How to show $ \sup \inf g(x,y) \leq \inf \sup g(x,y)$? Came across this little practice exercise, and I couldn't properly convince myself of this relation:
Let $X,Y \subset \mathbb{R}^n$ and $g:X\times Y \rightarrow \mathbb{R}$. Show that
$$\sup_{y \in Y} \inf_{x \in X}g(x,y) \leq \inf_{x \in X} \sup_{y \in Y} g(x,y).$$
My thinking was starting with $\inf_{x \in X}g(x,\overline{y}) \leq\sup_{y \in Y} g(\overline{x},y)$, but does that even hold for all $\overline{x} \in X$ and $\overline{y} \in Y$?

Well silly me, I think I've found a decent explanation. First note that
$$ g(\overline{x},\overline{y}) \leq \sup_{y \in Y}g(\overline{x},y)$$
for all $\overline{x} \in X$ and $\overline{y} \in Y$. Then take the infimum wrt $X$ on both sides, giving
$$ \inf_{x \in X}g(x,\overline{y}) \leq \inf_{x \in X}\sup_{y \in Y}g(x,y) $$
which now holds for all $\overline{y} \in Y$. Thus we can take the supremum over $Y$ on the lhs to give the desired result.
