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 [?], we 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 ?? we have the implication.For a complemented space we trivially find that each limit operator ofa Cauchy filter is supertight and hence by lemma ?? we 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 we have an approach frame associated with an approach space, thenξ f (g) = sup x∈X(g(x) − f(x)). We will use this 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 we prove that ˆf is an element of R ˆX, we have an isomorphism betweenRX 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 we findthatδ P(f(¯ɛ(F)), f(¯ɛ(Γ)))≤ ˆδ( F, Γ)+ 2ɛ
SOBER APPROACH SPACES 11We can now start the calculation of δ P( ˆf(F), ˆf(Γ)).δ P( ˆf(F), ˆf(Γ))≤ δP( ˆf(F), {f(¯ɛ(F))})+ δP(f(¯ɛ(F)), ˆf(Γ))( ) ( ) (≤ δ P ˆf(F), {f(¯ɛ(F))} + δP f(¯ɛ(F)), f(¯ɛ(Γ)) + sup δ P f(¯ɛ(F ′ )), ˆf(Γ) )F ′ ∈ΓWe also have that( ) ( )δ P ˆf(F), {f(¯ɛ(F))} = ˆf(F) − f(¯ɛ(F)) ∨ 0( ( ) )= sup f(x) − λF(x) − f(¯ɛ(F)) ∨ 0x∈X( ( ) )= sup f(x) − f(¯ɛ(F)) − λF(x) ∨ 0x∈X( ( ) )≤ δ(x, {¯ɛ(F)}) − λF(x) ∨ 0supx∈X≤ λF(¯ɛ(F)) ≤ ɛand(sup δ P f(¯ɛ(F ′ )), ˆf(Γ) ) = sup inf δ (F ′ ∈ΓF ′ ∈Γ F ′′ P f(¯ɛ(F ′ )), ˆf(F ′′ ) )∈Γ(= sup inf f(¯ɛ(F ′ ()) − sup f(x) − λF ′′ (x) )) ∨ 0F ′ ∈Γ F ′′ ∈Γ= sup inf infF ′′ ∈ΓF ′ ∈Γx∈X≤ sup inf infF ′′ ∈ΓF ′ ∈Γx∈Xx∈X= sup inf λF ′′ (¯ɛ(F ′ )) ≤ ɛF ′′ ∈ΓF ′ ∈ΓSo we finally find thatand hence ˆf is a contraction.()f(¯ɛ(F ′ )) − f(x) + λF ′′ (x)∨ 0(δ (¯ɛ(F ′ ), {x} ) + δ ( x, {¯ɛ(F ′ )} ) )+ λF ′′ (¯ɛ(F ′ )) ∨ 0δ P( ˆf(F), ˆf(Γ))≤ ɛ + ˆδ( F, Γ)+ 2ɛ + ɛ5.19. Corollary. (1) Take X metric, ΣRX is the completion of X.(2) Uniform approach spaces are sober iff they are complete and T 0 .Proof. (1) In a metric space the supertight maps can be expressed aslim n→∞ d(·, x n ) with (x n ) n a Cauchy sequence.(2) All uniform spaces are complemented.□5.20. Remark. There exist complemented approach spaces that are notuniform. Take for example all Hausdorff topological spaces that are notcompletely regular.6. SpatialityRemark that in frame theory spatiality can be defined in the same way aswe defined it for approach frames. Hence it seems logical that we can findresults analogous to the classical theory. The first two results are examplesof this.□
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