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Use the recursive definitions of union and intersection to prove the following general De Morgan's law: For all positive integers \(n\), if \(A_{1}, A_{2}, \ldots, A_{n}\) are sets, then $$ \left(\bigcap_{i=1}^{n} A_{i}\right)^{c}=\bigcup_{i=1}^{n}\left(A_{i}\right)^{c} $$

Short Answer

Expert verified
The proof for the general De Morgan's law for all positive integers \(n\) is established using the principle of mathematical induction. The base case is proven for \(n = 2\). The inductive hypothesis assumes that the De Morgan's law holds for \(n = k\) for arbitrary sets \(A_1, A_2, \ldots, A_k\). Using this, we prove the induction step where the law also holds for \(n = k+1\). This shows that the general De Morgan's law \(\left(\bigcap_{i=1}^{n} A_i\right)^c = \bigcup_{i=1}^{n} \left(A_i\right)^c\) holds true for all positive integers \(n\).

Step by step solution

01

Prove Base Case for n = 2

We will first demonstrate that the De Morgan's law holds for n = 2. Let \(A_1\) and \(A_2\) be two arbitrary sets, and let \(^c\) denote the complement of a set. We must prove that $$ \left(\bigcap_{i=1}^{2} A_i\right)^c = \bigcup_{i=1}^{2} \left(A_i\right)^c $$ This simplifies to showing that \((A_1 \cap A_2)^c = A_1^c \cup A_2^c\). Let \(x\) be an arbitrary element in the universe. By the definition of set complements, \(x \in (A_1 \cap A_2)^c\) if and only if \(x \notin (A_1 \cap A_2)\). This means that either \(x \notin A_1\) or \( x \notin A_2\), or both. Now, by the definition of union, we can conclude that \(x \in A_1^c \cup A_2^c\). Therefore, we have proved that \((A_1 \cap A_2)^c = A_1^c \cup A_2^c\).
02

Inductive Hypothesis for n = k

Now, we will assume that the De Morgan's law is true for n = k and try to prove that it also holds for n = k + 1. Let \(A_1, A_2, \ldots, A_k\) be arbitrary sets. We assume that $$ \left(\bigcap_{i=1}^{k} A_i\right)^c = \bigcup_{i=1}^{k} \left(A_i\right)^c $$ is true.
03

Prove for n = k + 1

Our goal is now to show that $$ \left(\bigcap_{i=1}^{k+1} A_i\right)^c = \bigcup_{i=1}^{k+1} \left(A_i\right)^c $$ Let's rewrite the left-hand side of the equation as: $$ \left(\bigcap_{i=1}^{k+1} A_i\right)^c = \left(\left(\bigcap_{i=1}^{k} A_i\right) \cap A_{k+1}\right)^c $$ By applying the base case of De Morgan's law at this stage (Step 1), we can write $$ \left(\left(\bigcap_{i=1}^{k} A_i\right) \cap A_{k+1}\right)^c = \left(\bigcap_{i=1}^{k} A_i\right)^c \cup A_{k+1}^c $$ Now, we use the inductive hypothesis (Step 2) to substitute the left-hand side of the equation: $$ \bigcup_{i=1}^{k} \left(A_i\right)^c \cup A_{k+1}^c = \bigcup_{i=1}^{k+1} \left(A_i\right)^c $$ Thus, we have proved that the statement also holds for n = k + 1.
04

Conclusion

By using the principle of mathematical induction, we have proven that the general De Morgan's law holds for all positive integers n. For each n in the domain, \(\left(\bigcap_{i=1}^{n} A_i\right)^c\) is equal to \(\bigcup_{i=1}^{n} \left(A_i\right)^c\).

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Key Concepts

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

Recursive Definitions
Recursive definitions play a fundamental role in mathematics for defining sequences, functions, and other mathematical constructs through recursion. A recursive definition specifies the initial term or terms of a sequence and then provides a rule for determining subsequent terms based on the preceding values.

For example, in the context of set operations, recursive definitions are used to define the union and intersection of a collection of sets indexed by positive integers. To recursively define the union of sets, one would start by defining the union of the first two sets, then the union of that result with the third set, and so on, until all sets are included. The same concept applies to intersections, starting with the intersection of the first two sets and then continuing to intersect with each next set in the sequence.

This method simplifies complex operations by breaking them down into a sequence of simpler operations applied iteratively, making it powerful for proving properties like De Morgan's laws.
Set Intersection
Set intersection is a basic operation in set theory where the intersection of two or more sets contains only the elements that are present in all of the sets. Symbolically, the intersection of sets A and B is represented as \( A \cap B \).

For instance, if set \( A = \{1,2,3\} \) and set \( B = \{2,3,4\} \), then the intersection \( A \cap B \) would be \( \{2,3\} \) as these are the elements common to both sets. Use of the set intersection is crucial in understanding the formulation of De Morgan's laws, especially when these laws are extended to more than two sets.
Set Union
In contrast to intersection, the set union operation combines all the unique elements from each set. The union of sets A and B is denoted as \( A \cup B \).

Following our previous example, for the sets \( A = \{1,2,3\} \) and \( B = \{2,3,4\} \), the union would be \( A \cup B = \{1,2,3,4\} \) since it includes every distinct element from both sets. In the recursive proof of De Morgan's laws, the union operation takes a significant place as it helps to create a single set from the complements of multiple sets.
Set Complement
The set complement concept involves elements not in a given set, relative to a universal set which contains all possible elements. If U is the universal set and A is a subset of U, the complement of A, denoted \( A^c \), contains all elements in U that are not in A.

Using the set complement is essential to De Morgan's laws, which establish a relationship between the complement of set intersections and unions. For an element x, saying \( x \in A^c \) is equivalent to \( x otin A \). This duality is at the heart of the proof showing that the complement of an intersection is the union of the complements, and vice versa.
Mathematical Induction
Mathematical induction is a powerful proof technique used in mathematics to prove statements about all positive integers. The process entails two main steps: proving the base case and establishing the inductive step.

Firstly, the base case confirms that the statement holds true for the initial integer, typically n=1 or n=2. Then, assuming the statement holds for an arbitrary positive integer k (the induction hypothesis), you must prove the statement for k+1. If successful, you can infer that the statement holds for all succeeding integers.

Induction is especially effective for recursive definitions, just as it is used in the proof of De Morgan's laws, because it validates the general step built upon the assumption of smaller, previously proven cases.

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Most popular questions from this chapter

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