Set of Reciprocals of Positive Integers is Nowhere Dense in Reals

Theorem
Let $N$ be the set defined as:
 * $N := \set {\dfrac 1 n: n \in \Z_{>0} }$

where $\Z_{>0}$ is the set of (strictly) positive integers.

Let $\R$ denote the real number line with the usual (Euclidean) metric.

Then $N$ is nowhere dense in $\R$.

Proof
Consider the set $\R \setminus N$, the relative complement of $N$.

By definition:
 * $\displaystyle \R \setminus N = \openint 1 \to \cup \paren {\bigcup_{n \mathop \in \N} \openint {\frac 1 {n + 1} } {\frac 1 n} }$

That is, $\R \setminus N$ is a (countable) union of open real intervals.

From Open Real Interval is Open Set and its corollary, each one of these is an open set of $\R$.

From open set axiom $\text O 1$, it follows that $\R \setminus N$ is itself an open set of $\R$.

So by definition, $N$ is closed in $\R$.

From Closed Set Equals its Closure:
 * $N = N^-$

where $N^-$ is the closure of $N$ in $\R$.

But $N$ contains no open real intervals and so has no subset which is open in $\R$.

Thus by definition $N$ is nowhere dense.