Two part question about $\sup_y\inf_xf(x,y)\leqslant\inf_x\sup_yf(x,y)$ Background Info:
Let $X,Y\neq\emptyset$
  and let $f:X\times Y\to\mathbb{R}$
  have a bounded range in $\mathbb{R}$
  . Also, let 
$f_{1}(x)=\sup\{f(x,y):\: y\in Y\}$ and $f_{2}(y)=\sup\{f(x,y):\: x\in X\}$
Establish the Principle of Iterated Suprema: 
\begin{align*}
\sup\{f(x,y):\: x\in X,y\in Y\} &=\sup\{f_{1}(x):\: x\in X\}\\
&=\sup\{f_{2}(y):\: y\in Y\}
\end{align*}
I proved the Principle of Iterated Suprema but now I am stuck on the proceeding question
The Question let $f$
  and $f_{1}$
  be as in the preceding exercise and let 
$$g_{2}(y)=\inf\{f(x,y):\: x\in X\}$$
Prove that $$\sup\{g_{2}(y):y\in Y\}\leqslant\inf\{f_{1}(x):\: x\in X\}$$
Show that strict inequality can hold. We sometimes express this inequality as $$\underset{y\quad x}{\sup\inf}f(x,y)\leqslant\underset{x\quad y}{\inf\sup}f(x,y)$$
I don't really know where to start. A hint would be greatly appreciated! Thanks
 A: Note that
$$
g_2(y) = \inf\{f(x,y) : x\in X\} \leq f(x,y) \quad\forall x \in X, y\in Y
$$
and similarly
$$
f(x,y) \leq f_1(x) \quad\forall x\in X, y\in Y
$$
Thus, for all $x,y$,
$$
g_2(y)\leq f_1(x)
$$
Taking sup over $y\in Y$, we get
$$
\sup \{g_2(y) : y\in Y\} \leq f_1(x)
$$
This is true for all $x\in X$, so
$$
\sup \{g_2(y) : y\in Y\} \leq \inf \{ f_1(x) : x\in X\}
$$
Now for the counterexample, let $X = \mathbb{N}$ and $Y = \mathbb{N}$. Then $f(x,y)$ can be represented by an infinite matrix; where the $x$ variable represents the rows, and $y$ represents the columns.
Let's say you want to construct such a matrix whose row infimum is zero for any row, and whose column supremum is 1 for any column. Just take $f$ to be the matrix
$$
f(0,j) = (1,0,1,0,1,\ldots )
$$
$$
f(1,j) = (0,1,0,1,0,\ldots )
$$
and $f(2n,j) = f(0,j)$, and $f(2n+1,j) = f(1,j)$ for all $n$,
A: Hints: 


*

*$g(y) \leq \sup\limits_y\ g(y)$

*$h(x) \leq k(x) \Rightarrow \inf\limits_x\ h(x) \leq \inf\limits_x\ k(x)$

*$l(y) \leq M \Rightarrow \sup\limits_y\ l(y) \leq M$

A: So, the claim is that 
\begin{align*}\sup\{g_{2}(y):y\in Y\} &\leqslant \inf\{f_{1}(x):\: x\in X\}\\
\sup\{\inf\{f(x,y):\: x\in X\}:y\in Y\} &\leqslant \inf\{\sup\{f(x,y):\: y\in Y\}:\: x\in X\}\end{align*}
Since we know that $\text{Ran}g_{2}$
  is a bounded subset of $\mathbb{R}$
  , we know that $\sup\{g_{2}(y):y\in Y\})$
  exists, likewise since $\text{Ran}f_{1}$
  is a bounded subset of $\mathbb{R}$
 , we know that $\inf\{f_{1}(x):\: x\in X\}$
  exists. Let us examine an arbitrary $x'\in X$
  and an arbitrary $y'\in Y$
  we have 
\begin{align*} g_{2}(y') \quad &? \quad f_{1}(x')\\
\inf\{f(x,y'):\: x\in X\} \quad &? \quad  \sup\{f(x',y):\: y\in Y\}
\end{align*}
Let $f(x_{0},y')=\inf\{f(x,y'):\: x\in X\}$
  and $f(x',y_{0})=\sup\{f(x',y):\: y\in Y\}$
 , then we have that $(\forall x)(f(x_{0},y')\leqslant f(x,y'))$
  and $(\forall y)(f(x',y)\leqslant f(x',y_{0}))$
 . In words, we have that is doesn’t matter what $y'$
  you choose $(f(x',y')\leqslant f(x',y_{0})$
 , and for not matter what $x'$
  you choose $(f(x_{0},y')\leqslant f(x',y')$
 . That is that $(f(x_{0},y')\leqslant f(x',y_{0})$
 , i.e. $g_{2}(y')\leqslant f_{1}(x')$
 . Since our selection of $x'$
  and $y'$
  was arbitrary we have that $(\forall x,y)(g_{2}(y)\leqslant f_{1}(x))$
 . So, it follows that $\sup\{g_{2}(y):y\in Y\}\leqslant\inf\{f_{1}(x):\: x\in X\}$
 . 
