Let $A$ and $B$ be nonempty sets of positive real numbers that are bounded above. Also let $AB = \{ab: a \in A, b \in B \}$. Prove that $AB$ is bounded above and $\sup(AB) = (\sup A) (\sup B)$.

So $\sup A$ and $\sup B$ exist by completeness. An upper bound for $AB$ is $(\sup A)(\sup B)$. Let $\alpha = \sup A$ and $\beta = \sup B$. We want to show that if $c$ is an upper bound for $AB$ then $\alpha \beta \leq c$. For $a \in A$, $ab \leq c$ for all $b \in B$. So $c/b$ is an upper bound for $A$. Thus $\alpha \leq c/b$. It follows that $\alpha \beta \leq c$.

Is this correct?

  • 1
    $\begingroup$ I think you should add some more details between the last two sentences since it's unclear whether you are mixing $\beta$ and $b$ or not. $\endgroup$
    – Apostolos
    Jun 21, 2011 at 16:42

3 Answers 3


Let $\alpha = \sup A$ and $\beta = \sup B$.

For every $a\in A$ and $b \in B$ we have

$$ab \leq \sup_{b\in B} a b = a \beta \leq \sup_{a\in A} a \beta = \alpha \beta,$$

so we have $\sup AB \leq \alpha\beta$.

Now let $(a_n)_{n\in \mathbb N} \subset A$ and $(b_n)_{n \in \mathbb N} \subset B$ be sequences such that $a_n \to \alpha$ and $b_n \to \beta$ as $n \to \infty$.

It is then clear, that $(a_nb_n)_{n \in \mathbb N} \subset AB$ and $a_nb_n \to \alpha\beta$ as $n \to \infty$, so $\sup AB \geq \alpha\beta$ and therefor we have $\sup AB = \alpha \beta$.

  • $\begingroup$ How could I salvage my proof? $\endgroup$
    – Damien
    Jun 21, 2011 at 18:35
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    $\begingroup$ @Damien: Your proof is correct, but ( the presentation of ) your use of $\alpha$ and $\beta$, is rather puzzling. While creativity surely is a good thing to cultivate, try to refine your proofs to present the arguments in a more clear way. It will help you correct your proofs by yourself, make them more accessible and easier to remember. Note how the second half of your proof reads like this: Fix an element $a$ of $A$. For every $b\in B$ we have $ab\leq c$. By rearrangement we get $a\leq c/b$ for all $b\in B$. Because $a$ was arbitrary we have $α\leq c/b$ for all $b\in B$. $\endgroup$ Jun 21, 2011 at 21:34
  • 2
    $\begingroup$ Again by rearrangement we get $b\leq c/\alpha$ for all $b\in B$, so $\beta\leq c/\alpha$, hence $\alpha\beta\leq c$. $\endgroup$ Jun 21, 2011 at 21:35

I provide the answer in a more general setting, which gives $\sup(A + B) = \sup A + \sup B$ as well.

Theorem: Let $G$ be a bi-ordered group with least upper bound property. Then for any $A, B\subseteq G$, we have $$ \sup(AB) = (\sup A)(\sup B)\text. $$

Proof. "$\le$": We show that the RHS is an upper bound of $AB$. Let $x\in A$ and $y\in B$. Then $x\le\sup A$ and $y\le \sup B$. Due to bi-invariance of $\le$, we have that $xy\le(\sup A)(\sup B)$, as required.
"$\ge$": Let $x\in A$ and $y\in B$. Then $xy\in AB$ and hence $xy\le \sup(AB)$, which implies $x\le \sup(AB)y^{-1}$. Since $x\in A$ was arbitrary, we have $\sup A\le\sup(AB)y^{-1}$, which gives, again using bi-invariance, $y\le (\sup A)^{-1}\sup(AB)$. Again, $y\in B$ was arbitrary, and hence $\sup B\le (\sup A)^{-1}\sup(AB)$, or $(\sup A)(\sup B)\le\sup(AB)$, as required.

Now, we just note that $(\mathbb R, +)$ and $(\mathbb R^+, \cdot)$ are bi-ordered (in fact abelian) groups (under the usual order). And boom! You get the additive and multiplicative facts in a single shot!

Note that the set of nonzero reals does not form an ordered group under multiplication. This gives a clearer picture of why we concern ourselves with only positive reals in the multiplicative case.


Let us first show that $ AB $ is bounded above. If we take an arbitrary element $ c \in AB $ it must be equal to a product: $ c = ab $ where $ a \in A $ and $ b \in B $. Since $ a \leq sup(A) $ and $ b \leq sup(B) $, $ c= ab \leq sup(A)sup(B) $ and hence $ AB $ is bounded above by $sup(A)sup(B)$. We know that any subset of real numbers bounded above should except a least upper bound, say it is $sup(AB)$.

To show $sup(AB) = sup(A)sup(B) $ we try to show that any number less than $sup(A)sup(B)$ could not be an upper bound for $ AB $ and because we know that $sup(A)sup(B)$ is an upper bound for $AB$ we can deduce that $sup(A)sup(B)$ is the smallest one of all upper bounds, ie the least upper bound.
Now lets take an arbitrary positive number $\epsilon$. If $ \epsilon < 2sup(A)sup(B) $ then in the following equation $$ (sup(A) - \frac{sup(A) \epsilon}{2sup(A)sup(B)-\epsilon})(sup(B) - \frac{\epsilon}{2sup(A)}) \tag{*} $$ we subtract positive numbers from $sup(A)$ and $sup(B)$ hence we can find $a \in A$ and $b \in B$ with $ab$ is greater than the above product $(*)$. Therefore $ (*) $ can not be an upper bound for $AB$.
Observe that: $$ (sup(A) - \frac{sup(A) \epsilon}{2sup(A)sup(B)-\epsilon})(sup(B) - \frac{\epsilon}{2sup(A)}) = sup(A)sup(B) - \epsilon $$ So If $ \epsilon < 2sup(A)sup(B) $ then $sup(A)sup(B) - \epsilon $ can not be an upper bound for $AB$.
For if $ \epsilon \geq 2sup(A)sup(B) $: $$ sup(A)sup(B) - \epsilon \leq sup(A)sup(B)-2sup(A)sup(B) = -sup(A)sup(B)$$ This means that $sup(A)sup(B) - \epsilon$ is a negative number. But it can not be an upper bound for a non-empty set consisting of positive numbers only.
As a result we can say that any number less then $sup(A)sup(B)$ can not be an upper bound for $AB$ and hence $$ sup(AB) = sup(A)sup(B) $$


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