Consider two subsets $A$ and $B$ of the interval $[0,2019]$ that are non-empty and bounded from above. Define a new set $ C =\{ \ ab \ | \ a\in A \ \wedge b\in B \ \ \} $. Prove that sup $C = $ sup$A$ $\cdot$ sup$B$.
Because $A$ and $B$ are non-empty subsets of $\mathbb{R}$ that are bounded from above, they will have a supremum according to completeness. It's fairly simple to see that $0\leq $ sup$A = S_A$ and $0\leq $ sup$B = S_B$. Consider now a $c \in C$. Then by definition $c = ab$ with $a \in A$ and $b\in b$. Because $a,b,S_A,S_B$ are all greater than or equal to $0$, it's also simple to conclude that $ab \leq S_A \cdot S_B$. Hence, $C$ will have a supremum because it is non-empty and bounded from above (by $S_A \cdot S_B$). This supremum will be less than or equal to any upper bound, thus: sup $C = S_C \leq S_A \cdot S_B$
So far it seems alright, but I can't manage to prove the reverse inequality. I first tried the "$\epsilon > 0$ " method but this leads to: $S_A \cdot S_B< ab \ + \ a\epsilon \ + \ b\epsilon \ + \ \epsilon^2$. For a set like $A + B$ this works fine because taking the sums of inequalities isn't messy.
Then I tried using the following argument. For all $a \in A$ and $b \in B$: $c = ab < S_C$ (because $c\in C$ by definition). Thus : $a \leq \frac{S_C}{b}$ and thus $\frac{S_C}{b}$ is an upper bound of $A$. So $S_A \leq \frac{S_C}{b}$ (because $S_A$ is the supremum of $A$ ). Then $ b \leq \frac{S_C}{S_A}$ and a similar argument for $B$ leads to $S_B \leq\frac{S_C}{S_A}$ and thus $S_A \cdot \ S_B \leq S_C$. This seems to work but since $a,b,S_A$ and $S_B$ can all be equal to $0$ this argument is invalid ( as you will then divide by $0$ ).
I can't seem to get past that $0$ singularity. There are duplicates but they all assume subsets without $0$.
Thanks in advance !
I would approach in this way:
Proof that $S_C = S_A \cdot S_B$:
Let $M$ be an upper bound of $C$. Then $M\ge ab$ for all $a,b \in A,B$.
(1) Show that $M\ge S_A \cdot b$ for any $b\in B$. (This can be done by contradiction. Assume that $M<S_A \cdot b$ for some $b$, and show why this is impossible)
(2) Similarly, show that $M\ge S_A \cdot S_B$. This can also be done by contradiction (assume that $M < S_A \cdot S_B$ and show this conflicts with the fact that $M\ge S_A \cdot b$ for all $b\in B$).
(3) Since $M$ was arbitrary, it has been shown that $S_A \cdot S_B$ is less than any upper bound of $C$. Therefore it is, by definition, the supremum of $C$.