10.30 Set
Workshop 3
Exercise 1
Decide if the following statements are true or false.
a) \(\{\emptyset\}\in\{\emptyset,\{\emptyset\}\}\).
True
b) \(\{\emptyset\}\subseteq\{\emptyset,\{\emptyset\}\}\).
True
c) \(\{1,2,3\}\in\{1,2,3,\{4\}\}\).
False
d) \(\{ \{ 4 \} \} \subseteq \{ 1, 2, 3, \{ 4 \} \}\).
True
Exercise 2
Let the universe be all real numbers. Let \(A = [3, 8)\), \(B = [2, 6]\), \(C = (1, 4)\), and \(D=(5,\infty)\). Find:
a) \(A \cup C\).
\(\left(1,8\right)\)
b) \(B \cap C\).
\([2,4)\)
c) \(A^c\).
\((-\infty,3)\cup[8,+\infty)\)
d) \((A \cup C) - (B \cap D)\).
\((1,8)-(5,6]=(1,5]\cup(6,8)\)
e) \(A \cap B \cap C\).
\([3,4)\)
Exercise 3
Let \(A\), \(B\), \(C\), and \(D\) be sets. Decide if the statement is true or false. If it is true, prove it. If it is false, give a counterexample.
a) If \(C \subseteq A\), \(D \subseteq B\), and \(A\) and \(B\) are disjoint (\(A \cap B = \emptyset\)), then \(C\) and \(D\) are disjoint.
Since \(A \cap B = \emptyset\), for any \(x \in A\), \(x \notin B\).
For any \(x \in C\subseteq A\), \(x \in A\). Thus \(x\in C\) and \(x\notin B\)
Since \(x\notin B\), and \(D \subseteq B\), then \(x \notin D\).
Now, \(\forall x \in C\), \(x \notin D\). So \(C \cap D = \emptyset\).
b) If \((A - B) \cap (A - C) = \emptyset\), then \(B \cap C = \emptyset\).
False. \(A=(1,10),B=(1,9),C=(2,10)\)
c) \(A - (B - C) = (A - B) - (A - C)\).
False. \(A=B=C\)
d) \(\mathcal{P}(A\cap B)=\mathcal{P}(A)\cap\mathcal{P}(B)\).
True: So let's prove it by double inclusion.
\(\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)\)
Take \(X \in \mathcal{P}(A \cap B)\), so \(X \subseteq A \cap B\).
Thus: \(X \subseteq A\) and \(X \subseteq B\).
Finally, \(X \in \mathcal{P}(A) \cap \mathcal{P}(B)\).
Another way is similar.
Exercise 4
Find \(\mathcal{P}(A \times B)\) for \(A = \{1, 2, \{1, 2\}\}\) and \(B=\{q,\{t\},\pi\}\).
\(A\times B=\{(1,q),(1,\{t\}),(1,\pi),(2,q),(2,\{t\}),\left(2,\pi\right),\left(\left\lbrace1,2\right\rbrace,q\right),\left(\left\lbrace1,2\right\rbrace,\left\lbrace t\right\rbrace\right),\left(\left\lbrace1,2\right\rbrace,\pi\right)\}\)
\(\mathcal{P}(A\times B)=\left\lbrace\ldots\right\rbrace\)
Exercise 5
Find the union and intersection of each of the following families, i.e., find \(\bigcup_{A \in \mathcal{A}} A\) and \(\bigcap_{A \in \mathcal{A}} A\), for each \(\mathcal{A}\).
a) \(\mathcal{A} = \{\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}, \{4, 5, 6, 7, 8\}\}\).
\(\bigcup_{A\in\mathcal{A}}A=\{1,2,3,4,5\}\cup\{2,3,4,5,6\}\cup\{3,4,5,6,7\}\cup\{4,5,6,7,8\}=\left\lbrace1,2,3,4,5,6,7,8\right\rbrace\)
\(\bigcap_{A\in\mathcal{A}}A=\{1,2,3,4,5\}\cap\{2,3,4,5,6\}\cap\{3,4,5,6,7\}\cap\{4,5,6,7,8\}=\left\lbrace4,5\right\rbrace\)
b) For each natural number \(n\), let \(A_n = \{5n, 5n + 1, 5n + 2, \dots, 6n\}\) and let \(\mathcal{A} = \{A_n : n \in \mathbb{N}\}\).
c) Let \(\mathcal{A}\) be the set of all sets of integers that contain 10.
\(\bigcup_{A \in \mathcal{A}} A = \mathbb{N}\)
\(\bigcap_{A \in \mathcal{A}} A = \{10\}\)
Exercise 6
Let \(\mathcal{A} = \{A_{\alpha} : \alpha \in \Delta\}\) be a family of sets and let \(B\) be a set. Prove that:
\(\begin{aligned}A \cup \left( \bigcap_{\alpha \in \Delta} B_\alpha \right) = \bigcap_{\alpha \in \Delta} (A \cup B_\alpha)\end{aligned}\)
Proof
\(\begin{aligned}A\cap\bigcup_{\alpha\in\Delta}B_{\alpha}=A\cap\{x:x\in B_{\alpha}\text{ for some }\alpha\in\Delta\}\end{aligned}\)
\(\subseteq\)) Then \(x\in A\text{ and }x\in B_{\alpha}\text{ for some }\alpha\in\Delta\Rightarrow x\in A\cap B_{\alpha}\text{ for some }\alpha\in\Delta\Rightarrow\bigcap_{\alpha\in\Delta}(A\cup B_{\alpha})\)
Thus \(\begin{aligned}A\cup\left(\bigcap_{\alpha\in\Delta}B_{\alpha}\right)\subseteq\bigcap_{\alpha\in\Delta}(A\cup B_{\alpha})\end{aligned}\)
\(\supseteq\)) Similarly
Exercise 7
Let \(\mathcal{A} = \{A_{\alpha} : \alpha \in \Delta\}\) and let \(\mathcal{B} = \{B_{\beta} : \beta \in \Gamma\}\) be families of sets.
Prove that: \(\left( \bigcup_{\alpha \in \Delta} A_{\alpha} \right) \cap \left( \bigcup_{\beta \in \Gamma} B_{\beta} \right) = \bigcup_{\beta \in \Gamma} \bigcup_{\alpha \in \Delta} (A_{\alpha} \cap B_{\beta}).\)
Since\(\begin{aligned}\left(\bigcup_{\alpha\in\Delta}A_{\alpha}\right)\cap\left(\bigcup_{\beta\in\Gamma}B_{\beta}\right)=\{x:x\in A_{\alpha}\text{ for some }\alpha\in\Delta\}\cap\{x:x\in B_{\beta}\text{ for some }\beta\in\Gamma\}\end{aligned}\)
\(\subseteq\)) Then we have \(x\in A_{\alpha}\text{ and }x\in B_{\beta}\) for some \(\alpha\in \Delta\) and \(\beta\in \Gamma\)
Thus we have \(x\in A_\alpha \cap B_\beta \subset \bigcup_{\beta\in\Gamma}\bigcup_{\alpha\in\Delta}(A_{\alpha}\cap B_{\beta})\)
\(\supseteq\)) Similarly