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Chapter 5: Problem 4

Prove Theorem 5.25, Part (4): \((A \cup B) \times C=(A \times C) \cup(B \times C)\).

Short Answer

Expert verified
To prove Theorem 5.25, Part (4): \((A \cup B) \times C = (A \times C) \cup (B \times C)\), we need to show that any elements in either side of the equation are also in the other side. By considering an arbitrary element \((x,y)\) in both sides and analyzing the two possible cases for \(x\) belonging to either set \(A\) or \(B\), we showed that \((A \cup B) \times C \subseteq (A \times C) \cup (B \times C)\) and \((A \times C) \cup (B \times C) \subseteq (A \cup B) \times C\). Thus, we can conclude that \((A \cup B) \times C = (A \times C) \cup (B \times C)\).

Step by step solution

01

Prove \((A \cup B) \times C \subseteq (A \times C) \cup (B \times C)\)

Let's take an arbitrary element \((x,y) \in (A \cup B) \times C\). By definition of the Cartesian product, this means that \(x \in A \cup B\) and \(y \in C\). Now, if \(x \in A \cup B\), then \(x\) must be in either set \(A\) or set \(B\). We have two cases: Case 1: If \(x \in A\), then \((x,y) \in A \times C\) because \(x \in A\) and \(y \in C\). Case 2: If \(x \in B\), then \((x,y) \in B \times C\) because \(x \in B\) and \(y \in C\). In either case, we find that \((x, y) \in (A \times C) \cup (B \times C)\), since an element belonging to at least one of the sets \(A \times C\) or \(B \times C\) is in their union. Thus, we've shown that any element of \((A \cup B) \times C\) is also an element of \((A \times C) \cup (B \times C)\), which means \((A \cup B) \times C \subseteq (A \times C) \cup (B \times C)\).
02

Prove \((A \times C) \cup (B \times C) \subseteq (A \cup B) \times C\)

Now, let's take an arbitrary element \((x,y) \in (A \times C) \cup (B \times C)\). By definition of the union of sets, this means that either \((x,y) \in A \times C\) or \((x,y) \in B \times C\). Again, we have two cases: Case 1: If \((x,y) \in A \times C\), then by definition of the Cartesian product, \(x \in A\) and \(y \in C\). Since \(x \in A\), it also belongs to the union \(A \cup B\). Hence, \((x,y) \in (A \cup B) \times C\). Case 2: If \((x,y) \in B \times C\), then by definition of the Cartesian product, \(x \in B\) and \(y \in C\). Since \(x \in B\), it also belongs to the union \(A \cup B\). Hence, \((x,y) \in (A \cup B) \times C\). In either case, we find that \((x, y) \in (A \cup B) \times C\). Thus, we've shown that any element of \((A \times C) \cup (B \times C)\) is also an element of \((A \cup B) \times C\), which means \((A \times C) \cup(B \times C) \subseteq (A \cup B) \times C\).
03

Conclusion

We've shown that \((A \cup B) \times C \subseteq (A \times C) \cup (B \times C)\) in Step 1 and \((A \times C) \cup (B \times C) \subseteq (A \cup B) \times C\) in Step 2. Therefore, we can conclude that \((A \cup B) \times C = (A \times C) \cup (B \times C)\).

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Key Concepts

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

Cartesian Product
The Cartesian product is a mathematical operation that returns a set from two given sets. This set consists of all possible ordered pairs where the first element is from the first set and the second element is from the second set.

For example, if we have two sets, Set 1: \( A = \{1, 2\} \) and Set 2: \( B = \{a, b\} \) then the Cartesian product of these sets, denoted \( A \times B \) will be \( \{ (1, a), (1, b), (2, a), (2, b) \} \).

Understanding this concept is critical in mathematics because it forms the basis for creating relations and functions between sets. It's also a foundational element in the fields of geometry, group theory, and other areas of higher mathematics.
Set Theory
Set theory is a branch of mathematical logic that deals with sets, which are collections of objects. In set theory, anything that can be thought of as a distinguished object can be considered an element of a set, whether it's a number, a letter, a symbol, or even another set.

Key concepts in set theory include union and intersection of sets, the notion of subsets, and the empty set. The language and notation of set theory are used throughout modern mathematics. In our example, the consideration of \( A \) and \( B \) as sets allows us to use operations like union and Cartesian products to combine and relate these sets in meaningful ways.

Being proficient in set theory enhances a student's mathematical literacy and problem-solving skills, which are vital in further education and future scientific endeavors.
Union of Sets
The union of sets is an operation that combines all unique elements from each set into a single set. It's denoted with the symbol \( \cup \). For instance, if we have Set 1: \( A \) and Set 2: \( B \) their union \( A \cup B \) includes every element that is in \( A \) or \( B \) or in both.

To put it in a simple context, if Set \( A \) has apples and Set \( B \) has bananas, the union \( A \cup B \) represents a fruit bowl with apples and bananas. There won't be duplicates in a set, just as you wouldn't count the same apple twice in the bowl.

This concept is particularly useful in our original exercise because the operation \( (A \cup B) \times C \) relies on combining all elements from both \( A \) and \( B \) before taking their Cartesian product with \( C \).
Proof Techniques
Mathematical proof writing is an essential skill in set theory and other areas of mathematics. Proofs provide a logical argument that establishes the truth of a mathematical statement beyond doubt. There are several techniques for proving theorems, such as direct proofs, contradiction, induction, and contrapositive.

In the context of our example, the direct proof technique was used. Direct proofs typically start with known information and use logical steps to arrive at the conclusion. As shown, we started with an element of one set, and through logical argument and the definitions of union and Cartesian product, we demonstrated that it must belong to another set.

Understanding proof techniques not only helps validate mathematical theories but also strengthens critical thinking and problem-solving skills, which are invaluable in academic and real-world situations alike.

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Most popular questions from this chapter

Let \(A, B,\) and \(C\) be subsets of a universal set \(U\). (a) Draw a Venn diagram with \(A \subseteq B\) and \(B \subseteq C .\) Does it appear that \(A \subseteq C ?\) (b) Prove the following proposition: If \(A \subseteq B\) and \(B \subseteq C,\) then \(A \subseteq C\) This may seem like an obvious result. However, one of the reasons for this exercise is to provide practice at properly writing a proof that one set is a subset of another set. So we should start the proof by assuming that \(A \subseteq B\) and \(B \subseteq C .\) Then we should choose an arbitrary element of \(A\).

Let \(A, B,\) and \(C\) be subsets of some universal set \(U\) (a) Draw two general Venn diagrams for the sets \(A, B,\) and \(C .\) On one, shade the region that represents \(A-(B-C),\) and on the other, shade the region that represents \((A-B)-C .\) Based on the Venn diagrams, make a conjecture about the relationship between the sets \(A-(B-C)\) and \((A-B)-C .\) (Are the two sets equal? If not, is one of the sets a subset of the other set?) (b) Prove the conjecture from Exercise (7a).

Intervals of Real Numbers. In previous mathematics courses, we have frequently used subsets of the real numbers called intervals. There are some common names and notations for intervals. These are given in the following table, where it is assumed that \(a\) and \(b\) are real numbers and \(aa\\} & \text { Open ray } \\ (-\infty, b)= & \\{x \in \mathbb{R} \mid x2\\}\) as the union of two intervals.

To prove the following set equalities, it may be necessary to use some of the properties of positive and negative real numbers. For example, it may be necessary to use the facts that: \- The product of two real numbers is positive if and only if the two real numbers are either both positive or both negative. \- The product of two real numbers is negative if and only if one of the two numbers is positive and the other is negative. For example, if \(x(x-2)<0,\) then we can conclude that either \((1) x<0\) and \(x-2>0\) or \((2) x>0\) and \(x-2<0 .\) However, in the first case, we must have \(x<0\) and \(x>2\), and this is impossible. Therefore, we conclude that \(x>0\) and \(x-2<0,\) which means that \(0

Use the definitions of set intersection, set union, and set difference to write useful negations of these definitions. That is, complete each of the following sentences (a) \(x \notin A \cap B\) if and only if \(\ldots\) (b) \(x \notin A \cup B\) if and only if \(\ldots\) (c) \(x \notin A-B\) if and only if \(\ldots\)
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