Question

Prove the distributive laws (i) \(A \cap \bigcup X_{i}=\bigcup\left(A \cap X_{i}\right)\); (ii) \(A \cup \bigcap X_{i}=\bigcap\left(A \cup X_{i}\right)\); \((\mathrm{iii})\left(\bigcap X_{i}\right)-A=\bigcap\left(X_{i}-A\right) ;\) (iv) \(\left(\bigcup X_{i}\right)-A=\bigcup\left(X_{i}-A\right)\); \((\mathrm{v}) \cap X_{i} \cup \bigcap Y_{j}=\bigcap_{i, j}\left(X_{i} \cup Y_{j}\right) ;^{4}\) (vi) \(\bigcup X_{i} \cap \bigcup Y_{j}=\bigcup_{i, j}\left(X_{i} \cap Y_{j}\right)\)

Short answer

All distributive laws are valid in set theory as demonstrated by step-by-step proofs.

Step-by-step solution

Understanding Distributive Laws in Set Theory

The distributive laws for sets are analogs of the polynomial distributive laws and involve intersection (\( \cap \)) and union (\( \cup \)). They generally express that these operations can be distributed over each other, similar to how multiplication distributes over addition. Our task is to prove the validity of these distributive laws.

Set Notation and Concepts Review

Let's recall that \( \bigcup \) denotes the union over a number of sets, while \( \bigcap \) denotes the intersection over a number of sets. Further, the set difference \( X - A \) means we take elements of \( X \) that are not in \( A \). With these definitions in mind, we can begin proving each law step-by-step.

Proving Distributive Law (i)

Proof for \( A \cap \bigcup X_{i} = \bigcup (A \cap X_{i}) \):**Left to Right (\( \subseteq \))**1. Take an element \( x \in A \cap \bigcup X_{i} \).2. Then \( x \) is in both \( A \) and \( \bigcup X_{i} \), which implies \( x \) is in some \( X_i \).3. Hence, \( x \) must be in \( A \cap X_i \) for some index \( i \).4. Therefore, \( x \in \bigcup (A \cap X_i) \), proving the left to right inclusion.**Right to Left (\( \supseteq \))**1. Take an element \( x \in \bigcup (A \cap X_i) \).2. Then \( x \in A \cap X_i \) for some index \( i \), implying \( x \in A \) and \( x \in X_i \).3. Consequently, \( x \in \bigcup X_i \), and thus \( x \in A \cap \bigcup X_i \).4. This proves the right to left inclusion.Both directions are proven; hence, (i) is true.

Proving Distributive Law (ii)

Proof for \( A \cup \bigcap X_{i} = \bigcap (A \cup X_{i}) \):**Left to Right (\( \subseteq \))**1. Assume an element \( x \in A \cup \bigcap X_{i} \).2. This means \( x \in A \) or \( x \in \bigcap X_{i} \).3. If \( x \in A \), then \( x \in A \cup X_i \) for all \( i \).4. If \( x \in \bigcap X_i \), \( x \) is in every \( X_i \), implying \( x \in A \cup X_i \) for all \( i \).5. Therefore, \( x \in \bigcap (A \cup X_{i}) \).**Right to Left (\( \supseteq \))**1. Assume \( x \in \bigcap (A \cup X_i) \).2. This means \( x \in A \cup X_i \) for every \( i \), implying \( x \in A \) or \( x \in X_i \).3. If \( x \in X_i \) for every \( i \), then \( x \in \bigcap X_i \).4. Thus, \( x \in A \cup \bigcap X_i \).All elements are proven in both directions; hence, (ii) is true.

Proving Distributive Law (iii)

Proof for \( \left( \bigcap X_{i} \right) - A = \bigcap (X_{i} - A) \):**Left to Right (\( \subseteq \))**1. Take an element \( x \in \left( \bigcap X_{i} \right) - A \).2. This means \( x \in \bigcap X_i \) and \( x otin A \).3. Therefore, \( x \in X_i \) for every \( i \) and not in \( A \), so \( x \in X_i - A \) for all \( i \).4. Consequently, \( x \in \bigcap (X_i - A) \).**Right to Left (\( \supseteq \))**1. Take an element \( x \in \bigcap (X_i - A) \).2. Then \( x \in X_i - A \) implies \( x \in X_i \) and \( x otin A \) for each \( i \).3. Hence, \( x \in \bigcap X_i \), and \( x otin A \), so \( x \in \left( \bigcap X_i \right) - A \).Verified, thus (iii) is true.

Proving Distributive Law (iv)

Proof for \( \left( \bigcup X_i \right) - A = \bigcup (X_i - A) \): **Left to Right (\( \subseteq \))**1. Take an element \( x \in \left( \bigcup X_i \right) - A \).2. This means \( x \in \bigcup X_i \) and \( x otin A \).3. Thus, \( x \in X_i \) for some \( i \) but not in \( A \), leading \( x \in X_i - A \) for some \( i \).4. Therefore, \( x \in \bigcup (X_i - A) \).**Right to Left (\( \supseteq \))**1. Take an element \( x \in \bigcup (X_i - A) \).2. Then \( x \in X_i - A \) for some \( i \), ensuring \( x \in X_i \) and \( x otin A \).3. Thus, \( x \in \left( \bigcup X_i \right) - A \).So, (iv) is confirmed true.

Proving Distributive Law (v)

Proof for \( \cap X_{i} \cup \bigcap Y_{j} = \bigcap_{i, j} (X_{i} \cup Y_{j}) \):**Left to Right (\( \subseteq \))**1. Take \( x \in \cap X_{i} \cup \bigcap Y_{j} \).2. It means \( x \) is in at least one \( X_{i} \) or in every \( Y_{j} \).3. If \( x \) is in every \( Y_{j} \), \( x \in X_{i} \cup Y_{j} \) holds for any \( i \), \( j \).4. If \( x \in X_i \), \( x \in X_{i} \cup Y_{j} \) for every \( j \).5. Thus, \( x \in \bigcap_{i, j}(X_{i} \cup Y_{j}) \).**Right to Left (\( \supseteq \))**1. Take \( x \in \bigcap_{i, j} (X_{i} \cup Y_{j}) \).2. Then \( x \) is in each \( X_i \cup Y_j \).3. If \( x otin X_i \) for any \( i \), then \( x \in Y_j \) for all \( j \), so \( x \in \bigcap Y_j \).4. If \( x \in X_i \) for all \( i \), \( x \in \cap X_i \).Thus, verified for (v).

Proving Distributive Law (vi)

Proof for \( \bigcup X_{i} \cap \bigcup Y_{j} = \bigcup_{i, j}(X_{i} \cap Y_{j}) \):**Left to Right (\( \subseteq \))**1. Suppose \( x \in \bigcup X_{i} \cap \bigcup Y_{j} \).2. This implies \( x \in X_i \) for some \( i \) and \( x \in Y_j \) for some \( j \).3. Therefore, \( x \in X_i \cap Y_j \), making \( x \in \bigcup_{i, j}(X_i \cap Y_j) \).**Right to Left (\( \supseteq \))**1. Assume \( x \in \bigcup_{i, j}(X_i \cap Y_j) \).2. This means \( x \in X_i \cap Y_j \) for some \( i \) and \( j \).3. Thus, \( x \in \bigcup X_i \) and \( x \in \bigcup Y_j \).4. Therefore, \( x \in \bigcup X_i \cap \bigcup Y_j \).Proven, showing (vi) is true.

Key concepts

Set Union

In set theory, a union of sets is a fundamental operation, denoted by \( \cup \). When you create a union of two sets, you are essentially merging them. You take all the distinct elements from each set and combine them to form a new set. This process keeps each element unique, so no duplicates are present.
In mathematical terms, for two sets \( A \) and \( B \), the union \( A \cup B \) will include any element that is either in \( A \), or in \( B \), or in both. This operation is quite similar to joining different groups together without losing any elements in the process.

• If \( x \in A \) or \( x \in B \), then \( x \in A \cup B \).
• If \( A = \{1, 2, 3\} \) and \( B = \{3, 4, 5\} \), then \( A \cup B = \{1, 2, 3, 4, 5\} \).

The union helps in proofs and problems involving distributive laws, as it is often paired with other operations like intersection.

Set Intersection

The set intersection operation is represented by the symbol \( \cap \). The intersection of two sets includes only the elements that are present in both sets. It's like finding the common ground between two different groups. When you intersect sets, you are identifying what's shared.
For any two sets \( A \) and \( B \), their intersection \( A \cap B \) contains all the elements that \( A \) and \( B \) have in common.

• If \( x \in A \) and \( x \in B \), then \( x \in A \cap B \).
• If \( A = \{1, 2, 3\} \) and \( B = \{3, 4, 5\} \), then \( A \cap B = \{3\} \).

In distributive laws, intersections can be distributed over unions, showcasing the flexibility and importance of this operation in set theory.

Set Difference

Set difference is another fundamental concept in set theory, noted for removing elements from a set. This operation is depicted by the symbol \( - \). Conceptually, the difference between two sets, \( A - B \), means we include every element that is in \( A \) but not in \( B \). This operation highlights the uniqueness of \( A \) relative to \( B \).
Understanding the difference between sets is crucial for operations that require discerning what's left when subtracting shared elements.

• If \( x \in A \) and \( x otin B \), then \( x \in A - B \).
• If \( A = \{1, 2, 3, 4\} \) and \( B = \{3, 4, 5\} \), then \( A - B = \{1, 2\} \).

This operation is essential for proving distributive laws that involve differences.

Subset Proof

Proving one set is a subset of another is a frequent exercise in set theory. A set \( A \) is a subset of a set \( B \) if every element of \( A \) is also in \( B \), indicated by \( A \subseteq B \).
To prove that \( A \subseteq B \), you need to show that if you take an arbitrary element from \( A \), it must always be part of \( B \) too.

• A formal proof involves arguing that for any element \( x \), if \( x \in A \), then \( x \in B \).
• For instance, if \( A = \{1, 2\} \) and \( B = \{1, 2, 3\} \), then \( A \subseteq B \).

The concept of subsets is pivotal when working with distributive laws, as they often depend on subset relationships between intersections and unions.

Set Notation

Set notation is the language we use to describe and work with sets. It includes a variety of symbols and conventions that help us clearly convey set-related concepts and operations.
Some key notations include:

• \( \{ \} \) for denoting a set, e.g., \( \{a, b, c\} \).
• \( \cup \) for union, which combines elements.
• \( \cap \) for intersection, representing common elements.
• \( - \) for difference, indicating elements that remain after exclusion.
• \( \subseteq \) for subset, describing inclusion of all elements.

These symbols and concepts are the backbone of discussing distributive laws in set theory, as they offer a clear and unambiguous way to express mathematical ideas.