statement is true for n=k+1 if it's true for n=k.

Key concepts

Proof by Induction

Proof by induction is a powerful technique typically used in mathematics to establish that a given statement is true for all natural numbers. The technique involves two crucial steps: the base case and the inductive step. With each step functioning as a crucial pillar, the overarching goal is to form an unbreakable chain of reasoning.

In the base case, one must demonstrate that the statement is valid for the initial natural number, usually 0 or 1. Returning to our exercise, we've focused on the base case of n=2, since our universal set starts from n=2, showcasing that the principle holds at the beginning of the number line.

The inductive step, which is where the actual induction occurs, requires assuming that the given statement is true for a natural number k (also known as the inductive hypothesis) and subsequently proving that the statement is also true for k+1. This logical step in our exercise connects the truth of the premises for k sets to the truth of the statement for k+1 sets, thus potentially extending the truth to all natural numbers.

When we use induction, we liken it to climbing a ladder – each rung depends on the firm placement of the previous one. If our initial step onto the ladder—the base case—is secure and each subsequent step up—the inductive step—is reliable, then we can confidently climb to any rung, securely knowing that all previous ones hold firm.

Proof by Smallest Counterexample

The proof by the smallest counterexample is a technique used to prove statements by contradiction. This method assumes that if the statement were false, then there would be a smallest counterexample that disproves the statement. By analyzing this supposed smallest counterexample, one often finds a contradiction that demonstrates the statement must indeed be true.

This approach has an inherent beauty, reflecting a detective's work by seeking out the ‘weakest link’ that would break the chain of a proposed theory. It creates a scenario where you would consider the contradiction arising from the existence of a set that does not satisfy the given condition. In our textbook exercise, if we were to implement this strategy, we would assume a smallest non-empty n for which the statement failed. Through meticulous analysis, our goal would be to show that this assumption leads to a logical inconsistency, thereby nullifying its very existence and proving the statement correct.

Set Theory

Set theory is the branch of mathematical logic that studies sets, which are collections of objects. In relation to our exercise, working with universal sets and subsets under the umbrella of set theory is akin to organizing different groups based on shared characteristics.

Key operations in set theory include the formation of the union (denoted by \( \cup \)) and intersection (denoted by \( \cap \)) of sets. The union represents the set of all elements belonging to at least one of the contributing sets, whereas the intersection comprises only those elements found in all contributing sets. The concept of complement (denoted by an overline) refers to the set of all elements in the universal set that are not in the specified subset.

In our textbook problem, the equality we aim to prove integrates these fundamental set operations. It demonstrates a quality of symmetry between the union of sets and their complements, providing evidence that the edge of a group's union can reflect the core intersection of separate identities, an intriguing juxtaposition within the realm of set theory.