By Bagaria J., Todorcevic S. (Editors)

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I then −bs ∈ uI which is a Conversely, if s ∈ θuI then bs ∈ uI . Now if s ∈ contradiction. Thus, s ∈ I. 5. θ is continuous. For let I = θu and take t1 , . . , tn ∈ I; s1 , . . , sm ∈ / I. p (T ) : t1 , . . , tn ∈ J and s1 , . . , sm ∈ / J}. −1 (V(I)), where s is the Stone Then b = ni=1 bti · m j=1 −bsj = 0. So, u ∈ s(b) ⊆ θ representation mapping see [9, p. 99]. p (T ). Thus, θ is a homeomorphism. Case 2. )-poset. 1. There is u0 ∈ Ult(B(T )) so that θu0 = ∅. Indeed, set V0 = {−bs : s ∈ T }.

Suppose b decomposes as, b = bt1 · · · btn = bs1 · · · bsm with n ≤ m, si and tj are pairwise distinct. ), there are i, j such that bti = bsj . Hence, b = bti l=i btl = bti k=j bsk . ), we have {btk : 1 ≤ k ≤ n and k = i} = 1 ≤ l ≤ m and l = j}. Case 1. m = n we are done. bsj2 · · · bsjp = 0 Case 2. m = n bsj1 {bsl : bsj1 = bsj2 · · · bsjp . ) there is k so that bsj1 = bsjk . Hence, sj1 = sjk with k = 1. This is a contradiction since all sj s are distinct. This finishes the proof of the theorem.

Proof. Let T be an upper semi-lattice, with a least element t0 , so that Prim(T ) is ∨-generating set of T , and set P := {↓ t : t ∈ Prim(T )}; then (P, ⊆) is a poset with a least element ↓ {t0 } = {t0 }. Define ϕ from Id(T ) into I(P ) by ϕ(I) = (↓ I) ∩ P , where ↓ I :=def {J ∈ Id(T ) : J ⊆ I}. Again it easy to see that Id(T ) ∼ =homeo I(P ). Let(T, ≤) be a poset with a least element. 14 give indeed, necessary and sufficient conditions on Id(T ) and I(P ) to be homeomorphic spaces. We state these conditions in the following corollary.

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