The problem, from the 'pigeon-hole principle' section of Bona's combinatorics book:
"One afternoon, a mathematics library had several visitors. A librarian noticed that it was impossible to find three visitors so that no two of them met in the library that afternoon. Prove that then it was possible to find two moments of time that afternoon so that each visitor was in the library at one of those two moments."
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My attempt at a solution:
pf.: Let $m_1,m_2,...,m_n$ be our mathematicians. We are given that it is true that for any three $m_i,m_j,m_k$ we have at least one of the following hold: $$m_i\text{ met }m_j,\quad m_i\text{ met }m_k,\quad\text{ or }\quad m_j\text{ met }m_k.\qquad(\text{I.})$$
$\quad$We first claim there can be no more than two moments. Note $$m_i \text{ met with }m_j\quad\text{ implies }\quad m_i,m_j \text{ were present during the same moment}.\qquad(\text{II.)}$$Now suppose for contradiction we have at least three moments. From the first moment take some $m_i$, from the second some $m_j$, and from the third some $m_k$. By (I.) above, either $m_i$ met with $m_j$, $m_i$ met with $m_k$, or $m_j$ met with $m_k$. But then (II.) would imply $m_i,m_j$, $m_i,m_k$, or $m_j,m_k$ share the same moment, which is a contradiction to the fact that the mathematicians were pulled from three separate moments.
$\quad$This shows there are at most two moments. If there are exactly two moments then we are done. Note one moment can trivially be extended to two moments, either by splitting it into two moments or by adding a second 'empty' moment (e.g. if it were true the $n$ mathematicians were in the library between 12:00-12:10 then it is also true they were in the library between 12:00-12:20). QED.
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Remarks:
I feel like this logic is pretty sound and would perhaps qualify as the most obvious solution. But I'm not sure where, if at all, the pigeon-hole principle is used in this particular solution. I'm thinking the pigeon-hole principle is encoded into the instructions themselves, particularly (I.) above - I'm thinking for any 2 mathematicians there are 2 possibilities, they met or did not meet. Then for 3 mathematicians, by the PHP it is true either two met or two did not meet. The problem is FORCING that the former is true, and asking us to study the consequences. Is this a good read at what's going on in this problem? Am I completely missing the mark? Just looking for some opinions, many thanks!
As I mentioned in the comments, I don't think the question is correct unless you add the condition that a visitor cannot reenter the library after leaving.
Let $m_i$ be the person that is the latest to arrive, and let $m_j$ be the person that is earliest to leave.
If $m_i$ and $m_j$ overlap, then choosing a single point in time when they are both in the library is sufficient.
Now suppose $m_i$ and $m_j$ don't overlap, i.e. $m_j$ left before $m_i$ arrived. By Condition I, each of the other visitors either met $m_i$ or $m_j$, so it is possible to divide the other visitors into two groups $I$ and $J$ such that all visitors in group $I$ met $m_i$, and all visitors in group $J$ met $m_j$. Let $m_k$ be the earliest to leave among visitors in group $I$, and let $m_\ell$ be the latest to arrive among visitors in group $J$. If we choose a point in time when both $m_i$ and $m_k$ are in the library, then all visitors in group $I$ are also in the library. Similarly if we choose a point in time where both $m_j$ and $m_\ell$ are in the library, then all visitors in group $J$ are also in the library.