Cofinite topology is compact

demonelite123 said:

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.