Let $A$ and $B$ be two sets of positive numbers bounded above, and let $a=\sup A$ and $b=\sup B$. Let $C$ be the set of all products of form $xy$. where $x\in A$ and $y\in B$. To prove that $\sup C=ab$.
My attempt
Now to prove existense of supremum is easy as it is based on lub axiom. I am writing about other part. Since $a_1 \leq a$ and $b_1 \leq b$ for all $a\in A$ and $b\in B$.
So multiplying two inequalities i get $a_1b_1 \leq ab$. This shows that $ab$ is upper bound for set $C$. Now to prove that it is least is main issue.
So i use contradiction by assuming that $K$ is least upper bound and so $K \leq ab$.
Since by assumption given in question i can write as there exists some $a_1 > a - \epsilon$ and $b_1 > b - \epsilon$ (Because $a$ and $b$ are given to be supremum), epsilon is bigger than $0$
So i have now $a < a_1 + \epsilon$ and $b < b_1 + \epsilon$. Multiplying these i get $ab < a_1b_1 + a_1\epsilon + b_1 \epsilon + \epsilon^2$. So $K < a_1b_1 + \epsilon(a_1b_1)$. I think this is a contradcition
Please cross check my proof. Thanks
In the last step of your proof you said $K\le ab <a_1b_1 + a_1\epsilon + b_1\epsilon + \epsilon^2$ But how far are you sure that $ab <a_1b_1 + a_1\epsilon + b_1\epsilon$ ? or even that $K< a_1b_1 + a_1\epsilon + b_1\epsilon$ since $\epsilon$ could be any positive real number,In real analysis it is sometimes useful to visualise your result and try various means to achieve it so I suppose that your idea was to show that $ab\le K$ and that $ab\ge K$ inorder to show that $ ab = K $
Lets assume that ab is not the supremum then there exist $ a'b' \in C $ such that $ab \le a'b'$ .But $a \ge a'$ and $b \ge b'$ thus $\frac{a}{a'} \ge 1$ multiply by $\frac{b}{b'}$ we have $\frac{ab}{a'b'} \ge \frac{b}{b'} \ge 1$ and as a result $ab \ge a'b'$
On the other hand suppose that $ab \ge a'b'$ this is a contradiction since $a'b'$ is the supremum of the set C so $ab \le a'b'$
We have the 02 results $ab \le a'b'$ and $ab \ge a'b'$ we conclude that $ab = a'b'=SupC $