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Given any $f: X \times Y \rightarrow \mathbb{R} \cup \{ \infty, -\infty\}$, I was wondering

  1. if it is true that $\inf_{x \in X} \sup_{y \in Y} f(x,y) \geq \sup_{y \in Y} \inf_{x \in X} f(x,y)$?
  2. when it is true that $\inf_{x \in X} \sup_{y \in Y} f(x,y) = \sup_{y \in Y} \inf_{x \in X} f(x,y)$?
  3. if the answers to the above will be different if $\inf$ and/or $\sup$ be replaced with $\min$ and/or $\max$?
  4. if the answers to the above will be different if the codomain of $f$ is any totally ordered set? Suppose all exist.

Thank!

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    @Arturo: Nice proof! Thanks!2011-04-05

3 Answers 3

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\mathop{\inf\vphantom{p}}_{x\in X}\sup_{y\in Y}f(x,y)\ge\mathop{\inf\vphantom{p}}_{x\in X}\sup_{y\in Y}\mathop{\inf\vphantom{p}}_{x'\in X}f(x',y)=\sup_{y\in Y}\mathop{\inf\vphantom{p}}_{x'\in X}f(x',y)=\sup_{y\in Y}\mathop{\inf\vphantom{p}}_{x\in X}f(x,y)\;.

This proves $1$, and the same proof works if you replace inf by min and/or sup by max and/or replace $\mathbb{R}$ by any totally ordered set.

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    @joriki: I think an important caveat is that if we take a different total order then the $\sup$ might exist and the $\inf$ might not, which renders the question somewhat moot.2011-04-29
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2 is not true in general, but under certain assumptions, it is true. This is called Minimax theorem. See, for instance, Minimax Theorem.

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For 1 and 2, they might not exist. Try $f(x,y)=\frac{1}{y}$. Even more so for 3 and 4.

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    Thanks! You are right about the existence issue. What about when they exist? I just modified my questions for this.2011-04-05