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


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. $$]$$