Why is the set of all integers not open?

A point $p$ is an interior point of a set $E$ if there exists a neighborhood $N_r(p)$ such that $N_r(p)\subseteq E$, and a set is open if all of its points are interior points.

Now my question is that since $r\in \mathbb{R^+}$ then why can't we take a $p\in \mathbb{Z}$ and let $r=\frac{3}{2}$ so that $N_{3/2}(p)=\{p-1,p,p+1\}$, which is a subset of $\mathbb{Z}$? Then all points of $\mathbb{Z}$ would be interior points.

• You are correct, $\mathbb Z$ is open as a subset of $\mathbb Z$. It is not open in the real line, because $\mathbb p-\frac{1}{2}$ is in that neighborhood, too. – Thomas Andrews Jul 18 '13 at 16:24

Openness of a set is only defined in terms of as a subset of some parent set.

It is true that $\mathbb Z$ is open as a subset of $\mathbb Z$ under the usual metric. But that's hardly surprising, since any set is open as a subset of itself. Indeed, you need not use $r=3/2$, you could just use $r=1/2$ in this case, and then $N_r(p)=\{p\}$ is a subset of $\mathbb Z$.

When considering $\mathbb Z$ as a subset of $\mathbb R$, however, $\mathbb Z$ is not open. You can find non-integers in any neighborhood of a point $p\in\mathbb Z$.

• Very clear. Thank you!! – Melanie Jul 18 '13 at 20:01

When I learn the analysis, the definitions are:

Set $\mathbb{E}$ is close if every limit point of $\mathbb{E}$ is a point of $\mathbb{E}$.

A point $p$ is a limit point of set $\mathbb{E}$ if every neighborhood of $p$ contains a point $q\neq p$ such that $q \in \mathbb{E}$.

So, I also got confused when I study it from the definitions above. But actually I found another question about the limit point in set of integers, which clear the confusion.

limit point of the set of integers

I quote here: "If a set is closed, then every one of its points are adherent points; but not necessarily limit points. "