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

The infimum of $A$ is its greatest lower bound: $\alpha =\inf A$ means that $\alpha \le x$ for every $x\in A$, and that every lower bound $l$ of $A$ satisfies $l\le \alpha$. The supremum is its least upper bound: $\beta =\sup A$ means that $x\le \beta$ for every $x\in A$, and every upper bound $u$ satisfies $\beta \le u$. The completeness of the real numbers ensures that both exist for a nonempty bounded set of real numbers.

Put $s=\sup (-A)$. For every $x\in A$, $-x\le s$, hence $-s\le x$, so $-s$ is a lower bound of $A$. If $l$ is any lower bound of $A$, then $-x\le -l$ for every $x\in A$, making $-l$ an upper bound of $-A$. The definition of supremum gives $s\le -l$, or $l\le -s$. Thus $-s$ is the greatest lower bound, and

Let $M=\max \{\sup A,\sup B\}$. Every element of the union $A\cup B$ is at most $M$, so $M$ is an upper bound. Conversely, every upper bound of $A\cup B$ is an upper bound of each of $A$ and $B$, hence is at least both suprema and therefore at least $M$. Consequently

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.