Zero is Limit Point 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 $0$ is the only limit point of $A$ in $\R$.

Proof

There are three cases to consider:

Points in $A$

Consider $x \in \R$ such that $x \in A$.

That is, $x = \dfrac 1 n$ for some $n \in \N$.

Let:

$d = \dfrac 1 n - \dfrac 1 {n + 1}$

Consider the open real interval:

$I := \openint {\dfrac 1 n - d} {\dfrac 1 n + d} \subseteq \R$

By definition, $I$ is an open set of $\R$.

Thus $I$ is an open set of $\R$ which contains no element of $A$ distinct from $x$.

Thus $x$ is not a limit point of $A$ in $\R$.

$\Box$

Non-Zero Points not in $A$

Let $x \in \R$ such that $x \ne 0$ and $x \notin A$.

Let:

$d := \min \set {\size {x - m}: m \in A}$

that is, the smallest distance from $x$ to an element of $A$.

Consider the open real interval:

$I := \openint {x - d} {x + d} \subseteq \R$

By definition, $I$ is an open set of $\R$.

Thus $I$ is an open set of $\R$ which contains no element of $A$ (distinct from $x$ or not).

Thus $x$ is not a limit point of $A$ in $\R$.

$\Box$

Zero

Finally, consider the point $0$.

Let $U$ be an open set of $\R$ which contains $0$.

From Open Sets in Real Number Line, there exists an open interval $I$ of the form:

$I := \openint {-a} b \subseteq U$

By the Archimedean Principle:

$\exists n \in \N: n > \dfrac 1 b$

and so:

$\exists n \in \N: \dfrac 1 n < b$

But $\dfrac 1 n \in A$.

Thus an open set $U$ which contains $0$ contains at least one element of $A$ (distinct from $0$).

Thus, by definition, $0$ is a limit point of $A$ in $\R$.

$\Box$

Thus the only limit point of $A$ in $\R$ is $0$.

$\blacksquare$