Finding Limit Points? I have two sets:
$ A = ({{ (-1)^n + 2/n : n = 1, 2, 3, ...}}) $ and 
$ B = ( x \in \mathbb{Q} : 0 < x < 1 ) $
How does one go about finding the limit points for these sets? Would I just do 
$ \lim_{n\to\infty} $ for each set? Could someone clarify what exactly limit points are, and how they relate to closed sets? Also, how do I know if a set given like this is open or closed? How do I find the isolated points? What about the closure? I don't necessarily want the answers, but rather the method to finding these. I can try them and post what answers I get later so they can be checked by the community! 
 A: A limit point is a point of a set $S$, is a point $x$, which may or may not be an element of the set $S$, such that for every possible real number $\epsilon > 0$.  There will exist an element $y \in S$, $y \ne x$ such that the distance between $x$ and $y$ is less than $\epsilon$.
In set $A$, $1$ is a limit point because for every $\epsilon > 0$ I can find an even $n$ so that $0 < 2/n < \epsilon$ (by finding and even $n > 2/\epsilon$) So $(-1)^n + 2/n = 1 + 2/n \in A$ and the distance between $1$ and $1 + 2/n$ is $2/n < \epsilon$.  I can find such an $n$ for every possible epsilon.
$-1$ is also a limit point because for every $\epsilon > 0$ I can find an odd $n$ so the $0 < 2/n < \epsilon$ (by finding an odd $n > 2/\epsilon$) so $(-1)^n + 2/n = -1 + 2/n \in A$ and the distance between $-1$ and $-1 + 2/n$ is $2/n < \epsilon$.
Furthermore these are the only two limit points.  This is a bit harder to show.  Basically if $x$ is between $(-1)^{n-1}+2/(n-1)$ and $(-1)^{n+1} + 2/(n+1)$ inclusively there are no other points of $A$ in that interval and we can find a very small distance that has no other points in it.  If $x$ is out of that range we can find a distance with no points in that range.
For set $B$, $0$, $1$ and every value in between are limit points.  For any distance $\epsilon > 0$ and any $x \in [0,1]$ we can find rational numbers in $(x-\epsilon, x + \epsilon)$ that are between $0$ and $1$ exclusively.  $[0,1]$ are the only numbers where that is true.
How do they relate to closed sets?  The definition of a closed set is that all the limit points are in $S$.  As $1 \not \in A$, $-1 \not \in A$, $0 \not \in B$,, and $1 \not \in B$, neither $A$ nor $B$ are closed.
How do you find the isolated points?  Okay, all the points in A are isolated and none of the points in B are.  An isolate point of $S$, is a point $x \in X$ is one in which there exists a distance $\epsilon > 0$ so that there are no other point of $S$ within that distance from $x$.  This is the exact opposite of a limit point.  In $A$ all points are isolated and are not limit points.  In $B$, none of the points are isolated as far any distance from any point, there will be a rational number in the set.  So no point in $B$ is isolated and in fact all points in $B$ are limit points of $B$.  (But $B$ is not closed because there are other limit points that are not in $B$).
So what is a closure?  The closure of $S$ is the smallest closed set containing $S$.  In other words $\overline{S}$ is the union of $S$ with its limit points.  (Because to be closed means that it contains its limit points.
So $\overline{A} = A \cup $ limit points of $A = A \cup {0} \cup {1}$.
$\overline{B} = B \cup $ limit points of $B = \{x \in \mathbb Q| 0< x < 1\} \cup [0,1] = [0,1]$. (Note.  This includes all the irrationals as well as just the rationals.  The irrationals between 0 and 1 are limit points as well.)
