Show that $\underset{x\in X}{\sup}\left(\underset{y\in Y}{\sup}f(x,y)\right) = \underset{(x,y)\in X\times Y}{\sup} f(x,y)$

Question

I have the following problem I don't know how to start.

Prove that:

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

always when $X\times Y\neq \emptyset$ and $f: X\times Y \rightarrow\mathbb{R}$ is a mapping bounded above.

How should I approach this problem?

Answer 1

For every $\tilde x\in X$ we have $$ \sup_{y\in Y}f(\tilde x,y)\leq \sup_{x\in X, y\in Y}f(x,y)\qquad \implies\qquad\sup_{x\in X}\bigg(\sup_{y\in Y}f(\tilde x,y)\Bigg)\leq \sup_{x\in X, y\in Y}f(x,y).$$ For every $\tilde x\in X$ and $\tilde y\in Y$ we have $$ f(\tilde x,\tilde y)\ \leq\ \sup_{y\in Y}f(\tilde x, y)\ \leq\sup_{x\in X}\bigg(\sup_{y\in Y}f(\tilde x,y)\Bigg)\qquad \implies\qquad\sup_{x\in X, y\in Y}f(x,y)\leq \sup_{x\in X}\bigg(\sup_{y\in Y}f(\tilde x,y)\Bigg).$$

Answer 2

Fix $x\in X$. Fix $y\in Y$. Then clearly:

$$f(x,y)\le \underset{(x,y)\in X\times Y}{\sup} f(x,y).$$

Because $y$ was arbitrary, you have $$\underset{y\in Y}{\sup} f(x,y)\le\underset{(x,y)\in X\times Y}{\sup} f(x,y).$$

Because $x$ was arbirtary, you have $$\underset{x\in X}{\sup}\left(\underset{y\in Y}{\sup}f(x,y)\right) \le \underset{(x,y)\in X\times Y}{\sup} f(x,y).$$

Note that we only used the definition of supremum as the least upper bound.

Try to do something similar to obtain the other inequality.