Exercise 0.4.6

Wed, 05 Apr 2023 00:00:00 +0000

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*}

Exercise 0.3.1

Fri, 17 Mar 2023 00:00:00 +0000

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*}

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Exercise 0.4.6

1 \( f^{-1} (A \cap B) = f^{-1} (A) \cap f^{-1} (B) \)

Exercise 0.3.1

Question

Answer