This question has only an answer using theorems relying on continuity of a non-decreasing function. While I (think I) can understand the answer, I have this same exercise, but we still not studied continuity yet, we are studying real numbers and preparing to study sequences. Perhaps because of seeing this answer, the only way I see to prove this is using continuity also, but there must be a way without using those theorems about continuity. Could anyone show me the way to prove this only with real numbers/supremum/infimum/etc properties?
Any help would be appreciated.
Hint
Suppose $S\subset [0,\infty )$. Let $c=\sup C$ and $s=\sup(S)$. Let $\varepsilon >0$. There is $x\in C$ s.t. $c-\varepsilon \leq x^2\leq s^2$. Since it holds for all $\varepsilon >0$, we $c\leq s^2$.
Suppose that $c<s^2$, i.e. there is $x\in S$ s.t. $c<x^2\leq s^2$. This contradict the fact that $c=\sup\{x^2\mid x\in S\}$.
Therefore $c=s^2$ as wished.
I let you adapt the proof in the case where $S\subset \mathbb R$ instead of $S\subset [0,\infty )$ only.