Set Proofs Continued

Inclusion

Basic principle: \(A\subseteq B\) is equivalent to the statement \(x\in A\implies x\in B\). One can prove this both directly and as \(x\not\in B\implies x\not\in A\).

Proposition: Let \(A=\{4x+2 : x\in\mathbb{Z}\}\). Let \(B=\{ 2x: x\in \mathbb{Z}\}\). Then \(A\subseteq B\).

More examples

Proposition: For all \(k\in\mathbb{Z}\), let \(A=\{n\in\mathbb{Z} : n|k\}\) and let \(B=\{n\in\mathbb{Z}: n|k^2\}\). Then \(A\subseteq B\). (Note: this is problem 3 on page 171.)

More examples

Proposition: Suppose \(A\), \(B\), and \(C\) are sets. If \(B\subseteq C\), then \(A\times B\subseteq A\times C\). (This is problem 7 on page 171.)

One more

Proposition: Let \(A\) and \(B\) be sets. Prove that \(A\subseteq B\) if and only if \(A-B=\emptyset\). (This is problem 21 on page 171.)

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