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 $X\notin P(A)$. $X$ was an arbitrary element, and

1. $X \in P(B)$
2. $X\notin P(A)$

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

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