Analysis, Limit Points of the alternating sequence (-1)^n 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.
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 $A$.
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.