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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?

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    $\begingroup$ 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). $\endgroup$ Nov 27 '21 at 17:42
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    $\begingroup$ 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 $\endgroup$
    – Henry
    Nov 27 '21 at 17:44
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    $\begingroup$ 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 $\mathbb R$. $\endgroup$
    – Lee Mosher
    Nov 27 '21 at 17:45
  • $\begingroup$ @LeeMosher Of course, I forgot that in the definition of an open ball the specific metric is included. Thank you, now I understood it! $\endgroup$
    – nipohc88
    Nov 27 '21 at 17:52
  • $\begingroup$ 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. $\endgroup$
    – Henry
    Nov 27 '21 at 17:53
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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 $\mathbb R$.

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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 $X$.

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