Suprema + Density Practice

1.3.1, 1.3.2, 1.3.3, 1.3.8, 1.3.11, 1.4.1, 1.4.2, 1.4.3, 1.4.4, 1.4.5, 1.4.8

1.3.1

Question

a. Write a formal definition in the style of the suprema for the infimum or greatest lower bound of a set.
b. Now, state and prove a version of suprema lemma for greatest lower bounds.

(a): Here it is:

Infimum, Greatest Lower Bounds

Some $s \in R$ is the greatest lower bound for a set $A \subseteq R$ if it meets:

$s$ is a lower bound for $A$ .
If $b$ is any lower bound for $A$ , then $s \geq b$ .
Denote $s = inf (A)$ .

(b):

Infima Lemma

Suppose $s \in R$ is a lower bound for a set $A \subseteq R$ . Then $s = inf (A)$ iff, for all $ε > 0$ there is an element $a \in A$ such that $s + ε > a$ .

Proof
( $\to$ ): Suppose $s = inf (A)$ . Then $s$ is obviously a lower bound for $A$ . Furthermore, we know for all other lower bounds $b$ for $A$ that $s \geq b$ . Now, let $ε > 0$ . Notice that $s + ε > s$ , so since $s$ is the greatest lower bound, then we know that $s + ε$ cannot be a lower bound for $A$ . Hence, there is some $a \in A$ such that $a < s + ε$ .

( $\leftarrow$ ): Suppose that $\forall ε > 0 \exists a \in A (s + ε > a)$ . We'll show the two properties of $inf$ :

It is given that $s$ is a lower bound.
Let $b$ be any lower bound for $A$ , so $\forall a \in A (a \geq b)$ . We want to show that $s \geq b$ . Notice that if we assume that $b > s$ is a greater lower bound, then clearly $b - s > 0$ so then plugging into our given then $\exists a \in A$ where $s + (b - s) > a$ So then $b > a$ . But we know that $b$ is a lower bound and found that $b > a$ ! That's a contradiction, so then our assumption is false. Hence $b \leq s$ as required.
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1.3.2

Question

Give an example of each of the following, or state that the request is impossible:
(a): A set $B$ with $inf (B) \geq sup (B)$
(b): A finite set that contains its infimum but not its supremum
(c): A bounded subset of $Q$ that contains its supremum but not its infimum.

Proof
(a): Simple. Have $B = {2}$ . Clearly $2 = inf (B) = sup (B)$ .

(b): Impossible. Finite sets with only finite elements must have to contain their elements (since then you can order the elements into an ordered list, and the infimum/suprema are the endpoints of the list).

(c): This is pretty easy. Chose $B = {q \in Q | 4 \geq q^{2} \geq 2}$ . Notice that $2 = inf (B) \in B$ but $\sqrt{2} = sup (B) \notin B$ since $\sqrt{2}$ is irrational.

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1.3.3

Question

a. Let $A$ be nonempty and bounded below, and define $B = {b \in R | b is a lower bound for A}$ . Show that $sup (B) = inf (A)$ .
b. Use (a) to explain why there is no need to assert that greatest lower bounds exist as part of the axiom of completeness.

Let's prove (a):

Proof
Since $A$ is nonempty and bounded below, then $\exists b$ as a lower bound such that $\forall a \in A (b \leq a)$ , and further $\exists a \in A$ . We immediately can say $b \leq a$ . Since $b$ is a lower bound for $A$ then clearly $b \in B$ .

Say $β = inf (A)$ . Then $\forall ε > 0 \exists a \in A (β + ε > a)$ using our infima lemma. Furthermore, we know that $β$ clearly is a lower bound for $A$ .

If we show $β$ is the suprema for $B$ then we are done:
(i) We should show $β$ is an upper bound for $B$ , or that $\forall b \in B (β \geq b)$ . Let $b \in B$ be arbitrary. Now assume that $β < b$ , for contradiction. Then clearly $ε = b - β > 0$ so then via our infima lemma given, then $\exists a \in A$ such that $β + (b - β) > a$ so then $b > a$ . But then that means $b$ is not a lower bound for $A$ , so our assumption is false. Thus $\forall b \in B (β \geq b)$ .
(ii). We should show $β$ is specifically the least upper bound for $B$ , or that $β \leq b$ for any upper bound $b$ for $B$ . Let $b$ be that arbitrary upper bound for $B$ . Assume that $β > b$ . Then $β \notin B$ since $b$ is an upper bound for $B$ . then $β$ is not a lower bound for $A$ , so $\exists a \in A$ where $β > a$ . But $β$ is a lower bound for $A$ since it's the infimum, so this is a contradiction. Thus $β \leq b$ so $β$ is the smallest upper bound.
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(b) Let $A$ be nonempty and bounded below. Then (a) suggests that $sup (B) = inf (A)$ . But clearly by the axiom of completeness, because $B$ is nonempty ( $A$ is bounded below so $b$ exists as a lower bound for $A$ so $b \in B$ exists) and bounded above (if there were always larger lower bounds for $A$ , then $A$ wouldn't be bounded below), then $sup (B)$ exists. Then clearly $inf (A)$ exists by consequence.

1.3.8

Question

Compute, without proofs, the suprema and infima (if they exist) of the following sets:
a. ${\frac{m}{n} | m, \in N; m < n}$
b. ${\frac{(- 1)^{m}}{n} | m, n \in N}$
c. ${\frac{n}{3 n + 1} | n \in N}$
d. ${\frac{m}{m + n} | m, n \in N}$

(treat them like 'limits')
a. $inf (A) = 0$ . $sup (A) = 1$ .
b. $inf (B) = - 1$ . $sup (A) = 1$ .
c. $inf (B) = \frac{1}{3}$ . $sup (A) = \frac{1}{4}$ .
d. $inf (B) = 0$ . $sup (A) = \frac{1}{2}$ .

1.3.11

Question

Decide if the following statements about suprema and infima are true or false. Give a short proof for those that are true. For any that are false, supply an example where the claim in question does not appear to hold.
a. If $A, B$ are nonempty, bounded, and satisfy $A \subseteq B$ , then $sup (A) \leq sup (B)$ .
b. If $sup (A) < inf (B)$ for sets $A, B$ , then $\exists c \in R$ satisfying $a < c < b$ for all $a \in A, b \in B$
c. If $\exists c \in R$ where $a < c < b$ for all $a \in A, b \in B$ , then $sup (A) < inf (B)$ .

a. True.

Proof
$A, B$ are bounded and nonempty so then both their suprema and infima exist. Let $α = sup (A), β = sup (B)$ . If we assume for contradiction that $sup (A) > sup (B)$ then $ε = α - β > 0$ so then $\exists a \in A$ where $α - (α - β) < a$ , or just $β < a$ . But since $a \in A$ and $A \subseteq B$ then $a \in B$ . Thus then clearly $β$ is not an upper bound for $B$ , which is a contradiction as $β$ is the suprema of $B$ . Therefore, we must have $sup (A) \geq sup (B)$ .
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b. True.

Proof
Suppose $sup (A) < inf (B)$ , denoted $α < β$ for simplicity here. Choose $c = \frac{α + β}{2} \in R$ . Let $a \in A, b \in B$ be arbitrary.

First let's show $a < c$ . Notice that $a < sup (A) < \frac{α + β}{2} = c$ . Similarly $c = \frac{α + β}{2} < inf (B) < b$ by the definitions of $sup$ and $inf$ .
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c. False. You can have equality. Let $A = {q \in Q | q^{2} < 2}$ and $B = {q \in Q | q^{2} > 2}$ . Here you can always choose some $c = \frac{a + b}{2}$ where $a \in A, b \in B$ and have $a < c < b$ . However, clearly $sup (A) = \sqrt{2} = inf (B)$ .

1.4.1

Question

Recall $I$ stands for the irrational numbers.
a. Show that if $a, b \in Q$ then $a b, a + b$ are both elements of $Q$ as well.
b. Show that if $a \in Q$ and $t \in I$ then $a + t \in I$ and $a t \in I$ as long as $a \neq 0$ .
c. (a) can be summarized by saying $Q$ is closed under addition and multiplication. Is $I$ closed under addition and multiplication? Given two irrationals $s, t \in I$ what can we say about $s + t$ and $s t$ ?

Proof
a. Here $a = \frac{m_{a}}{n_{a}}$ and $b = \frac{m_{b}}{n_{b}}$ where all the $m_{i} \in Z$ and $n_{i} \in N$ . Then:

a + b = \frac{m_{a}}{n_{a}} + \frac{m_{b}}{n_{b}} = \frac{m_{a} n_{b} + n_{a} m_{b}}{n_{a} n_{b}}

Now clearly $n_{a} n_{b} \in N$ by the closure of multiplication over $N$ . Similarly $m_{a} n_{b}, n_{a} m_{b} \in Z$ by the same idea, and by closure of addition on $Z$ thne $m_{a} n_{b} + n_{a} m_{b} \in Z$ . Therefore then $a + b$ has the correct form of a rational number so $a + b \in Q$ . Furthermore:

a b = \frac{m_{a}}{n_{b}} \cdot \frac{m_{b}}{n_{a}} = \frac{m_{a} m_{b}}{n_{a} n_{b}}

Here $m_{a} m_{b} \in Z$ by closure via multiplication so then $a b$ is also of the right form. $a b \in Q$ .

b. Use $a = \frac{m}{n}$ and $t \in I$ , assume for contradiction that $a + t Q$ . Then it's rational, so $a + t = \frac{m^{'}}{n^{'}}$ . But then:

a + t = \frac{m^{'}}{n^{'}} \Rightarrow t = \frac{m^{'}}{n^{'}} - \frac{m}{n} \in Q

because $Q$ is closed under addition and scalar multiplication. As such then $t \in Q$ which contradicts $t \in I$ . Hence $a + t \notin Q$ so $a + t \in I$ . Similarly for $a t$ if we suppose $a \neq 0$ then $m \neq 0$ . Thus:

a t = \frac{m^{'}}{n^{'}} \Rightarrow t = \frac{m^{'}}{a n^{'}} = \frac{m^{'} n}{n^{'} m} \in Q

which contradicts $t \notin Q$ . Thus $a t \in I$ .

c. $I$ is not closed under addition nor scalar multiplication. For multiplication we know that $\sqrt{2} \in I$ via Irrationality of sqrt(2). But:

(\sqrt{2}) (\sqrt{2}) = 2 \notin I

Thus $I$ isn't closed under scalar multiplication. Furthermore, we know $- \sqrt{2} + 1 \in I$ (without proof here, but it's easy to verify) but:

\sqrt{2} + (- \sqrt{2} + 1) = 1 \notin I

so $I$ isn't closed under addition neither.
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1.4.2 (check, didn't use s+1/n)

Question

Let $A \subseteq R$ be nonempty and bounded above, and let $s \in R$ have the property that for all $n \in N$ , $s + \frac{1}{n}$ is an upper bound for $A$ and $s - \frac{1}{n}$ is not an upper bound for $A$ . Show $s = sup (A)$ .

Proof
If we show $\forall ε > 0 \exists a \in A (s - ε < a)$ then we are done by the Suprema Lemma. Let $ε > 0$ . Notice that since $ε > 0$ then the archimedian property gives that $\exists n \in N$ where $\frac{1}{n} < ε$ . For later, this has:

\frac{1}{n} < ε \Rightarrow - ε < - \frac{1}{n} \Rightarrow s - ε < s - \frac{1}{n}

Then notice that since $s - \frac{1}{n}$ is not an upper bound for $A$ then $\exists a \in A$ where $s - \frac{1}{n} < a$ . But:

s - ε < s - \frac{1}{n} < a \Rightarrow s - ε < a

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1.4.3

Question

Prove that $⋂_{n = 1}^{\infty} (0, \frac{1}{n}) = \emptyset$ . Notice that this demonstrates that the intervals in the Nested Interval Property must be closed for the conclusion of the theorem to hold.

Proof
Assume for contradiction $\neq \emptyset$ . Then $\exists a \in ⋂_{n = 1}^{\infty} (0, \frac{1}{n})$ . That means that for all $n \in N$ then $a \in (0, \frac{1}{n})$ , so $0 < a < \frac{1}{n}$ . Notice that this is just the opposite of the The Density of Q in R#^46383a, where we've shown $\exists a \in R \forall n \in N (\frac{1}{n} > a)$ , so we've hit a contradiction (since we know the Archimedean property is true). Thus, the set must be the empty set.
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1.4.4 (check)

Question

Let $a < b$ be real numbers and consider $T = Q \cap [a, b]$ . Show $sup (T) = b$ .

Proof

( $b$ is an upper bound). We want to show that $\forall t \in T (b \geq t)$ . Let $t \in T$ and assume instead $b < t$ . Since $t \in Q \cap [a, b]$ then $\exists m \in Z, n \in N$ where $t = \frac{m}{n}$ . Furthermore, we know that $a \leq t \leq b$ . But look! $t \leq b$ cannot be true while $t > b$ ! Thus our assumption is false so then $b \geq t$ as required.
( $b$ is the least upper bound). We want to show that for all upper bounds $u$ for $T$ that $b \leq u$ . It's true that $u \geq b$ , because if we had assumed that $u < b$ then choose $b \in T$ to show that $u$ is not a valid upper bound.
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1.4.5

Question

Using Suprema + Density Practice#1.4.1, prove The Density of Q in R#^ca78a0 by considering the real numbers $a - \sqrt{2}$ and $b - \sqrt{2}$ .

Proof
Given two reals $a, b \in R$ , then since $Q$ is dense in $R$ and $a - \sqrt{2}, b - \sqrt{2} \in R$ then $\exists q \in Q$ where:

a - \sqrt{2} < q < b - \sqrt{2}

adding $\sqrt{2}$ to both sides:

a < q + \sqrt{2} < b

Now choose $t = q + \sqrt{2}$ . Notice that by Suprema + Density Practice#1.4.1 that $t \in I$ because $q \in Q$ .
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1.4.8

Question

Give an example of each or state that the request is impossible. When impossible, provide a compelling argument for why it's the case.
a. Two sets $A, B$ with $A \cap B = \emptyset$ and $sup A = sup B$ where $sup A \notin A$ and $sup B \notin B$
b. A sequence of nested open intervals $J_{1} \supseteq J_{2} \supseteq J_{3} \supseteq \dots$ with $⋂_{n = 1}^{\infty} J_{n}$ nonempty but containing only a finite number of elements.
c. A sequence of nested unbounded closed intervals $L_{1} \supseteq L_{2} \supseteq L_{3} \supseteq \dots$ with $⋂_{n = 1}^{\infty} L_{n} = \emptyset$ where the unbounded closed intervals are of the form $[a, \infty)$ where $a \in R$ .
d. A sequence of closed bounded (not necessarily nested) intervals $I_{1}, I_{2}, I_{3}, . . .$ with the property that $⋂_{n = 1}^{N} I_{n} \neq \emptyset$ for all $N \in N$ , but $⋂_{n = 1}^{\infty} I_{n} = \emptyset$

Proof
a. $A = Q \cap (0, 1)$ and $B = I \cap (0, 1)$ . Here $A \cap B = \emptyset$ , and $sup A = sup B = 1 \notin A, B$ .
b. Define $J_{i} = (a_{i}, b_{i})$ for all $i \in N$ where $A = {a_{j} | j \in N \land \forall i, j (a_{i} < b_{j}}$ and $B = {b_{j} | j \in N \land \forall i, j (a_{i} < b_{j})}$ . Notice that $sup A, inf B \in ⋂_{n = 1}^{\infty} J_{n}$ .
c. Have each $L_{i}$ be of the form $[i, \infty)$ . Clearly we have the nested interval property. Now assume for contradiction there was some $α \in ⋂_{n = 1}^{\infty} L_{n}$ to our contrary. Then since $α \in L_{1}$ then $1 \leq α$ . Indeed for any $j$ we have that $j \leq α$ . But since $α \in R$ then it must bounded between two integers by the Archimedian Property:

n \leq α < n + 1

But then $α < n + 1 \in N$ which implies that $α \notin L_{i + 1} \Rightarrow α \notin ⋂_{n = 1}^{\infty} L_{n}$ which is a contradiction.
d. This is impossible. Assume you could do such a thing. Then there is $α \in ⋂_{n = 1}^{N} I_{n}$ for any $n \in N$ . We have to have $α \notin ⋂_{n = 1}^{\infty} I_{n}$ so then $\exists I_{i}$ such that $α \notin I_{i}$ . But since $i \in N$ then clearly $α \in ⋂_{n = 1}^{i} I_{n}$ as our given, so then that's our contradiction.
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