Suppose $A$ and $B$ are bounded and nonempty subsets in $\Bbb{R}$. Define the set $A \cdot B$ as follows $$A \cdot B = \{ab\mid a \in A, b \in B\}$$

I tried proving the statement using inequalities but only managed to show $\sup(A \cdot B) \leq \sup A \sup B$.

I tried a different approach using the fact that

$$\sup S = u \iff \forall \varepsilon > 0 \exists s_{\varepsilon} \in S s.t. u -\varepsilon < s_{\varepsilon}$$

Here's my sort of proof

Proof. Let $\sup A = a$ and $\sup B = b$. Let $\varepsilon > 0$. Then $\exists x_{\varepsilon} \in A$, $\exists y_{\varepsilon} \in B$ s.t. $a - \varepsilon < x_{\varepsilon}$ and $b - \varepsilon < y_{\varepsilon}$. Then $(a - \varepsilon)(b - \varepsilon) = ab - a\varepsilon - b\varepsilon + {\varepsilon}^2 < x_{\varepsilon}y_{\varepsilon}$.

The goal is to show that $ab - \varepsilon < xy$ $\exists x \in A, y \in B$. I'm pretty sure that $x_{\varepsilon}$ and $y_{\varepsilon}$ are those numbers but I'm not sure if $ab - \varepsilon < ab - a\varepsilon - b\varepsilon + {\varepsilon}^2$

Feedback is appreciated. Have a nice day/night!

  • 3
    $\begingroup$ Equality may not hold if the sets contain negative numbers. $\endgroup$
    – Martin R
    Apr 17 at 12:52
  • 4
    $\begingroup$ What if $A = B = \{0,-1\}$? $\endgroup$
    – Hetebrij
    Apr 17 at 12:54
  • 1
    $\begingroup$ Is $\sup(A)$ really $-1$? Are both $-1$ and $0$ smaller than $-1$? $\endgroup$
    – Hetebrij
    Apr 17 at 12:58
  • 2
    $\begingroup$ @860009898987 no, $\sup A\sup B=0\cdot 0=0\leq 1=\sup(A)\sup(B)$. $\endgroup$
    – C Squared
    Apr 17 at 12:58
  • 1
    $\begingroup$ @860009898987 Yes, if both $A$ and $B$ only have positive elements, then you can show $\sup(A \cdot B) = \sup(A) \sup(B)$. $\endgroup$
    – Hetebrij
    Apr 17 at 13:02

For now, lets assume $A$ and $B$ only contain positive elements.

Then in your proof you have shown for all $\varepsilon > 0$ there exists $z_\varepsilon = x_\varepsilon y_\varepsilon \in A \cdot B$ such that \begin{align*} ab - a \varepsilon - b \varepsilon + \varepsilon^2 < z_\varepsilon. \end{align*}

Notice that it is not necessary to show that $ab - \varepsilon < ab - a \varepsilon - b \varepsilon + \varepsilon^2$.

For the definition of the supremum for $A \cdot B$ we will use $\delta > 0$ instead of $\varepsilon > 0$. I.e. we want to show that for all $\delta >0 $ there exists an $z \in A \cdot B$ such that $ab - \delta < z$.

Now, for $\delta > 0$ small enough there exists an $\varepsilon >0$ such that $\delta = a \varepsilon + b \varepsilon - \varepsilon^2$.

Hence, you have $ab - \delta < z_\varepsilon$ for $z_\varepsilon \in A \cdot B$.


As already mentioned in the comments this is true if both $A$ and $B$ have positive elements .

You have already shown that for all $\epsilon>0$ there exists $x\in A$ and $y\in B$ such that

$(a-\epsilon)(b-\epsilon)<xy\le \sup(AB)$

So $(a-\epsilon)(b-\epsilon)<\sup(AB)$ for all $\epsilon>0$

So $ab\le sup(AB)$ and you’ve already proven the reverse inequality .


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