Exercise 3.4.3

Let \(A,B\) be two subsets of a set \(X\), and let \(f:X\to Y\) be a function. Show that \(f(A\cap B)\subseteq f(A)\cap f(B)\), that \(f(A)\setminus f(B)\subseteq f(A\setminus B)\), \(f(A\cup B)=f(A)\cup f(B)\). For the first two statements, is it true that the \(\subseteq\) relation can be improved to \(=\)?

  • \(\bullet\) Show that \(f(A\cap B)\subseteq f(A)\cap f(B)\)

  • \(\Vdash\) \(f(A\cap B)\subseteq f(A)\cap f(B)\)

  • \(\equiv\) { Definition 3.4.1 }

    • \(\{f(x):x\in A\cap B\}\subseteq\{f(x):x\in A\}\cap \{f(x):x\in B\}\)

  • \(\equiv\) { Definition 3.1.23 }

    • \(\{f(x):x\in\{a\in A:a\in B\}\}\subseteq\{f(x):x\in A\}\cap \{f(x):x\in B\}\)

  • \(\equiv\) { Definition 3.1.23 }

    • \(\{f(x):x\in\{a\in A:a\in B\}\}\subseteq\{y\in\{f(x):x\in A\}: y\in \{f(x):x\in B\}\}\)

  • \(\equiv\) { \(x\in\{a\in A:a\in B\}\equiv x\in A\land x\in B\) }

    • \(\{f(x):x\in A\land x\in B\}\subseteq\{y\in\{f(x):x\in A\}: y\in \{f(x):x\in B\}\}\)

  • \(\equiv\) { \(\{a\in A: a\in B\}=\{a: a\in A\land a\in B\}\) }

    • \(\{f(x):x\in A\land x\in B\}\subseteq\{y:y\in\{f(x):x\in A\}\land y\in\{f(x):x\in B\}\}\)

  • \(\equiv\) { \(\{a: a\in A\}=\{a\in A\}\) }

    • \(\{f(x):x\in A\land x\in B\}\subseteq\{f(x):x\in A\}\land \{f(x):x\in B\}\)

  • \(\equiv\) { Definition 3.1.15 }

    • True

  • \(\square\)

  • \(\bullet\) Show that \(f(A\cap B)\neq f(A)\cap f(B)\)

  • – H1: \(A=\{1\}\)

  • – H2: \(B=\{-1\}\)

  • – H3: \(f(1)=1\)

  • – H4: \(f(-1)=1\)

  • \(\Vdash\) \(f(A\cap B)\neq f(A)\cap f(B)\)

  • \(\equiv\) { By H1 and H2, \(A\cap B=\emptyset\) }

    • \(f(\emptyset)\neq f(A)\cap f(B)\)

  • \(\equiv\) { By Definition 3.4.1 \(f(\emptyset)=\{f(x):x\in\emptyset\}=\emptyset\) }

    • \(\emptyset\neq f(A)\cap f(B)\)

  • \(\equiv\) { By Definition 3.4.1 and H1~4, \(f(A)=\{1\},f(B)=\{1\}\) }

    • \(\emptyset\neq \{1\}\)

  • \(\square\)


  • \(\bullet\) Show that \(f(A)\setminus f(B)\subseteq f(A\setminus B)\)

  • \(\Vdash\) \(f(A)\setminus f(B)\subseteq f(A\setminus B)\)

  • \(\equiv\) { Definition 3.4.1 }

    • \(\{f(x):x\in A\}\setminus \{f(x):x\in B\}\subseteq\{f(x):x\in A\setminus B\}\)

  • \(\equiv\) { Definition 3.1.27 }

    • \(\{y\in\{f(x):x\in A\}:y\notin\{f(x):x\in B\}\}\subseteq\{f(x):x\in A\setminus B\}\)

  • \(\equiv\) { Definition 3.1.27 }

    • \(\{y\in\{f(x):x\in A\}:y\notin\{f(x):x\in B\}\}\subseteq\{f(x):x\in \{a\in A:a\notin B\}\}\)