Why is minimax always greater or equal to maximin?

Question

Let $f(x,y)$ be a function representing a game, we define the maximin : $\underline{f} = \max _{x∈X} \min _{y∈Y} f(x, y)$ and the minimax $\overline{f} = \min _{y∈Y} \max _{x∈X} f(x, y)$. It is said that we clearly have that $\underline{f}\leq \overline{f}$ for all games $f$. I saw a few examples to agree on the fact that we don't always have the equality but I don't really see how this is clear in general and why cannot we have $\underline{f}\geq \overline{f}$ for some game. Can I have an intuitive explanation or even a rigorous proof of it maybe ?

Answer 1

If max, min are attained in all cases, the following argument seems intuitive enough.

There exists a function $\alpha:Y\rightarrow X$ such that $$\max_{x\in X}f(x,y)=f(\alpha(y),y)\geq f(x,y), \forall (x,y)\in X\times Y .$$ Similarly, there exists a function $\beta:X\rightarrow Y$ such that $$f(x,y)\geq f(x,\beta(x))=\min_{y\in Y}f(x,y), \forall (x,y)\in X\times Y .$$ It follows that $$f(\alpha(y),y)\geq f(x,y)\geq f(x,\beta(x)),\forall (x,y)\in X\times Y$$ $$\Rightarrow f(\alpha(y),y)\geq f(x,\beta(x)),$$ where each side of the inequality depends only on one variable, hence $$\min_{y\in Y}f(\alpha(y),y)\geq \max_{x\in X}f(x,\beta(x)),$$ i.e. $$\min_{y\in Y}\max_{x\in X}f(x,y)\geq \max_{x\in X}\min_{y\in Y}f(x,y),$$ as required.