# Proof of $\sup ST = \sup S \sup T$

Claim

Let $S, T\subset$ ordered field $F$

Let $P_F=\{a\in F: a \gt 0\}$

Let $ST = \{st \in P_F\mid s \in S, t\in T\}$ Then,

$\forall S,T \in P_F\;\;\;$$\sup ST = \sup S \sup T$

Proof

Let $\sup S=\alpha,\; \sup T=\beta$ Then

$\forall s \in S$ and $\forall t \in T$, $\;st \le \alpha\beta$ Thus

$\alpha \beta$ is an upper bound of $ST$

Now, take$\;\;\gamma\in P_F$ s.t. $\gamma \lt \alpha \beta$

Let $e = \alpha \beta - \gamma \gt 0$ Then

${e \over 2}\gt 0$ Thus

$\alpha - {e \over 2} \lt \alpha$ and $\alpha - {e \over 2}$ is not least upper bound of $S$.

Thus $\exists s \in S$ s.t. $\alpha - {e \over 2} \lt s$

Similar to the case of $\alpha$,

$\exists t \in T$ s.t. $\beta - {e \over 2} \lt t \;$ Then

I am stuck in here.

I would like to show that $\gamma \lt st$ with above inequalities.

• What happens when you multiply the inequalities $\alpha-\frac{e}{2}<s$ and $\beta-\frac{e}{2}<t$? (You may beed to be careful about signs.). Perhaps you don't want to use $\frac{e}{2}$, but some other value? May 23 '17 at 1:48
• @MichaelBurr maybe.. that is because I had constructed similar inequalities from supS+supT case May 23 '17 at 1:55
• Right, $\frac{e}{2}$ works great for addition, not so for multiplication. May 23 '17 at 1:56

At the OP's request, here is an updated version which does not use square roots (if you want a version that works in the reals, check the previous edits). We must also follow the comments above and assume that $S$ and $T$ are positive sets!

The problem that this answer is facing is that $y_1=\alpha-\frac{e}{2}$ and $y_2=\beta-\frac{e}{2}$ is well-suited to addition since $y_1+y_2=\alpha+\beta-e$, so the $\frac{e}{2}$'s are balanced. This doesn't work for multiplication because the $\frac{e}{2}$'s don't combine to become $e$.

• Starting from $\gamma=\alpha\beta-e$. Our goal is to construct $x_1$ and $x_2$ so that $x_1<\alpha$, $x_2<\beta$, and $x_1x_2=\alpha\beta-e$. If this is true, then there are $s$ and $t$ so that $x_1<s\leq \alpha$ and $x_2<t\leq \beta$. Hence, $\alpha\beta-e=x_1x_2<st$, so $\alpha\beta-e$ is not an upper bound.

• Now, what we must do is to construct this $x_1$ and $x_2$. Suppose that we can find $x_1$ so that $x_1<\alpha$ and $x_1\beta>\alpha\beta-e$. Then, if we let $x_2=\frac{\alpha\beta-e}{x_1}$, then all we must check is that $x_2<\beta$. This is true by cross multiplying.

• So, how do we find such an $x_1$. What I'm going to use is the average between $\alpha$ and $\frac{\alpha\beta-e}{\beta}$. In this case, try $x_1=\alpha-\frac{e}{2\beta}$. Then $x_1<\alpha$ and $x_1\beta=\alpha\beta-\frac{e}{2}>\alpha\beta-e$. Therefore, this $x_1$ has the required properties.

I've skipped a few steps in here, so you might need additional details, but this sketch works.

• Any hint for finding appropriate $x_1$ and $x_2$ without usung squreroot? May 23 '17 at 4:37
• I rewrote my answer to avoid square roots. May 24 '17 at 3:17