# Trouble understanding why natural numbers are not open

general-topologymetric-spacesreal-analysis

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?

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 $$\mathbb R$$.