Calculate $\sup$ (Supremum) and $\inf$ (Infimum) of the following set:
$A=\{x\in\mathbb{R}:x|x|>x+2\}$
My solution.
$A=\{x\in\mathbb{R}:x^2>x+2 \ \ (x>0)\ \cup\ -x^2>x+2\ \ (x<0) \}$
The inequality $-x^2>x+2$ is satisfied on the empty set, and the
$x^2>x+2 \ \ (x>0)$ is satisfied for $x>2$. Then $\inf(A)=2$ and $\sup(A)=+\infty$.
Is my procedure right? I made a mistake?
Thank you very much
Best Answer
CW answer to push it from unanswered queue:
It is correct.