Quoting: Find the supremum and infimum of $S=\{x|x^2 \leq 7\}$. State whether they are in S.
Given $$x^2 \leq 7 $$
$$ \sqrt{x^2} \leq \sqrt{7} $$
$$ |x| \leq \sqrt{7}$$
$$- \sqrt{7} \leq x \leq \sqrt{7}$$
It shows that $- \sqrt{7}$ and $\sqrt{7}$ are a lower bound and upper bound of S respectively
As $- \sqrt{7}$ is the greatest lower bound and $\sqrt{7}$ the least upper bound,
Infimum of S: $$inf(S)=- \sqrt{7}$$
supremum of S:$$ sup(S)= \sqrt{7}$$
Both are contained in S
Is this enough to declare that both inf and sup are contained? Is there a more efficient approach?
Any input is much appreciated
Best Answer
You can give an argument that they are contained, instead of just claiming it: Since $(\sqrt7)^2 \le 7$ and $(-\sqrt7)^2 \le 7$, both are contained in S.
The argument for finding the infimum and the supremum is correct.