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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.

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    $\begingroup$ Let $A=B=(-\infty,0)$. Both sets have supremums. But $AB$ has no supremum since it is not bounded from above. $\endgroup$
    – Mark
    Commented Apr 22, 2019 at 9:18
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    $\begingroup$ For (3), consider $A=B=[-1,0]$. $\endgroup$ Commented Apr 22, 2019 at 9:21
  • $\begingroup$ @Mark Your comment rests on product of $-\infty,0$ to yield undefined value. Please provide a theoretical approach also. $\endgroup$
    – jiten
    Commented Apr 22, 2019 at 9:23
  • $\begingroup$ @YuiToCheng Thanks for insight in bounded sets based example. Please provide a theoretical approach also. $\endgroup$
    – jiten
    Commented Apr 22, 2019 at 9:25
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    $\begingroup$ 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. $\endgroup$
    – Mark
    Commented Apr 22, 2019 at 9:29

1 Answer 1

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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 $\alpha^2$.

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  • $\begingroup$ 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. $\endgroup$
    – jiten
    Commented Apr 22, 2019 at 10:03
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    $\begingroup$ 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 $(\frac{1}{2})^2$. $\endgroup$
    – Mark
    Commented Apr 22, 2019 at 10:07

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