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I've recently learnt the definition of limit point in the context of metric spaces. I'm comfortable with the definition geometrically, but am having trouble showing that $$A=(-1)^{n}, n \in \mathbb{N} $$

Doesn't have any limit points.

I initally thought it would do, because for example, could I not take $$U(-1,4)$$ where $$U(a,r)$$ denotes the open ball center a radius r, r>0. Then under the usual metric this would give an open interval from (-5, 3). Then looking at all points in A not including -1, we only have the point 1. But the point 1 is also in the open interval, so the intersection is non-empty...

I'm clearly misunderstanding the concept.

Any help appreciated,

Thanks.

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    Any two points in $A$ has a minimal distance separating them. The radius of $U$ be less then half this minimal distance. – 2017-02-21
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    Should there not be exactly two limit points -- $1$ and $-1$? You can certainly find infinitely many points that are arbitrarily close (read: exactly equal) to each. – 2017-02-21
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    Interestingly, the accepted answer confuses the notions of limit point of *a set* and limit point of a *sequence* and is, as a consequence, quite wrong. This confusion is already present in the question itself. Consulting [WP on the subject](https://en.wikipedia.org/wiki/Limit_point#Types_of_limit_points) might help. – 2017-05-10

2 Answers 2

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The range of the sequence is the set E={1,-1} and it does not converge to some point. 1 and -1 are not limit point because the ball $U(1,1/4)$ does not contain a point of E different from 1, and the ball $U(-1,1/4)$ does no contain a point of $E$ different from -1. So the sequence have no Limits Points.

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You need to show that it holds for all balls centered at the supposed limit point, not just one.

For this problem, you must let $x$ be a real number, and find a deleted ball around it which does not intersect $A$.