Co-separability of filter objects (Solved)

Importance: Medium ✭✭
Author(s): Porton, Victor
Subject: Unsorted
Keywords: filters
Recomm. for undergrads: no
Posted by: porton
on: November 29th, 2009
Solved by: Porton, Victor
Conjecture   Let $ a $ and $ b $ are filters on a set $ U $ and $ a\cap b = \{U\} $. Then $$\exists A\in a,B\in b: (\forall X\in a: A\subseteq X \wedge \forall Y\in b: B\subseteq Y \wedge A \cup B = U).$$

See here for some equivalent reformulations of this problem.

This problem (in fact, a little more general version of a problem equivalent to this problem) was solved by the problem author. See here for the solution.

Maybe this problem should be moved to "second-tier" because its solution is simple.

* indicates original appearance(s) of problem.

A counterexample?

From the link it seems you have proved the result. What about the following what seems to be a counterexample?

$ U = \mathbb N $ (the set of integers), $ a = \{ U \} $, $ b = $ the set of all infinite sets of integers

Now there is no set $ B $ that would be minimal in $ b $ ...

Your example is wrong

The set of all infinite sets of integers is not a filter. For example $ \{0,2,4,\dots\} \cap \{1,3,5,\dots\} = \emptyset $.

I haven't read your comment further.


Sorry, I was too hasty. What I meant is that $ b $ is a "nontrivial ultrafilter" (wikipedia page ultrafilter) calls this "non-principal ultrafilter".

No counterexamples, it is proved

Then take $ A=U $ and $ B=\emptyset $ (I do not require filters to be proper).

Robert, why you are trying to find a counter-example for a proved theorem?

Here is the proof.

Victor Porton -

Is it really?

You require that $ B \in b $, and my filter $ b $ does not contain empty set. I'm trying to find a counter-example because either I misunderstand the statement of the theorem, or the theorem is false.

Some proofs just happen to have mistakes. Unfortunately, I don't understand yours, it apparently uses lot of notation (up, down, Cor, ...) that I'm unfamiliar with.

Oh, my mistake

I made a mistake in the statement of the conjecture as published on OPG. I corrected the problem statement both on OPG and on my blog. It should be $ \exists A,B\in\mathscr{P}U $ rather than $ \exists A\in a,B\in b $.

Indeed the equivalent reformulations of the theorem are correct and my proof (of a more general statement than this theorem) is not affected by the above mentioned error.

Robert, you do not understand me because I introduced new notations (that up, down, Cor, etc.) You may wish to read my preprint about these things (filters on posets and generalizations).

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