2.4.E: Problems on Upper and Lower Bounds (Exercises)
Complete the proofs of Theorem 2 and Corollaries 1 and 2 for infima.
Prove the last clause of Note \(4 .\)
Prove that \(F\) is complete iff each nonvoid left-bounded set in \(F\) has an infimum.
Prove that if \(A_{1}, A_{2}, \ldots, A_{n}\) are right bounded (left bounded) in \(F,\) so is
\[
\bigcup_{k=1}^{n} A_{k}
\]
Prove that if \(A=(a, b)\) is an open interval \((a<b),\) then
\[
a=\inf A \text { and } b=\sup A.
\]
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.
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. \(]\)
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 \(\inf A \leq \inf B\).
\([\text { Hint for }(\mathrm{i}) : \text { By Corollary } 1,(\forall y \in B) \sup A \leq y, \text { so } \sup A \leq \inf B .(\text { Why? })]\)
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. \(]\)
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!
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 \(a \leq b\).
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.,
\[
p \in\left[a_{n}, b_{n}\right] . ]
\]
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."
\([\text { Hint: Take } a=\inf A, b=\sup A]\).
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. \(]\)