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