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SOBER APPROACH SPACES 1. Introduction In this article we will ...

SOBER APPROACH SPACES 1. Introduction In this article we will ...

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<strong>SOBER</strong> <strong>APPROACH</strong> <strong>SPACES</strong> 5Proof. For the first claim, put a := ∨ {x|ξ(x) = 0}, then obviously ξ(a) = 0.If b ∧ c ≤ a, then by monotonicity ξ(b) ∧ ξ(c) = ξ(b ∧ c) = 0, hence eitherξ(b) = 0 or ξ(c) = 0. Finally, if α > 0 is such that α ≤ a, then ξ(α) ≤ξ(a) = 0 which is a contradiction.To prove the second claim, note that ξ is clearly order-preserving and thatξ a (0) = 0. Suppose that ξ a (∞) ≠ ∞, then there exists an α ∈ ]0, ∞[ suchthat ∞ = A α a. Consequently 2α ≤ A α a, and thus α = S α 2α ≤ a which isa contradiction.If S ⊂ L then <strong>we</strong> find that∨ ∨S ≤ A ξa(s)a = A W ξ a(S)as∈Sand hence that ξ a ( ∨ S) ≤ ∨ ξ a (S). Since ξ is monotone, <strong>we</strong> have the otherinequality and so <strong>we</strong> find that ξ commutes with arbitrary joins.ξ a (x ∧ y) = ξ a (x) ∧ ξ a (y): since ξ is order-preserving, <strong>we</strong> find ξ(x ∧ y) ≤ξ(x) ∧ ξ(y). For the other inequality: suppose that x ∧ y ≤ A α a. Bylemma ??, <strong>we</strong> find that x ≤ A α a or y ≤ A α a.ξ a (A α x) = ξ a (x) + α: take ξ a (x) = β, then x ≤ A β a ⇔ A α x ≤ A β+α a ifα < ∞. With α = ∞, the result is immediate.ξ a (S α x) = (ξ a (x) − α) ∨ 0: S α x ≤ A β a ⇔ x ≤ A α+β a.□4.6. Corollary. The space of points of L can be expressed as the set ofapproach prime elements. We denote <strong>this</strong> set by aprim(L).4.7. Definitions and Notations. We <strong>will</strong> use p ξ for the approach primeelement associated with the homomorphism ξ : L → J.For the homomorphism associated with the approach prime element a, <strong>we</strong><strong>will</strong> use h a .Using <strong>this</strong> definition <strong>we</strong> can easily express the approach structure onaprim(L). This must still be L, but the construction is slightly differentnow:∀f ∈ L, a ∈ aprim(L) : f(a) = h a (f)The distance δ on the spectrum can be constructed explicitly. First rememberthe relation bet<strong>we</strong>en the distance and the regular function frame inan approach space X:δ(x, A) = sup{ρ(x)|ρ ∈ R, ρ| A = 0}4.8. Proposition. On ΣL, the distance is defined as:δ(ξ, A) = ∨ {ξ(a)|a ∈ L, ∀ζ ∈ A : ζ(a) = 0}If <strong>we</strong> work with aprim(L), <strong>this</strong> givesδ(a, A) = h a ( ∧a i )a i ∈AProof. The formula on ΣL is an easy translation of the relation bet<strong>we</strong>enregular function frames and the distance.For aprim(L), note that h ai (b) = 0 iff b ≤ a i . It now follows that thelargest element of L that has evaluation 0 for all h ai must be ∧ a i . □

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