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From: Saviola on 23 Jul 2010 14:52
(AC): Given a non-empty family A={A_i}_(i belongs to I) of non-empty sets, there exists a choice function for A.

(BPI): Given a proper ideal J of a boolean lattice B, there exists a prime ideal I of B such that I contains J.

(DPI): Given a distributive lattice L, an ideal J of L and a filter G of L such that J and G are disjoint, there exists a prime ideal I of L such that I contains J and L\I contains G.

Prove that (BPI) ===> (DPI) and (DMI) ==> (AC).

Supposedly (BPI) ===> (DPI) can be proved by constructing an embedding of a given distributive lattice into a Boolean lattice, to which (BPI) is applied.

Similarly, (DMI) ==> (AC) can be proved by applying (DMI) to a suitable lattice of sets.

Can anyone please shed some light on this?

If you'd rather write on paper and scan it, please PM me so I can give you my e-mail address.

Thank you in advance.
From: William Elliot on 25 Jul 2010 04:30
On Fri, 23 Jul 2010, Saviola wrote:

> (AC): Given a non-empty family A={A_i}_(i belongs to I) of non-empty
> sets, there exists a choice function for A.

> (BPI): Given a proper ideal J of a boolean lattice B, there exists a
> prime ideal I of B such that I contains J.

> (DPI): Given a distributive lattice L, an ideal J of L and a filter G of
> L such that J and G are disjoint, there exists a prime ideal I of L such
> that I contains J and L\I contains G.

> Prove that (BPI) ===> (DPI) and (DMI) ==> (AC).

Where you write (DMI), do you mean (DPI)?

> Supposedly (BPI) ===> (DPI) can be proved by constructing an embedding
> of a given distributive lattice into a Boolean lattice, to which (BPI)
> is applied.

> Similarly, (DMI) ==> (AC) can be proved by applying (DMI) to a suitable
> lattice of sets.

Let C be a collection of pairwise disjoint, not empty sets.
I'd suppose the lattice would be the lattice of subsets of \/C.
Perhaps the ideal would be the ideal generated by C.

> Can anyone please shed some light on this?

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