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*}
Best Answer
The proof you give is correct.