The back of the book gave a proof similar to the proof here Proving principle of the Iterated Suprema, but I proved it following way before I checked the back of the book. Could some one verify this proof?
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 $f_{2}(y)=\sup\{f(x,y):\: x\in X\}$
Prove that \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*}
proof. Since $\text{Ran}f$ is a bounded subset of $\mathbb{R}$ we have that the set $\{f(x,y):\: x\in X,y\in Y\}$ indeed has a supremum (and an infimum). If we let $u=\sup\{f(x,y):\: x\in X,y\in Y\}$ then we have that $(\forall x\in X)(\forall y\in Y)(f(x,y)\leqslant u)$ , this implies that $(\forall x\in X)(f_{1}(x)\leqslant u)$ and $(\forall y\in Y)(f_{2}(y)\leqslant u)$ , hence $\sup f_{1}(x)\leqslant u$ and $\sup f_{2}(y)\leqslant u$ . Now suppose $\sup f_{1}(x)<u$ Then there exist a $x'$ such that $f(x',y)\leqslant u'<u$ for all $x$ and $y$ hence $u'$ is an upper bound on $f(x,y)$ , but this contradicts our selection of $u$ as the supremum of $\{f(x,y):\: x\in X,y\in Y\}$ , thus $(u\leqslant\sup f_{1}(x)\leqslant u)\implies\sup f_{1}(x)=u$ . Likewise, if we suppose that $\sup f_{2}(y)<u$ then $(\exists y')(\forall x,y)(f(x,y)\leqslant f(x,y')<u)$ which means that $f(x,y')$ is an upper bound on $f(x,y)$ again this contradicts our selection of $u$ as $\sup\{f(x,y):\: x\in X,y\in Y\}$ . Consequetly, we have that $(u\leqslant\sup f_{2}(y)\leqslant u)\implies(\sup f_{2}(y)=u)$ . So indeed we have that \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*}
