Unformatted text preview: You can assume that this gives a well-deﬁned topology. (a) Show that the standard topology on R 2 n ∼ = C n is ﬁner than the Zariski topology. ( Note: By “ﬁner”, I always mean “at least as ﬁne”; see the Deﬁnition in Munkres, p. 77.) (b) Consider the subset A ⊂ C n consisting of all points of the type ( s, , . . . , 0) with s ∈ Z . Prove that A is not closed in the Zariski topology. Conclude that the standard topology is strictly ﬁner than the Zariski topology. Challenge (extra credit): Munkres, Ch.2 § 17, exercise 21 on p. 102....
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- Spring '05
- Math, Empty set, Munkres, Zariski topology