# Closure of Integer Reciprocal Space

## Theorem

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:

$A := \set {\dfrac 1 n: n \in \Z_{>0} }$

Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.

Then:

$A^- = A \cup \set 0$

where $A^-$ denotes the closure of $A$ in $\R$.

## Proof

By definition, the closure of $A$ is:

$A \cup A'$

where $A'$ is the derived set of $A$.

By definition of derived set, $A'$ consists of all the limit points of $A$ in $\R$.

From Zero is Limit Point of Integer Reciprocal Space, the only limit point of $A$ is $0$.

Hence the result:

$A^- = A \cup \set 0$

$\blacksquare$