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

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?

• Don't you mean "$M^2$ is an upper bound of $B$"? I don't particularly like the way you're going about it. I suggest you prove that $(\sup(A))^2$ is the smallest upper bound of $B$. Let $s$ be $\sup(A)$. A similar argument to what you've done for "$M$" proves that $s^2$ is an upper bound of $B$. Suppose there is a smaller (than $s$) upper bound $u$ of $B$. It follows that $\forall a\in A(a^2\leq u)$ and $u$ is positive, therefore $\forall a\in A(a\leq \sqrt u)$, which means that $\sqrt u$ is a smaller upper bound of $A$ than $s$ is, a contradiction. – Git Gud Jan 31 '17 at 21:08

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…
• @Ufomammut Hence $\sqrt{t}$ is an upper bound for $A$. Therefore $\sqrt{t}\ge s$, which implies $t\ge s^2$. Thus $s^2$ is the least upper bound for $B$. – egreg Jan 28 '18 at 23:04