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May 11 at 18:55 comment added bof You can disprove this conjecture with $23$ sets if you don't mind $\varnothing$ being one of the sets.
May 11 at 8:19 comment added Fabius Wiesner This version of the conjecture was disproved with $25$ sets, so maybe it can be shown that attempt 3 needs more than $53$ sets, the current lower bound for a counterexample of the union-closed sets conjecture.
May 11 at 8:02 vote accept Fabius Wiesner
May 11 at 8:02 comment added Fabius Wiesner Thank you! I have started now attempt 3. I think it can't be restricted more than that. I think also that the requirement that the empty set is not in the family should not be relevant.
May 11 at 7:46 history edited bof CC BY-SA 4.0
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May 10 at 22:31 history edited bof CC BY-SA 4.0
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May 10 at 22:23 history answered bof CC BY-SA 4.0