Textbook-Solutions

To show \(A=B\), show \(A\subseteq B\) and \(B\subseteq A\). Suppose \(x\in B\), then we know by definition that \(x\in (A\cup B)\) if \(x\in B\) or \(x\in A\). Which then implies that \(x\in A\) from rule 1. Thus \(B\subseteq A\). Now suppose \(x\in A\) which implies \(x\in (A\cap B)\) (rule 2). The definition of this means that \(x\in A\) and \(x\in B\). \(\therefore x\in A \implies x\in B\), i.e. \(A\subseteq B\), so \(A=B\).

This only holds for \(B\subseteq A\). We proceed by counterexample. Let \(A={1,2}, B={3,4}\). Then \((A-B) = {1,2}\) and \((A-B)\cup B = {1,2,3,4} \neq {1,2}\).

\[ A\cap B = {6n \mid n\in \mathbb{N}} \] \[ A - B = {2n \mid n\in \mathbb{N}, 2n \not\in {6m \mid m\in \mathbb{N}}} \]