1
$\begingroup$

(Adapted) (Mathematical Circles) Eleven students formed five study groups. Prove that we can find two students, say A and B, that every study group that includes student A also includes student B

One solution is to name every group as $1,\dots ,5$ and then say that every student must be in a subset of $\{1,\dots, 5\}$. Then rearrange these subsets in other sets. Say that set $A$ is one of those other sets. $A$ is such that if two students are in the same set, they must be in the same groups (every set element is a subset of a element of $A$). Example: $A = \{ \{1\}, \{1,2\}, \{1,2,3\}, \dots\}$. This way, we can apply the pidgeonhole principle and show that at least two students are in the same $A$.

However, this solution is not very intuitive. Is there any other solution?

  • 0
    To be clear, a group that includes student $B$ but not student $A$ would not disqualify the pairing? – 2017-01-19
  • 0
    @Joffan iff *all* groups for *all* students A and B – 2017-01-20
  • 0
    Let me ask again a different way. If we have such a pairing $A$ and $B$ as per the question, is it possible that there is a group that contains student $B$ but does not contain student $A$? – 2017-01-20
  • 0
    There is. But only if we're talking about students A, ..., (student 11). There *might* be students with the property you said, but there are *at least two* that are in the same group – 2017-01-21
  • 1
    My motivation is that it is relatively easy to make sure that no two students are always together - that is, they are always both in a group or both out of it. You can do that with only four groups. So I was trying to find out how the one-way relation works, exactly. – 2017-01-21

2 Answers 2

1

So we have $11$ sets $A_1,...,A_{11}$ which are subsets of the set $S= \{1,2,3,4,5\}$. ($A_i$ represents i-th student and $j\in A_i$ iff the student $i$ is in the $j$-th group). We have to prove that there exist $A_m=:A$ and $A_n=:B$ such that if $k\in A$ then $k\in B$ i.e. $A\subset B$.

Say there is no such pair. But then we have an antichain $A_1,...,A_{11}$ which wide is $11$. But this is impossible by Sperner theorem which tell's us that in $S$ we have an antichain which is wide at most ${5\choose 2} = 10$. A contradiction.

0

I would look at it as a function $f$ from $[[11]]$ to $P([[5]])$ such that for all $a,b \in [[11]]$ $f(a) \not \subset f(b)$. Then with Dilworth's theorem you can see that there are at most $\binom{5}{3}=10$ anti-chains.

  • 0
    I never saw anything about antichains, just read about it today. However, this is basically the same solution for the problem but using set and equivalence abstraction – 2017-01-21
  • 0
    I think the idea is basically the same, yes, but imo it makes for a way more intuitive proof if you write it out. – 2017-01-22