Show $\sup B=(\sup A)^2$

Question

Let $A$ be a non-empty set with positive elements $a\in \mathbb{R}$ which is bounded above. Let $B=\{a^2\mid a \in A\}$. I want to show that $\sup B=(\sup A)^2$. If $M$ is an upper bound of A then $a \leq M $ so $a^2\leq M^2$ for all $a$ so $M^2$ is an upper bound of B. I think of arguing by contradiction, so assume that $\sup B<(\sup A)^2$. I am not sure how to arrive at the contradiction. We have $a^2 \leq \sup B$ for all $a$ and $\sup B<(\sup A)^2$. Can we then use the completeness of the real number so that there exists an element in $(\sup B, (\sup A)^2)$ which gives a contradiction?

Answer 1

Let $s=\sup A$; note first that $s>0$, because $A$ contains positive elements by assumption. Then, for every $a\in A$, $a^2\le s^2$, which implies $s^2$ us an upper bound for $B$.

Let $t$ be an upper bound of $B$, that is, $a^2\le t$, for every $a\in A$. Then $t>0$, so $a\le\sqrt{t}$, for every $a\in A$. Hence…