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I figured someone would have asked the question here, but I could not find it.

I know it is not open, because $ \forall n \in \mathbb{N}$, $V_\epsilon (n) \notin \mathbb{N}$. In other words, it is made up of a bunch of isolated points.

But I keep reading that it is closed, and I'm having trouble thinking about why, except that perhaps the complement is open and thus $\mathbb{N}$ is closed? Or is it closed vacuously like $\mathbb{Z}$, it contains all its limit points because it has no limit points.

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    $\begingroup$ You are right: the complement of $\mathbb{N}$ in $\mathbb{R}$ is open, hence, by definition, $\mathbb{N}$ is a closed set. $\endgroup$ – Catalin Zara Apr 3 '16 at 2:10
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    $\begingroup$ The terms "open" and "closed" are not absolute terms, one has to reference a topology (a system of open sets). In the (usual) relative topology of $\Bbb N$ in $\Bbb R$, the set $\Bbb N$ is open (since it is $\Bbb R \cap \Bbb N$, and $\Bbb R$ is open), in the (usual) topology of $\Bbb R$ it is closed (since its complement is a union of open intervals, which is open). $\endgroup$ – David Wheeler Apr 3 '16 at 2:14
  • $\begingroup$ ^perhaps you meant N is open in N? (using the inheritance property what you have said was true). In R, N should not be open since no neighborhood of maximal distance r around any natural number should have only natural numbers in it (i.e. take r = 0.5) $\endgroup$ – Charlie Tian Sep 21 '17 at 16:25
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$ N$ is closed for either, or both, reasons.(1).Each interval $(n-1,n)$ is open , and $(-\infty,0)$ is open, so $\mathcal R\backslash N=(-\infty,0)\cup (\cup_{n\in N}(n-1,n)$, a union of open sets, is open. (2) A subset of $\mathcal R$ with no limit points is a closed set. Note that the sentence $\forall x\;(( x$ is a limit point of $ S)\implies x\in S)$ means "No $x$ can be a limit point of $S$ without belonging to $S$", which is certainly true if there aren't any limit points of $S$.

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It is closed vacuously.

Another reasoning can be derived as follows: If $x \in \mathbb{N}^c$, then we can let $r = \min(x - \lfloor x \rfloor, \lceil x \rceil - x)$, then $B(x;r)$ is contained in $\mathbb{N}^c$, and hence $\mathbb{N}^c$ is open.

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Let $\mathbb{N}$ be the set of natural numbers. $\overline{\mathbb{N} }=\mathbb{N} \cup \partial \mathbb{N} .$ Clearly, $\partial{N}=\emptyset .$ Hence, $\overline{\mathbb N}=N .$ This means that the set of natural numbers are closed in the usual topology in $\mathbb{R}.$

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