Prove $\sup \{x^2|x \in S \}$ exists, and equal to $s^2$, given $S \subset \Bbb{R}, S \neq \emptyset$, with sup $s$, inf $t$ and $s \geq -t$

Question

Question

Suppose $S$ is a non-empty set of real numbers, with supremum $s$ and infimum $t$, and also that $s \geq -t$.

a) Show that $-s \leq x \leq s, \forall x \in S.$

b) Show that $\sup \{x^2|x \in S \}$ exists, and it is equal to $s^2$.

Answer

a) is fully answered, but please verify ?

We know that, given $s=\sup(S), t=\inf(S)$, then $\forall x \in S \; t \leq x \leq s$, but since $$s \geq -t, \text{ then } \\ -s \leq t, \text{ hence } \\ -s \leq t \leq x \leq s \\ \Rightarrow -s \leq x \leq s$$

The answer to b) is only partly answered and I require assistance

We know that $\forall x \in S, x \leq s=\sup(S)$, which implies $x^2 \leq s^2$, hence $s^2$ is an upper bound for $x^2$.

However the "least" upper bound for $x^2$ is $\sup(x^2) \; \Rightarrow x^2 \leq \sup(x^2) \leq s^2$. **

I need to show that $s^2 \leq \sup(x^2)$ , which is what I think is required, since combined with (**) one can say $\sup(x^2)=s^2$. Need assistance with this?

Answer 1

Did you try by contradiction for part b?

Hint: Suppose there exists a $b^2$ such that $x^2\le b^2 < s^2$

Answer 2

Part a) looks correct.
For second part of part b):
suppose $s^2 < \sup(x^2)\implies\exists p \in S : p^2 > s^2\implies |p| > |s|$ so it will contradict a)

Answer 3

$"x\le s\implies x^2\le s^2"$ In general this is not true if $s\lt 0$. Ho nwever, if $-t\le s\lt 0$, then $t\gt 0$, but we need to have $s\ge t$, so $s$ must be non-negative.

Since $s=\sup S,$ any upper bound of S (say $b$) must satisfy $b\ge s$. Now, $b^2\ge s^2$ is an arbitrary upper bound for $T=\{x^2|x \in S \}$, so by definition $\sup T=s^2$.