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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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 Fall '09
 Robinson
 Topology, Empty set, Metric space, Closed set, Bδ, disjoint open subsets

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