# $\sup(A \cdot B) = \sup A \sup B$

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!

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

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

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