\begin{align*}
A & = \{ x : x \in \mathbb N \text{ and } x \text{ is even} \},\\
B & = \{x : x \in \mathbb N \text{ and } x \text{ is prime}\},\\
C & = \{ x : x \in \mathbb N \text{ and } x \text{ is a multiple of } 5\}\text{.}
\end{align*}
If \(A = \{ a, b, c \}\text{,}\)\(B = \{ 1, 2, 3 \}\text{,}\)\(C = \{ x \}\text{,}\) and \(D = \emptyset\text{,}\) list all of the elements in each of the following sets.
Observe that \(x \in A \cup B\) if and only if \(x \in A\) or \(x \in B\text{.}\) Equivalently, \(x \in B\) or \(x \in A\text{,}\) which is the same as \(x \in B \cup A\text{.}\) Therefore, \(A \cup B = B \cup A\text{.}\)
\((A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}\)
\(A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}\)
Which of the following relations \(f: {\mathbb Q} \rightarrow {\mathbb Q}\) define a mapping? In each case, supply a reason why \(f\) is or is not a mapping.
(a) \(f\) is one-to-one but not onto. \(f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}\text{.}\) (c) \(f\) is neither one-to-one nor onto. \(f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}\)
Let \(f :A \rightarrow B\) and \(g : B \rightarrow C\) be invertible mappings; that is, mappings such that \(f^{-1}\) and \(g^{-1}\) exist. Show that \((g \circ f)^{-1} =f^{-1} \circ g^{-1}\text{.}\)
(a) Let \(x, y \in A\text{.}\) Then \(g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))\text{.}\) Thus, \(f(x) = f(y)\) and \(x = y\text{,}\) so \(g \circ f\) is one-to-one. (b) Let \(c \in C\text{,}\) then \(c = (g \circ f)(x) = g(f(x))\) for some \(x \in A\text{.}\) Since \(f(x) \in B\text{,}\)\(g\) is onto.
(a) Let \(y \in f(A_1 \cup A_2)\text{.}\) Then there exists an \(x \in A_1 \cup A_2\) such that \(f(x) = y\text{.}\) Hence, \(y \in f(A_1)\) or \(f(A_2) \text{.}\) Therefore, \(y \in f(A_1) \cup f(A_2)\text{.}\) Consequently, \(f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)\text{.}\) Conversely, if \(y \in f(A_1) \cup f(A_2)\text{,}\) then \(y \in f(A_1)\) or \(f(A_2)\text{.}\) Hence, there exists an \(x\) in \(A_1\) or \(A_2\) such that \(f(x) = y\text{.}\) Thus, there exists an \(x \in A_1 \cup A_2\) such that \(f(x) = y\text{.}\) Therefore, \(f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\text{,}\) and \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)
Determine whether or not the following relations are equivalence relations on the given set. If the relation is an equivalence relation, describe the partition given by it. If the relation is not an equivalence relation, state why it fails to be one.
(a) The relation fails to be symmetric. (b) The relation is not reflexive, since \(0\) is not equivalent to itself. (c) The relation is not transitive.
Define a relation \(\sim\) on \({\mathbb R}^2\) by stating that \((a, b) \sim (c, d)\) if and only if \(a^2 + b^2 \leq c^2 + d^2\text{.}\) Show that \(\sim\) is reflexive and transitive but not symmetric.
Find the error in the following argument by providing a counterexample. “The reflexive property is redundant in the axioms for an equivalence relation. If \(x \sim y\text{,}\) then \(y \sim x\) by the symmetric property. Using the transitive property, we can deduce that \(x \sim x\text{.}\)”
Define a relation on \({\mathbb R}^2 \setminus \{ (0,0) \}\) by letting \((x_1, y_1) \sim (x_2, y_2)\) if there exists a nonzero real number \(\lambda\) such that \((x_1, y_1) = ( \lambda x_2, \lambda y_2)\text{.}\) Prove that \(\sim\) defines an equivalence relation on \({\mathbb R}^2 \setminus (0,0)\text{.}\) What are the corresponding equivalence classes? This equivalence relation defines the projective line, denoted by \({\mathbb P}({\mathbb R}) \text{,}\) which is very important in geometry.
