Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 1 / b

True. Write $s=\sup A$ and $u=\sup \{\sup R_m:m\ge 1\}$. All these suprema exist in $[0,1]$. Since $R_m\subseteq A$, we have $\sup R_m\le s$ for every $m$, hence $u\le s$. Conversely, every $x_{m,n}\le \sup R_m\le u$, so $u$ is an upper bound of $A$ and $s\le u$. Thus the iterated extrema of an array satisfy

True. Set $l=\inf A$ and $v=\inf \{\inf R_m:m\ge 1\}$. Because $R_m\subseteq A$, every row infimum satisfies $\inf R_m\ge l$, so $v\ge l$. But $x_{m,n}\ge \inf R_m\ge v$ for all $m,n$, making $v$ a lower bound of $A$. Therefore $v\le l$, and

True. For every $m,n$,Fix $m$. The number $\sup R_m$ is an upper bound for every column infimum, so ${\sup }_n\inf C_n\le \sup R_m$. This holds for every $m$, making ${\sup }_n\inf C_n$ a lower bound of the row suprema. Taking their infimum gives the weak minimax inequality

False. Take $x_{1,n}=0$ for every $n$, and $x_{m,n}=1$ for every $m\ge 2$ and every $n$. Each row is constant, so its infimum and supremum coincide. Nevertheless,The two outer extrema can select different rows.