Prove $\inf_{x \in X} \sup_{y \in Y} f(x,y) \ge \sup_{y \in Y}\inf_{x \in X} f(x,y)$ Let $f : Z \rightarrow \mathbb{R}$ where $Z = X \times Y$, with both $X, Y \subset \mathbb{R}$ and $f$ is bounded, that is, $\exists$ $M >0$ such that $|f(z)|\le M$ $\forall$ $z \in Z$. Prove that $\inf_{x \in X} \sup_{y \in Y} f(x,y) \ge \sup_{y \in Y}\inf_{x \in X} f(x,y)$.
First of all, I've only dealt with the supremum and infimum of sets and not with functions yet. So I want to confirm whether my understanding of the notation is correct first. Is it true that $\inf_{x \in X} \sup_{y \in Y} f(x,y)$ means: First, we fix $x \in X$, then find $\sup \{f(x,y) : y \in Y\}$, so the supremum essentially becomes a function of $x$, which we can denote $f^*(x) = \sup \{f(x,y) : y \in Y\}$, then we find $\inf \{f^*(x) : x \in X\}$. Is my understanding correct? If so, how can I complete the proof?
 A: That is the meaning.
For all $x$ and $y$, $f(x,y)\le f^*(x)$. Thus $$\forall y\in Y,\qquad f_*(y)=\inf_{x\in X} f(x,y)\le \inf_{x\in X}f^*(x)$$
Notice that the RHS is just a fixed extended real number. Thus, $$\sup_{y\in Y} f_*(y)\le\inf_{x\in X} f^*(x)$$
The hypothesis of boundedness is unnecessary. Meaning: if you are working with functions which take values in the appropriate version of the extended real numbers (namely, $[-\infty,\infty]$), the inequality still holds in the natural way.
A: First things first, for every $x \in X$ and $y \in Y$ (to avoid confusion, I intentionally used $y'$ on the left hand side to distinguish $y$ and $y'$ in this step), 
$$\sup_{y' \in Y} f(x, y') \geq f(x, y) \quad \tag{1}$$
Now taking $\inf$ over the set $X$ on both sides of $(1)$ gives that
$$\inf_{x \in X} \sup_{y' \in Y} f(x, y') \geq \inf_{x \in X} f(x, y). \quad \tag{2} $$
Remember that $(2)$ holds for every $y \in Y$, therefore a $\sup$ over $Y$ can be safely appended, gives:
$$\inf_{x \in X} \sup_{y' \in Y} f(x, y') \geq \sup_{y \in Y}\inf_{x \in X} f(x, y). \quad \tag{3} $$
Now you can replace the $y'$ with $y$ as it is a dummy index, and this gives you the desired result.
The boundedness assumption is needed so that every step in $(1)-(3)$ will not cause abnormal cases.
