1. ## Iterated sup

Let $f:A\times B\to\mathbb R$ bounded above, prove that $\underset{(x,y)\in A\times B}{\mathop{\sup }}\,f(x,y)=\underset{x\in A}{\mathop{\sup }}\,\left(\underset{y\in B}{\mathop{\sup }}\,f(x,y)\right).$

How to prove this? I think is not that easy.

2. Originally Posted by Connected
Let $f:A\times B\to\mathbb R$ bounded above, prove that $\underset{(x,y)\in A\times B}{\mathop{\sup }}\,f(x,y)=\underset{x\in A}{\mathop{\sup }}\,\left(\underset{y\in B}{\mathop{\sup }}\,f(x,y)\right).$

How to prove this? I think is not that easy.
Let $(x_0,y_0)\in A\times B$ be fixed. We clearly then that that $\displaystyloe f(x_0,y_0)\leqslant \sup_{y\in B}f(x_0,y)$ and moreover it's clear that $\displaystyle \sup_{y\in B}f(x_0,y)\leqslant \sup_{x\in A}\sup_{y\in B}f(x,y)$ putting these two together gives $\displaystyle f(x_0,y_0)\leqslant \sup_{x\in A}\sup_{y\in B}f(x,y)$ and since $(x_0,y_0)\in A\times B$ was arbitrary it follows that $\displaystyle \sup_{(x,y)\in A\times B}f(x,y)\leqslant \sup_{x\in A}\sup_{y\in B}f(x,y)$. Conversely, fix $y_0\in B$ and $x_0\in A$ then evidently $\displaystyle \sup_{(x,y)\in A\times B}f(x,y)\geqslant f(x_0,y_0)$. Since $y_0$ was arbitrary we have that $\displaystyle \sup_{(x,y)\in A\times B}f(x,y)\geqslant \sup_{y\in B}f(x_0,y)$ and since $x_0$ was arbitrary $\displaystyle \sup_{(x,y)\in A\times B}f(x,y)\geqslant \sup_{x\in A}\sup_{y\in B}f(x,y)$. The conclusion follows.

3. Originally Posted by Drexel28
and moreover it's clear that $\displaystyle \sup_{y\in B}f(x_0,y)\leqslant \sup_{x\in A}\sup_{y\in B}f(x,y)$
Okay, not much clear for me, haha, wonder why is this true? I can't see it.

4. Originally Posted by Connected
Okay, not much clear for me, haha, wonder why is this true? I can't see it.
Why? If I said to you consider the function $g:\mathbb{R}\to\mathbb{R}$ one would say that $g(x_0)\leqslant \sup_{x\in\mathbb{R}}g(x)$. Let $g(x)=\sup_{y\in B}f(x,y)$