Geometry And Topology

A second course in general topology by Heikki Junnila

By Heikki Junnila

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Since G consists of τd - continuous functions, we have that τ ⊂ τd . To show that τd ⊂ τ , let x ∈ X and ǫ > 0. Choose n ∈ N so that 2 n < ǫ and choose i ∈ In so that fi (x) > 0. Let z ∈ X be such that Supp(fi ) ⊂ Bd (z, n1 ). Then we have that x ∈ Bd (z, n1 ) and hence that Bd (z, n1 ) ⊂ Bd (x, n2 ). It follows from the foregoing that we habve x ∈ Supp(gi ) ⊂ Bd (x, ǫ) for the the τ -open set Supp(gi ). We have shown that τd ⊂ τ . Sufficiency. We assume that the topology τ of a space X is the weak topology induced by a partition of unity G = {gi : i ∈ I}.

Proof. To show that θ is lsc, let G ⊂◦ Y and let x be a point of the set {z ∈ X : θ(z) ∩ G = ∅}. Let u ∈ θ(x) ∩ G. Then u ∈ Bd (ϕ(x), r) and hence there exists δ > 0 such that ϕ(x) ∩ Bd (u, r − δ) = ∅. Since ϕ is lsc, the set V = {z ∈ X : ϕ(z) ∩ Bd (u, r − δ) = ∅} is a nbhd of x. We also have that u ∈ ψ(x), and it follows, since ψ is lsc, that the set W = {z ∈ X : ψ(z) ∩ G ∩ Bd (u, δ)} is a nbhd of x. We show that V ∩ W ⊂ {z ∈ X : θ(z) ∩ G = ∅}. Let z ∈ V ∩ W . Since z ∈ V , there exists v ∈ ψ(z) ∩ G such that d(v, u) < δ.

Assume on the contrary that there exist α ∈ A fα (z) > 1 4 h(x) 1 h(z), 2 fαx (z) > > fα (z) > 1 2 h(z) 3 h(x) 4 B and z ∈ Uα ∩ V ∩ W . Then we have that and fα (z) < and h(z) ≥ fαx (z) > 1 h(x). 4 3 4 h(x), As a consequence, we have that but this is a a contradiction. It follows from the foregoing that U is locally finite (as an indexed family). For every α ∈ A, the function kα = 0 ∨ (fα − 12 h) is continuous and Supp(kα ) = Uα . Hence the collection {kα : α ∈ A} is locally finitely supported, and it follows that the function k = α∈A kα is continuous.

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