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All spaces that have the cofinite topology are sequentially compact

  • #1
i want to show that given any space X with the cofinite topology, the space X is sequentially compact.

i have already shown that any space X with the cofinite topology is compact since any open cover has a finite subcover on X. i know that if we are dealing with metric spaces, then the notions of compactness and sequentially compactness are equivalent but in general topological spaces it is not true.

so the cofinite topology defines all the open sets as subsets of X whose complement is finite. and a sequentially compact set is a set which has the property that all sequences in the set have a convergent subsequence in the set.

these are the definitions i have but i am not sure how to use them together to show that the set X is sequentially compact. can someone give me some pointers in the right direction? thanks.
 

Answers and Replies

  • #2
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Think about whether the sequence takes only finitely many or infinitely many distinct values.

Also, remember that a sequence can converge to more than one point in sufficiently bad topological spaces.
 
  • #3
doesn't that depend on whether X is finite or infinite? if X is finite then the sequence will just be finitely many terms and i'm not sure how to talk about convergence in that case but if X is infinite then a sequence will have infinitely many terms right?
 
  • #4
352
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doesn't that depend on whether X is finite or infinite? if X is finite then the sequence will just be finitely many terms and i'm not sure how to talk about convergence in that case but if X is infinite then a sequence will have infinitely many terms right?
A sequence in [tex]X[/tex] is a function [tex]n \mapsto x_n[/tex] from [tex]\mathbb{N}[/tex] to [tex]X[/tex]; it has infinitely many terms regardless of the cardinality of [tex]X[/tex]. The range or image of the sequence is the set [tex]\{x_n \mid n\in\mathbb{N}\}[/tex] of values taken by the sequence; it is a subset of [tex]X[/tex].

If [tex]X[/tex] is a finite set, then the range (set of values) of any sequence in [tex]X[/tex] must be finite, but if [tex]X[/tex] is infinite, the range of a sequence in [tex]X[/tex] can be finite or infinite. An example of a sequence in [tex]\mathbb{R}[/tex] with infinite range is [tex](x_n)[/tex] where [tex]x_n = n[/tex]; an example with finite range is [tex](y_n)[/tex] where [tex]y_n = (-1)^n[/tex].

My hint above was meant to tell you that you should try to construct your convergent subsequence differently depending on whether the range of your sequence is finite or infinite.
 
  • #5
so if a sequence in X has a finite range, does it have to be in one of the closed sets? if it is then i can take away finitely many terms so that the subsequence will converge. if the sequence has an infinite range then would it be contained in one of the open sets? based on that i don't know if it has a convergent subsequence. i'm still unsure on how to continue. is there still something i may be missing?
 
  • #6
in order to talk about the convergence of a sequence don't you need a metric space? or at least the only definition of convergence of sequences that i know uses a distance in its definition. if i'm trying to prove this fact for all topological spaces in general how would i go about this?
 

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