Let A and B be bounded set of nonnegative real numbers Let $$C=\{ab:a\in A,b\in B\}$$

Prove that $\sup(C)=\sup(A)\sup(B)$

I got idea for the solution but I am not sure with my solution.

Let $m=\sup(A),n=\sup(B),p=\sup(C)$

It is easy to prove that $p\leq mn$. Now is the problem on proving $p \geq mn$

Here, given any $\epsilon>0$ $$\exists x\in A:m-\epsilon<x$$ $$\exists y\in B:n-\epsilon<y$$ $$\implies(m-\epsilon)(n-\epsilon)<xy$$ $$\implies mn-(m+n)\epsilon<(m-\epsilon)(n-\epsilon)<xy\leq p$$ Since the result holds for any $\epsilon>0$, we get $$mn\leq p$$ So combining both inequalities, $mn=p$

Please help me to check whether my way of proving is correct or not.

  • $\begingroup$ I've fixed a minor mistake in your existential statement about $y$. It's a fine proof, but you should take away your "$\Rightarrow$"s as they are not really logically correct. $\endgroup$
    – user21820
    Commented Dec 12, 2014 at 6:53

1 Answer 1


Your solution looks fine. If you want to be even more rigorous, notice that you can choose $\epsilon$ to be $\frac{1}{m+n}$ times something that goes to $0$.


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