# Zero is Limit Point of Integer Reciprocal Space Union with Closed Interval

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

Let $B$ be the uncountable set:

- $B := A \cup \closedint 2 3$

where $\closedint 2 3$ is a closed interval of $\R$.

$2$ and $3$ are to all intents arbitrary, but convenient.

Then $0$ is a limit point of $B$ in $\R$.

## Proof

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 $\dfrac 1 n \in B$.

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

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

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 32: \ 1 \ \text{(b)}$