Zero is Omega-Accumulation 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 an $\omega$-accumulation 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:
 * $I := \left({- a \,.\,.\, b}\right) \subseteq U$

By the Archimedean Principle:
 * $\exists n \in \N: n > \dfrac 1 b$

and so:
 * $\exists n \in \N: \dfrac 1 n < b$

Let:
 * $M := \left\{{m \in \N: m \ge n}\right\}$

Then:
 * $\forall m \in M: 0 < \dfrac 1 m < b$

Thus:
 * $\forall m \in \N, m \ge n: \dfrac 1 m \in I \cap B$

Thus an open set $U$ which contains $0$ contains a countably infinite number of elements of $B$ (distinct from $0$).

Thus, by definition, $0$ is an $\omega$-accumulation point of $B$ in $\R$.