TITLE: The set of integers is not open or is open
QUESTION [9 upvotes]: Baby Rudin gives the example of the set of all integers being not open if it is a subset of $\mathbb{R}^2$.
If we consider the set of integers in $\mathbb{R}$, is this set also not open? I can find a neighbourhood which will contain any point, $p$, however is it a requirement that a neighbourhood contains more than one point?
I'm trying to understand this fully and have searched through the various posts that have a slight relation and can not find out specifically how these take interior and isolated points into account and how these relate to openess.

REPLY [14 votes]: A set $U\subset \mathbb R$ is open if and only if for every $x\in U$, there exists some $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon)$ is a subset of $U$.
For $U=\mathbb Z$, this is clearly not the case:

Take $x=0$
Take any $\epsilon > 0$.
Then, $\min\{x+\frac\epsilon2, x+\frac12\}$ is an element of $(x-\epsilon, x+\epsilon)$, but it is not an element of $\mathbb Z$. 
Therefore, $(x-\epsilon, x+\epsilon)$ is not a subset of $\mathbb Z$ for any value of $\epsilon$
Therefore, $\mathbb Z$ is not open.