(Real) Set Theory: Part 4.1

Sample Proofs of Set Theory Statements

1. Proof that A ∪ B = B ∪ A

• Objective: To prove that the union of two sets A and B is commutative, that is, the equation A ∪ B is equal to B ∪ A holds for all sets A and B.

• Set Containment: To establish set equality, we must demonstrate set containment in both directions:

  • Direction 1: Show that A ∪ B ⊆ B ∪ A

  • Direction 2: Show that B ∪ A ⊆ A ∪ B

• Proof for A ∪ B ⊆ B ∪ A:

  • Assume x ∈ A ∪ B. By the definition of union, this implies that x must be an element of either A or B.

  • Case 1: If x ∈ A, then by the logical implication (if x is in A, then x is also in B ∪ A), it must follow that x ∈ B ∪ A.

  • Case 2: If x ∈ B, by the implication similarly, x is also in B ∪ A.

  • Thus, regardless of which set x belongs to, we conclude that x ∈ B ∪ A.

  • Hence, we have demonstrated that if x ∈ A ∪ B, then x must also belong to B ∪ A.

• Conclusion of Direction 1: From both cases, we conclude A ∪ B ⊆ B ∪ A is established.

• Proof for B ∪ A ⊆ A ∪ B:

  • Armed with similar reasoning, let x be an element in B ∪ A.

  • By the definition of union, this either means x ∈ B or x ∈ A.

  • Case 1: If x ∈ B, then clearly x meets the condition for A ∪ B, so we can conclude that x ∈ A ∪ B.

  • Case 2: If x ∈ A, it follows directly that x must also be in A ∪ B.

• Final Conclusion: Since both containments are proven, we conclude that A ∪ B = B ∪ A, confirming the commutative property of set union. QED.

2. Proof that A ∩ B ≠ A ∪ B

• Objective: This statement requires a careful examination of the relationship between intersection and union of sets A and B, with a particular focus on their definitions and properties.

  • To establish that A ∩ B equals A ∪ B necessitates the explicit demonstration of two important inclusions:

    • Inclusion 1: A ∩ B ⊆ A ∪ B

    • Inclusion 2: A ∪ B ⊆ A ∩ B

Step 1: Prove A ∩ B ⊆ A ∪ B

• Consider an element x ∈ A ∩ B. By definition of intersection, this means that x must satisfy:

  • x ∈ A and x ∈ B

• Consequently, as both conditions are fulfilled, it follows by the definition of union that:

  • x ∈ A ∪ B.

Step 2: Prove A ∪ B ⊆ A ∩ B

• Let x be an arbitrary element in A ∪ B.

• By the definition of union, it implies:

  • x is in A or x is in B.

• Case Analysis: Here comes the critical point of failure:

  • Case 1: Suppose x ∈ A, then it does not necessitate that x is also in B; therefore this does not imply x ∈ A ∩ B.

  • Case 2: In a similar manner, if x ∈ B, it does not automatically infer that x is in A, making it evident that x cannot be established as an element of A ∩ B.

• Therefore, we discover that A ∪ B is not a subset of A ∩ B.

Conclusion

• This indicates that the intersections and unions do not contain the same elements collectively. Hence, while we have validated the inclusion from intersection to union, the reverse case fails. Thus, this reveals the distinct nature of these operations, confirming contradictions that arise in our attempt to equate them. QED.

3. Proving A ∩ B ⊆ B (where A = {1, 2, 3} and B = {2, 4})

• General Objective: Demonstrating that for any two sets, A and B, the intersection A ∩ B is contained within set B emphasizes the allocation of shared elements.

• Assume:

  • Let x ∈ A ∩ B.

• Given the definition of intersection, this condition leads us to:

  • x must be an element of both sets A and B.

Example Counterargument

• Exploring the sets in context, we know that A ∩ B = {2}, given that this is the only common element.

• We now establish:

  • B = {2, 4}, confirming that the element 2 is indeed part of B.

  • While x must belong to A, we verify that our intersection A ∩ B = {2} does not include every element of A.

• Conclusion of Containment: This vital finding shows A ∩ B = {2} is entirely contained within B = {2, 4} confirming that B remains valid as a containing structure for the intersection of the two sets. Here, the containment holds true, with no counter-example invalidating the assertion.

4. Proof that A ∪ Ø = A

• Goal: To demonstrate that the union of any set A with the empty set Ø remains equal to A itself; formal proof entails demonstrating two containment directions:

• For A ∪ Ø ⊆ A:

  • Let x ∈ A ∪ Ø. By definition of union:

  • x could either belong to A or Ø, and since Ø possesses no elements, we deduce that:

    • x must be in A.

• For A ⊆ A ∪ Ø:

  • For any x ∈ A, it is clear by the definition of union that:

    • Since x is in A, it follows that x is also in A ∪ Ø.

Final Summary

• Thus, we affirmatively conclude both containment directions are valid, confirming the statement A ∪ Ø = A holds true with certainty. QED.