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Given a 2-D array $\{a(i,j)\}$ of real numbers indexed by $i\in I$ and $j\in J$, do we have $\sup\limits_{i\in I}\inf\limits_{j\in J}a(i,j)=\inf\limits_{j\in J}\sup\limits_{i\in I}a(i,j)$? I can only prove $\sup\limits_{i\in I}\inf\limits_{j\in J}a(i,j)\le\inf\limits_{j\in J}\sup\limits_{i\in I}a(i,j)$ but could not establish the converse. Does the converse, i.e., $\sup\limits_{i\in I}\inf\limits_{j\in J}a(i,j)\ge\inf\limits_{j\in J}\sup\limits_{i\in I}a(i,j)$ holds in gereral? Thanks a lot.

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up vote 2 down vote accepted

Define $a(i,j)=1$ if $i=j$ and $0$ otherwise. Then $$\sup_{i\in\Bbb N}\inf_{j\in\Bbb N}a(i,j)=0<1=\inf_{j\in\Bbb N}\sup_{i\in\Bbb N}a(i,j).$$

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Since the statement is not true. Let us cook up an counterexample. $$ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} $$

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