# Show that sup$AB$=(sup$A$)(sup$B$)

Where $AB$ is the product of the sets and $A,B \in \mathbb{R^+}$.

Since $A,B$ are bounded above sup $A$ and sup $B$ exist. Let $\alpha =$ sup $A$ and $\beta =$ sup $B$. This implies $\forall a \in A$ and $\forall b \in B$ $a \leq \alpha$ and $b \leq \beta$. Then $ab \leq \alpha\beta$ because $a,b > 0$. Thus $ab$ is bounded above and sup $AB$ exists and sup $AB \leq \alpha\beta$. \ We now show sup $AB \geq \alpha\beta$. \ Let $\varepsilon > 0$ then $\exists a \in A$ s.t. $\alpha - \varepsilon < a \leq \alpha$ and $\exists b \in B$ s.t. $\beta - \varepsilon < b \leq \beta$. So: \begin{equation*} (\alpha-\varepsilon)(\beta-\varepsilon) < ab \leq \alpha\beta \text{ since } a,b,\varepsilon > 0 \end{equation*} \begin{equation*} = \alpha\beta-\varepsilon(\alpha+\beta-\varepsilon) < ab \leq \alpha\beta \end{equation*}

I spoke with my professor today about this and he suggested I show that $\varepsilon$ is sufficiently small to proceed. I'm not sure exactly how to write this detail.

EDIT: I was meant to show what $\varepsilon$ was bounded by to proceed. The proof below realizes this idea. Feedback is welcome and appreciated.

Since $A,B$ are bounded above sup $A$ and sup $B$ exist. Let $\alpha =$ sup $A$ and $\beta =$ sup $B$. This implies $\forall a \in A$ and $\forall b \in B$ $a \leq \alpha$ and $b \leq \beta$. Then $ab \leq \alpha\beta$ because $a,b > 0$. Thus $ab$ is bounded above, sup $AB$ exists and sup $AB \leq \alpha\beta$. Let $\varepsilon > 0$ then $\exists a \in A$ s.t. $\alpha - \varepsilon < a \leq \alpha$ and $\exists b \in B$ s.t. $\beta - \varepsilon < b \leq \beta$. So: \begin{equation*} (\alpha-\varepsilon)(\beta-\varepsilon) < ab \leq \alpha\beta \end{equation*} \begin{equation*} = \alpha\beta-(\varepsilon\alpha+\varepsilon\beta-\varepsilon^2) < ab \leq \alpha\beta \end{equation*} Since $ab$ is bounded above by $\alpha\beta$ we have $ab \leq \text{ sup}(AB)$. We let $\varepsilon' = \varepsilon\alpha+\varepsilon\beta-\varepsilon^2 > 0$ so $\forall(0 < \varepsilon' < \alpha+\beta)$ we have $\alpha\beta-\varepsilon'< ab < \text{ sup}(AB) \implies \alpha\beta \leq \text{ sup}(AB) + \varepsilon' \implies \alpha\beta \leq \text{ sup}(AB)$ by elbow room''.

I will try to be as clearly as posible.

For $$A, B \subseteq \mathbb{R}^+$$, we will prove that $$\sup(A)\cdot \sup(B) = \sup(AB)$$. For this, we will prove that $$\sup(A)\cdot \sup(B) \leqslant \sup(AB)$$ and $$\sup(AB) \leqslant \sup(A)\cdot \sup(B)$$. Of course, $$AB=\{xy:x\in A~\wedge~y\in B\}$$.

Let, $$a=\sup(A), b=\sup(B), M=\sup(AB)$$

Proof ($$\sup(A)\cdot \sup(B) \leqslant \sup(AB)$$).

By definition, $$\forall x\in A(x\leqslant a)$$ and $$\forall y\in B(y\leqslant b)$$. Since $$A, B \subseteq \mathbb{R}^+$$, $$x, y>0$$ thus $$0 and $$0 where it follows that $$xy\leqslant ab$$, where $$xy\in AB$$. This means that $$AB\leqslant ab$$ which implies that $$M\leqslant ab$$.

Now, by definition, $$\forall z\in AB (z\leqslant M)$$. Since $$z\in AB$$, $$z=xy$$ where $$x\in A$$ and $$y\in B$$ thus $$xy\leqslant M$$. Since $$x, y>0$$, $$xy>0$$, this way we can define the quotient $$x\leqslant \frac{M}{y}$$ which means that $$\frac{M}{y}$$ is an upper bound for $$A$$ which implies that $$a\leqslant \frac{M}{y}$$. It follows that $$ay\leqslant M$$ thus $$y\leqslant \frac{M}{a}$$ which means that $$\frac{M}{a}$$ is an upper bound for $$B$$ which implies that $$b\leqslant \frac{M}{a}$$. Therefore, $$ab\leqslant M$$.

$$\therefore \sup(A)\cdot \sup(B) = \sup(AB)$$

• What if $a$ (you divided by) is $0$. ?! Feb 17 at 13:08

You are almost there, from the first inequality of your last line $$\alpha\beta- \epsilon (\alpha + \beta - \epsilon) < ab \leq \sup AB$$ Observe that the right hand side $\sup AB$ is independent of $\epsilon$, send $\epsilon$ to zero. Then $$\alpha\beta \leq \sup AB.$$

• Right I understand that part but don't I need to show the $\epsilon$ is "small enough"? Oct 7, 2015 at 2:15
• @Chris An $\epsilon$-argument works like the following, if for each $\epsilon>0$, we have $$a-\epsilon \leq b$$, then we can conclude $$a\leq b.$$ What do you mean by $\epsilon$ is small enough?
– Xiao
Oct 7, 2015 at 2:19
• Right so in this case ϵ(α+β−ϵ) becomes our $\epsilon$. I'm not sure, my professor said I needed to show that it was small enough and that my proof was lacking something. Oct 7, 2015 at 2:23
• @Chris I believe the important part here is moving from the term $ab$ to $\sup AB$, because by the way you choose $a$ and $b$, $ab$ depends on $\epsilon$.
– Xiao
Oct 7, 2015 at 2:25

$$\alpha \beta -\varepsilon(\alpha+\beta -\varepsilon)<ab\le \alpha \beta$$

and this is true for every $\varepsilon>0$. Now let $\eta >0$ be arbitrary, and put $\varepsilon = \min \left \{\frac{2\eta}{\alpha+\beta} , \frac{\alpha+\beta}{2}\right \}$. It follows that:

$$\alpha+\beta-\varepsilon\ge\frac{\alpha+\beta}{2}\implies -\varepsilon(\alpha+\beta-\varepsilon)\ge-\varepsilon \left (\frac{\alpha+\beta}{2}\right )$$

Hence:

$$\alpha\beta -\varepsilon \left (\frac{\alpha+\beta}{2}\right )<ab$$

By our choice of $\varepsilon$:

$$\alpha\beta <\varepsilon \left (\frac{\alpha+\beta}{2}\right )+ab \le \eta +ab$$

Since $\eta >0$ is arbitrary, the conclusion follows.

• Why $\alpha +\beta -\epsilon \geq {\alpha + \beta \over 2} \implies -\epsilon(\alpha + \beta -\epsilon) \geq - \epsilon ({{\alpha + \beta } \over 2})$ ?? you are multiplying by a negative number $- \epsilon$, right? Aug 24, 2017 at 19:22