I find the following thread If $\overline{x}$ is a point such that $f(x)=f(\overline{x}) \Rightarrow$ $x$ is a local minimizer, then $\overline{x}$ is global minimizer
I am trying to understand the solution, I understand that the target is prove that the set $A=\emptyset$ and use that $\mathbb{R}^{n}$ is connected. but I do not understand, why the set $B$ is open? and the affirmation: And if $f(x) > f(\overline{x})$ then the same is true because of the continuity of $f$.
Could someone help me?. Thanks!