Can I switch the order of taking minimums? Suppose I have some function $F(x,y):\mathbb{R}^2 \rightarrow \mathbb{R}$. 
Is it always the case that
$$
\min_{1 \leq x \leq N} (\min_{1 \leq y \leq M} |F(x,y)|) = \min_{1 \leq y \leq M}
(\min_{1 \leq x \leq N} |F(x,y)|)
$$
holds?
Here I am interested in situations where $x,y \in \mathbb{R}$ or $\mathbb{N}$. 
If it's not always the case, under what conditions are they the same? Thank you!
 A: Clearly, for any $y$ in $[1,M]$, we have that 
$$
\min_{1 \leq x \leq N} (\min_{1 \leq y \leq M} |F(x,y)|) \le 
(\min_{1 \leq x \leq N} |F(x,y)|)
$$
since for any $y$ in $[1,M]$ we have that $
\min_{1 \leq y \leq M} |F(x,y)| \le |F(x,y)|
$. In particular for the values of $y$ attaining the minimum of the right  hand side. Hence $$
\min_{1 \leq x \leq N} (\min_{1 \leq y \leq M} |F(x,y)|) \le \min_{1 \leq y \leq M}
(\min_{1 \leq x \leq N} |F(x,y)|).
$$
The same argument interchanging $x$  and $y$ proves the reversed inequality.  
A: This is an explanation for my comment. I will show the following 

Claim: For any nonempty set $X,Y$ and a function $f:X\times Y\to \mathbb{R}$, we have 
  $$\inf_{(x,y)\in X\times Y}f(x,y)=\inf_{x\in X}\left(\inf_{y\in Y}f(x,y)\right).$$

Let $L,R$ be the left and right hand side respectively. It is clear that for any $x\in X$, we have $L\le \inf_{y\in Y}f(x,y)$, and thus 
$$L\le\inf_{x\in X}\left(\inf_{y\in Y}f(x,y)\right)=R.$$ 
To show the converse, we first assume that $L\ne -\infty$. Then for any $\epsilon>0$, we can find $(x_0,y_0)\in X\times Y$ such that $f(x_0,y_0)<L+\epsilon$. Thus we have 
$$R\le f(x_0,y_0)<L+\epsilon.$$
Since this is true for any $\epsilon>0$, it must be that $R\le L$. 
If $L=-\infty$, then for any $l\in\mathbb{R}$, we can find $(x_0,y_0)\in X\times Y$ such that $f(x_0,y_0)<l$, and thus $R\le f(x_0,y_0)<l$. This implies that $R=-\infty$ as well. QED.
Finally, by symmetry we also have 

$$\inf_{(x,y)\in X\times Y}f(x,y)=\inf_{y\in Y}\left(\inf_{x\in X}f(x,y)\right).$$

In particular, for compact sets $X=[1,N],Y=[1,M]$, if we assume that $F$ is continuous, then the infima are attained so we can write 
$$\min_{1 \leq x \leq N} (\min_{1 \leq y \leq M} |F(x,y)|) = \min_{1 \leq y \leq M}
(\min_{1 \leq x \leq N} |F(x,y)|)$$
