Trouble understanding why natural numbers are not open I have a real analysis course based on the Principles of Real Analysis by Aliprantis and Birkenshaw. I have done previous analysis courses and know how to show that the set of natural numbers is a closed set.
They define the open ball of radius $r$ in the typical way. However, in the section on metric spaces, the authors define a point $x_{0}$ to be an interior point of a subset $A$ if there exists an open ball $B(x_{0},r)$ such that $B(x_{0},r)\subseteq A$. Here lies my lack of understanding. If we consider $\mathbb{N}$ to be a metric space itself, not a subset of $\mathbb{R}$, with the usual metric, for any $r>0$ the open ball only includes the number itself and other natural numbers. Then clearly, $B(r,x_{0})\subseteq \mathbb{N}$. Therefore, it is an open set.
Where am I making a fallacy?
 A: I'll turn my comment into an answer.
Your fallacy is treating "open set" as an absolute notion, independent of any metric space.
Instead, "open subset" is a notion defined relative to a particular metric space of which the given set is a subset. So it is possible for one and the same set, for example $\mathbb N$, to be an open subset of one space such as $\mathbb N$ itself, and to be a non-open subset of another space such as $\mathbb R$.
A: To expand upon the previous answer, consider the example $x_0 = 3$, $r = 0.5$. If regarded as a subset of $\mathbb N$, then $B(r, x_0) = \{3\}$ is an open subset of $\mathbb N$. But if regarded as a subset of $\mathbb R$, then $B(r, x_0) = (2.5, 3.5)$ which is an uncountable open interval containing all the points between 2.5 and 3.5, and is not a subset of $\mathbb N$.
It is all relative to the parent space. If $X$ is a metric space then
$$B(r, x_0) = \{x \in X : d(x,x_0) < r\}$$
is the definition of the open ball of radius $r$ centered at $x_0$ in $X$. Notice the ball contains all the points from $X$ which are a distance $r$ from $x_0$. So in order for a subset $A\subseteq X$ to have an interior point, it must contain an open ball from $X$.
Any subset can be considered "open" relative to itself (that is, in the inherited topology), but this is different from being an open subset of $X$.
