Rational Number Space is Totally Separated

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Theorem

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.


Then $\struct {\Q, \tau_d}$ is totally separated.


Proof

Let $x, y \in \Q$.

From Between two Rational Numbers exists Irrational Number:

$\exists \alpha \in \R \setminus \Q: x < \alpha < y$

Consider the unbounded open real intervals:

$A := \openint \gets \alpha$, $B := \openint \alpha \to$

Let:

$U := A \cap \Q, V := B \cap \Q$

Let $\beta \in \Q$.

Then either:

$(1): \quad \beta < \alpha$

in which case:

$\beta \in U$

or:

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

in which case:

$\beta \in V$

Thus $U \cup V = \Q$.


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 $\Q$ such that $x \in U$ and $y \in V$.

Hence the result by definition of totally separated.

$\blacksquare$


Sources