More Suprema and Density Practice

1.3.4, 1.3.5, 1.3.7, 1.3.9, 1.3.10, 1.4.6, 1.4.7

1.3.4 (check)

Question

Let $A_{1}, A_{2}, A_{3}, . . .$ be a collection of nonempty sets, each of which is bounded above.
a. Find a formula for $sup (A_{1} \cup A_{2})$ and extend to $⋃_{k = 1}^{n} A_{k}$
b. Consider $sup (⋃_{k = 1}^{\infty} A_{k})$ . Does the formula in (a) extend to the infinite case?

Proof
a.

sup (A_{1} \cup A_{2}) = max (sup A_{1}, sup A_{2})

Proof
Let $β = max (sup A_{1}, sup A_{2})$ . Clearly the claim holds if $A_{1} = A_{2}$ so say $A_{1} \neq A_{2}$ . Then either $A_{1} \subset A_{2}$ or $A_{1} \supset A_{2}$ , which are argued in similar ways.

Using the former then $A_{1} \cup A_{2} = A_{2}$ so then we just have to show $sup A_{2} \geq sup A_{1}$ . Notice if they equal then we are done since $sup (A_{1} \cup A_{2}) = sup A_{2} = sup A_{1} = max (\dots)$ . Now consider the case they don't equal. Then $sup A_{2} - sup A_{1} > 0$ so then choose $ε > 0$ to be that positive number. Since $sup A_{2}$ then there is a $a \in A_{2}$ where $sup A_{2} - ε = sup A_{2} - (sup A_{2} - sup A_{1}) = sup A_{1} < a$ . But since $sup A_{2}$ then $\forall a^{'} \in A_{2} (a^{'} \leq sup A_{2})$ , thus giving that $sup A_{1} < a \leq sup A_{2} \Rightarrow sup A_{1} < sup A_{2}$ as we wanted.
☐

Using an inductive proof shows the following with ease:

sup (⋃_{k = 1}^{n} A_{k})) = {max}_{k = 1}^{n} sup A_{k}

b. It doesn't. Consider the collection $A = {A_{i} \subseteq R | A_{i} = [i, i + 1] \forall i \in N}$ . If you say that $α = sup (⋃_{k = 1}^{\infty} A_{i})$ exists then then $\exists A_{i}$ such that $α \in A_{i}$ . But then that means that $α \geq sup A_{j}$ for any $j \in N$ . But then that means that $α \geq j$ for all $j$ by the way we constructed $A_{j}$ . This is a contradiction since that breaks the Cantor's Approach#^5404f9.
☐

1.3.5

Question

Let $A \in R$ be nonempty and bounded above, let $c \in R$ . This time define the set $c A = {c a | a \in A}$ .
a. If $c \geq 0$ then show that $sup (c A) = c sup (A)$
b. Postulate a similar type of statement for $sup (c A)$ for the case that $c < 0$ .

Proof
a. Clearly $sup (c A)$ exists since $A$ is nonempty and bounded above. Start from $c α = c sup (A)$ and we'll show that it is $sup (c A)$ . For the $c = 0$ case notice that then $c A = {0}$ so clearly $sup (c A) = 0 = 0 \cdot sup (A)$ . Thus, take $c \neq 0$ :

( $c α$ is an upper bound for $c A$ ): Let $a^{'} \in c A$ be arbitrary. Then $\exists a \in A$ where $a^{'} = c a$ . Then $\frac{a^{'}}{c} = a$ . Since $α = sup (A)$ then $a \leq α$ so consequently $a^{'} \leq c α$ as desired.
( $c α$ is the least upper bound for $c A$ ): Let $α^{'}$ be another upper bound for $c A$ . Assume that $α^{'} < c α$ . Then $\frac{α^{'}}{c} < α$ so then since $α = sup A$ then $\frac{α^{'}}{c} \in A$ . Thus then $\exists a \in A$ where $\frac{α^{'}}{c} = a \Rightarrow α^{'} = c a$ which contradicts $α^{'}$ being an upper bound for $c A$ .

For

c < 0

, then

sup (c A) = - inf (A)

☐

1.3.7

Question

Prove that if $a$ is an upper bound for $A$ , and if $a$ is also an element of $A$ , then it must be that $a = sup (A)$ .

Proof
Clearly the upper bound for $A$ property is satisfied so we just have to show that $a$ is the least upper bound for $A$ . Let $a^{'}$ be another upper bound for $A$ . Assume for contradiction that $a^{'} < a$ . Then since $a \in A$ that contradicts $a^{'}$ being an upper bound for $A$ , so then clearly $a^{'} \geq a$ as required.
☐

1.3.9 (check)

Question

a. If $sup A < sup B$ , then $\exists b \in B$ that is an upper bound for $A$ .
b. Give an example to show that this is not always the case if we only assume $sup A \leq sup B$ .

Proof
a.
Since $sup A < sup B \Rightarrow sup B - sup A = ε > 0$ so then by the Suprema Lemma for $sup (B)$ then $\exists b \in B$ such that $sup B - (sup B - sup A) = sup A < b$ . Thus $b$ is clearly an upper bound for $A$ .

b. If $A = {q \in I | q^{2} < 2}$ and $B = {q \in Q | q^{2} < 2}$ then clearly $sup A = sup B = \sqrt{2}$ but if there was some $b \in B$ that's an upper bound for $A$ then we know that $b$ is irrational and $\sqrt{2}$ is irrational, so since The Density of Q in R#^ca78a0, then $\exists q \in I$ where $b < q < \sqrt{2}$ , so clearly $q \in A$ so clearly $b$ is not an upper bound for $A$ .
☐

1.3.10 (FINISH, don't forget the other parts)

(Cut Property)

If $A, B$ are nonempty, disjoint sets with $A \cup B = R$ and $a < b$ for all $a \in A, b \in B$ then $\exists c \in R$ such that $x \leq c$ whenever $x \in A$ and $x \geq c$ whenever $x \in B$ .

Proof

☐

1.3.4 (check)

1.3.5

1.3.7

1.3.9 (check)

1.3.10 (FINISH, don't forget the other parts)

1.4.7 (TODO)