$\mathbb{Q}$ has closure $\mathbb{R}$, empty interior, and boundary $\mathbb{R}$

Write ${\mathbb{Q}}_{\mathbb{R}}$ for the copy of $\mathbb{Q}$ inside $\mathbb{R}$ (The rationals embed densely in the reals). Then

$\overline{{\mathbb{Q}}_{\mathbb{R}}}=\mathbb{R},({\mathbb{Q}}_{\mathbb{R}}{)}^{\circ }=\mathrm{\varnothing },\mathrm{\partial }{\mathbb{Q}}_{\mathbb{R}}=\mathbb{R},$

with closure, interior and boundary as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$. So the rationals are as large as possible for the closure operator and as small as possible for the interior operator at once, and their boundary is everything.

• The same computation applies verbatim to the irrationals. $X$ is dense by [L2] and its complement ${\mathbb{Q}}_{\mathbb{R}}$ is dense too, so $\bar{X}=\mathbb{R}$, $X^{\circ }=\mathrm{\varnothing }$ and $\mathrm{\partial }X=\mathbb{R}$. Two complementary sets can therefore both have boundary everything, which is what the density of each of them forces.

• Empty interior is not smallness in any counting sense. $X$ has empty interior and is uncountable (The irrationals are uncountable), while ${\mathbb{Q}}_{\mathbb{R}}$ has empty interior and is countable ($\mathbb{Q}$ is countably infinite). The interior measures whether the set contains an interval, and nothing else.

• Where the density of the irrationals comes from. It is proved in Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable by counting: an interval is uncountable and the rationals are not, so an interval cannot consist of rationals alone. No explicit irrational is needed for this computation, though one is available (Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$, FALSE: some rational number squares to 2).