Proof: Hausdorffness is simple, as C(X,Y) inherits a topology at least as fine as that of a subspace under the product topology of YX, which is Hausdorff when Y is. So suppose Y is regular, let f∈C(X,Y), and let K⊆C(X,Y) be a closed subset not containing f. Then there exists compact C1,...,Cn⊆X and open U1,...,Un⊆Y such that f(Ci)⊆Ui and for all g∈K there exists ig such that g(Cig)⊈Uig. Since f(Ci) is compact for each i, by regularity of Y choose neighborhoods Vi of these sets such that ¯Vi⊆Ui. Then ∩B(Ci,Vi) is a hood of f, and since ¯B(Ci,Vi)⊆B(Ci,¯Vi) we observe ¯∩B(Ci,Vi)⊆∩¯B(Ci,Vi)⊆∩B(Ci,Ui) is disjoint from K. ◻
Thursday, January 8, 2015
Hausdorffness and Regularity of Compact-Open C(X,Y) (7.46.6)
James Munkres Topology, chapter 7.46, exercise 6:
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Let C(X,Y) be under the compact-open topology. Show C(X,Y) is Hausdorff if Y is Hausdorff, and regular if Y is regular.
Proof: Hausdorffness is simple, as C(X,Y) inherits a topology at least as fine as that of a subspace under the product topology of YX, which is Hausdorff when Y is. So suppose Y is regular, let f∈C(X,Y), and let K⊆C(X,Y) be a closed subset not containing f. Then there exists compact C1,...,Cn⊆X and open U1,...,Un⊆Y such that f(Ci)⊆Ui and for all g∈K there exists ig such that g(Cig)⊈Uig. Since f(Ci) is compact for each i, by regularity of Y choose neighborhoods Vi of these sets such that ¯Vi⊆Ui. Then ∩B(Ci,Vi) is a hood of f, and since ¯B(Ci,Vi)⊆B(Ci,¯Vi) we observe ¯∩B(Ci,Vi)⊆∩¯B(Ci,Vi)⊆∩B(Ci,Ui) is disjoint from K. ◻
Proof: Hausdorffness is simple, as C(X,Y) inherits a topology at least as fine as that of a subspace under the product topology of YX, which is Hausdorff when Y is. So suppose Y is regular, let f∈C(X,Y), and let K⊆C(X,Y) be a closed subset not containing f. Then there exists compact C1,...,Cn⊆X and open U1,...,Un⊆Y such that f(Ci)⊆Ui and for all g∈K there exists ig such that g(Cig)⊈Uig. Since f(Ci) is compact for each i, by regularity of Y choose neighborhoods Vi of these sets such that ¯Vi⊆Ui. Then ∩B(Ci,Vi) is a hood of f, and since ¯B(Ci,Vi)⊆B(Ci,¯Vi) we observe ¯∩B(Ci,Vi)⊆∩¯B(Ci,Vi)⊆∩B(Ci,Ui) is disjoint from K. ◻
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