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im struggling to prove the following problem.

Let $A \subseteq \Bbb R$ be a non empty set, (multipication is for example $A\cdot B= \{a \cdot b\ |\ a \in A,\ b \in B\}$)

Show: If $A$ is bounded, then $A\cdot A$ is bounded too!

I know that from the completenes axiom the set A has an infimum and a supremum. what should I do next . I'm new to this material and i dont understand it to it's fullest.

pls help, thanks

Calvin Khor
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Aviv Barel
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1 Answers1

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If a $A \subseteq \mathbb{R}$ is bounded then there exists a positive number $r \in \mathbb{R}_{+}$ such that $A$ is contained in the ball $B_r(0)$ with radius $r$ centered in $0$. Clearly, $$ A\cdot A \subseteq B_r(0) \cdot B_r(0) \subseteq B_{r^2}(0). $$ For the last inclusion take $x,y \in B_r(0)$. Then we have $|x|, |y| < r$, which implies $$|x\cdot y| = |x|\cdot |y| < r\cdot r.$$ So $A\cdot A$ is bounded by $r^2$.

This proof can be easily reused for bounded sets in $\mathbb{C}$.

Nathanael Skrepek
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  • I Don' t understand the thing with the ball . – Aviv Barel Nov 22 '19 at 09:20
  • Which thing? The definition of a ball is $B_r(0) := {x \in \mathbb{R} : |x| < r }$. This may seem like an overkill because you can also just write $(-r,r)$. – Nathanael Skrepek Nov 22 '19 at 09:27
  • Calling the interval a ball is incredibly common in literature because it generalizes nicely to higher dimensions. In $\Bbb R^3$ you would have a ball of radius $r$ centered around a point $p$ would be the set ${x\in\Bbb R^3:|x-p|<r}$. Yes, in the case of $\Bbb R^1$ balls and intervals are the same thing, but the naming comes from how it appears in $\Bbb R^2$ and $\Bbb R^3$ and higher. – JMoravitz Nov 22 '19 at 12:58
  • can someone explain to me the thing with the absolute value ? – Aviv Barel Nov 23 '19 at 22:23
  • @AvivBarel again which thing? you should try to ask your question more precisely! – Nathanael Skrepek Nov 26 '19 at 08:29