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?
Best Answer
$"x\le s\implies x^2\le s^2"$ 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 $\sup T=s^2$.