Given a finite collection of of sets, closed under union (that is, if \(A\) and \(B\) are sets belonging to the collection, then \(A\cup B\) is also a set in the collection.)

Prove or disprove: There exists an element that belongs to at least half the sets in the collection.

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## Comments

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TopNewestI'll be impressed if someone comes up with a proof/disproof here, as it is presently an open question in mathematics, (first posed by Peter Frankl in 1979). :)

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I don't know about what does this "closed under union" means exactly !! No theoretical knowledge about it. But I think It can be proved using the

Principle of Mathematical Induction. One can easily show it for only sets (that is A and B). Assume it true for \(n\) sets and then try to prove it for \(n+1\)sets.Log in to reply

As explained, a collection of sets is

closed under unionif given any 2 sets \(A\) and \(B\) in the collection, then \( A \cup B \) must also be in the collection.Standard Induction on the number of sets in the collection would unlikely work. Strong Induction might offer a possibility of approach.

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Any hints, @Calvin Lin?

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Not really. Brian gave it away by stating that it is an open question.

Sometimes, open questions are solved by "amateur mathematicians" who had no idea that the problem was really difficult. This is a question that has "Olympiad Combinatorics" roots, so I felt was interesting to the L4/5's, and so I posted it. I was hoping that someone had an ah-ha moment, and could figure out a way to solve this.

I am interested in seeing approaches that people tried, and how far they pushed it out.

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Sorry for spoiling the fun, but I thought it would help others to know the background of the question and what approaches have been taken before.

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I was going to mention the empty set, but then I realized that you mentioned that "there exists an ELEMENT", not a set. If that were the case, this conjecture would have been proven a long time ago.

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Well, avoid trivial cases by adding "non-empty collection" into the conditions.

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