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?
Set theory puzzles - chess players and mathematicians
2 Answers
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.
-
0@Asaf: Right next to the $C$ on my keyboard. – 2013-03-18
(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.