limit point of the set of integers What are the limit point of the set of integers? I know that the closure of this set is the set of integers, so is this set of limit points empty?
 A: The confusion here is that the definition of a limit point $x$ in a set $X$ requires that every neighborhood of $x$ contains a point of $X$ different from $x$. A limit point is also known as an accumulation point.
http://en.wikipedia.org/wiki/Limit_point
This is in contrast to the definition of an adherent point, also known as a contact point, which is a point whose every neighborhood intersects $X$. 
http://en.wikipedia.org/wiki/Adherent_point
If a set is closed, then every one of its points are adherent points; but not necessarily limit points. For example the set $[0, 1] \cup \{2\}$ is a closed set in $\mathbb{R}$. Every point is an adherent point, but $2$ is not a limit point.
This terminology a common point of confusion. 
To answer the original question, the integers have no limit points in the reals, since all integers are isolated; that is, each integer has a neighborhood that does not contain any other integers.
A: I'm going to assume your space is $\Bbb{R}$ with the standard topology. You claimed that $\overline{\Bbb{Z}} = \Bbb{Z}$ (which is correct, assuming you are indeed using the standard topology.) Since $\overline{\Bbb{Z}} = \Bbb{Z}$ we know $\Bbb{Z}$ is closed with respect to $\Bbb{R}$. Hence if $\space \Bbb{Z}$ has a limit point, say $p$, then $p \in \Bbb{Z}$. However, for all $p \in \Bbb{Z}$, you can take the open set $\left(p-\frac{1}{2},p+\frac{1}{2}\right)$ and use the definition of a limit point $$\Bbb{Z}\cap \left[\left(p-\frac{1}{2},p+\frac{1}{2}\right) - \{p\} \right] =\emptyset$$ to see that $\Bbb{Z}$ contains no limit points.
