# Why is the set of integers in $\mathbb{R}$ closed?

A point $p$ is a limit point of the set $E$ if every neighbourhood of $p$ contains a point $q \not= p$ such that $q \in E$

So there are no limit points in $\mathbb{Z} \in \mathbb{R}$

Is that why it's closed?

A point $p$ is an interior point of the set $E$ if there is a neighbourhood $N$ of $p$ such that $N \subset E$

So every point in $\mathbb{Z} \in \mathbb{R}$ is an interior point.

So shouldn't $\mathbb{Z}$ be open as well?

• You have proved that it's closed. But check your logic about interior points again! – Slade May 7 '15 at 3:37
• @Slade Oh right, the neighbourhood N can't be a subset of the Integers because it contains reals – Jack May 7 '15 at 3:51

By definition, a set $F$ is closed in a topological space $X$ if its complement $X\setminus F$ is open. Notice that with the usual topology on $\mathbb{R}$, $\mathbb{R}\setminus\mathbb{Z}=\bigcup_{n\in\mathbb{Z}}(n,n+1)$, which is the union of open intervals (open sets).
Another way to look at it is, as you pointed out, by showing that $\mathbb{Z}$ contains all its limit points in $\mathbb{R}$. Let $\mathbb{Z}'$ be the set af all limit point of $\mathbb{Z}$ in $\mathbb{R}$. As you already noticed, $\mathbb{Z}'=\emptyset\subset\mathbb{Z}$. The conclusion follows from the obvious fact that $\emptyset\subset A$ for any $A\subset\mathbb{R}$.
limit points are used to describe the boundary. The easiest answer is that $\mathbb Z$ is closed in $\mathbb R$ because $\mathbb R \backslash \mathbb Z$ is open.
Note that $\mathbb{Z}$ is a discrete subset of $\mathbb{R}$. Thus every converging sequence of integers is eventually constant, so the limit must be an integer. This shows that $\mathbb{Z}$ contains all of its limit points and is thus closed.
• Do you mean by discrete a subset with pairwise distances greater than a constant $\delta >0\,$? I think the argument works. – Orest Bucicovschi May 7 '15 at 4:19
• @orangeskid: Discrete as having the discrete topology. I.e. every singleton is an open set. The metric that gives this topology, for instance the so called discrete metric, is defined by $d(i,j)=1-\delta_{i,j}$. – T. Eskin May 7 '15 at 4:21