10.30 Set

1. Let \(A = \{ \{1, 3, 8\}, \{4, 5\}, 10 \}\).

   a) What are the elements of \(A\)?

   The elements are \(\{1, 3, 8\}\), \(\{4, 5\}\), \(10\).

   b) Write the power set of \(A\).
   ​\(\mathcal{P}(A) = \{ \emptyset, A, \{ \{1, 3, 8\} \}, \{ \{4, 5\} \}, \{ 10 \}, \{ \{1, 3, 8\}, \{4, 5\} \}, \{ \{1, 3, 8\}, 10 \}, \{ \{4, 5\}, 10 \}\)

   c) Write T or F.

   i) \(3 \in A\) \(\quad F\)

   ii) \(\left\lbrace\{10\,\right\rbrace\}\subseteq\mathcal{P}(A)\) \(\quad T\)

   iii) \(\left\lbrace\{10\right\rbrace\}\subseteq A\) \(\quad F\)

   iv) \(\{\left\lbrace10\right\rbrace\}\in\mathcal{P}(A)\) \(\quad F\)

   v) \(\{\emptyset, \{1, 3, 8\} \} \subseteq \mathcal{P}(A)\) \(\quad F\)

   vi) \(\{\emptyset,\{\left\lbrace1,3,8\right\rbrace\}\}\subseteq\mathcal{P}(A)\) \(\quad T\)

2. Let \(A = \{ x \in \mathbb{R} \mid -4 \leq x^2 \leq 4 \}\) and \(B = [-2, 2]\). Prove that \(A = B\).

   Let \(x \in A \Rightarrow -4 \leq x^2 \leq 4 \Leftrightarrow 0 \leq x^2 \leq 4 \Leftrightarrow -2 \leq x \leq 2\). So \(x \in B\).

3. Let \(A, B\) be sets. Prove that \(A \subseteq B \Leftrightarrow A \setminus B = \emptyset\).

   (\(\Rightarrow\)) We have \(x\in A\Rightarrow x\in B\) and we want to prove \(x\in A \cap x \notin B\) is empty

   Which means \(\neg(x\in A \cap x \notin B)\equiv x\notin A\cup x\in B\) (N.T.P.)

   Since \(P\Rightarrow Q\equiv\neg P\vee Q\), then \(x\in A\Rightarrow x\in B\equiv x\notin A\cup x\in B\)​

   (\(\Leftarrow\)) Similarly, we have \(A\setminus B=\emptyset\), then \(x\notin A\cup x\in B\)​

   Which is equivalent to \(x\in A\Rightarrow x\in B\)

4. Prove: \(A \times B = B \times A \Leftrightarrow A = B\).

   Proof

   (\(\Leftarrow\)) Obviously

   (\(\Rightarrow\))N.T.P. \(A \subseteq B \text{ and } B \subseteq A.\)

   \((\subseteq\)) Let \(a \in A\), and \(b\) be any element. \(\Rightarrow(a,b)\in A\times B=B\times A\Rightarrow a\in B\Rightarrow A\subseteq B\)

   (\(\supseteq\)) Let \(b \in B\) and \(a\) be any element \((b,a)\in B\times A=A\times B\Rightarrow b\in A\Rightarrow B\subseteq A\)

   Therefore \(A\times B=B\times A\Leftrightarrow A=B\)​

5. Let \(A_n = \{n, n+1, 2n\}\) and \(\Delta = \mathbb{N}\). Prove that \(\bigcup_{n \in \Delta} A_n = \mathbb{N}\).
   ​\(A_1 = \{1, 2\}\)
   ​\(A_2 = \{2, 3, 4\}\)
   ​\(A_{10} = \{10, 11, 20\}\)

   (\(\subseteq\)​) It is obvious because every \(A_n \subseteq \mathbb{N}\). So the union of all of them will be \(\subseteq \mathbb{N}\).

   (\(\supseteq\)​) Let \(m \in \mathbb{N} \Rightarrow m \in A_m \Rightarrow m \in A_m \subseteq \bigcup_{n \in \Delta} A_n\).

6. For each natural number \(n\), let \(B_n = \mathbb{N} \setminus \{1, 2, 3, \dots, n\}\) and let \(\mathcal{B}=\{B_{n}:n\in\mathbb{N}\}\).

   Find \(\bigcup_{n \in \mathbb{N}} B_n = \bigcup_{B \in \mathcal{B}} B\) and \(\bigcap_{n \in \mathbb{N}} B_n = \bigcap_{B \in \mathcal{B}} B\).

   \(B_1 = \{2, 3, 4, \dots\} = \mathbb{N} \setminus \{1\}\)

   \(B_{10} = \{11, 12, \dots\}\)

   \(B_{100} = \{101, 102, \dots\}\)

   I will prove that \(\bigcup_{n\in\mathbb{N}}B_{n}=\mathbb{N}\setminus\left\lbrace1\right\rbrace\)​ \(\bigcup_{n\in\mathbb{N}}B_{n}=\bigcup_{n\in\mathbb{N}}\mathbb{N}\setminus\{1,2,3,\dots,n\}=\bigcup_{n\in\mathbb{N}}\{1,2,3,\dots,n\}^{c}=\left(\bigcap_{n\in\mathbb{N}}\{1,2,3,\dots,n\}\right)^{c}=\left\lbrace1\right\rbrace^{c}=\mathbb{N}\setminus\left\lbrace1\right\rbrace\).

   \(\bigcap_{n \in \mathbb{N}} B_n = \bigcap_{n \in \mathbb{N}} \mathbb{N} \setminus \{1, 2, 3, \dots, n\}\) \(=\bigcup_{n\in\mathbb{N}}\{1,2,3,\ldots,n\}^{c}=\left(\bigcup_{n\in\mathbb{N}}\left\lbrace1,2,3,...,n\right\rbrace\right)^{c}=\mathbb{N}^{c}=\emptyset\).

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