Could anyone help me to show that the sets $\{(x,y)|f(x,y)\le \gamma, x>0, y>0\}$ are compact for all scalars $\gamma$, for the function $f(x,y)=xy+\frac{1}{x}+\frac{1}{y}$?
I think it is easy to show that the set is bounded at $0<x<\gamma, 0<y<\gamma$, because above function is actually the surface area of a rectangular parallelepiped with volumne 1 and $x, y, z$ as length, height and depth and replace $z$ with $\frac{1}{xy}$. But how to prove the set is closed?
My understanding of open set is that pick any point in the set, if all the neighbors of this point are also in this set then the set is open, like (2,3) is open, while [2,3] is closed, is this true?
How $f(x,y)$ is continous in that set (because $x,y>0$). The set of the $(x,y)$ with $f(x,y)\leq \gamma$ is the preimage of the set $(-\infty,\gamma]$ (which is closed) and therefore is closed.