# 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?

• As a metric space with the usual metric, $\Bbb N$ is discrete. That makes every subset of $\Bbb N$ (of course, including itself) is open (and closed at the same time). Nov 27 '21 at 17:42
• There is a distinction between not open and closed. With the usual $d(m,n)=|m-n|$ metric on $\mathbb N$, any set is both closed and open Nov 27 '21 at 17:44
• 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, e.g. $\mathbb N$, to be an open subset of one space, e.g. of $\mathbb N$ itself, and to be a non-open subset of another space, e.g. of $\mathbb R$. Nov 27 '21 at 17:45
• @LeeMosher Of course, I forgot that in the definition of an open ball the specific metric is included. Thank you, now I understood it! Nov 27 '21 at 17:52
• Meanwhile for any set $S$ and any metric on $S$, both $S$ and $\emptyset$ are both open and closed. For example on $\mathbb R$ and the usual metric, $\mathbb R$ and and $\emptyset$ are the only sets both open and closed; $\mathbb N$ and the interval $[0,1]$ are closed but not open; $\mathbb N^c$ and the interval $(0,1)$ are open but not closed; the interval $(0,1]$ is neither open nor closed. Nov 27 '21 at 17:53

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 $$\mathbb R$$.

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 $$X$$.