Final answer:
To prove that p(A) ⊆ p(B) if and only if A ⊆ B, we consider elements in the power sets and show that elements of A being subsets of B implies p(A) is a subset of p(B), and conversely, elements of p(A) being subsets of p(B) implies A is a subset of B.
Step-by-step explanation:
Let's prove that p(A) ⊆ p(B) if and only if A ⊆ B. Here, p(A) and p(B) represent the power sets of A and B, respectively, which are the sets of all subsets of A and B.
→ Direction: Assume that A ⊆ B. Consider any element x in p(A). Since x is a subset of A, and A is a subset of B, x must also be a subset of B. Hence, x belongs to p(B). This shows that every element of p(A) is also an element of p(B), thus p(A) ⊆ p(B).
← Direction: Conversely, assume that p(A) ⊆ p(B). Take any element y in A. The set containing y only, {y}, is a subset of A, and since p(A) is a subset of p(B), {y} must be a subset of B. Therefore, y must be an element of B. Since this is true for any arbitrary element y in A, it follows that A ⊆ B.
Through this proof, we have established the bidirectional conditional statement confirming the relationship between the subsets of sets and their power sets.