Question

Prove that if A intersect B = empty set then Power set(B) - {empty set} subset Power set(A union B) - Power set(A)

Answer 1

\[\text{If }A\cap B=\emptyset,\text{ then }(\mathcal{P}(B)-\emptyset)\subseteq(\mathcal{P}(A\cup B)-\mathcal{P}(A)).\]
Let \(X\in(\mathcal{P}(B)-\emptyset)\) Since \(A\cap B=\emptyset\), then \(X\not\in\mathcal{P}(A)\). However, \(X\in\mathcal{P}(A\cup B)\), since the power set of the union will contain both subsets of \(A\) as well as \(B\). By the defintion of the set complement, \(X\in\mathcal{P}(A\cup B)\) and \(X\not\in\mathcal{P}(A)\) means \(X\in(\mathcal{P}(A\cup B)-\mathcal{P}(A))\). Therefore, \((\mathcal{P}(B)-\emptyset)\subseteq(\mathcal{P}(A\cup B)-\mathcal{P}(A)).\)