- #1
YamiBustamante
- 17
- 0
There's what I have so far.
We assume that P(A) ⊆ P(B). This means that every element x that exists in P(A), also exits in P(B). By definition of a power set, x∈P(A) if x ⊆ A. Therefore, A∈P(A). Since P(A) ⊆ P(B), A∈P(B), meaning all x ⊆ A, x ∈ P(B). Furthermore, B∈P(B), meaning all x ⊆ B, x ∈ P(B). Since x ⊆ A and x ⊆ B and P(A) ⊆ P(B), A ⊆ B.
Is my proof correct?
We assume that P(A) ⊆ P(B). This means that every element x that exists in P(A), also exits in P(B). By definition of a power set, x∈P(A) if x ⊆ A. Therefore, A∈P(A). Since P(A) ⊆ P(B), A∈P(B), meaning all x ⊆ A, x ∈ P(B). Furthermore, B∈P(B), meaning all x ⊆ B, x ∈ P(B). Since x ⊆ A and x ⊆ B and P(A) ⊆ P(B), A ⊆ B.
Is my proof correct?