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In case of an infinite set $S = \{(-1)^n /\; n \in N\}$, it has two limit points -1 and +1.

Similarly, in case of a finite set $S = \{(-1)^n /\; n \in N \;\&\; n<5\}$ shouldn't it have limit points +1 and -1, because neighborhood of every point in $S$ has at least one point other than itself?

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    finite set is given in second line, please see below infinite set example. – 2017-02-08
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    And I quote "In case of an infinite set $S=\{(-1)^n:n\in\Bbb N\}$". Who says this? Well, in this case, we get $S=\{-1,1\}$ which is finite. – 2017-02-08
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    "Because every neighborhood of every point in S has at least one point other than it". Uh, no they don't. The points are -1,1/2,-1/3,1/4, and -1/5. Take the neighborhoods with radius .000000001. Those neighbors don't have any other points of S in them. – 2017-02-08
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    so infinite set should have infinite number of **distinct** elements? Also in definition of limit points does _other than itself_ means distinct element? please clarify. – 2017-02-08
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    "It has two limit points -1 and +1." Uh, no it doesn't. It has one limit point; 0. – 2017-02-08
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    By the way, are you referring your first as $$S=\{(-1)^n:n\in\Bbb N\}$$ or $$\left\{\frac{(-1)^n}{n}:n\in\Bbb N\right\}?$$ – 2017-02-08
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    What do you think limit point means. Why would every neiborhood of ever point having a point other than itself (which doesn't happen btw) mean that the limit points are 1 and -1? – 2017-02-08
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    @fleablood I'm considering every element distinct here even though they have same value (e.g. -1^2, -1^4 are distinct from each other). – 2017-02-08
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    If the have the same values they aren't distinct. Why on earth did you think you could count them distinctly? *sheesh* So those weren't fractions? Then those two sets are both finite with exactly two elements. And no limit points. – 2017-02-08
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    Thank you sir. I was looking at them as abstract elements not as numbers. – 2017-02-08

3 Answers 3

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Your post was even wronger than I thought. $S=\{(-1)^n|n\in \mathbb N\} $ is not an infinite set and it doesn't have any limit points. $S=\{-1,1\} $. You absolutely can NOT treat $(-1)^3$ and $(-1)^5$ as distinct elements. $(-1)^3=(-1)^5=-1$ so they are one element; just two notations for the same element. Neither $1$ nor $-1$ is a limit point as $(.9,1.1) $ has no point of $S $ other than $1$ and same is true for $(-1.1,-.9) $ and $-1$.

Meanwhile $\{(-1)^n|n\in\mathbb N\}=\{(-1)^n|n\in \mathbb N, n\le 5\}=\{-1,1\} $.

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Everything you say in your post is wrong.

$S=\{\frac {(-1)^n}n|n\in \mathbb N\} $ has only one limit point. That limit point is $0$ because $0$ is the only point in which every neighborhood contains a point of $S $ that isn't the point itself.

If $T=\{\frac {(-1)^n}n|n\le 5\} $ has no limit points. There is no point $x $ in $\mathbb R $ that every neighborhood has a point of $T $ that isn't $x$.

Let $x$ be a real number. Let $\epsilon= \min\{|x- y|: y\in T, y\ne x\}>0$. The neighborhood $\{z:|x-z|<\epsilon\} $ has no points of $T $ other than (possibly) $x$. If $z$ is in the neighborhood than $|x-z|< \min\{|x- y|: y\in T, y\ne x\}$. But if $z \in T$ and $z\ne x $, then $|x-z| \ge \min\{|x- y|: y\in T, y\ne x\}$ so that's impossible.

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Not at all. The point $-1$ has no neighbor in a neighborhood of radius $\frac12$.

In your definition you seem to be missing an important point: every neighborhood of every point has at lest a neighbor. If you drop this every, and assume that a single neighborhood suffices, the definition makes no sense: in a set of at least two points, every point has a neighbor in some neighborhood.

Actually, no finite set can have an accumulation point, because for points to be arbitrarily close you need an infinity of them.

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It's unclear what you're asking for. Every non-empty subset of a topological space has limit points since every point in the set is a limit point. The question is whether there are limit points outside the set, which they don't in the case of $\mathbb R$, if the set is being finite there's a minimum distance to the points being larger than $0$ - so given a point outside the set there's a neighborhood of it not intersecting the set.