Exercises §4 — Ross, Elementary Analysis, 2nd Edition
Exercise 4.9. Complete the proof of the Corollary — every nonempty subset \(S\) of \(\mathbb{R}\) that is bounded below has a greatest lower bound \(\inf S\) — by proving \(\inf S = -\sup(-S)\), where \(-S = \{-s : s \in S\}\). Let \(s_0 = \sup(-S)\).
Proof.
(1) \(-s_0\) is a lower bound of \(S\). Suppose for contradiction that there exists \(s \in S\) such that \(-s_0 > s\). Then \(-s > s_0\). But \(-s \in -S\) by definition of \(-S\), so this means \(-S\) contains an element exceeding \(s_0\), which contradicts the fact that \(s_0 = \sup(-S)\). Hence \(-s_0 \le s\) for all \(s \in S\).
(2) \(-s_0\) is the greatest lower bound. Consider \(s_0\) from (1). We prove that if \(t \le s\) for all \(s \in S\), then \(t \le -s_0\). Suppose for contradiction that \(t > -s_0\). Note this means \(-t < s_0\). Since \(t \le s\) for all \(s \in S\), we have \(-t \ge -s\) for all \(s \in S\); that is, \(-t\) is an upper bound of \(-S\). But then \(-t\) is an upper bound smaller than \(s_0\) for the set \(-S\), which contradicts the fact that \(s_0 = \sup(-S)\). Hence \(t \le -s_0\).
By (1), \(-s_0\) is a lower bound of \(S\), and by (2) it is greater than or equal to every lower bound of \(S\). Therefore \[ \inf S = -s_0 = -\sup(-S). \qquad\square \]
Exercise 4.14. Let \(A\) and \(B\) be nonempty bounded subsets of \(\mathbb{R}\), and let \(A + B\) be the set of all sums \(a + b\) where \(a \in A\) and \(b \in B\).
Exercise 4.14. (a) Prove \(\sup(A + B) = \sup A + \sup B\).
Proof. We prove this by first proving \(\sup(A+B) \le \sup A + \sup B\) and then proving \(\sup(A+B) \ge \sup A + \sup B\).
(\(\le\)) Note \(a \le \sup A\) and \(b \le \sup B\) for all \(a \in A\), \(b \in B\). Hence \(a + b \le \sup A + \sup B\) for all \(a \in A\), \(b \in B\). Since every element of \(A + B\) has the form \(a + b\), the number \(\sup A + \sup B\) is an upper bound of \(A + B\), so \[ \sup(A + B) \le \sup A + \sup B. \]
(\(\ge\)) Now we show \(\sup(A+B) - b \ge \sup A\) for each \(b \in B\). Fix \(b \in B\) and assume for contradiction that there exists \(a \in A\) such that \(a > \sup(A + B) - b\). Then \(a + b > \sup(A + B)\). But \(a + b \in A + B\), which contradicts the fact that \(\sup(A+B)\) is an upper bound of \(A + B\). Hence \(\sup(A + B) - b \ge \sup A\) for the fixed \(b\), and therefore for all \(b \in B\). This means \(b \le \sup(A + B) - \sup A\) for all \(b \in B\), so \(\sup(A+B) - \sup A\) is an upper bound of \(B\), giving \(\sup B \le \sup(A + B) - \sup A\). That is, \[ \sup A + \sup B \le \sup(A + B). \]
Combining the two inequalities, \(\sup(A + B) = \sup A + \sup B\). \(\square\)
Exercise 4.14. (b) Prove \(\inf(A + B) = \inf A + \inf B\).
Proof. Consider the sets \(-A = \{-a : a \in A\}\), \(-B = \{-b : b \in B\}\), and \(-(A+B) = \{-(a + b) : a \in A,\, b \in B\}\). Note that \(-(A + B) = (-A) + (-B)\), so by part (a) applied to \(-A\) and \(-B\), \[ \sup\bigl(-(A + B)\bigr) = \sup(-A) + \sup(-B). \] From Exercise 4.9 we proved \(\inf S = -\sup(-S)\). Thus \[ \inf(A + B) = -\sup\bigl(-(A + B)\bigr) = -\bigl(\sup(-A) + \sup(-B)\bigr) = -\sup(-A) + \bigl(-\sup(-B)\bigr) = \inf A + \inf B. \qquad\square \]
Exercise 4.15. Let \(a, b \in \mathbb{R}\). Prove that if \(a \le b + \frac{1}{n}\) for all \(n \in \mathbb{N}\), then \(a \le b\).
Proof. Note \(\frac{1}{n} > 0\) for all \(n \in \mathbb{N}\). Suppose for contradiction that \(a > b\). Then \(a - b > 0\). By the Archimedean property, there exists \(n_0 \in \mathbb{N}\) such that \(n_0(a - b) > 1\). This implies \(a - b > \frac{1}{n_0}\), i.e. \(a > b + \frac{1}{n_0}\). But that contradicts the hypothesis \(a \le b + \frac{1}{n}\) taken at \(n = n_0\). Thus \(a \le b\). \(\square\)
Exercise 4.16. Show \(\sup\{r \in \mathbb{Q} : r < a\} = a\) for each \(a \in \mathbb{R}\).
Proof. Let \(S = \{r \in \mathbb{Q} : r < a\}\). By definition every \(r \in S\) satisfies \(r < a\), so \(a\) is an upper bound of \(S\).
Suppose for contradiction that \(a\) is not the least upper bound; then there exists an upper bound \(a_0\) of \(S\) with \(a_0 < a\). Since \(a_0 \in \mathbb{R}\) and \(a \in \mathbb{R}\) with \(a_0 < a\), the denseness of \(\mathbb{Q}\) gives a rational \(r_0 \in \mathbb{Q}\) such that \(a_0 < r_0 < a\). This implies \(r_0 \in S\) (as \(r_0 < a\)), yet \(r_0 > a_0\), which contradicts the assumption that \(a_0\) is an upper bound of \(S\). Hence no upper bound smaller than \(a\) exists, so \(a\) is the least upper bound, and thus \[ a = \sup\{r \in \mathbb{Q} : r < a\}. \qquad\square \]