Set of Reciprocals of Positive Integers is Nowhere Dense in Reals

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Theorem

Let $N$ be the set defined as:

$N := \left\{{\dfrac 1 n: n \in \Z_{>0}}\right\}$

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


Then $N$ is nowhere dense in the set of real numbers $\R$ considered as a metric space.


Proof

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

By definition:

$\displaystyle \R \setminus N = \left({1 \,.\,.\, \infty}\right) \cup \left({\bigcup_{n \mathop \in \N} \left({\frac 1 {n+1} \,.\,.\, \frac 1 n}\right)}\right)$



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 $O1$, 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.

$\blacksquare$


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