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

Theorem
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
 * $A := \left\{{\dfrac 1 n : n \in \Z_{>0}}\right\}$

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

Let $B$ be the uncountable set:
 * $B := A \cup \left[{2 \,.\,.\, 3}\right]$

where $\left[{2 \,.\,.\, 3}\right]$ 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 $I$ be an open set of $\R$ which contains $0$.

Then $I$ is a of the form:
 * $I := \left({- a \,.\,.\, b}\right) \subseteq \R$

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 real interval $I$ which contains $0$ contains at least one element of $B$ (distinct from $0$).

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