Question

Either prove or disprove (by giving a counterexample) this claim re- garding powersets and set containment: For any sets A and B, if A ⊆ B, then P(A) ⊆ P(B)

Answer 1

Answer :

For all sets A and B, A ⊆ B then P(A) ⊆ P(B).

Proof (→):

Suppose that A and B are sets and A ⊆ B. Suppose that S is an element of P(A). By the definition of power set, S
must be a subset of A. Since S ⊆ A and A ⊆ B, we must have that S ⊆ B . Since S ⊆ B, the definition of power set implies that S ∈ P(B). Since we’ve shown that any element of P(A) is also an element of P(B), we have that P(A) ⊆ P(B).
A really common mistake is to stop at this point, thinking you are done.But we’ve only done half the job. We need to show that the implication works in the other direction:

Proof (←): Suppose that A and B are sets and P(A) ⊆ P(B).

By the definition of power set, A ∈ P(A). Since A ∈ P(A) and P(A) ⊆ P(B), we know that A ∈ P(B) (definition of subset). So,
by the definition of power set, A ⊆ B.

Example :

Suppose first that p(A)⊆p(B). A⊆A,

so A∈p(A)⊆p(B),

so A∈p(B),

hence A⊆B

.Now suppose that A⊆B

. Then for any X∈p(A) we have X⊆A⊆B,

so X⊆B, and therefore X∈p(B).

Thus, p(A)⊆p(B).