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?
Best Answer
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 $\mathbb R$.