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Answers, Assignment #10Exercises 1.41.(a) True. Ifs∈intS,thensdoes not belong tobdS.(b) False. LetS= (0,1). ThenbdS={0,1}which is not a subset ofS.(c) False.S= [0,1], bdS={0,1},S= bdS.(d) True. Supposes∈S. Ifs /∈intS,then every neighborhood ofsmust intersectScwhich impliess∈bdS.(e) True.s∈bdSimpliesN∩S=∅andN∩Sc=∅for every neighborhoodNofs.t∈bdScimpliesN∩Sc=∅andN∩(Sc)c=N∩S=∅for every neighborhoodNoft. ThereforebdS= bdSc.(f) False.S= [0,1], bdS={0,1} ⊂S.2.(a) True.Letz∈N(x,).Letδ= min{|z-x|,|z-(x+)|,|z-(x-)|}.ThenN(z, δ)⊂N(x,).(b) True. Theorem 10 (a).(c) False.∞n=11n,1-1n= (0,1).(d) False.∞n=1(-1n,1 +1n)= [0,1].(e) True. LetTbe a collection of closed sets. Then[∩TT∈T]c=∪Tct∈T.Tclosed impliesTcis open and the union of a collection of open sets is open. Therefore,[∩TT∈T]cisopen and∩TT∈Tis closed.(f) False. The set of real numbers is both open and closed.
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Fall '08
Staff
Topology, Metric space, Sn, Closed set, General topology, ∩TT ∈T