From Minimax theorems, we roughly know that if $f(x,y)$ is convex on $X$ and concave $Y$ for both compact $X,Y$, then: $$ \max_{x\in X}\min_{y\in Y} f(x,y)=\min_{y\in Y}\max_{x\in X} f(x,y). $$ We can define the region $\mathcal{R}(x,y)=(-\infty,f(x,y))$ and change minimum and maximum with union and intersection as follows: $$ \bigcup_{x\in X}\bigcap_{y\in Y} \mathcal{R}(x,y)=\bigcap_{y\in Y}\bigcup_{x\in X} \mathcal{R}(x,y). $$ So Minimax theorems give us the condition for interchanging union and intersection operator applied to an interval in $\mathbb R$.
Now consider a bounded region $\mathcal{R}(x,y)\subset\mathbb R^d$ parametrized by $x$ and $y$.
Question: What is the condition under which we can similarly state that: $$ \bigcup_{x\in X}\bigcap_{y\in Y} \mathcal{R}(x,y)=\bigcap_{y\in Y}\bigcup_{x\in X} \mathcal{R}(x,y). $$ This looks like a generalization of Minimax results to multidimensional space and I expect that we need some convexity and concavity assumption at the end. We can also assume that regions are some kind of geometric objects like polytopes.
