Irrational Number Space is Dense-in-itself

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

Then $\struct {\R \setminus \Q, \tau_d}$ is dense-in-itself.

Proof
Let $x \in \R \setminus \Q$.

Let $U \subseteq \R$ be an open set of $\struct {\R \setminus \Q, \tau_d}$ such that $x \in U$.

From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau_d}$.

By Basis for Topological Subspace, the set of all intersections of $\R \setminus \Q$ and open real intervals of $\R$ form a basis for $\struct {\R \setminus \Q, \tau_d}$.

So there exists $V = \paren {\R \setminus \Q} \cap \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V$.

By Between two Real Numbers exists Irrational Number we have:


 * $\exists r \in \R \setminus \Q: x < r < x + \epsilon$.

Then $r \in \paren {\R \setminus \Q} \cap \openint {x - \epsilon} {x + \epsilon}$.

Therefore $x$ is not isolated.

Hence $\struct {\R \setminus \Q, \tau_d}$ is dense-in-itself.

Also See

 * Rational Number Space is Dense-in-itself