Assignment 2
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw2/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Exercise 1.1.1
Prove part (iii) of Proposition 1.1.8. That is, let $ F $ be an ordered field and $ x, y, z \in F $. Prove if $ x < 0 $ and $ y < z $, then $ xy > xz $.
$x < 0 \Rightarrow -x > 0, \, y < z \Rightarrow 0 = y + (-y) < z + (-y) = z - y$. For $ x, y, z \in F $, $-x > 0$, $ z - y > 0$, $\Rightarrow -x (z - y) > 0 \Rightarrow xy - xz > 0$. Thus $ xy > xz $.
Exercise 1.1.2
Let $ S $ be an ordered set. Let $ A \subset S $ be a nonempty finite subset. Then $ A $ is bounded. Furthermore, in $ A $ exists and is in $ A $ and sup $ A $ exists and is in $ A $. Hint: Use induction.
Let $ A = {a_1, \ldots, a_n} \subset S $. Since $ A $ is finite, there exists $ \min A $, $ \max A \in A $.
Lemma. If $ \min A $, $ \max A $ exist, then $ \inf A = \min A $, $ \sup A = \max A $.
Proof: Let $ a = \min A \in S $. $\forall x \in A$, $ x \geq a \Rightarrow a $ is a lower bound of $ A $. $\forall \epsilon > 0$, $ a > a - \epsilon $, then $ a - \epsilon \notin A $, otherwise $ a \neq \min A $ which causes a contradiction. Thus $ a = \inf A $, $\Rightarrow \inf A = \min A $. The supremum is proved similarly. $\square$
Continue the proof by induction.
(i) $ n = 1 $, $ A = {a_i}, \, i \in {1, \ldots, N} $. $ \min A = \max A = a_i $. Therefore $ \inf A = \min A = a_i \in A $, $ \sup A = \max A = a_i \in A $.
(ii) By induction hypothesis, assume when $ n = k $, $ \vert A\vert = k $, then $\inf A = \min A \in A, \, \sup A = \max A \in A.$ For $ n = k + 1 $, consider $ A’ = A \cup {x} $ with $ \vert A’\vert = k + 1 $, $ x \in S \setminus A $.
Let $ m = \min A $, $ M = \max A $, then $ \min A’ = \min {m, x} \in {m, x} \subset A’ $, $\max A’ = \max {M, x} \in {M, x} \subset A’$. So $ \inf A’ = \min A’ \in A’$, $\sup A’ = \max A’ \in A’$. Then $ \forall n \in \mathbb{N}, \vert A\vert = n $, $ \inf A $ and $ \sup A $ exist and are both in $ A $.
Exercise 1.1.5
Let $ S $ be an ordered set. Let $ A \subset S $ and suppose $ b $ is an upper bound for $ A $. Suppose $ b \in A $. Show that $ b = \sup A $.
By definition of supremum, $ \sup A $ is the least upper bound of $ A $. Since $ b $ is an upper bound for $ A $ and $ b \in A $, we need to show it is the least upper bound. Suppose for contradiction that $ b \neq \sup A $, then $ \exists h \neq 0 $ such that $ b + h = \sup A $.
(Case 1) $ h < 0 $. Then $ \exists x_0 = b - \frac{\vert h\vert }{2} \in A $ such that $ b \geq x_0 > b + h $. This contradicts $ b + h $ being an upper bound since $ x_0 \in A $ and $ x_0 > b + h $.
(Case 2) $ h > 0 $. Then $ \forall x \in A $, $ x \leq b < b + h $. But $ b \in A $ and $ b < b + h $ implies $ b + h $ is not the least upper bound.
Therefore $ b = \sup A $.
Exercise 1.1.6
Let $ S $ be an ordered set. Let $ A \subset S $ be nonempty and bounded above. Suppose $ \sup A $ exists and $ \sup A \notin A $. Show that $ A $ contains a countably infinite subset.
To show $ A $ contains a countably infinite subset, we prove $ A $ cannot be finite. Assume for contradiction that $ A $ is finite. Then $ A $ has a maximum element $ a = \max A $. Since $ a \in A $ and $ a \geq x $ for all $ x \in A $, $ a $ is an upper bound for $ A $.
By definition of supremum, $ \sup A \leq a $. But since $ a \in A $, the least upper bound must satisfy $ \sup A \geq a $. Thus $ \sup A = a \in A $, which contradicts the assumption that $ \sup A \notin A $.
Therefore $ A $ must be infinite, and any infinite set contains a countably infinite subset.
Exercise 1.2.7
\[0 \leq \left( \sqrt{x} - \sqrt{y} \right)^2 = x + y - 2\sqrt{xy} \Rightarrow \frac{x + y}{2} \geq \sqrt{xy}\]Prove the arithmetic-geometric mean inequality. For two positive real numbers $ x, y $,
\(\sqrt{xy} \leq \frac{x + y}{2}.\)Furthermore, equality occurs if and only if $ x = y $.
Now show that equality occurs if and only if $ x = y $.
($\Rightarrow$) If $\frac{x + y}{2} = \sqrt{xy}$, then $x + y - 2\sqrt{xy} = 0$, which implies:
\[\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} - 2 = 0\]Let $t = \sqrt{\frac{x}{y}} \in \mathbb{R}$, then:
\[t + \frac{1}{t} - 2 = 0 \Rightarrow t^2 - 2t + 1 = 0 \Rightarrow (t-1)^2 = 0 \Rightarrow t = 1\]Thus $\sqrt{\frac{x}{y}} = 1 \Rightarrow x = y$.
($\Leftarrow$) If $x = y$, then:
\[\frac{x + y}{2} = x = \sqrt{x^2} = \sqrt{xy}\]Exercise 1.2.9
Let $A$ and $B$ be two nonempty bounded sets of real numbers. Let $C := {a + b : a \in A, b \in B}$. Show that $C$ is a bounded set and that
\[\sup C = \sup A + \sup B \quad \text{and} \quad \inf C = \inf A + \inf B.\]
First, $C$ is bounded because:
\[\inf A + \inf B \leq c \leq \sup A + \sup B \quad \forall c \in C\]By the least upper bound property, $\sup C$ and $\inf C$ exist in $\mathbb{R}$. $\sup A + \sup B$ is an upper bound of $C$, so $\sup C \leq \sup A + \sup B$.
To show $\sup C \geq \sup A + \sup B$, suppose for contradiction that $\sup C < \sup A + \sup B$. Then $\exists \varepsilon > 0$ such that $\sup C \leq \sup A + \sup B - \varepsilon$.
By definition of supremum, $\exists a \in A$ with $a > \sup A - \varepsilon/2$, and $\exists b \in B$ with $b > \sup B - \varepsilon/2$. Then $a + b > (\sup A - \varepsilon/2) + (\sup B - \varepsilon/2) = \sup A + \sup B - \varepsilon \geq \sup C$. But $ c:=a + b \in C, \sup C < c $, contradicting $\sup C$ being an upper bound. Thus $\sup C \geq \sup A + \sup B$.
Therefore $\sup C = \sup A + \sup B$. (The proof for $\inf C = \inf A + \inf B$ is analogous.)
Problem 7
Let \(E = \{x \in \mathbb{R} : x > 0 \text{ and } x^3 < 2\}.\)
(a) Prove that $ E $ is bounded above.
For any $ x \in E $, we have $ x^3 < 2 < 8 $, which implies $ x < 2 $.
Therefore, 2 is an upper bound of $ E $, and $ E $ is bounded above.
(b) Let $ r = \sup E $ (which exists by part (a)). Prove that $ r > 0 \text{ and } r^3 = 2 $.
First, note that $ 1 \in E $ since $ 1^3 = 1 < 2 $, so $ r \geq 1 > 0 $.
We prove $ r^3 = 2 $ by showing both $ r^3 \geq 2 $ and $ r^3 \leq 2 $.
(Step 1) Suppose $ r^3 < 2 $. We want to find $ h > 0 $ such that $ (r + h)^3 < 2 $, which would imply $ r + h \in E $ and contradict $ r $ being the supremum.
\[\begin{aligned}(r + h)^3 - r^3 &= 3r^2h + 3rh^2 + h^3 \\&= h(3r^2 + 3rh + h^2) \\&< h(3r^2 + 3r + 1) \quad (\text{since } h < 1) \\&< 2 - r^3 \quad (\text{for } h < \frac{2 - r^3}{3r^2 + 3r + 1})\end{aligned}\]Choose \(h = \min\left\{\frac{1}{2}, \frac{2 - r^3}{2(3r^2 + 3r + 1)}\right\}\). Then $ (r + h)^3 < r^3 + (2 - r^3) = 2 $, contradicting $ r = \sup E $. Thus $ r^3 \geq 2 $.
(Step 2) Suppose $ r^3 > 2 $. We find $ h > 0 $ such that $ (r - h)^3 > 2 $, showing $ r - h $ is a smaller upper bound.
\[\begin{aligned}r^3 - (r - h)^3 &= 3r^2h - 3rh^2 + h^3 \\&< 3r^2h + h^3 \\&< h(3r^2 + 1) \quad (\text{since } h < 1) \\&< r^3 - 2 \quad (\text{for } h < \frac{r^3 - 2}{3r^2 + 1})\end{aligned}\]Choose \(h = \min\left\{\frac{1}{2}, \frac{r^3 - 2}{2(3r^2 + 1)}\right\}\). Then $ (r - h)^3 > r^3 - (r^3 - 2) = 2 $, contradicting $ r $ being the least upper bound. Thus $ r^3 \leq 2 $.
Combining both results, we conclude $ r^3 = 2 $.
(Note: The solution above demonstrates how to come up with a proof. It should be reorganized to give a formal proof.)