Integer Reciprocal Space with Zero is Totally Separated

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Theorem

Let $A \subseteq \R$ be the set of all points on $\R$ defined as:

$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$

Let $\struct {A, \tau_d}$ 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 two Rational Numbers exists Irrational Number:

$\exists \alpha \in \R \setminus \Q: a < \alpha < b$

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


Consider the half-open intervals $S = \hointr 0 \alpha$ and $T = \hointl \alpha 1$

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 $U \ne \O$ and $V \ne \O$.


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.

$\blacksquare$


Sources

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