How to show this $\sup_{x} [ \sup_{y} f(x,y)] = \sup_{y} [ \sup_{x} f(x,y)]$?

Question

Let $X,Y$ non empty sets and $f: X \times Y \to R$ a bounded function. For each $x_0 \in X$ and for each $y_0 \in Y$, let $s_1 (x_0)=\sup\{f(x_0,y): y \in Y\}$ and $s_2 (y_0) = \sup\{f(x,y_0): x \in X\}$ . This defines functions $s_1 : X \to R$ and $s_2 Y \to R$. Show that $\sup_{x \in X} s_1 (x) = \sup_{y \in Y} s_2 (y)$. In other words: $$ \sup_{x} [ \sup_{y} f(x,y)] = \sup_{y} [ \sup_{x} f(x,y)] $$

Answer 1

By definition of the sup you have :

$$f(x,y) \leq s_2(y), \quad \forall x \in X, \forall y \in Y$$

You can pass in this inequality to the sup on $y \in Y$ :

$$\underset{y}{sup} \ f(x,y) \leq \underset{y}{sup} \ s_2(y), \quad \forall x \in X$$

Finally you can pass to the sup on $x \in X$ in this last inequality (the right term doesn't depend on $x$ and is a regular number in $\mathbb{R}$)

$$\underset{x}{sup} \ \underbrace{\underset{y}{sup} \ f(x,y) }_{s_1(x)}\leq \underset{y}{sup} \ s_2(y)$$ which is $\underset{x}{sup} \ s_1(x) \leq \underset{y}{sup} \ s_2(y)$.