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.
$\endgroup$ 34 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$$ 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 $S$.
$\endgroup$ $\begingroup$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 $\mathbb{R}.$
$\endgroup$ 2 $\begingroup$- The set $\mathbb{N}^c$ can be written as a union of open intervals:$(-\infty, 1), (1, 2), (2, 3), \dots, (n, n+1), \dots$.
- Each open interval is an open set.
- The arbitrary union of open sets is open.
- Thus, $\mathbb{N}^c$ is open.
- Thus, $\mathbb{N}$ is closed.
