Singleton Set is Nowhere Dense in Rational Space

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

Then every singleton subset of $\Q$ is nowhere dense in $\struct {\Q, \tau_d}$.

Proof
Let $x \in \Q$.

By definition of nowhere dense, we need to show that:
 * $\paren {\set x^-}^\circ = \O$

where $S^-$ denotes the closure of a set $S$ and $S^\circ$ denotes its interior.

By Real Number is Closed in Real Number Space, $\set x$ is closed in $\struct {\R, \tau_d}$.

$\struct {\Q, \tau_d}$ is a subspace of $\struct {\R, \tau_d}$.

Thus by Closed Set in Topological Subspace, $\set x$ is closed in $\struct {\Q, \tau_d}$.

From Closed Set Equals its Closure, it follows that:
 * $\set x^- = \set x$

From Interior of Singleton in Real Number Space is Empty:
 * $\set x^\circ = \O$

Hence:
 * $\paren {\set x^-}^\circ = \set x^\circ = \O$