How to Prove that if $A$ is bounded, then $A \cdot A$ is bounded? [closed]

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Definition: Let $\varnothing$ $\neq$ $A,B$ be a subset of $\Bbb R$, then we define $A \cdot B$ $:=$ $\lbrace$ $a\cdot b$ | $a \in A$ , $b \in B$ $\rbrace$

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Let $\varnothing$ $\neq$ $A$ be a subset of $\Bbb R$.

Question: Prove or disprove the following statement: ''If $A$ is bounded, then $A \cdot A$ is bounded''.

Can someone help me with the proof?

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2 Answers

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If A is bounded, $\forall x \in A$, $\exists M \in \mathrm{R}^+$ such that $|x| \le M$.

For any $a \in A\cdot A$, there exists $x,y \in A$ such that $xy=a$, hence $|xy|=|x||y|\le M\cdot M=M^2$.

Hence is bounded.

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Hint: A set $A$ is bounded, if there exist two numbers $l, h$ such that for all $x \in A$ we have $l \le x \le h$. Now if $A$ is bounded, can you find the bounds for the set $A\cdot A$?

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