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# HW04 - Modern Analysis 1 Homework 04 1 For x ∈ M and A...

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Unformatted text preview: Modern Analysis 1 Homework 04 1. For x ∈ M and A ⊂ M deﬁne d(x, A) = inf {d(x, a) : a ∈ A}. Show that if δ > 0 then Bδ (A) = {x ∈ M : d(x, A) < δ } is an open subset of M . 2. Let the compact subset K ⊂ M and the closed subset F ⊂ M be disjoint. Prove that there exist disjoint open subsets U ⊂ M containing K and V ⊂ M containing F . Solutions (1) Let p ∈ Bδ (A): by deﬁnition, d(p, A) < δ so δ is not a lower bound for {d(p, a) : a ∈ A}; choose a ∈ A with d(p, a) < δ and set ε = δ − d(p, a). If x ∈ Bε (p) then the triangle inequality yields d(x, a) d(x, p) + d(p, a) < (δ − d(p, a)) + d(p, a) = δ ; thus d(x, A) < δ and so x ∈ Bδ (A). This proves Bε (p) ⊂ Bδ (A). (2) For each x ∈ M − F we may choose δx > 0 such that Bδx (x) ⊂ M − F and for each y ∈ M − K we may choose εy > 0 such that Bεy (y ) ⊂ M − K ; in particular, these apply when x ∈ K and when y ∈ F . Now let U= Bδx /2 (x) x∈ K and V= Bεy /2 (y ). y ∈F Plainly, U is an open set containing K and V an open set containing F . Moreover, U ∩ V = ∅. Let z ∈ U ∩ V : if z ∈ Bδx /2 (x) ∩ Bεy /2 (y ) then the ∆ inequality implies that d(x, y ) < (δx + εy )/2 δx ∨ ε y whence either Bδx (x) meets F or Bεy (y ) meets K (depending on which of δx and εy is the greater); this is contrary to choice. 1 Remark : This proof applies when K and F are closed. In case K is actually compact, one can proceed diﬀerently: for each x ∈ K let Ux be the open ball with centre x and radius d(x, F )/2 > 0; extract a ﬁnite subcover {Ux1 , . . . , Uxn } of K and put U = Bd(x1 ,F )/2 (x1 ) ∪ · · · ∪ Bd(xn ,F )/2 (xn ) V = Bd(x1 ,F )/2 (F ) ∩ · · · ∩ Bd(xn ,F )/2 (F ). Indeed, one can go further still and show (conveniently using sequences) that δ := inf {d(x, y ) : x ∈ K, y ∈ F } > 0 whence we may simply take U = Bδ/2 (K ) and V = Bδ/2 (F ). 2 ...
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HW04 - Modern Analysis 1 Homework 04 1 For x ∈ M and A...

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