Question

Prove that (i) \(\left(\bigcup A_{i}\right) \times B=\bigcup\left(A_{i} \times B\right) ;\) (ii) \(\left(\cap A_{i}\right) \times B=\bigcap\left(A_{i} \times B\right) ;\) (iii) \(\left(\bigcap_{i} A_{i}\right) \times\left(\bigcap_{j} B_{j}\right)=\bigcap_{i, j}\left(A_{i} \times B_{i}\right) ;\) (iv) \(\left(\bigcup_{i} A_{i}\right) \times\left(\bigcup_{j} B_{j}\right)=\bigcup_{i, j}\left(A_{i} \times B_{j}\right)\).

Short answer

Each of the given set operations in the Cartesian product holds as stated: (i), (ii), (iii), and (iv) are proven true by set inclusion arguments.

Step-by-step solution

Understanding the Problem Statement (i)

To prove \( (\bigcup A_i) \times B = \bigcup (A_i \times B) \), we need to show that any element \((x, y)\) belongs to the left-hand side if and only if it belongs to the right-hand side.

Proof of (i) Left to Right Inclusion

Assume \((x, y) \in (\bigcup A_i) \times B\). This means \(x \in \bigcup A_i\) and \(y \in B\). Hence, there exists an \(i\) such that \(x \in A_i\), which implies \((x, y) \in A_i \times B\). Thus, \((x, y) \in \bigcup (A_i \times B)\).

Proof of (i) Right to Left Inclusion

Assume \((x, y) \in \bigcup (A_i \times B)\). This means there exists an \(i\) such that \((x, y) \in A_i \times B\), implying \(x \in A_i\) and \(y \in B\). Therefore, \(x \in \bigcup A_i\), and hence \((x, y) \in (\bigcup A_i) \times B\).

Conclusion for Statement (i)

We have shown both inclusions, hence \((\bigcup A_i) \times B = \bigcup (A_i \times B)\), proving statement (i).

Understanding the Problem Statement (ii)

For \((\bigcap A_i) \times B = \bigcap (A_i \times B)\), we need to show that \((x, y)\) belongs to the left-hand side if and only if it belongs to the right-hand side.

Proof of (ii) Left to Right Inclusion

Assume \((x, y) \in (\bigcap A_i) \times B\). This means \(x \in \bigcap A_i\) (i.e., \(x \in A_i\) for all \(i\)) and \(y \in B\). Hence, \((x, y) \in A_i \times B\) for each \(i\), so \((x, y) \in \bigcap (A_i \times B)\).

Proof of (ii) Right to Left Inclusion

Assume \((x, y) \in \bigcap (A_i \times B)\). Then \((x, y) \in A_i \times B\) for every \(i\), hence \(x \in A_i\) for every \(i\), implying \(x \in \bigcap A_i\) and \(y \in B\), thus \((x, y) \in (\bigcap A_i) \times B\).

Conclusion for Statement (ii)

Both inclusions are shown, thus \((\bigcap A_i) \times B = \bigcap (A_i \times B)\), proving statement (ii).

Setting Up for Problem Statement (iii)

For \((\bigcap_{i} A_{i}) \times (\bigcap_{j} B_{j}) = \bigcap_{i,j} (A_{i} \times B_{j})\), show any element \((x, y)\) in the left-hand side is in the right-hand side and vice versa.

Proof of (iii) Left to Right Inclusion

If \((x, y) \in (\bigcap A_i) \times (\bigcap B_j)\), then \(x \in \bigcap A_i\) and \(y \in \bigcap B_j\). Therefore, for each \(i, j\), \(x \in A_i\) and \(y \in B_j\), thus \((x, y) \in A_i \times B_j\), implying \((x, y) \in \bigcap (A_i \times B_j)\).

Proof of (iii) Right to Left Inclusion

Assume \((x, y) \in \bigcap (A_i \times B_j)\). It means for all \(i, j\), \((x, y) \in A_i \times B_j\), implying \(x \in A_i\) and \(y \in B_j\) for all \(i, j\). Thus, \(x \in \bigcap A_i\) and \(y \in \bigcap B_j\). Hence, \((x, y) \in (\bigcap A_i) \times (\bigcap B_j)\).

Conclusion for Statement (iii)

With both inclusions validated, \((\bigcap A_i) \times (\bigcap B_j) = \bigcap (A_i \times B_j)\). Hence, statement (iii) is proven.

Set Up for Problem Statement (iv)

For \((\bigcup A_i) \times (\bigcup B_j) = \bigcup (A_i \times B_j)\), show each element \((x, y)\) in the left side is in the right side and vice versa.

Proof of (iv) Left to Right Inclusion

Assume \((x, y) \in (\bigcup A_i) \times (\bigcup B_j)\). This implies \(x \in \bigcup A_i\) and \(y \in \bigcup B_j\). Then \(\exists i, j\) such that \(x \in A_i\) and \(y \in B_j\), indicating \((x, y) \in A_i \times B_j\) and hence, \((x, y) \in \bigcup (A_i \times B_j)\).

Proof of (iv) Right to Left Inclusion

Assume \((x, y) \in \bigcup (A_i \times B_j)\). There exist \(i, j\) such that \((x, y) \in A_i \times B_j\), meaning \(x \in A_i\) and \(y \in B_j\). Therefore, \(x \in \bigcup A_i\) and \(y \in \bigcup B_j\), so \((x, y) \in (\bigcup A_i) \times (\bigcup B_j)\).

Conclusion for Statement (iv)

Both inclusions have been confirmed, thus \((\bigcup A_i) \times (\bigcup B_j) = \bigcup (A_i \times B_j)\). Therefore, statement (iv) is proven.

Key concepts

Union of Sets

In set theory, the concept of the union of sets is fundamental. The union of a collection of sets refers to the set containing all the elements from each of the sets in the collection. It is a way to gather all elements without repetition. In mathematical notation, the union of a collection of sets \( \{A_i\} \) is expressed as \( \bigcup A_i \). This notation indicates all elements that are in at least one of the sets \( A_i \).

To better understand, consider you have three sets: \(A_1 = \{1, 2\}\), \(A_2 = \{2, 3\}\), and \(A_3 = \{3, 4\}\). The union of these sets \( \bigcup A_i = \{1, 2, 3, 4\}\) includes every element from each set, but only counts each unique item once. This set theory operation is crucial when working with Cartesian Products, as it allows us to combine elements from various sets into a single overarching set.

Intersection of Sets

The intersection of sets is another core operation within set theory, focusing on common elements shared among sets. When two or more sets are involved, their intersection comprises only the elements found in every set. Represented mathematically by the symbol \( \cap \), the intersection between sets \( \{A_i\} \) is denoted as \( \bigcap A_i \). This means identifying elements that are present in each set \( A_i \).

For instance, if you have \(A_1 = \{1, 2\}\), \(A_2 = \{2, 3\}\), and \(A_3 = \{2, 4\}\), their intersection \( \bigcap A_i = \{2\}\) contains only the number 2, since it is the only element common to all three sets. The concept of intersection is particularly relevant in the context of proving Cartesian product properties, as it helps in outlining elements that strictly belong to specified conditions across different datasets.

Proof Techniques in Mathematics

Proof techniques are essential tools in mathematics that help validate statements, identities, or properties. Proving mathematical statements ensures that they are universally accepted and logically sound. Some key proof methods include direct proofs, proof by contradiction, proof by induction, and combinatorial proofs. In the context of Cartesian products and set theory, direct proofs and contradictions often play a vital role.

In direct proofs, you assume the premises of a statement to demonstrate its conclusions directly. For the exercises above, demonstrating equivalences between various expressions involving Cartesian products involves showing both directions of inclusion. For example, to prove \((\bigcup A_i) \times B = \bigcup (A_i \times B)\), we need to show that each side is a subset of the other, thereby proving equality.

By leveraging these proof techniques, one can rigorously establish the correctness of mathematical theorems and foster a deeper comprehension of their underlying principles.

Distributive Laws for Cartesian Products

The distributive laws in set theory are analogous to those in arithmetic, helping to simplify and manipulate expressions involving union and intersection within Cartesian products. These laws outline how Cartesian products interact with union and intersection operations, providing valuable tools for dealing with complex set expressions.

The distributive law for Cartesian products states that:

• \((\bigcup A_i) \times B = \bigcup (A_i \times B)\)
• \((\bigcap A_i) \times B = \bigcap (A_i \times B)\)
• \((\bigcap A_i) \times (\bigcap B_j) = \bigcap (A_i \times B_j)\)
• \((\bigcup A_i) \times (\bigcup B_j) = \bigcup (A_i \times B_j)\)

These laws are essential for simplifying complex expressions and play an instrumental role in formal proofs, as illustrated in the original exercise. By understanding and applying these distributive properties, you can effectively navigate and solve problems that involve set operations and Cartesian products.