Question

Prove that ( P(A-B) subseteq(P(A)-P(B)) cup{emptyset} ).

Answer 1

To prove the statement, we need to show that every element in P(A-B) is also in (P(A)-P(B)) cup {emptyset}.

Step-by-step explanation:

To prove that ( P(A-B) ⊆ (P(A)-P(B)) ∪{emptyset} ), we can start by showing that every element in P(A-B) is also in (P(A)-P(B)) ∪{emptyset}.

Let x be an arbitrary element in P(A-B). This means that x is in A but not in B. Therefore, x is in P(A) but not in P(B), so x is in (P(A)-P(B)). Thus, x is in (P(A)-P(B)) ∪ {emptyset}.

Since x was arbitrary, this shows that every element in P(A-B) is also in (P(A)-P(B)) ∪ {emptyset}. Hence, we have proven that P(A-B) ⊆ (P(A)-P(B)) ∪ {emptyset}.