$(-1)^n=\{-1,1,-1,1,...\}$ Now according to the definition of limit point A point $x$ in $X$ is a limit point or cluster point or accumulation point of a set of $S$ if every neighborhood of $x$ contains at least one point of $S$ different from $x$ itself Now if I take deleted nbd of $1$ I don't have any element in the set. So how $1$ or $-1$ would be its limit point.
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4$\begingroup$ The limit point of a sequence and the limit point of a set are two different concepts. $\endgroup$– jjagmathCommented Sep 28, 2021 at 10:49
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$\begingroup$ Thank you @jjagmath $\endgroup$– VivekCommented Sep 28, 2021 at 10:56
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$\begingroup$ I have downvoted the question because it isn't communicated well in my opinion. $\endgroup$– Adam RubinsonCommented Sep 28, 2021 at 11:21
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$\begingroup$ The equation in line 1 does not make sense. On the LHS you have a number, on the RHS a set of numbers. $\endgroup$– Paul FrostCommented Sep 28, 2021 at 17:35
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1 Answer
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Just as jjagmath says in his comment, the concept of a limit point is different for sequences and sets.
If you were to consider $\{-1,1,-1,...\}$ as a set, then it would actually be $\{-1,1\}$ since sets do not contain the same element more than once (unless it is a multiset). In this case you are correct, this set does not have limit points.
However, if you consider the sequence $-1,1,-1,...$, then a limit point is one that for every neighbourhood around the point, there are infinitely many terms that lie within the neighbourhood. So $-1$ and $1$ are limit points of the sequence because they occur infinitely often.
