Exercises §13 — Ross, Elementary Analysis, 2nd Edition
Exercise 13.4. Let \((S, d)\) be a metric space.
(iii) Prove the union of any collection of open sets is open.
(iv) Prove the intersection of finitely many open sets is again an open set.
Proof. (iii) Let \(D\) be the union of a (possibly infinite) collection of open sets in \(S\). Let \(x_0 \in D\). Then there exists \(r > 0\) with open ball \(B(x_0, r) \subseteq U\), where \(U\) is one of the open sets in the union. But note \(B(x_0, r) \subseteq D\) by definition of \(D\), so \(D^\circ = D\), meaning \(D\) is open. \(\square\)
(iv) Let \[ D = \bigcap_{k=1}^{n} B_k, \] where \(n \ge 1\) and each \(B_k\) is an open set. Let \(x_0 \in D\). Since \(x_0 \in \bigcap_{k=1}^{n} B_k\), there exist \(r_k > 0\), \(k = 1, \ldots, n\), such that \(B(x_0, r_k) \subseteq B_k\). Let \[ r = \min(r_1, \ldots, r_n). \] Then \[ B(x_0, r) \subseteq \bigcap_{k=1}^{n} B_k, \] which means \(D^\circ = D\), so \(D\) is open. \(\square\)
Exercise 13.5. Let \((S, d)\) be a metric space. Prove the intersection of any collection of closed sets is a closed set.
Proof. Note first the De Morgan identity \[ \bigcap \{\, S \setminus U : U \in \mathcal{U} \,\} = S \setminus \bigcup \{\, U : U \in \mathcal{U} \,\}. \]
Let \(D = \bigcap_\alpha S_\alpha\), where each \(S_\alpha\) is a closed set in \(S\). Note each \(S \setminus S_\alpha\) is open. We need to show \(S \setminus D\) is open. Note \[ D = S \setminus \bigcup_\alpha (S \setminus S_\alpha). \] Note \(\bigcup_\alpha (S \setminus S_\alpha)\) is open since each \(S \setminus S_\alpha\) is open [union of open sets, Exercise 13.4(iii)]. Since \(\bigcup_\alpha (S \setminus S_\alpha)\) is open, then \[ S \setminus \Bigl( \bigcup_\alpha (S \setminus S_\alpha) \Bigr) \] is closed, thus \(D\) is closed. \(\square\)
Exercise 13.6. Let \(E\) be a subset of a metric space \((S, d)\).
(a) Prove \(E\) is closed if and only if \(E = E^-\).
(b) Prove \(E\) is closed if and only if it contains the limits of every convergent sequence of points in \(E\).
(c) Prove \(x \in E^-\) if and only if \(x\) is the limit of some sequence of points in \(E\).
(d) Prove a point \(x \in E^- \setminus E^\circ\) if and only if \(x \in E^- \cap (S \setminus E)^-\).
Proof. (a) (\(\Rightarrow\)) Suppose \(E\) is closed. Note \(E \subseteq E^-\) by definition of \(E^-\). Since \(E^-\) is the smallest closed subset of \(S\) containing \(E\), and \(E\) is closed, \(E^- \subseteq E\), so \(E = E^-\).
(\(\Leftarrow\)) Suppose \(E = E^-\). Then \(S \setminus E = S \setminus E^-\). By definition \(S \setminus E^-\) is open since \(E^-\) is closed. Thus \(S \setminus E\) is open, so \(E\) is closed. \(\square\)
(b) (\(\Rightarrow\)) Suppose \(E\) is closed. Suppose for the sake of contradiction that there exists a convergent sequence of points in \(E\) whose limit is not in \(E\); call this limit \(x\). Note \(x \in S \setminus E\), and \(S \setminus E\) is open since \(E\) is closed. Thus there exists \(r > 0\) such that \(B(x, r) \subseteq S \setminus E\). Let \((x_n)\) denote the convergent sequence mentioned above. There exists \(N\) such that \(d(x_n, x) < r\) for all \(n > N\) — this is true by definition of convergence. Note this implies \[ x_n \in B(x, r) \subseteq S \setminus E \quad \text{for all } n > N, \] but this contradicts \(x_n \in E\). Thus the limit \(x \in E\).
(\(\Leftarrow\)) Suppose \(E\) contains the limits of every convergent sequence of points in \(E\). Let \(x \in E^-\). Note for all \(n \in \mathbb{N}\), \(B(x, \tfrac{1}{n}) \cap E \neq \varnothing\) since \(x \in E^-\). So choose \[ x_n \in B(x, \tfrac{1}{n}) \cap E. \] Then \(d(x_n, x) < \tfrac{1}{n}\) for all \(n \in \mathbb{N}\) by definition of \(B(x, \tfrac{1}{n})\). Note \(\lim_{n \to \infty} d(x_n, x) = 0\), so \(x_n \to x\), meaning \(x\) is the limit of the sequence \((x_n)\) of points in \(E\), since \(x_n \in E\) for all \(n \in \mathbb{N}\). Note \(E\) contains the limit of every convergent sequence of points in \(E\), so \(x \in E\). Thus \(E^- \subseteq E\). Since \(E \subseteq E^-\), this implies \(E = E^-\), so \(E\) is closed [part (a)]. \(\square\)
(c) (\(\Rightarrow\)) Suppose \(x \in E^-\). For each \(n \in \mathbb{N}\) there exists \(x_n \in E\) such that \(x_n \in B(x, \tfrac{1}{n}) \cap E\), since \(B(x, \tfrac{1}{n}) \cap E \neq \varnothing\) because \(x \in E^-\). Note \((x_n)\) is a sequence of points in \(E\). Further, note \[ d(x_n, x) < \tfrac{1}{n} \] by definition of \(B(x, \tfrac{1}{n})\), and \(\lim_{n \to \infty} d(x_n, x) = 0\), so \((x_n)\) converges to \(x\).
(\(\Leftarrow\)) Suppose \(x\) is the limit of some sequence \((x_n)\) of points in \(E\). Suppose for contradiction that \(x \notin E^-\). Since \(E \subseteq E^-\), \(x \notin E\). There exists \(r > 0\) such that \(B(x, r) \neq \varnothing\) and \(B(x, r) \subseteq S \setminus E^-\), and this is possible because \(E^-\) is closed so \(S \setminus E^-\) is open. By definition of convergence there exists \(N\) such that \(d(x_n, x) < r\) for all \(n > N\). But this implies \(x_n \notin E^-\) for all \(n > N\), so \(x_n \notin E\) since \(E \subseteq E^-\), which is a contradiction. Thus \(x \in E^-\). \(\square\)
(d) (\(\Rightarrow\)) Suppose \(x \in E^- \setminus E^\circ\). Clearly \(x \in E^-\). Since \(x \notin E^\circ\), for every \(r > 0\) we have \(B(x, r) \cap (S \setminus E) \neq \varnothing\). Thus \(x \in (S \setminus E)^-\). Thus \[ x \in E^- \cap (S \setminus E)^-. \]
(\(\Leftarrow\)) Suppose \(x \in E^- \cap (S \setminus E)^-\). Since \(x \in (S \setminus E)^-\), there is no \(r > 0\) such that \(B(x, r) \subseteq E\), thus \(x \notin E^\circ\). But \(x \in E^-\), so then \(x \in E^- \setminus E^\circ\). \(\square\)
Exercise 13.14. Let \(E\) be a compact nonempty subset of \(\mathbb{R}^k\), and let \(\delta = \sup\{\, d(x, y) : x, y \in E \,\}\). Show \(E\) contains points \(x_0, y_0\) such that \(d(x_0, y_0) = \delta\).
Proof. Since \(E\) is compact, it is closed and bounded, by the Heine–Borel Theorem.
Case: \(E\) is finite. Choose \(x_0, y_0 \in E\) such that \(d(x_0, y_0) \ge d(x, y)\) for all \(x, y \in E\). Then \[ \delta = d(x_0, y_0) = \sup\{\, d(x, y) : x, y \in E \,\}. \]
Case: \(E\) is infinite. For all \(n \in \mathbb{N}\) there exist \(x_n, y_n \in E\) such that \[ \delta - \tfrac{1}{n} < d(x_n, y_n) \le \delta. \] Thus \(d(x_n, y_n) \to \delta\).
Since \(E\) is compact, there exists \(x_0\) such that \((x_n)\) has a convergent subsequence \((x_{n_k}) \to x_0 \in E\). Similarly, there exists \(y_0\) such that \((y_{n_k})\) has a convergent subsequence \((y_{n_{k_j}}) \to y_0 \in E\).
Note \(x_{n_{k_j}} \to x_0\), so by the triangle inequality \[ \bigl| d(x_{n_{k_j}}, y_{n_{k_j}}) - d(x_0, y_0) \bigr| \le d(x_{n_{k_j}}, x_0) + d(y_{n_{k_j}}, y_0). \] Since the right side goes to \(0\), that implies \[ d(x_{n_{k_j}}, y_{n_{k_j}}) \to d(x_0, y_0) \quad \text{and} \quad d(x_{n_{k_j}}, y_{n_{k_j}}) \to \delta. \] So \(d(x_0, y_0) = \delta\). \(\square\)