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Hi, I read the following theorem in a book:
If {[tex]A_n[/tex]} is a sequence of nowhere dense sets in a complete metric space X, then there exists a point in X which is not in any of the [tex]A_n[/tex]'s.
But what if I say X={1,1/2,1/3, ...} [tex]\cup[/tex] {0} with the regular metric d(x,y)=|x-y|, and [tex]A_n[/tex]={1/n,0}.
Why wouldn't this be a counterexample?
As I see it, X is a complete metric space, each [tex]A_n[/tex] is nowhere dense, and their union equals X.
If {[tex]A_n[/tex]} is a sequence of nowhere dense sets in a complete metric space X, then there exists a point in X which is not in any of the [tex]A_n[/tex]'s.
But what if I say X={1,1/2,1/3, ...} [tex]\cup[/tex] {0} with the regular metric d(x,y)=|x-y|, and [tex]A_n[/tex]={1/n,0}.
Why wouldn't this be a counterexample?
As I see it, X is a complete metric space, each [tex]A_n[/tex] is nowhere dense, and their union equals X.