Appendix C Hints and Answers to Selected Even Exercises

<hint>
    <p>(a) <m>A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}</m>; (d) <m>A \times D = \emptyset</m>.</p>
</hint>

<hint>
    <p>If <m>x \in A \cup (B \cap C)</m>, then either <m>x \in A</m> or <m>x \in B \cap C</m>.  Thus,   <m> x \in A \cup B</m> and <m>A \cup C</m>.  Hence, <m> x \in (A \cup B) \cap (A \cup C)</m>. Therefore, <m>  A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)</m>.  Conversely, if <m>x \in (A \cup B) \cap (A \cup C)</m>, then <m>x \in A \cup B</m> and <m>A \cup C</m>. Thus, <m>x \in A</m> or <m>x</m> is in both <m>B</m> and <m>C</m>. So <m>x \in A \cup (B \cap C)</m> and therefore <m>(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)</m>. Hence, <m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</p>
</hint>

<hint>
    <p><m>(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</m>.</p>
</hint>

<hint>
    <p><m>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)</m>.</p>
</hint>

<hint>
    <p>(a) <m>f</m> is one-to-one but not onto. <m>f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}</m>.  (c) <m>f</m> is neither one-to-one nor onto.  <m>f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}</m>.</p>
</hint>

<hint>
    <p>(a) <m>f(n) = n + 1</m>.</p>
</hint>

<hint>
    <p>(a) Let <m>x, y \in A</m>. Then <m>g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))</m>.  Thus, <m>f(x) = f(y)</m> and <m>x = y</m>,  so <m>g \circ f</m> is one-to-one. (b) Let <m>c \in C</m>, then <m>c = (g \circ f)(x) = g(f(x))</m> for some <m>x \in A</m>. Since <m>f(x) \in B</m>, <m>g</m> is onto.</p>
</hint>

<hint>
    <p>(a) Let <m>y \in f(A_1 \cup A_2)</m>.  Then there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Hence, <m> y \in f(A_1)</m> or <m>f(A_2) </m>.  Therefore, <m> y \in f(A_1) \cup f(A_2)</m>. Consequently, <m> f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)</m>. Conversely, if <m>y \in f(A_1) \cup f(A_2)</m>, then <m> y \in f(A_1)</m> or <m>f(A_2)</m>.  Hence, there exists an <m>x \in A_1</m> or there exists an <m>x \in A_2</m> such that <m>f(x) = y</m>.  Thus, there exists an <m>x \in A_1 \cup A_2</m> such that <m>f(x) = y</m>.  Therefore, <m> f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)</m>, and <m>f(A_1 \cup A_2) = f(A_1) \cup f(A_2)</m>.</p>
</hint>

<hint>
    <p>Let <m>X = {\mathbb N} \cup \{ \sqrt{2}\, \}</m> and define <m>x \sim y</m> if <m>x + y \in {\mathbb N}</m>.</p>
</hint>