数学 – 集合と写像 - 写像に関する諸概念(特徴関数, 定義関数とその性質)

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集合・位相入門(松坂和夫(著)、岩波書店)の第1章(集合と写像)、4(写像に関する諸概念)、問題15.を解いてみる。

問題15.

$\begin{array}{l} x \in A \cap B \\ χ_{A \cap B} (x) = 1, χ_{A} (x) χ_{B} (x) = 1 \cdot 1 = 1 \\ x \in {(A \cap B)}^{c} \\ χ_{A \cap B} (x) = 0, \\ χ_{A} (x) = 0 \lor χ_{B} (x) = 0 \\ χ_{A} (x) χ_{B} (x) = 0 \end{array}$
$\begin{array}{l} x \in A \cup B \\ χ_{A \cup B} (x) = 1 \\ x \in A \land x \in B \\ 1 + 1 - 1 = 1 \\ x \in A \land x \notin B \\ 1 + 0 - 0 = 1 \\ x \notin A \land x \in B \\ 0 + 1 - 0 = 1 \\ x \notin A \cup B \\ χ_{A \cup B} (x) = 0 \\ 0 + 0 - 0 = 0 \end{array}$
$\begin{array}{l} x \in A^{c} \\ χ_{A^{c}} (x) = 1 \\ 1 - 0 = 1 \\ x \in A \\ χ_{A^{c}} (x) = 0 \\ 1 - 1 = 0 \end{array}$
$\begin{array}{l} x \in A - B \\ χ_{A - B} (x) = 1 \\ 1 \cdot (1 - 0) = 1 \\ x \notin A - B \\ χ_{A - B} (x) = 0 \\ x \in A \cap B \\ 1 (1 - 1) = 0 \\ x \notin A \land x \in B \\ 0 (1 - 1) = 0 \\ x \notin A \land x \notin B \\ 0 (1 - 0) = 0 \end{array}$
$\begin{array}{l} x \in A △ B \\ χ_{A △ B} = 1 \\ x \in A \land x \notin B \\ | 1 - 0 | = 1 \\ x \notin \land x \in B \\ | 0 - 1 | = 1 \\ x \notin A △ B \\ χ_{A △ B} (x) = 0 \\ x \in A \cap B \\ | 1 - 1 | = 0 \\ x \notin A \cup B \\ | 0 - 0 | = 0 \end{array}$

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