Question

Given nonempty sets A and B, prove that every set in P(B − A) ⊆ P(B) − P(A).

Answer 1

Proof:

Let
`X \in P(B-A)`. As we chose
`X` in
`P(B-A)` we know that
`X \subseteq B-A`. Since
`B-A \subseteq B` by transitivity we get:

`X \subseteq B \quad \implies X \in P(B)`.

If
`X` is the empty set, we already have that
`X = \emptyset \in P(B) - P(A)`. But if
`X` is not empty, that means that it can't be subset of
`A`, because
`X` is already subset of
`B-A`, and those sets do not share any element. In other words:

`X \subseteq A \cup (B-A) = \emptyset`

`\Rightarrow X = \emptyset`

As
`X` can't be subset of
`A`, then
`X\\otin P(A)`.
`X` was an arbitrary element, and

1. 
   `X \in P(B)`
2. 
   `X\\otin P(A)`

Thus,
`X\in P(B)-P(A)`, where we conclude that

`P(B-A) \subseteq P(B) - P(A)`