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$AB= \{\ x \in (0, \infty) : \exists a \in A, \exists b \in B \ s.t. x=ab\} $

Then I have to show that $supAB \leq supAsupB$

My own attempt is that

$\forall a,b \in A,B$

$sup(AB) \leq ab \iff sup(AB)$ is bounded above by AB. But this can only happen if

$sup(AB) \leq sup(A)sup(B)$

But I feel like my proof is lacking..

2 Answers2

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To answer the (implicit) question: Yes, your proof is certainly "lacking". In fact it makes no sense: Saying something is bounded above by $AB$ makes no sense, because $AB$ is not a number.

What you say you want to prove is not true - you're leaving out an important hypothesis. If $A=B=[-2,-1]$ then $\sup(AB)=4$ and $\sup(A)\sup(B)=1$.

Hint: Once you get the hypotheses straight this is very easy: Since $\sup(AB)$ is the least upper bound of $AB$, to show that $\sup(AB)\le\sup(A)\sup(B)$ you just need to show that $\sup(A)\sup(B)$ is an upper bound for $AB$.

In other words, you need to show that if $x\in AB$ then $x\le\sup(A)\sup(B)$. And that's obvious: If $x\in AB$ then $x=ab$ for some $a\in A$ and $b\in B$. Hence $a\le\sup(A)$ and $b\le\sup(B)$, so $x=ab\le\sup(A)\sup(B)$.

Except that that's wrong; in fact knowing that $a\le\sup(A)$ and $b\le\sup(B)$ does not imply that $ab\le\sup(A)\sup(B)$! Why not? Annd what could you assume about $A$ and $B$ to fix this?

Note that the counterexample $A=B=[-2,-1]$ gives a hint regarding the "why not?" above... $-2<-1$ but $(-2)(-2)>(-1)(-1)$.

David C. Ullrich
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  • thanks, any suggestion on how I should look at this problem? –  Apr 22 '18 at 14:44
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    First you have to state the problem correctly! Then note this: Since the sup is the least upper bound, saying $\sup(S)\le\alpha$ is the same as saying that $x\le\alpha$ for every $x\in S$. – David C. Ullrich Apr 22 '18 at 14:50
  • I edit' my original post I believe that should state the problem correctly now –  Apr 22 '18 at 16:53
  • No. This seems like you didn't bother to look at the counterexample I gave: If $A=B=[-2,-1]$ and we use your curious redefinition of $AB$ it's still true that $\sup(AB)=4$ and $\sup(A)\sup(B)=1$. – David C. Ullrich Apr 22 '18 at 16:58
  • Changing the definition of $AB$ is changing the problem; you're not allowed to do that. The problem with the question is not the definition of $AB$, the problem is that there is a hypothesis missing. – David C. Ullrich Apr 22 '18 at 17:08
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    See the Hint I added... – David C. Ullrich Apr 22 '18 at 17:16
  • thank you very much for your time I am reading it now –  Apr 22 '18 at 17:48
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Your proof is not correct. What makes you think that $\sup(AB)\leqslant ab$? It is not true in general.

José Carlos Santos
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  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – AlexR Apr 22 '18 at 13:54
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    @AlexR Of course it provides an answer to the question! Of course there's no "?" in the post - the question is implicit in "But I feel like my proof is lacking". The sentence "Your proof is not correct" is an answer to that. – David C. Ullrich Apr 22 '18 at 14:02
  • @DavidC.Ullrich I agree this isn‘t a clear-cut case but as-is I think it should be a comment. Note that the above comment is auto-generated so the wording is definitely off. With a little elaboration it might become a helpful answer, though. – AlexR Apr 22 '18 at 14:06
  • @AlexR I think that the (implicit) question was whether the proof was correct or not. What I wrote answers that question. – José Carlos Santos Apr 22 '18 at 14:08
  • @JoséCarlosSantos The proof can‘t be correct anyway because the proposition is false. Such a remark would be more useful to the OP and future readers... – AlexR Apr 22 '18 at 14:10
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    @AlexR Saying you "agree" iit's not clear-cut comes close to suggesting that I feel it's not\ clear-cut. Just for the record: No, it's perfectly clear-cut. The OP (almost) asked a question, and the Answer answers it. Tell us: When someone asks whether a proof is correct, and it's not, what would constitute an acceptable Answer? – David C. Ullrich Apr 22 '18 at 14:21
  • @DavidC.Ullrich An answer saying: No it‘s not, and here‘s why / Here‘s what you might have missed / Here‘s a working counter-example. Much like your second revision. – AlexR Apr 22 '18 at 14:26