Hubbard

Prove Proposition 0.4.11 (Inverse image of intersection, union) 1 \( f^{-1} (A \cap B) = f^{-1} (A) \cap f^{-1} (B) \) Proof: 1 To show \(f^{-1}(A \cap B) \subset f^{-1} (A) \cap f^{-1} (B)\) \(x \in f^{-1}(A \cap B)\) を任意にとる。 すると、 \begin{align*} & f(x) \in A \cap B \qquad (\because \text{逆像の定義})\\ \Leftrightarrow & f(x) \in A \text{ and } f(x) \in B \\ \Leftrightarrow & x \in f^{-1}(A) \text{ and } x \in f^{-1}(B) \qquad (\because \text{逆像の定義}) \\ \Leftrightarrow & x \in f^{-1}(A) \cap f^{-1}(B) \quad \square \end{align*}...

Question a Make up a nonmathematical function that is one to one but not onto. b Make up a mathematical function that is one to one but not onto. Answer a “The student number of” from Keio University students to integers. b \(f(x) = 2^x\)

Question \(E\) を\(A \subset E\) and \(B \subset E\)を満たす集合とし、 演算子\(\ast \) を以下で定義する。 \[ A \ast B = (E - A) \cap (E-B) \] このとき、 以下の a,b,c を、 \(A, B\), and \(\ast\)を用いて表せ。 a \(A \cup B\) b \(A \cap B\) c \(E - A\) Answer a \begin{align*} A \cup B &= E - ( \bar{A} \cap \bar{B} ) \\ &= E - \left\{ ( E - A) \cap (E-B) \right\} \\ &= E - A \ast B \\ &= (E - A \ast B) \cap (E - A \ast B) \\ &= (A \ast B) \ast (A \ast B) \end{align*}...

Question Explain Why one of these statements is true and the other is false: \begin{align*} (\forall \text{ man } M)(\exists \text{ woman } W) \mid W \text{ is the mother of } M \\ (\exists \text{ woman } W)(\forall \text{ man } M) \mid W \text{ is the mother of } M \end{align*} Answer The first statement is true, and the second is false because every man has a mother but there doesn’t exist a woman who is the mother of all men....