Closure of Irrational Interval is Closed Real Interval

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Theorem

Let $\left({\R, \tau_d}\right)$ be the real numbers under the usual (Euclidean) topology.

Let $\left({\R \setminus \Q, \tau_d}\right)$ be the irrational number space under the same topology.

Let $a, b \in \R$ such that $a < b$.

Let $\Bbb I \subseteq \R$ be an interval of $\R$


Then the closure of the set:

$\Bbb I \cap \left({\R \setminus \Q}\right)$

is the closed real interval $\left[{a \,.\,.\, b}\right]$.


Proof

Let $\Bbb I$ be an open real interval.

From Closure of Real Interval is Closed Real Interval:

$\Bbb I^- = \left[{a \,.\,.\, b}\right]$

From Closure of Irrational Numbers is Real Numbers:

$\left({\R \setminus \Q}\right)^- = \R$

From Closure of Intersection is Subset of Intersection of Closures:

$\left({\Bbb I \cap \left({\R \setminus \Q}\right)}\right)^- \subseteq \Bbb I^- \cap \left({\R \setminus \Q}\right)^-$

From Intersection with Subset is Subset:

$\Bbb I^- \cap \left({\R \setminus \Q}\right)^- = \left[{a \,.\,.\, b}\right]$

and so:

$\left({\Bbb I \cap \left({\R \setminus \Q}\right)}\right)^- \subseteq \left[{a \,.\,.\, b}\right]$


From Irrationals are Everywhere Dense in Reals:

$\left[{a \,.\,.\, b}\right] \subseteq \left({\Bbb I \cap \left({\R \setminus \Q}\right)}\right)^-$

and the result follows.

$\blacksquare$