2.4.E: Problems on Upper and Lower Bounds (Exercises)

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- 2.4: Upper and Lower Bounds. Completeness
- 2.5: Some Consequences of the Completeness Axiom

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Exercise $\PageIndex{1}$

Complete the proofs of Theorem 2 and Corollaries 1 and 2 for infima.
Prove the last clause of Note $4 .$

Exercise $\PageIndex{2}$

Prove that $F$ is complete iff each nonvoid left-bounded set in $F$ has an infimum.

Exercise $\PageIndex{3}$

Prove that if $A_{1}, A_{2}, \ldots, A_{n}$ are right bounded (left bounded) in $F,$ so is
$\bigcup_{k=1}^{n} A_{k}$

Exercise $\PageIndex{4}$

Prove that if $A=(a, b)$ is an open interval $(a<b),$ then
$a=\inf A \text { and } b=\sup A.$

Exercise $\PageIndex{5}$

In an ordered field $F,$ let $\emptyset \neq A \subset F .$ Let $c \in F$ and let $c A$ denote the set of all products $c x(x \in A) ;$ i.e.,
$c A=\{c x | x \in A\}.$
$\begin{array}{l}{\text { (i) if } c \geq 0 \text { , then }} \\ {\qquad \begin{array}{l}{\sup (c A)=c \cdot \sup A \text { and } \inf (c A)=c \cdot \inf A} \\ {\text { (ii) if } c<0 \text { , then }} \\ {\qquad \sup (c A)=c \cdot \inf A \text { and } \inf (c A)=c \cdot \sup A}\end{array}}\end{array}$
In both cases, assume that the right-side sup $A$ (respectively, inf $A )$ exists.

Exercise $\PageIndex{6}$

From Problem 5 $(\text { ii })$ with $c=-1,$ obtain a new proof of Theorem 1.
[Hint: If $A$ is left bounded, show that $(-1) A$ is right bounded and use its supremum. $]$

Exercise $\PageIndex{7}$

Let $A$ and $B$ be subsets of an ordered field $F .$ Assuming that the required lub and glb exist in $F,$ prove that
(i) if $(\forall x \in A)(\forall y \in B) x \leq y,$ then $\sup A \leq \inf B$ ;
(ii) if $(\forall x \in A)(\exists y \in B) x \leq y,$ then $\sup A \leq \sup B$ ;
(iii) if $(\forall y \in B)(\exists x \in A) x \leq y,$ then $\inf A \leq \inf B$ .
$[\text { Hint for }(\mathrm{i}) : \text { By Corollary } 1,(\forall y \in B) \sup A \leq y, \text { so } \sup A \leq \inf B .(\text { Why? })]$

Exercise $\PageIndex{8}$

For any two subsets $A$ and $B$ of an ordered field $F,$ let $A+B$ denote the set of all sums $x+y$ with $x \in A$ and $y \in B ;$ i.e.,
$A+B=\{x+y | x \in A, y \in B\}.$
Prove that if $\sup A=p$ and $\sup B=q$ exist in $F,$ then
$p+q=\sup (A+B);$
similarly for infima.
[Hint for sup: By Theorem $2,$ we must show that
(i) $(\forall x \in A)(\forall y \in B) x+y \leq p+q(\text { which is easy })$ and
(ii') $(\forall \varepsilon>0)(\exists x \in A)(\exists y \in B) x+y>(p+q)-\varepsilon$ .
Fix any $\varepsilon>0 .$ By Theorem 2,
$(\exists x \in A)(\exists y \in B) \quad p-\frac{\varepsilon}{2}<x \text { and } q-\frac{\varepsilon}{2}<y .(\mathrm{Why} ?)$
Then
$x+y>\left(p-\frac{\varepsilon}{2}\right)+\left(q-\frac{\varepsilon}{2}\right)=(p+q)-\varepsilon,$
as required. $]$

Exercise $\PageIndex{9}$

In Problem 8 let $A$ and $B$ consist of positive elements only, and let
$A B=\{x y | x \in A, y \in B\}.$
Prove that if $\sup A=p$ and $\sup B=q$ exist in $F,$ then
$p q=\sup (A B);$
similarly for infima.
[Hint: Use again Theorem 2 $\left(\mathrm{ii}^{\prime}\right) .$ For $\sup (A B),$ take
$0<\varepsilon<(p+q) \min \{p, q\}$
and
$x>p-\frac{\varepsilon}{p+q} \text { and } y>q-\frac{\varepsilon}{p+q};$
show that
$x y>p q-\varepsilon+\frac{\varepsilon^{2}}{(p+q)^{2}}>p q-\varepsilon.$
For inf $(A B),$ let $s=\inf B$ and $r=\inf A ;$ choose $d<1,$ with
$0<d<\frac{\varepsilon}{1+r+s}.$
Now take $x \in A$ and $y \in B$ with
$x<r+d \text { and } y<s+d,$
and show that
$x y<r s+\varepsilon.$
Explain!

Exercise $\PageIndex{10}$

Prove that
(i) if $(\forall \varepsilon>0) a \geq b-\varepsilon,$ then $a \geq b$ ;
(ii) if $(\forall \varepsilon>0) a \leq b+\varepsilon,$ then $a \leq b$ .

Exercise $\PageIndex{11}$

Prove the principle of nested intervals: If $\left[a_{n}, b_{n}\right]$ are closed intervals in a complete ordered field $F,$ with
$\left[a_{n}, b_{n}\right] \supseteq\left[a_{n+1}, b_{n+1}\right], \quad n=1,2, \ldots$
then
$\bigcap_{n=1}^{\infty}\left[a_{n}, b_{n}\right] \neq \emptyset.$
[Hint: Let
$A=\left\{a_{1}, a_{2}, \ldots, a_{n}, \ldots\right\}.$
Show that $A$ is bounded above by each $b_{n}$ .
Let $p=\sup A .$ (Does it exist?)
Show that
$(\forall n) \quad a_{n} \leq p \leq b_{n},$
i.e.,
$p \in\left[a_{n}, b_{n}\right] . ]$

Exercise $\PageIndex{12}$

Prove that each bounded set $A \neq \emptyset$ in a complete field $F$ is contained in a smallest closed interval $[a, b]$ (so $[a, b]$ is contained in any other $[c, d] \supseteq A )$ .
Show that this fails if "closed" is replaced by "open."
$[\text { Hint: Take } a=\inf A, b=\sup A]$ .

Exercise $\PageIndex{13}$

Prove that if $A$ consists of positive elements only, then $q=\sup A$ iff
(i) $(\forall x \in A) x \leq q$ and
(ii) $(\forall d>1)(\exists x \in A) q / d<x$ .
[Hint: Use Theorem 2. $]$

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Exercise 2.4.E.1\PageIndex{1}

Exercise 2.4.E.2\PageIndex{2}

Exercise 2.4.E.3\PageIndex{3}

Exercise 2.4.E.4\PageIndex{4}

Exercise 2.4.E.5\PageIndex{5}

Exercise 2.4.E.6\PageIndex{6}

Exercise 2.4.E.7\PageIndex{7}

Exercise 2.4.E.8\PageIndex{8}

Exercise 2.4.E.9\PageIndex{9}

Exercise 2.4.E.10\PageIndex{10}

Exercise 2.4.E.11\PageIndex{11}

Exercise 2.4.E.12\PageIndex{12}

Exercise 2.4.E.13\PageIndex{13}

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