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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $9$