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Mathematics LibreTexts

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

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    22255
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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. \(]\)