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Taken from sec. 1.4.1 of the book by Mary Hart, titled: Guide to Analysis.

Let $A, B$ be two non-empty sets of real numbers with supremums $\alpha, \beta$ respectively, and let the sets $A + B$ and $AB$ be defined by :
$A + B = {a + b: a\in A, b\in B}$,
$AB= {ab:a\in A, b\in B}$.

The first question in the sequence is stated here (and was found answered earlier).

  1. Give an example to show that AB need not have a supremum.

  2. Prove also that even if AB has a supremum, this supremum need not be equal to $\alpha \beta$.

  3. Show that if $A$ be set of positive reals with supremum $\alpha$, & let $Y = {x^2 : x\in X}$; then $\alpha^2$ is supremum of Y.

My attempts:

Q.#2 : The possible way seems not clear, as if sets $A, B$ do have valid supremum, then why their product cannot have. I hope that the only way to not have a valid supremum is to have an unbounded value ($+/- \infty$), which I hope cannot be formed by product of two valid values., i.e. if $a,b \lt \infty$, (or, $a,b \gt - \infty$) then $a.b$ is also $\lt \infty$ ($\gt - \infty$).

I am just elaborating by below the statement above, to substantiate it & is based on material here.
If take the sets $A,B$ as $A = \{1,2,3\}, B=\{4,5\}$;
then the set $AB= \{4,5,8,10,12,15\}$.

Q.#3 : There are two approaches by which the attempt is planned. First, theoretical one; & second using an example (as given in book as hint).

1st appr.: Unable to develop anything. Need help. Request help for providing minimum ground to develop upon.

2nd appr.: As per the book that gives hint by stating :
The set S is equal to ${x \in R: \frac13 \lt x \lt 3}$, since $3x^2 -10x +3 = (3x-1)(x-3) \lt 0$ if $\frac13 \lt x \lt 3$.

My understanding of the hint:
$3x^2-10x+3$ has roots $x=\frac13, 3$. The set of values taken by $x$ in $R$, in which the value of function is not - positive is in range ${x \in R: \frac13 < x < 3}$. So, the given function in bounded domain is having no maximum, but has supremum of $0$ apart from having minimum, infimum.

The individual linear components are: $(3x - 1), (x - 3)$, with supremum: $8,0$ respectively at $x=3$. While at the other end of the domain, $x=\frac13$, supremum are : $0, \frac{-2}3$.

Supremum of quadratic function is $0$, & linear factors' supremum product is also $0$ at both ends.

Q. 4: First, take an example of finite small set. If take the set $A$ as $A = \{1,2,3\}$ with supremum $3$, then $Y=\{1,2, 3,4, 6,9\}$; or alternately take set defined by a function, with domain limits specified to make it a bounded set. Let the set $A= 3x-1, 1 \le x \le 3$. The values in $A= \{0,1,2\}$. And similar multiplication can be done.

But, unable to develop theoretical basis.

jiten
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    Let $A=B=(-\infty,0)$. Both sets have supremums. But $AB$ has no supremum since it is not bounded from above. – Mark Apr 22 '19 at 09:18
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    For (3), consider $A=B=[-1,0]$. – YuiTo Cheng Apr 22 '19 at 09:21
  • @Mark Your comment rests on product of $-\infty,0$ to yield undefined value. Please provide a theoretical approach also. – jiten Apr 22 '19 at 09:23
  • @YuiToCheng Thanks for insight in bounded sets based example. Please provide a theoretical approach also. – jiten Apr 22 '19 at 09:25
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    How does it? We are not multiplying infinite numbers here. All the numbers in the set $(-\infty, 0)$ are finite real numbers, just you can take them as small as you wish. Let $M>0$. We know that $-2\sqrt{M}\in A$ and $-2\sqrt{M}\in B$, and hence $(-2\sqrt{M})(-2\sqrt{M})=4M\in AB$. So there is an element in $AB$ which is bigger than $M$, so $M$ is not an upper bound of $AB$. This is true for all $M>0$, hence $AB$ is not bounded from above. – Mark Apr 22 '19 at 09:29
  • @Mark Sorry, for taking values near $- \infty$ as that. I hope you have given theoretical basis. Please provide answer too, if possible. – jiten Apr 22 '19 at 09:38
  • Jiten. Mark's example. Show that there is no real s s.t. s=sup (AB).By def. sup is a least upper bound.We show that AB is not bounded above: Take b=-1, and a in A. Then C={-a| a \in (-\infty,0)= \mathbb{R_{>0}} \subset AB. \mathbb {R_{>0}} is not bounded , hence AB is not bounded, no real s . – Peter Szilas Apr 22 '19 at 09:52
  • @PeterSzilas I hope you meant to have two sets $A,B$, with set $B$ having $b=-1$, and set $A$ with elements $\in (-\infty,0)$. This is to show that the product is comprising only of positive elements. This I hope is conveyed by the product to be a subset of positive reals, i.e. $C =AB= {-a | a \in (-\infty,0)} \subset \mathbb {R_{>0}} }$. But, cannot understand why $AB$ is unbounded, if $A$ is? – jiten Apr 22 '19 at 18:55
  • jiten.C=(0,+\infty))( \subset AB) is the set of positive reals is not bounded above. If C as a subset of AB is not bounded above then AB is not bounded above.Is this your question?Took only one element of B, b=-1, then C=A{-1} \subset AB, and C are the positive reals unbounded above. – Peter Szilas Apr 22 '19 at 21:44
  • @PeterSzilas Sorry, but I meant that as $AB$ is bounded above due to not taking $+ \infty$ value, so $C$ (which, I took as $AB$, rather than a subset of) too is. Your statement seems to take $C$ as a valid subset of $AB$ and yet showing that $C$ is unbounded. I request elaboration as an answer, if correct; else kindly correct me as it is confusing how a valid subset of a bounded set is unbounded. – jiten Apr 23 '19 at 01:38
  • jiten.C is the set of positive reals , C is a subset of AB, correct? If C is not bounded above , then AB as a superset is not bounded above.Note: \infty is not a real number. Is this your question? – Peter Szilas Apr 23 '19 at 06:57

1 Answers1

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$2.$ Let $A=B=(-\infty,0)$. Then both sets have a supremum, but as I proved in the comments $AB$ is not bounded from above and hence has no supremum.

$3.$ Let $A=B=[-1,0]$. Then $\alpha=\beta=0$. But the supremum of $AB$ is $1$ (it is even a maximum) which is not $\alpha\beta$.

$4. $ If $A$ is a set of positive real numbers then $\alpha>0$. First we will show that $\alpha^2$ is an upper bound of $Y$. Let $y\in Y$. By the definition of $Y$ there is some $x\in A$ such that $y=x^2$. But $x\leq\alpha$ and hence $y=x^2\leq\alpha^2$. This is true for all $y\in Y$, so $\alpha^2$ is an upper bound of $Y$.

Now we have to show that $\alpha^2$ is the least upper bound. Let $\epsilon>0$ be small enough such that $\alpha-\frac{\epsilon}{\alpha}>0$. Since $\alpha$ is the least upper bound of $A$ there is some $x\in A$ such that $x>\alpha-\frac{\epsilon}{\alpha}$. Then $x^2>\alpha^2-2\epsilon+\frac{\epsilon^2}{\alpha^2}>\alpha^2-2\epsilon$. So we showed that there is an element in $Y$ which is greater than $\alpha^2-2\epsilon$. Since it is true for any small enough $\epsilon$ we conclude that there can't be an upper bound of $Y$ which is smaller than $\alpha^2$.

Mark
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  • In case of $A$ being not restricted to positive reals, with $\alpha \gt 0$, the logic should change as if infimum is negative (got due to lower bound in domain, if take a function) & its absolute value can be greater than $\alpha$. Please provide some thoughts on this too. – jiten Apr 22 '19 at 10:03
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    We use not only the fact that $\alpha>0$, but also that the elements in $A$ are positive. Otherwise we could take $A=[-1,\frac{1}{2}]$. The supremum of $A$ is positive, but the supremum of $A^2$ is $1$, which is not $(\frac{1}{2})^2$. – Mark Apr 22 '19 at 10:07