Integer Reciprocal Space with Zero is Totally Separated

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

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

Then $A$ is totally separated.

Proof
Let $a, b \in A$ such that $a < b$.

From Between Every Two Rationals Exists an Irrational:
 * $\exists \alpha \in \R \setminus \Q: a < \alpha < b$

Because $\forall x \in A: x \in \Q$ it follows that $\alpha \notin A$.

Consider the half-open intervals $S = \left[{0 \,.\,.\, \alpha}\right)$ and $T = \left({\alpha \,.\,.\, 1}\right]$

Let:
 * $U := S \cap A, V := T \cap A$

Let $\beta \in A$.

Then either:
 * $(1): \quad \beta < \alpha$

in which case:
 * $\beta \in U$

or:
 * $(2): \quad \beta > \alpha$

in which case:
 * $\beta \in V$

Thus $U \cup V = A$.

Let $a \in U$.

Then $a < \alpha$ and so $a \notin V$.

Similarly, let $b \in V$.

Then $b > \alpha$ and so $b \notin U$.

Then note that $x \in U$ and $y \in V$.

Thus $U \ne \varnothing$ and $V \ne \varnothing$.

Thus, by definition, $U$ and $V$ constitute a separation of $A$ such that $x \in U$ and $y \in V$.

Hence the result by definition of totally separated.