Finding limit points and isolated points of a set I'm trying to solve the following problem:
find limit points and isolated points of the following set:
$$A=\left\{(x,y)\in\mathbb{E}^2: x=(-1)^n\frac{n}{n+1}, y=3, n\in \mathbb{N}\right\}$$
How should I proceed? I know that limit point is a point, that where if I put a sphere of arbitrary length, I should always get some point that belongs to the set. However, I fail to draw that function. (Still learning functions of two variables). How should I proceed without drawing the function?
Thanks
 A: It seems as though the primary issue here is that of understanding the nature of the set $A$.  Note that we would normally read the definition presented as "$A$ is the set of all points $(x,y)$ (in $\Bbb E^2$) such that $x=(-1)^n\frac{n}{n+1}$ and $y=3$ for some $n \in \Bbb N$."  $\Bbb N$ here means the natural numbers $\Bbb N = \{0,1,2,3,\dots\}$.
Because we have a point for each $n \in \Bbb N$, we can list the points.  We have
$$
n = 0: x = 0,y=3\\
n = 1: x = - \frac 12, y=3\\
n = 2: x = \frac 23, y = 3\\
n = 3: x = - \frac 34,y=3\\
\vdots
$$
In other words, our set contains the points $(0,3),(-1/2,3),(2/3,3),(-3/4,3),$ and so on.  In order to draw this set: make an $x$-axis and $y$-axis, then put a dot at each of the coordinates listed above.  Note that unlike a set associated with a function such as $\{(x,y): y = 2x - 3\}$, the set $A$ will consist of many separate dots instead of a line or a region.
Once you have drawn several of these coordinates for the set $A$, consider the following question: which circles can you draw that will capture infinitely many points? Which circles can you draw that will capture exactly one point?
A: For even $n$, the term $(-1)^n \frac{n}{n+1}$ is just $\frac{n}{n+1}= 1- \frac{1}{n+1}$. For odd $n$ we get the negative $-\frac{n}{n+1} = -1 + \frac{1}{n+1}$. 
All these points are isolated but tend arbitarily close to $1$ (for even large values) and $-1$ (for odd large values).
In your case all these points lie on $y=3$, I just analysed their $x$-coordinates.
Conclusion: all points in the sequence are isolated and the two limit points are $(-1,3)$ and $(1,3)$.
A: HINT.-$$\pm\dfrac{n}{n+1}=\pm\dfrac{1}{1+\dfrac1n}\to \pm1$$ thus $(1,3)$ and $(-1,3)$ are limit points. 
All the other are isolated because of $\left]\dfrac{n}{n+1},\dfrac{2n}{2n+1}\right[$ is a non empty open interval. 
Thus all the $\left(-\dfrac{2n+1}{2n+2},3\right)$ and $\left(\dfrac{2n}{2n+1},3\right)$ are isolated points.
