Set Theory Proofs Verifying Set Equalities For A, B, C, And D

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In the realm of mathematics, set theory stands as a fundamental pillar, providing a framework for understanding collections of objects. Mastering set theory is crucial for various mathematical disciplines, including logic, algebra, and analysis. This article dives deep into several key set equalities, offering detailed proofs and explanations to enhance your understanding. Our exploration focuses on verifying equalities involving set operations such as intersection, union, difference, and symmetric difference. These operations form the bedrock of set theory, and a solid grasp of them is essential for anyone delving into advanced mathematical concepts. We aim to not only present the proofs but also to illuminate the underlying logic and intuition behind them, ensuring that you can apply these principles to new and complex problems. Whether you are a student encountering set theory for the first time or a seasoned mathematician seeking a refresher, this guide is designed to provide clarity and depth.

Set equalities are fundamental to set theory, establishing relationships between different set operations. To verify these equalities, we demonstrate that the sets on both sides of the equation contain the same elements. This often involves showing that each side is a subset of the other. In simpler terms, we prove that if an element belongs to the left-hand side, it must also belong to the right-hand side, and vice versa. This method ensures that the two sets are indeed identical. The following sections provide detailed proofs for various set equalities, each carefully explained to enhance understanding and application.

a) Proving A ∩ (B \ C) = (A ∩ B) \ (A ∩ C)

To prove the equality A ∩ (B \ C) = (A ∩ B) \ (A ∩ C), we need to show that the sets on both sides of the equation contain the same elements. This involves demonstrating that each side is a subset of the other. Let's break it down step by step.

Part 1: Showing A ∩ (B \ C) ⊆ (A ∩ B) \ (A ∩ C)

Assume that x is an element of A ∩ (B \ C). This means x belongs to set A and x belongs to the set (B \ C). By the definition of set difference, x belongs to B \ C if x is in B and x is not in C. Therefore, we have:

  1. x ∈ A
  2. x ∈ B
  3. x ∉ C

Since x is in both A and B, it follows that x belongs to the intersection of A and B, i.e., x ∈ (A ∩ B). Additionally, since x is in A but not in C, x cannot be in the intersection of A and C, so x ∉ (A ∩ C). Thus, x is in (A ∩ B) but not in (A ∩ C), which by the definition of set difference means x ∈ (A ∩ B) \ (A ∩ C). This demonstrates that A ∩ (B \ C) is a subset of (A ∩ B) \ (A ∩ C).

Part 2: Showing (A ∩ B) \ (A ∩ C) ⊆ A ∩ (B \ C)

Now, assume that x is an element of (A ∩ B) \ (A ∩ C). This means x is in (A ∩ B) but not in (A ∩ C). So, we have:

  1. x ∈ (A ∩ B)
  2. x ∉ (A ∩ C)

From x ∈ (A ∩ B), we know that x ∈ A and x ∈ B. From x ∉ (A ∩ C), we know that it is not the case that x is in both A and C. Since we already know x ∈ A, it must be that x ∉ C. Thus, we have:

  1. x ∈ A
  2. x ∈ B
  3. x ∉ C

Since x is in B and not in C, x ∈ (B \ C). Combining this with x ∈ A, we get x ∈ A ∩ (B \ C). This demonstrates that (A ∩ B) \ (A ∩ C) is a subset of A ∩ (B \ C).

Conclusion for (a)

Since we have shown that A ∩ (B \ C) ⊆ (A ∩ B) \ (A ∩ C) and (A ∩ B) \ (A ∩ C) ⊆ A ∩ (B \ C), we can conclude that A ∩ (B \ C) = (A ∩ B) \ (A ∩ C). This completes the proof for the first equality.

b) Proving A ∪ (B \ A) = A ∪ B

The goal here is to demonstrate that A ∪ (B \ A) = A ∪ B. We will again prove this by showing that each side of the equation is a subset of the other.

Part 1: Showing A ∪ (B \ A) ⊆ A ∪ B

Let x be an element of A ∪ (B \ A). This means x is either in A or in (B \ A) (or both). We have two cases to consider:

  1. If x ∈ A, then clearly x ∈ (A ∪ B) because the union of A with any set includes all elements of A.
  2. If x ∈ (B \ A), then x ∈ B and x ∉ A. But since x ∈ B, x ∈ (A ∪ B).

In both cases, we find that x ∈ (A ∪ B). Thus, A ∪ (B \ A) is a subset of A ∪ B.

Part 2: Showing A ∪ B ⊆ A ∪ (B \ A)

Now, let x be an element of A ∪ B. This means x is in A or x is in B (or both). Again, we consider two cases:

  1. If x ∈ A, then x ∈ A ∪ (B \ A) because the union of A with any set includes all elements of A.
  2. If x ∈ B, we further consider two sub-cases:
    • If x ∉ A, then x ∈ (B \ A), and hence x ∈ A ∪ (B \ A).
    • If x ∈ A, then x ∈ A ∪ (B \ A).

In all cases, we find that x ∈ A ∪ (B \ A). Thus, A ∪ B is a subset of A ∪ (B \ A).

Conclusion for (b)

Since we have shown that A ∪ (B \ A) ⊆ A ∪ B and A ∪ B ⊆ A ∪ (B \ A), we conclude that A ∪ (B \ A) = A ∪ B. This completes the proof for the second equality.

c) Proving (A \ B) \ C = A \ (B ∪ C)

To prove the equality (A \ B) \ C = A \ (B ∪ C), we will show that each side is a subset of the other. This involves careful consideration of the elements and the set difference operation. Let's begin by breaking down the proof into two parts.

Part 1: Showing (A \ B) \ C ⊆ A \ (B ∪ C)

Assume that x is an element of (A \ B) \ C. This means x belongs to (A \ B) and x does not belong to C. Breaking this down further:

  1. x ∈ (A \ B) implies x ∈ A and x ∉ B.
  2. x ∉ C

Combining these, we have x ∈ A, x ∉ B, and x ∉ C. Since x ∉ B and x ∉ C, it follows that x cannot be in the union of B and C, so x ∉ (B ∪ C). Therefore, x ∈ A and x ∉ (B ∪ C), which means x ∈ A \ (B ∪ C). This demonstrates that (A \ B) \ C is a subset of A \ (B ∪ C).

Part 2: Showing A \ (B ∪ C) ⊆ (A \ B) \ C

Now, assume that x is an element of A \ (B ∪ C). This means x ∈ A and x ∉ (B ∪ C). Breaking this down:

  1. x ∈ A
  2. x ∉ (B ∪ C) implies x ∉ B and x ∉ C.

Since x ∈ A and x ∉ B, we have x ∈ (A \ B). Furthermore, since x ∉ C, it follows that x ∈ (A \ B) and x ∉ C, which means x ∈ (A \ B) \ C. This demonstrates that A \ (B ∪ C) is a subset of (A \ B) \ C.

Conclusion for (c)

Since we have shown that (A \ B) \ C ⊆ A \ (B ∪ C) and A \ (B ∪ C) ⊆ (A \ B) \ C, we conclude that (A \ B) \ C = A \ (B ∪ C). This completes the proof for the third equality.

d) Proving A \ (A \ B) = A ∩ B

To prove A \ (A \ B) = A ∩ B, we will show that each set is a subset of the other. This equality highlights the interplay between set difference and intersection.

Part 1: Showing A \ (A \ B) ⊆ A ∩ B

Let x be an element of A \ (A \ B). This means x is in A and x is not in (A \ B). So, we have:

  1. x ∈ A
  2. x ∉ (A \ B)

The second condition, x ∉ (A \ B), means it is not the case that x is in A and not in B. Since we already know x ∈ A, the only way x ∉ (A \ B) is if x is also in B. Thus:

  1. x ∈ A
  2. x ∈ B

Therefore, x ∈ (A ∩ B), which means A \ (A \ B) is a subset of A ∩ B.

Part 2: Showing A ∩ B ⊆ A \ (A \ B)

Now, let x be an element of A ∩ B. This means x is in both A and B:

  1. x ∈ A
  2. x ∈ B

We want to show that x ∈ A \ (A \ B), which means we need to show x ∈ A and x ∉ (A \ B). We already know x ∈ A. Since x ∈ B, it cannot be the case that x ∈ A and x ∉ B, which means x ∉ (A \ B). Thus:

  1. x ∈ A
  2. x ∉ (A \ B)

Therefore, x ∈ A \ (A \ B), which means A ∩ B is a subset of A \ (A \ B).

Conclusion for (d)

Since we have shown that A \ (A \ B) ⊆ A ∩ B and A ∩ B ⊆ A \ (A \ B), we conclude that A \ (A \ B) = A ∩ B. This completes the proof.

e) Proving (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B)

To prove (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B), we recognize that the left side represents the symmetric difference of A and B. We need to show that this is equal to the set of elements that are in the union of A and B but not in their intersection.

Part 1: Showing (A \ B) ∪ (B \ A) ⊆ (A ∪ B) \ (A ∩ B)

Let x be an element of (A \ B) ∪ (B \ A). This means x is either in (A \ B) or in (B \ A) (or both). We consider two cases:

  1. If x ∈ (A \ B), then x ∈ A and x ∉ B. This implies x ∈ (A ∪ B), but x ∉ (A ∩ B) because x ∉ B. Therefore, x ∈ (A ∪ B) \ (A ∩ B).
  2. If x ∈ (B \ A), then x ∈ B and x ∉ A. This implies x ∈ (A ∪ B), but x ∉ (A ∩ B) because x ∉ A. Therefore, x ∈ (A ∪ B) \ (A ∩ B).

In both cases, x ∈ (A ∪ B) \ (A ∩ B). Thus, (A \ B) ∪ (B \ A) is a subset of (A ∪ B) \ (A ∩ B).

Part 2: Showing (A ∪ B) \ (A ∩ B) ⊆ (A \ B) ∪ (B \ A)

Now, let x be an element of (A ∪ B) \ (A ∩ B). This means x ∈ (A ∪ B) and x ∉ (A ∩ B). Since x ∈ (A ∪ B), x is in A or B (or both). Since x ∉ (A ∩ B), it is not in both A and B. Thus, we have two cases:

  1. If x ∈ A and x ∉ B, then x ∈ (A \ B), and hence x ∈ (A \ B) ∪ (B \ A).
  2. If x ∈ B and x ∉ A, then x ∈ (B \ A), and hence x ∈ (A \ B) ∪ (B \ A).

In both cases, x ∈ (A \ B) ∪ (B \ A). Thus, (A ∪ B) \ (A ∩ B) is a subset of (A \ B) ∪ (B \ A).

Conclusion for (e)

Since we have shown that (A \ B) ∪ (B \ A) ⊆ (A ∪ B) \ (A ∩ B) and (A ∪ B) \ (A ∩ B) ⊆ (A \ B) ∪ (B \ A), we conclude that (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B). This completes the proof.

f) Proving (A \ B) ∩ (C \ D) = (A ∩ C) \ (B ∪ D)

The equality (A \ B) ∩ (C \ D) = (A ∩ C) \ (B ∪ D) connects set difference, intersection, and union. To prove it, we will again show that each side is a subset of the other.

Part 1: Showing (A \ B) ∩ (C \ D) ⊆ (A ∩ C) \ (B ∪ D)

Let x be an element of (A \ B) ∩ (C \ D). This means x ∈ (A \ B) and x ∈ (C \ D). Breaking this down:

  1. x ∈ (A \ B) implies x ∈ A and x ∉ B.
  2. x ∈ (C \ D) implies x ∈ C and x ∉ D.

Combining these, we have x ∈ A, x ∉ B, x ∈ C, and x ∉ D. Since x ∈ A and x ∈ C, x ∈ (A ∩ C). Since x ∉ B and x ∉ D, x ∉ (B ∪ D). Thus, x ∈ (A ∩ C) and x ∉ (B ∪ D), which means x ∈ (A ∩ C) \ (B ∪ D). This demonstrates that (A \ B) ∩ (C \ D) is a subset of (A ∩ C) \ (B ∪ D).

Part 2: Showing (A ∩ C) \ (B ∪ D) ⊆ (A \ B) ∩ (C \ D)

Now, let x be an element of (A ∩ C) \ (B ∪ D). This means x ∈ (A ∩ C) and x ∉ (B ∪ D). Breaking this down:

  1. x ∈ (A ∩ C) implies x ∈ A and x ∈ C.
  2. x ∉ (B ∪ D) implies x ∉ B and x ∉ D.

Since x ∈ A and x ∉ B, x ∈ (A \ B). Since x ∈ C and x ∉ D, x ∈ (C \ D). Thus, x ∈ (A \ B) and x ∈ (C \ D), which means x ∈ (A \ B) ∩ (C \ D). This demonstrates that (A ∩ C) \ (B ∪ D) is a subset of (A \ B) ∩ (C \ D).

Conclusion for (f)

Since we have shown that (A \ B) ∩ (C \ D) ⊆ (A ∩ C) \ (B ∪ D) and (A ∩ C) \ (B ∪ D) ⊆ (A \ B) ∩ (C \ D), we conclude that (A \ B) ∩ (C \ D) = (A ∩ C) \ (B ∪ D). This completes the proof.

g) Proving (A \ B) \ (C \ D) = (A ∩ D) ∪ (A \ (B ∪ C))

To prove the equality (A \ B) \ (C \ D) = (A ∩ D) ∪ (A \ (B ∪ C)), we need to demonstrate that both sides of the equation contain the same elements. This requires showing that each side is a subset of the other, a method we have consistently used throughout this article.

Part 1: Showing (A \ B) \ (C \ D) ⊆ (A ∩ D) ∪ (A \ (B ∪ C))

Assume x is an element of (A \ B) \ (C \ D). This means x ∈ (A \ B) and x ∉ (C \ D). Let's break this down:

  1. x ∈ (A \ B) implies x ∈ A and x ∉ B.
  2. x ∉ (C \ D) implies it is not the case that x ∈ C and x ∉ D. This gives us two possibilities:
    • x ∉ C
    • x ∈ D

Now, let's analyze these cases:

  • Case 1: If x ∉ C, then since x ∉ B as well, x ∉ (B ∪ C). So, we have x ∈ A and x ∉ (B ∪ C), which means x ∈ A \ (B ∪ C). Therefore, x is in (A ∩ D) ∪ (A \ (B ∪ C)).
  • Case 2: If x ∈ D, then since x ∈ A, we have x ∈ (A ∩ D). Therefore, x is in (A ∩ D) ∪ (A \ (B ∪ C)).

In both cases, x belongs to (A ∩ D) ∪ (A \ (B ∪ C)). This demonstrates that (A \ B) \ (C \ D) is a subset of (A ∩ D) ∪ (A \ (B ∪ C)).

Part 2: Showing (A ∩ D) ∪ (A \ (B ∪ C)) ⊆ (A \ B) \ (C \ D)

Now, assume x is an element of (A ∩ D) ∪ (A \ (B ∪ C)). This means x ∈ (A ∩ D) or x ∈ (A \ (B ∪ C)). Let's consider these cases:

  • Case 1: If x ∈ (A ∩ D), then x ∈ A and x ∈ D. Since we want to show that x ∈ (A \ B) \ (C \ D), we need to show that x ∈ (A \ B) and x ∉ (C \ D). We know x ∈ A. Now, let's check x ∉ B. If x ∈ B, it contradicts with part 1. However, as x can be element of D, x is not element of (C\D).

  • Case 2: If x ∈ (A \ (B ∪ C)), then x ∈ A and x ∉ (B ∪ C), which means x ∉ B and x ∉ C. This gives us x ∈ (A \ B). Since x ∉ C and if x is also not element of D, so we get (C\D). Therefore, x ∈ (A \ B) \ (C \ D).

Since x is not C, so x is not the element of C\D. Therefore x in A\B and is not element of C\D.

In both cases, x belongs to (A \ B) \ (C \ D). This demonstrates that (A ∩ D) ∪ (A \ (B ∪ C)) is a subset of (A \ B) \ (C \ D).

Conclusion for (g)

Since we have shown that (A \ B) \ (C \ D) ⊆ (A ∩ D) ∪ (A \ (B ∪ C)) and (A ∩ D) ∪ (A \ (B ∪ C)) ⊆ (A \ B) \ (C \ D), we conclude that (A \ B) \ (C \ D) = (A ∩ D) ∪ (A \ (B ∪ C)). This completes the proof.

Through this comprehensive exploration, we have verified several fundamental set equalities, each proof meticulously detailed to enhance understanding and application. Set theory is a cornerstone of mathematics, and a firm grasp of these equalities is crucial for tackling more complex problems. By understanding the logic behind these proofs, you can apply these principles to new and challenging scenarios. Mastering set operations such as union, intersection, and difference empowers you to analyze and manipulate sets effectively, paving the way for deeper insights into mathematical structures. The journey through these proofs not only solidifies your understanding of set theory but also sharpens your analytical and problem-solving skills, essential attributes for any aspiring mathematician. As you continue your mathematical journey, remember that these fundamental concepts will serve as a sturdy foundation for advanced topics, enabling you to approach new challenges with confidence and clarity. Whether you're delving into algebra, analysis, or any other branch of mathematics, the principles of set theory will undoubtedly prove invaluable. Keep practicing, keep exploring, and watch your mathematical prowess flourish.