How to prove this real analysis problem?

Question

I cannot seem to prove this because the supremum is A is positive infinity, this also goes for B.

Answer 1

Well, since $(\sup (A))^2 \geq a^2$ for all $a \in A$, we have $(\sup (A))^2 \geq \sup (B)$.

The converse is similar: since $\sup (B) \geq a^2$ means $\sqrt{\sup (B)} \geq a$ for all $a$ because $x \mapsto \sqrt{x}$ is montonically increasing on its domain; so in turn it's greater than $\sup (A)^2$ and so on.