Assignment 3
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw3/
Problem 1
Suppose $ x, y \in \mathbb{R} $ and $ x < y $ . Prove that there exists $ i \in \mathbb{R} \setminus \mathbb{Q} $ such that $ x < i < y $ .
$ y - x > 0 $ . Pick an irrational number $ \theta = \frac{\sqrt{2}}{2} \in (0,1) $ .
Define $ i := x + (y - x)\theta $ . $ i \in (x, y) $ , and $ i \in \mathbb{R} \setminus \mathbb{Q} $ since $ \theta \notin \mathbb{Q} $ .
Problem 2
Let $ E \subset (0,1) $ be the set of all real numbers with decimal representation using only the digits 1 and 2:
\[E := \{ x \in (0,1) : \forall j \in \mathbb{N}, \exists d_{-j} \in \{1,2\} \text{ such that } x = 0.d_{-1}d_{-2}\ldots \}\]Prove that $ \vert E\vert = \vert \mathcal{P}(\mathbb{N})\vert $ . Hint: Consider the function $ f : E \to \mathcal{P}(\mathbb{N}) $ such that if
$ x \in E, x = 0.d_{-1}d_{-2}\ldots $ ,
We want to show that $ f $ is bijective.
(Step 1) If $ f(x_1) = f(x_2) $ , then $ { j \in \mathbb{N} : d_{-j}^{(1)} = 2 } = { j \in \mathbb{N} : d_{-j}^{(2)} = 2 } =: A$ , where $ d_{-j}^{(1)} $ denotes the $ j $ -th digit of $ x_1 $ . Then $ \forall j \in \mathbb{N}, d_{-j}^{(1)} = d_{-j}^{(2)} \Rightarrow x_1 = x_2.$ So $ f $ is 1-1.
(Step 2) Define a number $ y := 0.e_{-1}e_{-2}\ldots $ , where
\[e_j = \begin{cases}2 & \text{if } j \in A \\1 & \text{if } j \notin A\end{cases}\]$ \forall A \in \mathcal{P}(\mathbb{N}) $ , exists $ y \in E $ with $ f(y) = A $ . Thus $ f $ is onto.
Since $ f $ is bijective, $ \vert E\vert = \vert \mathcal{P}(\mathbb{N})\vert $
Problem 3
(a) Given that $ \vert A\vert = \vert B\vert = \vert \mathbb{N}\vert $ , we have to show that $ \vert A \cup B\vert = \vert \mathbb{N}\vert $ .
There exist bijections $ f: \mathbb{N} \to A$ , $ g: \mathbb{N} \to B$ .
Define $ h: \mathbb{N} \to A \cup B$ ,
\[h(n) = \begin{cases} f(n/2) & \text{if } n \text{ is even} \\ g((n-1)/2) & \text{if } n \text{ is odd} \end{cases}\]This means when $ n = 2k$ , we map to $ f(k) \in A$ ; when $ n = 2k + 1$ , we map to $ g(k) \in B$ . Therefore $ h$ is bijective from $ \mathbb{N}$ to $ A \cup B$ .
(b) Since $ \mathbb{R} \setminus \mathbb{Q}$ is infinite, assume that $ \mathbb{R} \setminus \mathbb{Q}$ is countable, thus $ \vert \mathbb{R} \setminus \mathbb{Q}\vert = \vert \mathbb{N}\vert $ .
By Assignment 1 Problem 6, $ \vert \mathbb{Q}\vert = \vert \mathbb{N}\vert $ . By (a), $ \vert \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})\vert = \vert \mathbb{R}\vert = \vert \mathbb{N}\vert $ , which is a contradiction. So $ \mathbb{R} \setminus \mathbb{Q}$ is uncountable.
Problem 4
Let $ A $ be a subset of $ \mathbb{R} $ which is bounded above, and let $ a_0 $ be an upper bound for $ A $ . Prove that $ a_0 = \sup A $ if and only if for every $ \epsilon > 0 $ , there exists $ a \in A $ such that $ a_0 - \epsilon < a $ .
By the least-upper-bound property of $ \mathbb{R} $ , $ \sup A $ exists.
$ (\Rightarrow)$ Since $ a_0 = \sup A $ , then $ \forall \epsilon > 0 $ , if $ \forall a \in A, a \leq a_0 - \epsilon $ , then $ a_0 - \epsilon $ is an upper bound of $ A $ . $ a_0 - \epsilon < a_0 $ , thus $ a_0 \neq \sup A $ , which contradicts with the assumption. $ \therefore \exists a \in A $ , s.t. $ a > a_0 - \epsilon $ .
$ (\Leftarrow)$ Given that $ \forall \epsilon > 0, \exists a \in A $ , s.t. $ a_0 - \epsilon < a $ . Suppose $ a_0 \neq \sup A $ . Then $ \exists u < a_0 $ , $ u $ is an upper bound of $ A $ . Let $ \epsilon = a_0 - u > 0 $ , then $ a_0 - \epsilon = u < a $ , which is a contradiction.
So $ a_0 = \sup A $ .
Problem 5
We say a set $ U \subset \mathbb{R} $ is open if for every $ x \in U $ there exists $ \epsilon > 0 $ such that
\[(x - \epsilon, x + \epsilon) \subset U.\]Since the definition is vacuous for $ U = \emptyset $ , it follows that the empty set is open. It is also clear from the definition that $ U = \mathbb{R} $ is open.
(a) Let $ a, b \in \mathbb{R} $ with $ a < b $ . Prove that the sets $ (-\infty, a)$ , $ (a, b)$ , and $ (b, \infty)$ are open.
When $ U = (-\infty, a) $ , $ \forall x \in (-\infty, a) $ , $ x < a $ , $ \exists \epsilon = \frac{a - x}{2} > 0 $ , s.t. $ x + \epsilon = \frac{a + x}{2} < a $ . $ U = (b, +\infty) $ is proved similarly.
$ \forall x \in (a, b) $ , $ a < x < b $ . \(\exists \epsilon = \min \left\{ \frac{x - a}{2}, \frac{b - x}{2} \right\}\) , s.t.
\(a < \frac{x + a}{2} = x - \epsilon < x + \epsilon = \frac{x + b}{2} < b.\)
(b) Let $ \Lambda $ be a set (not necessarily a subset of $ \mathbb{R} $ ), and for each $ \lambda \in \Lambda $ , let $ U_\lambda \subset \mathbb{R} $ . Prove that if $ U_\lambda $ is open for all $ \lambda \in \Lambda $ , then the set
\(\bigcup_{\lambda \in \Lambda} U_\lambda = \{ x \in \mathbb{R} : \exists \lambda \in \Lambda \text{ such that } x \in U_\lambda \}\) is open.
$ \forall \lambda \in \Lambda, \exists \epsilon_\lambda > 0 $ , s.t. $ (x - \epsilon_\lambda, x + \epsilon_\lambda) \subset U_\lambda, \forall x \in U_\lambda $ .
For $ \bigcup_{\lambda \in \Lambda} U_\lambda $ , let $ \epsilon = \min_{\lambda \in \Lambda} \epsilon_\lambda > 0 $ . If $ x \in U_\gamma $ ($ \gamma \in \Lambda $ ), then
\[(x - \epsilon, x + \epsilon) \subset (x - \epsilon_\gamma, x + \epsilon_\gamma) \subset U_\gamma \subset \bigcup_{\lambda \in \Lambda} U_\lambda.\]
(c) Let $ n \in \mathbb{N} $ , and let $ U_1, \ldots, U_n \subset \mathbb{R} $ . Prove that if $ U_1, \ldots, U_n $ are open, then the set
\(\bigcap_{m=1}^n U_m = \{ x \in \mathbb{R} : x \in U_m \text{ for all } m = 1, \ldots, n \}\) is an open set.
Prove by induction:
(i) $ k = 1 $ , $ U_1 $ is open
(ii) Suppose when $ k = n - 1 $ , $ \bigcap_{m=1}^{n-1} U_m $ is open, thus $ \exists \epsilon_1 > 0 $ s.t. $ \forall x \in \bigcap_{m=1}^{n-1} U_m $ , $ (x - \epsilon_1, x + \epsilon_1) \subset \bigcap_{m=1}^{n-1} U_m $ . For $ k = n $ , $ U_n $ is open. $ \forall x \in U_n, \exists \epsilon_2 > 0 $ s.t. $ (x - \epsilon_2, x + \epsilon_2) \subset U_n $ .
Let $ \epsilon = \min { \epsilon_1, \epsilon_2 } $ , then
\((x - \epsilon, x + \epsilon) \subset \left( \bigcap_{m=1}^{n-1} U_m \right) \cap U_n = \bigcap_{m=1}^n U_m\) So by induction, $ \bigcap_{m=1}^n U_m $ is an open set.
(d) Is the set of rationals $ \mathbb{Q} \subset \mathbb{R} $ open? Provide a proof to substantiate your claim.
No. If $ \mathbb{Q} $ is an open set, then $ \forall x \in \mathbb{Q}, \exists \epsilon > 0 $ , s.t. $ (x - \epsilon, x + \epsilon) \subset \mathbb{Q} $ .
Case 1: $ \epsilon \in \mathbb{R} \setminus \mathbb{Q} $ , then $ x + \frac{\epsilon}{2} \in (x - \epsilon, x + \epsilon) $ , but $ x + \frac{\epsilon}{2} \notin \mathbb{Q} $ , which causes a contradiction.
Case 2: $ \epsilon \in \mathbb{Q} $ , then $ x + \frac{\sqrt{2}}{2} \in (x - \epsilon, x + \epsilon) $ , since $ 0 < \frac{\sqrt{2}}{2} < 1 $ . $ \therefore x + \frac{\sqrt{2}}{2} \notin \mathbb{Q} $ , also a contradiction.
Therefore $ \mathbb{Q} $ is not open.
Problem 6
Prove that
\[\lim_{n \to \infty} \frac{1}{20n^2 + 20n + 2020} = 0.\]Let $ \epsilon > 0 $ and $ M = \frac{2}{\epsilon} $ . Thus, $ \forall n \geq M $ ,
\[\left\vert \frac{1}{20n^2 + 20n + 2020} - 0 \right\vert = \left\vert \frac{1}{20} \cdot \frac{1}{n^2 + n + 101} \right\vert < \left\vert \frac{1}{n(n+1)} \right\vert = \left\vert \frac{1}{n} - \frac{1}{n+1} \right\vert < \frac{2}{n} < 2 \cdot \frac{\epsilon}{2} = \epsilon\]