Assignment 4
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw4/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Problem 1
We say a set $ F \subset \mathbb{R} $ is closed if its complement $ F^c := \mathbb{R} \setminus F $ is open. Since $ \emptyset $ and $ \mathbb{R} $ are open, it follows that $ \emptyset $ and $ \mathbb{R} $ are closed as well.
(a) Let $ a, b \in \mathbb{R} $ with $ a < b $ . Prove that $ [a,b]$ is closed.
Proving $ [a, b]$ is closed is equivalent to prove $ (-\infty, a) \cup (b, +\infty)$ is open. See Assignment 3 Problem 5(a).
(b) Is the set $ \mathbb{Z} \subset \mathbb{R} $ closed? Provide a proof to substantiate your claim.
Yes. We show that $ \mathbb{R} \setminus \mathbb{Z} $ is open. $ \forall x \in \mathbb{R} \setminus \mathbb{Z} $ , let $ z_0 $ be the closest $ z \in \mathbb{Z} $ , i.e., $ z_0 = \arg\min_{z \in \mathbb{Z}} \vert x - z\vert $ . Then $ \exists \ \varepsilon = \frac{\vert x - z_0\vert }{2} $ , s.t. $ (x - \varepsilon, x + \varepsilon) \subset \mathbb{R} \setminus \mathbb{Z} $ .
(c) Is the set of rationals $ \mathbb{Q} \subset \mathbb{R} $ closed? Provide a proof to substantiate your claim.
Yes, because $ \mathbb{Q} \subset \mathbb{R} $ is not open. See A3P5(d).
Problem 2
(a) Let $ \Lambda $ be a set (not necessarily a subset of $ \mathbb{R} $ ), and for each $ \lambda \in \Lambda $ , let $ F_\lambda \subset \mathbb{R} $ . Prove that if $ F_\lambda $ is closed for all $ \lambda \in \Lambda $ , then the set
\[\bigcap_{\lambda \in \Lambda} F_\lambda = \{ x \in \mathbb{R} : x \in F_\lambda \text{ for all } \lambda \in \Lambda \}\]is closed.
Since $ F_\lambda $ is closed, then $ F_\lambda^c $ is open. By A3P5(b), $ \bigcup_{\lambda \in \Lambda} F_\lambda^c$ is open. Thus $ \left( \bigcup_{\lambda \in \Lambda} F_\lambda^c \right)^c = \bigcap_{\lambda \in \Lambda} F_\lambda$ is closed.
(b) Let $ n \in \mathbb{N} $ , and let $ F_1, \ldots, F_n \subset \mathbb{R} $ . Prove that if $ F_1, \ldots, F_n $ are closed then the set $ \bigcup_{m=1}^n F_m $ is closed.
$ F_1^c, \ldots, F_n^c $ are open, by A3P5(c), $ \bigcap_{m=1}^n F_m^c$ is an open set. Thus $ \left( \bigcap_{m=1}^n F_m^c \right)^c = \bigcup_{m=1}^n F_m$ is closed.
Problem 3
Let $ F \subset \mathbb{R} $ be a closed set, and let $ {x_n} $ be a sequence of elements of $ F $ converging to $ x \in \mathbb{R} $ . Prove that $ x \in F $ .
Hint: Assume that $ x \notin F $ and arrive at a contradiction.
Assume $ x \in F^c $ . Given $ \lim_{n \to \infty} x_n = x $ , then $ \forall \varepsilon > 0, \exists M \in \mathbb{N}, \forall n > M $ ,
\[\vert x_n - x\vert < \varepsilon \Rightarrow x - \varepsilon < x_n < x + \varepsilon\]Since $ F \subset \mathbb{R} $ is closed, so $ F^c $ is open. $ x \in F^c \Rightarrow x_n\in (x - \varepsilon, x + \varepsilon) \subset F^c $ , which contradicts with $ x_n \in F $ . Thus $ x \in F $ .
Exercise 2.2.3
Prove that if $ {x_n}_{n=1}^\infty $ is a convergent sequence, $ k \in \mathbb{N} $ , then
\[\lim_{n \to \infty} x_n^k = \left( \lim_{n \to \infty} x_n \right)^k\]Hint: Use induction.
Let $ x := \lim_{n \to \infty} x_n $ .
(i) $ k = 1 $ , $ \lim_{n \to \infty} x_n = x$
(ii) Suppose when $ k = m \in \mathbb{N} $ , $ \lim_{n \to \infty} x_n^m = x^m $ . That is, let $ \varepsilon > 0 $ , $ \exists M_1 \in \mathbb{N}, \text{ for all } n > M_1, \ \vert x_n^m - x^m\vert < \frac{\varepsilon}{2(1 + \vert x\vert )}$
Since $ \lim_{n \to \infty} x_n = x $ , $ \exists M_2 \in \mathbb{N}, \forall n > M_2, \ \vert x_n - x\vert < \frac{\varepsilon}{2(1 + \vert x_n^m\vert )}$
\[\begin{aligned} \vert x_n^{m+1} - x^{m+1}\vert &= \vert x_n^m x_n - x_n^m x + x_n^m x - x^m x\vert \\ & \leq \vert x_n^m\vert \cdot \vert x_n - x\vert + \vert x\vert \cdot \vert x_n^m - x^m\vert \\ & < \vert x_n^m\vert \cdot \frac{\varepsilon}{2(1 + \vert x_n^m\vert )} + \vert x\vert \cdot \frac{\varepsilon}{2(1 + \vert x\vert )} < \varepsilon \end{aligned}\]$ \therefore \lim_{n \to \infty} x_n^{m+1} = x^{m+1} $
Exercise 2.2.5
Let $ x_n := \frac{n - \cos(n)}{n} $ . Use the squeeze lemma to show that $ {x_n}_{n=1}^{\infty} $ converges and find the limit.
$ x_n = \frac{n - \cos(n)}{n} = 1 - \frac{\cos(n)}{n} $ . Since $ -1 \leq \cos(n) \leq 1 $ , we get: $ 1 - \frac{1}{n} \leq x_n \leq 1 + \frac{1}{n} $
Since $ \lim_{n \to \infty} (1 - \frac{1}{n}) = 1 $ and $ \lim_{n \to \infty} (1 + \frac{1}{n}) = 1 $ , by the squeeze lemma, $ \lim_{n \to \infty} x_n = 1 $
Problem 6
Let $ A \subset \mathbb{R} $ be bounded above, and let $ a_0 $ be an upper bound for $ A $ . Prove that
$ a_0 = \sup A $ if and only if there exists a sequence $ {a_n} \subset A $ such that $ \lim_{n \to \infty} a_n = a_0 $ .
Hint: By Assignment 3, if $ a_0 = \sup A $ , then for all $ n \in \mathbb{N} $ , there exists $ a_n \in A $ such that
\[a_0 - \frac{1}{n} < a_n \leq a_0\](⇒) Given $ a_0 = \sup A $ , construct $ a_n = a_0 - \frac{1}{n} $ , then $ \lim_{n \to \infty} a_n = a_0 $
(⇐) If \(\exists \{a_n\}_{n \in \mathbb{N}}\), $a_n \in A$ , such that $\lim_{n \to \infty} a_n = a_0$ , then $\forall \varepsilon > 0, \exists M \in \mathbb{N}$ , for $n > M$ ,
$\vert a_n - a_0\vert < \varepsilon \Rightarrow a_0 - \varepsilon < a_n < a_0 + \varepsilon$ .Since $a_0 $ is an upper bound of $ A $ , and $ a_n \leq a_0 $ , it follows that $ a_0 - \varepsilon < a_n \leq a_0 $
By A3P4, $ a_0 = \sup A $ .
Problem 7
Let $ E \subset \mathbb{R} $ be a nonempty set of real numbers. We say $ x \in \mathbb{R} $ is a cluster point of $ E $ if for every $ \varepsilon > 0 $ ,
\((x - \varepsilon, x + \varepsilon) \cap (E \setminus \{x\}) \neq \emptyset\)
Said less formally, $ x $ is a cluster point of $ E $ if every interval containing $ x $ contains at least one element of $ E $ other than $ x $ .
(a) Prove that $ x $ is a cluster point of $ E $ if and only if there exists a sequence $ {x_n} $ of elements of $ E \setminus {x} $ such that $ \lim_{n \to \infty} x_n = x $
Hint: If $ x $ is a cluster point of $ E $ , then for all $ n \in \mathbb{N} $ there exists $ x_n \in E $ with $ x_n \neq x $ such that
\(x - \frac{1}{n} < x_n < x + \frac{1}{n}\)
Denote $ S := (x - \varepsilon, x + \varepsilon) \cap (E \setminus {x}) $
(⇒) If $ x $ is a cluster point, then $ \forall \varepsilon > 0, S \neq \emptyset $ . For each $ n \in \mathbb{N} $ , choose $ x_n \in E \setminus {x} $ such that $ \vert x_n - x\vert < \frac{1}{n} $ . So $ \lim_{n \to \infty} x_n = x $ .
(⇐) If $ x_n \in E \setminus {x} $ with $ \lim_{n \to \infty} x_n = x $ , let $ \varepsilon > 0 $ , then $ \exists M \in \mathbb{N} $ such that $ n > M \Rightarrow \vert x_n - x\vert < \varepsilon \Rightarrow x_n \in (x - \varepsilon, x + \varepsilon) $ . Then $ x_n \in S $ , so $ S \neq \emptyset $ .
(b) Prove that the set of all cluster points of $ E $ is closed.
Define $ A := { x \in \mathbb{R} : x \text{ is a cluster point of } E } $ . To show $ A $ is closed, we show $ \mathbb{R} \setminus A $ is open.
Let $ x \in \mathbb{R} \setminus A $ , then $ \exists \varepsilon > 0 $ such that $ (x - \varepsilon, x + \varepsilon) \cap (E \setminus {x}) = \emptyset $ . Let $ y \in (x - \varepsilon, x + \varepsilon) $ , then $ \exists \delta > 0 $ such that $ (y - \delta, y + \delta) \subset (x - \varepsilon, x + \varepsilon) $ .
If $ y \neq x $ , then $ x \notin {y} $ , so $ (y - \delta, y + \delta) \cap E \subset {x} \Rightarrow (y - \delta, y + \delta) \cap (E \setminus {y}) \subset {x} $ . Let $ \delta = \frac{\vert y - x\vert }{2} $ , then $ x \notin (y - \delta, y + \delta) $ , so $ (y - \delta, y + \delta) \cap (E \setminus {y}) = \emptyset $ . Thus $ y $ is not a cluster point $ \Rightarrow y \notin A \Rightarrow y \in \mathbb{R} \setminus A $
We proved $ \mathbb{R} \setminus A$ is an open set, then $ A $ is closed.