When $\min_{x \in X,y \in Y} f(x,y) = \min_{x \in X} \min_{y \in Y} f(x,y)$? When $$ \min_{x \in X,y \in Y} f(x,y) = \min_{x \in X} \min_{y \in Y} f(x,y) \qquad? $$ 
I mean when we are minimizing a function with respect to two variables, under what conditions we are allowed to minimize over one variable and then minimize over the other one?
 A: For each $x$,
$$
\min_yf(x,y)\ge\min_{x,y}f(x,y)\tag{1}
$$
Taking the minimum over $x$,
$$
\min_x\min_yf(x,y)\ge\min_{x,y}f(x,y)\tag{2}
$$
However, there is an $(x_0,y_0)$ so that $f(x_0,y_0)=\min\limits_{x,y}f(x,y)$. Then
$$
\min_yf(x_0,y)=\min_{x,y}f(x,y)\tag{3}
$$
Therefore, we must have that the minimum over all $x$ is at most the value at one $x_0$:
$$
\min_x\min_yf(x,y)\le\min_{x,y}f(x,y)\tag{4}
$$
$(2)$ and $(4)$ show that
$$
\min_x\min_yf(x,y)=\min_{x,y}f(x,y)\tag{5}
$$
A: I compile the answer here for clarity.
First I reformulate the question : 
given a function of two variables $f(x,y)$, its minimum can be written $\min_x y^*(x)$, where $y^*(x)=\min_y f(x,y)$.
Question: under which condition $y^*(x)$ is a constant function, so is independent of $x$?
Answer:
Let $f'(x,y)$ be the $y$-derivative of $f(x,y)$.
$y^*(x)$ is exprimable as a solution of $f'(x,y)=0$.
A sufficient condition for $y^*$ to be a constant is if $f'(x,y)$ is of the form $h(x,y)g(y)$, where $y^*$ is a root of $g$.
I don't know about a proof of the converse, but this should cover most cases in practice...
