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

Question

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''.

Answer 1

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 $0<x\leqslant a$ and $0<y\leqslant b$ 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, $ab\leqslant M$.

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

Answer 2

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 $$\alpha\beta \leq \sup AB.$$

Answer 3

Proceeding with your argument:

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

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.