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

12 B. BANASCHEWSKI, R. LOWEN, C. VAN OLMEN6.1. Proposition. An approach frame is spatial iff each element is the meetof prime elements.Proof. ⇒ If the approach frame comes from an approach space, then weknow that each δ(·, {x}) is prime and since f − f(x) ≤ δ(·, {x}) for allcontractions, we find that f = ∧ A f(x) δ(·, {x})⇐ Take S = {approach prime elements a}, as approach structure take Land see the elements as functions. This set of functions has all the propertiesof an regular function frame and it is clear that this approach space givesthe approach frame we started with: if a ≠ b ∈ L then there must exist anapproach prime element e such that h e (a) ≠ h e (b), suppose not, thena = ∧ e∈SA he(a)e = ∧ e∈SA he(b)e = b.Hence an approach frame is spatial iff the underlying frame is spatial.6.2. Proposition. A T 0 approach space X is a dense subspace of ΣRX.Proof. If X is T 0 then for all x ≠ y we find δ(x, {y}) > 0 or δ(y, {x}) > 0.Hence δ(·, {y}) ≠ δ(·, {x}) and since all δ(·, {y}) are prime elements, ɛ X :X → ΣRX : x ↦→ δ(·, {x}) is injective. To show that ɛ X is a subspaceembedding, we remark that the initial structure on X by ɛ X is just R sincef ◦ɛ X (x) = f(x) (f first considered as element of RX, then seen as functionon X).For the density, consider a ∈ L such that ∀x ∈ X : a(ɛ X (x)) = 0, thenobviously a = 0. Hence we find that δ(·, ɛ X (X)) = ∨ {a(·)|a ∈ L, ∀x ∈ X :a(ɛ X (x)) = 0} = 0.□6.3. Definition. An approach frame L is said to be topological iff L can beobtained as the regular function frame of a topological approach space.6.4. Proposition. L topological iff it it spatial and S λ a = a for all λ < ∞and all approach primes a.Proof.⇒ The approach prime elements a of the approach frame are:{0 y ∈ Uθ U (y) =∞ y ∉ Ufor every closed and join-irreducible (with regard to the closed sets) U.This is because an approach prime a has values below every ɛ > 0 andfor each n ∈ N there exists an x ∈ X such that a(x) > n: suppose not,∀x ∈ X : a(x) < N and ∃x ∈ X : a(x) > N − ɛ > 0, then take{a ′ a(y) a(y) ≤ N − ɛ(y)) =∞ a(y) > N − ɛ ,then a ′ ∧ (N − ɛ) ≤ a, but a ′ and N − ɛ are not smaller than a.Finally a must be of the form θ Y : suppose a takes a value r > 0 otherthan ∞, then consider T = a −1 ([0, r 2 ]), then θ T ∧ (a ∨ r 2 ) ≤ a, but θ T ≰ aand a ∨ r 2 ≰ a.⇐ Using the isomorphism between aprim(L) and ΣL, we have an approachspace, but it remains to prove that this space is topological.□