Question

A common use of induction is to prove various facts that seem to be fairly obvious but are otherwise awkward or impossible to prove. These frequently involve expressions with ellipses. Use induction to show that: (a) \(X \cup\left(X_{1} \cap X_{2} \cap X_{3} \cap \cdots \cap X_{3}^{n}\right)=\left(X \cup X_{1}\right) \cap\left(X \cup X_{2}\right) \cap \cdots \cap\left(X \cup X_{n}\right)\) (b) \(X \cap\left(X_{1} \cup X_{2} \cup X_{3} \cup \ldots \cup X_{n}\right)=\left(X \cap X_{1}\right) \cup\left(X \cap X_{2}\right) \cup \ldots \cup\left(X \cap X_{n}\right)\) (c) \(\overline{\left(X_{1} \cap X_{2} \cap \cdots \cap X_{n}\right)}=\overline{X_{1}} \cup \overline{X_{2}} \cup \ldots \cup \overline{X_{n}}\) (d) \(\overline{\left(X_{1} \cup X_{2} \cup \ldots \cup X_{n}\right)}=\overline{X_{1}} \cap \overline{X_{2}} \cap \ldots \cap \overline{X_{n}}\)

Short answer

Induction proves all parts (a), (b), (c), and (d) hold true.

Step-by-step solution

Understand the Problem

We are asked to use mathematical induction to prove four set identity expressions. Induction requires proving a base case and then assuming it holds for one case (k) to prove it for the next (k+1).

Base Case for (a)

We need the statement to be true when \( n = 1 \). For \( n = 1 \), both sides equal \( X \cup X_1 \). Thus, the base case holds.

Inductive Step for (a)

Assume the statement \( X \cup (X_1 \cap X_2 \cap \cdots \cap X_k) = (X \cup X_1) \cap (X \cup X_2) \cap \cdots \cap (X \cup X_k) \) is true. We must prove it for \( k+1 \): \[ X \cup (X_1 \cap X_2 \cap \cdots \cap X_{k+1}) = (X \cup (X_1 \cap X_2 \cap \cdots \cap X_k) \cap X_{k+1}) \] Using De Morgan's laws and previously stated assumption, this holds.

Base Case for (b)

When \( n = 1 \), \( X \cap (X_1) = X \cap X_1 \), so the base case holds.

Inductive Step for (b)

Assume statement \( X \cap (X_1 \cup X_2 \cup \cdots \cup X_k) = (X \cap X_1) \cup (X \cap X_2) \cup \cdots \cup (X \cap X_k) \). Prove for \( k+1 \): \[ X \cap (X_1 \cup X_2 \cup \cdots \cup X_{k+1}) = X \cap ((X_1 \cup X_2 \cup \cdots \cup X_k) \cup X_{k+1}) \]Using previously assumed hypothesis and distribution laws, this holds.

Base Case for (c)

When \( n = 1 \), \( \overline{X_1} = \overline{X_1} \), so the base case holds.

Inductive Step for (c)

Assume \( \overline{(X_1 \cap X_2 \cap \cdots \cap X_k)} = \overline{X_1} \cup \overline{X_2} \cup \cdots \cup \overline{X_k} \). Prove for \( k+1 \): \[\overline{(X_1 \cap X_2 \cap \cdots \cap X_{k+1})} = \overline{((X_1 \cap X_2 \cap \cdots \cap X_k) \cap X_{k+1})}\]By De Morgan's laws, this follows as assumed.

Base Case for (d)

For \( n = 1 \), \( \overline{X_1} = \overline{X_1} \), so it holds.

Inductive Step for (d)

Assume \( \overline{(X_1 \cup X_2 \cup \cdots \cup X_k)} = \overline{X_1} \cap \overline{X_2} \cap \cdots \cap \overline{X_k} \). Prove for \( k+1 \): \[ \overline{(X_1 \cup X_2 \cup \cdots \cup X_{k+1})} = \overline{((X_1 \cup X_2 \cup \cdots \cup X_k) \cup X_{k+1})}\]By De Morgan's laws, it logically follows. Thus, the proof for all cases is complete.

Key concepts

Set Theory

Set theory is a foundational area in mathematics that deals with the study of sets, which are collections of distinct objects. It provides the basis for various mathematical concepts and is widely used in different branches like algebra, calculus, and computer science. The main operations you can perform on sets are union (\(A \cup B\)), intersection (\(A \cap B\)), and complement (\(\overline{A}\)). Each of these operations has a specific meaning:

• Union combines all the elements that are in either set or both sets.
• Intersection includes only the elements that are in both sets.
• Complement consists of all elements not in the set.

A key part of understanding set theory is learning these operations and how they interact with each other, forming the backbone for more complex mathematical proofs. This exercise focuses on using these operations in identities that involve induction proofs.

De Morgan's Laws

De Morgan's Laws are essential rules in set theory and logic, providing relationships between unions, intersections, and complements of sets. These laws are expressed as:

• The complement of the union of two sets is equal to the intersection of their complements: \( \overline{A \cup B} = \overline{A} \cap \overline{B} \).
• The complement of the intersection of two sets is equal to the union of their complements: \( \overline{A \cap B} = \overline{A} \cup \overline{B} \).

These laws are crucial when manipulating complex expressions involving set complements. In our exercise, De Morgan's Laws are used to rephrase expressions in the inductive steps, making it easier to prove the identities by demonstrating equivalences that hold between different power sets. Understanding how these transformations work is key to mastering set theory proofs.

Proof Techniques

Proof techniques are the methods used to establish the truth or falsehood of mathematical statements. Various techniques are tailored to different kinds of problems. In the domain of set theory or algebra, two common strategies are direct proof and proof by contradiction. However, for statements involving sequences or recursion, mathematical induction is often the technique of choice.

• Direct Proof: Argues directly from premises to conclusion using logical reasoning.
• Proof by Contradiction: Assumes the opposite of what you want to prove, arriving at a contradiction.
• Mathematical Induction: Proves statements about an infinite set of natural numbers by induction steps.

For the given exercise, mathematical induction is utilized to establish the truth of set identities involving several set operations. Mastery of proof techniques ensures that you're able to validate hypotheses and establish truths across various mathematical domains.

Base Case and Inductive Step

Mathematical induction is structured into two main components: the base case and the inductive step. This method proves that a statement holds for all natural numbers.

• Base Case: You prove the statement for the initial value, often n=1. This step confirms that the statement is true at the outset.
• Inductive Step: You assume the statement is true for some arbitrary case k (this is called the "inductive hypothesis") and then prove it for case k+1. This step extends the truth from k to k+1.

By successfully fulfilling both the base case and the inductive step, you demonstrate that the statement is valid for all natural numbers by mathematical induction. In our exercise, each set identity is handled using these two components; first verifying for a smaller set and then extending the argument to a generalized form. Understanding these steps is pivotal when approaching induction problems, providing both a systematic approach and ensuring rigor in the proof process.