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I'm looking at "Basic Set Theory" by A. Shen. The very first 2 problems are: 1) can the oldest mathematician among chess players and the oldest chess player among mathematicians be 2 different people? and 2) can the best mathematician among chess players and the best chess player among mathematicians be 2 different people? I think the answers are no, and yes, because a person can only have one age, but they can have separate aptitudes for chess playing and for math. Is this correct?

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up vote 10 down vote accepted

Yes, it’s correct. If $M$ is the set of mathematicians, and $C$ is the set of chess players, you’re looking rankings of the members of $M\cap C$. If for $x\in M\cap C$ we let $m(x)$ be $x$’s ranking among mathematicians, $c(x)$ be $x$’s ranking among chess players, and $a(x)$ be $x$’s age, then there is a unique $x_a\in M\cap C$ such that $$a(x_a)=\max\{a(x):x\in M\cap C\}\;,$$ but there can certainly be distinct $x_m,x_c\in M\cap C$ such that $$m(x_m)=\max\{m(x):x\in M\cap C\}$$ and $$c(x_c)=\max\{c(x):x\in M\cap C\}\;.$$

All of which just says what you said, but a bit more formally.

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Thanks very much for the help you guys! – Peter Jun 11 '12 at 5:06
So what's $V$ in the first displayed equation? – Asaf Karagila Mar 18 '13 at 22:07
@Asaf: Right next to the $C$ on my keyboard. – Brian M. Scott Mar 18 '13 at 22:08

(1) Think of it in terms of sets. Let $M$ be the set of mathematicians, $C$ the set of chess players. Both are asking for the oldest person in $C\cap M$.

(2) Absolutely fantastic reasoning, though perhaps less simply set-theoretically described.

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