Comments on $\sup(A\cdot B) = (\sup A)(\sup B)$ where $A$ and $B$ bounded sets of positive real numbers
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$\sup(A\cdot B) = (\sup A)(\sup B)$ where $A$ and $B$ bounded sets of positive real numbers
Problem. Suppose $A$ and $B$ are two subsets of positive real numbers. In addition, assume that $A$ and $B$ are both bounded. Show that $$ (\sup A)(\sup B) = \sup A\cdot B$$ where the set of the right-hand side is defined as $$ A\cdot B = \{ ab\mid a\in A, b\in B\} $$
The problem above is a typical exercise in real analysis manipulating the definition of supremum. This is an excellent example of what Gowers called a “fake difficulty” in his blog post for the undergraduate real analysis course at Cambridge. Solving the problem really only requires one to know the definition of the supremum and basic logic. I will share my own answer below.
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Let $L:=(\sup A)(\sup B)$. By definition of "supremum", one needs to show the following two things,
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$L$ is an upper bound for $A\cdot B$;
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$L$ is the smallest upper bound for $A\cdot B$.
The first statement is very easy to prove: since for every $a\in A$ and every $b\in B$, one has $a\le \sup A$ and $b\le \sup B$, which immediately implies that $ab\le (\sup A)(\sup B)$.
To prove the second statement, assume that $L'$ is an upper bound of $A\cdot B$. We want to show that $L\le L'$. But the assumption of $L'$, we have for every $a\in A$,
$$\text{for every } b\in B: ab\le L'$$which is equivalent to
$$\text{for every } b\in B: b\le \frac{1}{a}\cdot L'$$So it follows that $\sup B\le L'/a$. Thus for every $a\in A$, one has
$$ a\le \frac{L'}{\sup B} $$and thus $\sup(A)\le \frac{L'}{\sup B}$, which implies that $L\le L'$.
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