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In a classic proof that the set of sequence points together with its limit is compact, we use the statement that any neighbourhood of a limit point contains infinite number of sequence points, hence we need only finite number of open sets to cover the rest of the sequence. That is, we state,, that countable minus countable is finite. On the other hand, if we subtract from countable set of natural numbers all odd numbers, then the set of all even numbers is countable. Thus, countable minus countable is countable. How to deal with cardinality in this case? What am I missing? Marina

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Any neighborhood of a limit point contains all but finitely many of the points of the sequence. That is a stronger condition than containing infinitely many points of the sequence, which is what makes an accumulation point. Once you say all but finitely many it is clear that we need only finitely many open sets to cover the rest. You are correct that deleting a countable infinity from a countably infinite set can leave none, finitely many, or countably infinitely many.

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  • $\begingroup$ Thank you very much, Ross! Could you, please, explain, how did we get " all but finite"? $\endgroup$ – user314849 Dec 2 '16 at 4:21
  • $\begingroup$ That is the definition of a limit point. if the sequence is $a_n$, given any neighborhood you can find an $N$ such that for all $n \gt N, $ the $a_n$ is within the neighborhood. That means there are at most $N$ that are outside. $\endgroup$ – Ross Millikan Dec 2 '16 at 4:54

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