# Supremum of product of sets $A,B$.

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.

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

$$2.$$ 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 $$\alpha^2$$.
• 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. Apr 22, 2019 at 10:03
• 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 $(\frac{1}{2})^2$.