Construct a set of real numbers whose limit points comprise the set of integers $\mathbb{Z}$ My thought process is the following: Let $S=\{ m + \frac{1}{n}| m \in \mathbb{Z},n \in N \}$.  Then I need to show that the limit points of $S$ are indeed the integers and that these are the only limit points.  I don't know where to go from here.  
 A: Your example is correct because the $\lim_{n\to \infty} m+\frac{1}{n} = m\in \mathbb{Z}$ for all $m\in \mathbb{Z}$, however your trick here is that you use that $m\in\mathbb{Z}$.  
A: Your example works, and it actually does not contain any integers if you check it closely.  Another trivial example would be the integers themselves.  But for your set, it's not hard to show that any integer $m \in \mathbb{Z}$ is a limit point:
Take any arbitrary neighborhood $N_r(m)$ around $m$ (meaning that $r$ is an arbitrary, positive real number).  By definition, this will be of the form $N_r(m) = \{s \in S : |s - m| < r \}$.  Now find an element of $S$ that is inside of this set.  Thus any neighborhood of $m$ contains an element of $S$, so $m$ is a limit point.
To show that any other point is not a limit point, fix a noninteger point $m + \frac{1}{n} \in S$.  Now find some real number $r > 0$ such that $N_r(m + \frac{1}{n})$ contains no other points of $S$.  Thus some neighborhood of any noninteger point contains no elements of $S$, so no noninteger point is a limit point.
For example, take the number $5 + \frac{1}{10}$.  The closest point in $S$ to this number is $5 + \frac{1}{11}$, and the distance between them is $\frac{1}{10} - \frac{1}{11}$.  So any neighborhood around $5 + \frac{1}{10}$ of radius $r = \frac{1}{10} - \frac{1}{11}$ or less will not include this point.  You can actually take the radius to be $r = \frac{1}{10} - \frac{1}{11}$ because a neighborhood is defined to be all points of distance strictly less than $r$, thus it will not include a point that is exactly distance $r$ away.
