Singleton Set is Nowhere Dense in Rational Space
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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 Line, $\set x$ 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 Line is Empty:
- $\set x^\circ = \O$
Hence:
- $\paren {\set x^-}^\circ = \set x^\circ = \O$
$\blacksquare$