Prove that this set is bounded. Let $A \subseteq(0, \infty)$ be a bounded subset of positive numbers. We define the set $A^2:=\{a^2:a \in A\}$ which is the set of squares of the elements of $A$.
Prove that $A^2$ is also bounded.
How would I start an approach to this question? 
 A: 
Let A⊆(0,∞) be a bounded subset of positive numbers. 

That means there there is an $M$ so that $M \ge x$ for all $x \in A$.  As all $x > 0$ for $x\in A$ we know that $M > 0$.  And all for all $x \in A$, $0 < x \le M$.

We define the set $A^2:=\{a^2| a∈A\}$ which is the set of squares of the elements of A.

You had a typo. $A^2 :=\{a^2|a\in A\} \ne \{a^2 \in A\}$ which would be "the set of elements of $A$ that happen to be squares".  But as all positive real numbers are squares, that would just be $A$.

Prove $A^2$ is also bounded.

Well for all $x \in A$ we know $0 < x \le M$ so $0 < x^2 \le M^2$ for all $x^2 \in A^2$.  
That's it.  We're done. We can go home and eat an early lunch.
A: Since $A$ is bounded, there exists $m,M \in \mathbb{R}^+$ such that $m\leq a\leq M$ for all $a\in A$. Since all those numbers are positive, we can square that double inequality to get $m^2\leq a^2\leq M^2$, for any $a^2\in A^2$. This implies that $m^2$ is a lower bound of $A^2$ and $M^2$ is a upper bound of $A^2$, hence $A^2$ is also bounded.
A: If $A$ is bounded, there is some $M \in \mathbb{R}_+$, such that for every $a \in A$, $|a| \le M$. But this implies $a^2 \le M^2$. This precisely says for every $b \in A^2$, $|b| \,(\, = a^2 \text{ for some $a \in A$ }) \le M^2$.
