1. each Uiis finite, 2. the elements of X are pairwise inclusion-incompatible, and

3. for every set of atoms M, p ∈ GLP (M) if and only if for some Ui ∈ X ,
Ui ∩ M = ∅.
We will now define two sequences:
1. a sequence hKnin∈N of infinite sets of integers and 2. a sequence hpnin∈N\{0} of atoms.

We define K0 = N, and we define p1 as the first element of U1 such that
{j : p /∈ Uj}
is infinite. Clearly, K0 is well-defined. We need to show that p1 is well-defined.

If p1 is not well-defined, then for every p ∈ U1 there is an integer ip such that for all **m > i**p, p ∈ Um. But U1 is finite so taking n **= max**p∈U1 ip, we find that for all **m > n**, U1 ⊆ Um - which contradicts the fact that the sets in X are pairwise inclusion-incompatible. Thus p1 is well-defined. We now set K1 = {n ∈ K0 : p1 ∈/ Un} = {n ∈ K0 : {p1} ∩ Un = ∅}.

Clearly. K1 is infinite.

Now, let us assume that we already defined pl and Kl so that Kl = {n : Un ∩
{p1**, . . . , p**l} = ∅} is an infinite subset of N. We select pl+1 as the first element p ∈ Ul+1 so that
{j : j ∈ Kl and p /∈ Uj}
is infinite. Clearly, by an argument as above, there is such p, and so pl+1 is welldefined. We then set Kl+1 = {j ∈ Kl: pl+1 ∈/ Uj}.

Since {p1, . . . , pl} ∩ Uj = ∅ for all j ∈ Kl, {p1, . . . , pl+1} ∩ Uj = ∅ for all j ∈ Kl+1. By construction, the set Kl+1 is infinite.

Now, we complete the argument as follows. We set Xn = {p1**, . . . , p**n}. The sequence hXnin∈N is monotonically increasing. For each n there is j (in fact infinitely many j's) so that Xn ∩ Uj = ∅. Therefore, for each n, p ∈ GLP (Xn).

Hence p ∈Tn∈N GLP (Xn).

On the other hand, let X =Sn∈N Xn. Then X = {p1, p2**, ...**}.