Prove $\sup(A^2) = (\sup(A))^2$

Question

So we let A be non empty set of R that’s bounded. Define $A^2=\{a^2\mid a \in A\}.$ How do we prove that the $\sup(A^2) = (\sup(A))^2.$ I've been approaching this by showing that $(\sup A)^2$ is an upper bound of $A^2$ and then showing that it's the least among the set of upper bounds. That is to say, I've been approaching it from the definition of $\sup A^2.$

Answer 1

I will assume that $A$ is a bounded set of nonnegative real numbers. Let us prove that $\sup(A^2) = (\sup(A))^2$.

First we prove $\sup(A^2) \leq (\sup(A))^2$. If $a \in A$ then we have $0 \leq a \leq \sup(A)$. Since the square fonction is increasing on nonnegative numbers, this yields $0 \leq a^{2} \leq (\sup(A))^{2}$, for any $a \in A$. Hence $\sup(A^2) \leq (\sup(A))^2$.

Now we prove $\sup(A^2) \geq (\sup(A))^2$. Let $a \in A$. We have $a^{2} \leq \sup(A^{2})$. By taking the root, since everything is nonnegative, we get $a \leq \sqrt{\sup(A^{2})}$. Since this is true for any $a \in A$, we get $\sup(A) \leq \sqrt{\sup(A^{2})}$. Since this is nonnegative and the square function is increasing on nonnegative numbers, we get $(\sup(A))^{2} \leq \sup(A^{2})$.