Duals of filters, an explicit formula for meet?

Fix some set $U$. Recall that filters on $U$ are nonempty sets $F$ such that $A\cap B\in F \Leftrightarrow A\in F\land B\in F$.

Replacing every element of $F$ with its complement and simultaneously replacing every $\mathscr{P}U$ with its dual, I get definition of objects which I call free stars. (It is easy to show that free stars bijectively correspond to filters by this double "replacement".)

Free stars on $U$ are sets whole complement is not empty such that $A\cup B\in F \Leftrightarrow A\in F\lor B\in F$.

I denote the lattice of free stars ordered by set-theoretic inclusion as $\mathfrak{S}$.

It is easy to prove that union of any (possibly infinite) number of free stars is also a free star. Thus follows that $\bigcup^{\mathfrak{S}} S = \bigcup S$ for every set $S$ of free stars (where I denote $\bigcup^{\mathfrak{S}}$ the supremum on the set of free stars).

My question: Help to find an explicit formula for infimum of any number free stars (both finite of infinite number of free stars).

The following formulas for filters ordered reverse to set theoretic inclusion (I denote them $\mathfrak{F}$) proved in my draft book may help:

$F_0 \cap^{\mathfrak{F}} \dots \cap^{\mathfrak{F}} F_n = \{ K_0\cap\dots\cap K_n \mid K_i\in F_i \text{ for } i=0,\dots,n \}$.

$\bigcap^{\mathfrak{F}} S = \{ K_0\cap\dots\cap K_n \mid K_i\in\bigcup S \text{ where } i=0,\dots,n \text{ for } n\in\mathbb{N} \}$.

• The dual of a filter is called an ideal. – Asaf Karagila Jun 27 '15 at 21:20
• @AsafKaragila You've misunderstood. Ideal is a filter in dual order. But I take both dual order (that is replacing every element of the filter with its dual) and complement of the filter (considered as a set) – porton Jun 27 '15 at 21:22
• After some thinking, I conclude that it seems that there is no explicit formula in this case – porton Jun 28 '15 at 13:52