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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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10 B. BANASCHEWSKI, R. LOWEN, C. VAN OLMEN5.15. Corollary. Take X a pseudometric space. A regular function f isapproach prime iff it is supertight.From [?], <strong>we</strong> recall that an approach space X is called complemented iffit satisfies∀F ∈ F(X) : δ(x, {y}) ≤ λF(x) + λF(y)5.16. Lemma. Take X a complemented approach space. f is approach primeiff f can be written as the limit operator of a Cauchy filter.Proof. By proposition ?? <strong>we</strong> have the implication.For a complemented space <strong>we</strong> trivially find that each limit operator ofa Cauchy filter is supertight and hence by lemma ?? <strong>we</strong> find that suchfunctions are approach prime.□5.17. Proposition. Take X a complemented approach space. Then X issober iff it is complete and T 0 .Proof. ⇒ We know that ɛ X : X → ΣRX = {λF|F Cauchy} is an isomorphism.Hence every ξ λF can be written as ˜x for a unique x ∈ X. HenceλF = δ {x} , so all Cauchy filters converge. X is also T 0 since ɛ X is injective.⇐ Since X is complete every λF with F Cauchy is equal to a function δ {x}for a unique (T 0 ) x ∈ X. Hence ɛ X is a bijection. It is even an isomorphismsince the approach structure of ΣRX is evidently the same as that of X. □If <strong>we</strong> have an approach frame associated with an approach space, thenξ f (g) = sup x∈X(g(x) − f(x)). We <strong>will</strong> use <strong>this</strong> evaluation as the basis foran extension of f ∈ RX to a function ˆf in R ˆX.5.18. Theorem. Take X complemented and T 0 , then ΣRX is isomorphicto the completion of X.Proof. Take f ∈ RX and construct ˆf : ˆX → P by settingˆf(F) ={ f(x) if λF(x) = 0sup x∈X(f(x) − λF(x))if ̸ ∃y ∈ X : λF(y) = 0If <strong>we</strong> prove that ˆf is an element of R ˆX, <strong>we</strong> have an isomorphism bet<strong>we</strong>enRX and R ˆX.Take ɛ > 0 and make ¯ɛ : ˆX → X a fixed map for whichTake F ∈ ˆX and Γ ∈ 2 ˆX:¯ɛ ( V(x) ) = xλF (¯ɛ(F) ) ≤ ɛ∀x ∈ X∀F ∈ ˆXδ P(f(¯ɛ(F)), f(¯ɛ(Γ)))≤ δ(¯ɛ(F), ¯ɛ(Γ))= ˆδ(¯ɛ(F), ¯ɛ(Γ))≤ ˆδ ( V(¯ɛ(F)), {F} ) + ˆδ ( F, Γ ) + supF ′ ∈Γˆδ ( F ′ , ɛ X (¯ɛ(Γ)) )From the definition of ¯ɛ it immediately follows that both ˆδ ( V(¯ɛ(F)), {F} )and sup F ′ ∈Γ ˆδ ( F ′ , ɛ X (¯ɛ(Γ)) ) are smaller than or equal to ɛ. Hence <strong>we</strong> findthatδ P(f(¯ɛ(F)), f(¯ɛ(Γ)))≤ ˆδ( F, Γ)+ 2ɛ

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