7/25/2019 sec41no7b_scibd

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(b) We will show that if A is an open cover ofX, there exists a -locally finite open refinement D ofA

that covers X. Since Xis regular, the equivalence of (1) and (4) in Lemma 41.3 then implies thatX is paracompact.

Suppose X=nZ+

IntXn, where Xn is a closed subset ofXwhich is paracompact in the subspacetopology, for every nZ+. Since A is an open cover ofX, for every nZ+

An

U Xn : U A

is an open cover ofXn. Since Xn is paracompact, there exists a refinement Bn of An that covers Xn,is locally finite in Xn, and is open in Xn. For each D Bn in Xn, choose an open subset V(D) Xsuch that

D= V(D) Xn.

LetDn=

V(D)IntXn :D Bn

nZ+ and D=

nZ+

Dn.

All elements of D are open in X. Since Bn refines An (and thus A) and

V(D) IntXn V(D) Xn= D D B n,

Dn refines A. Since Bn covers Xn and X=nZ+

IntXn,

nZ+

DBn

V(D) IntXn= nZ+

DBn

V(D) IntXn nZ+

DBn

D IntXn=

nZ+

Xn IntXn

=

nZ+

Xn= X.

Thus, D is an open refinement of A that covers X. By the last paragraph of part (a), the collectionBn is locally finite in X. Since

V(D) IntXn V(D) Xn= D D B n,

the collection Dn is also locally finite in X. Thus, D is -locally finite.

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solution, sec 41 no 7b, munkres, topology, 2e