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8.15 Sets Tutorial

Subset question

\(A \subseteq B~and~C\subseteq D. Prove~that~(A\cup C)\subseteq(B\cup D)\)

Proof:

Pick \(x\), due to definition, we need to prove \(x\in(A\cup C)\Rightarrow x\in (B\cup D)\).

\(x\in(A\cup C)\Rightarrow (x\in A)\cup (x \in C)\Rightarrow (x\in B)\cup (x \in D)\Rightarrow x\in (B\cup D)\)

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Complement set question

\(If~A\subseteq B,~then~B^{C}\subseteq A^{C}\)​

Proof

Pick \(x\), due to definition, we need to prove \(x\in B^{C}\Rightarrow x\in A^{C}\)

\(x\in B^{C}\Rightarrow x\notin B\Rightarrow x\notin A \Rightarrow x\in A^{C}\)

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Relative complement question

\(A \setminus B=(A\cap B^{C})\)​

Proof

By definition, we need to prove \((A \setminus B)\subseteq (A\cap B^{C})~and~(A \setminus B)\supseteq (A\cap B^{C})\)

1.\(A \setminus B\Rightarrow x\in A:x\notin B\Rightarrow x\in A:x\in B^{C}\Rightarrow x\in (A\cap B^{C})\)​

2.Similarly

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Trick:

\[ For~all~sentences~P~and~Q~the~following\\P\Rightarrow P\vee Q~~~P\wedge Q\Rightarrow P~~~P\vee Q\Leftrightarrow Q\vee P\\Q\Rightarrow P\vee Q~~~P\wedge Q\Rightarrow Q~~~P\wedge Q\Leftrightarrow Q\wedge P\\are~true. \]

All can prove by truth table.

If A, B are sets, prove that

  1. \(A\subseteq A\cup B~and~B\subseteq A\cup B\)

    Proof

    \[ \begin{aligned}(a)&\quad x\in A~\Rightarrow x\in A~~or~~x\in B\quad\mathrm{since~}P\Rightarrow P\vee Q\\&\quad~~~~~~~~~~~\Rightarrow x\in A\cup B\quad\\&\quad\mathrm{thus~A\subseteq A\cup B\quad Analogously,~we~have~B\subseteq A\cup B.}\end{aligned} \]

  2. \(A\cap B\subseteq A~and~A\cap B\subseteq B\)

    Proof

\[ \begin{aligned}(b)~&x\in A\cap B\Rightarrow x\in A\mathrm{~and~}x\in B\mathrm{~~~since~}P\cap Q\Rightarrow P\\&~~~~~~~~~~~~~~~~~\Rightarrow x\in A\\&thus~A\cap B\subseteq A.~Analogously,we~have A\cap B\subseteq B\\\end{aligned} \]

  1. \(A\cup B=B\cup A\)

    Proof

    \[ \begin{aligned}(c)~&x\in A\cup B\Leftrightarrow x\in A\mathrm{~or~}x\in B~~~since~P\vee Q\Leftrightarrow Q\vee P\\&~~~~~~~~~~~~~~~~~~\Leftrightarrow x\in B\mathrm{~or~}x\in A\\&thus~A\cup B=B\cup A.\end{aligned} \]

  2. \(A\cap B=B\cap A\)

    Proof

    \[ \begin{aligned}(d)~&x\in A\cap B\Leftrightarrow x\in A\mathrm{~and~}x\in B~~~since~P\wedge Q\Leftrightarrow Q\wedge P\\&~~~~~~~~~~~~~~~~~~\Leftrightarrow x\in B\mathrm{~and~}x\in A\\&thus~A\cap B=B\cap A.\end{aligned} \]

Fun

Since \((A^{C})^{C}=A\), Prove \(A\setminus B=B^{C}\setminus A^{C}\)

\(A\setminus B=A\cap B^{C}=B^{C}\cap A=B^{C}\cap(A^{C})^{C}=B^{C}\setminus A^{C}\)

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