Part A: To show that if A, B ∈ F, then A \ B ∈ F, we need to prove that A \ B satisfies the properties of being an element of the set F.
1. A\B is a subset of the universal set: Since A and B are subsets of the universal set, their intersection A ∩ ( \ B) is also a subset of the universal set. Hence, A \ B is also a subset of the universal set.
2. A\B is a set: A \ B is defined as the set of all points that are in A but not in B. Since both A and B are sets, their difference A \ B is also a set.
3. A\B is closed under set operations: Let x be an arbitrary point in A \ B. This means that x is in A but not in B. Since A and B are in F, they satisfy the properties of being elements of F. Thus, x is in A and not in B, which implies that x is in A \ B. Therefore, A \ B is closed under set operations.
By satisfying these properties, A \ B is an element of F. Hence, if A, B ∈ F, then A \ B ∈ F.
Part B: To show that if A, B ∈ F, then A △ B ∈ F, we need to prove that A △ B satisfies the properties of being an element of the set F.
1. A △ B is a subset of the universal set: Since A and B are subsets of the universal set, A △ B is also a subset of the universal set.
2. A △ B is a set: A △ B is defined as the set of points that are in either A or B but not both. Since both A and B are sets, their symmetric difference A △ B is also a set.
3. A △ B is closed under set operations: Let x be an arbitrary point in A △ B. This means that x is either in A or in B, but not in both. Since A and B are in F, they satisfy the properties of being elements of F. Thus, x is either in A or in B (but not both), which implies that x is in A △ B. Therefore, A △ B is closed under set operations.
By satisfying these properties, A △ B is an element of F. Hence, if A, B ∈ F, then A △ B ∈ F.