Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable

Write ${\mathbb{Q}}_{\mathbb{R}}$ for the image of $\mathbb{Q}$ in $\mathbb{R}$ under the canonical embedding $q\mapsto \widehat{q}$ (The rationals embed densely in the reals), the set usually written $\mathbb{Q}$ once the identification is made, and put $X:=\mathbb{R}\setminus {\mathbb{Q}}_{\mathbb{R}}$ for the irrationals. Then:

1. ${\mathbb{Q}}_{\mathbb{R}}$ is dense in $\mathbb{R}$, that is, $\overline{{\mathbb{Q}}_{\mathbb{R}}}=\mathbb{R}$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$);
2. $X$ is dense in $\mathbb{R}$;
3. every nonempty open subset of $\mathbb{R}$ is uncountable (Finite, countably infinite, countable, uncountable).

Claim 2 is not a symmetry of claim 1: the rationals are dense because they are constructed to approximate, whereas the irrationals are dense because there are too many points in any interval for a countable set to exhaust it, which is why claim 3 is proved alongside and used for it.

The rationals embed densely in the reals

Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$

Finite, countably infinite, countable, uncountable

• Two dense sets can be disjoint. ${\mathbb{Q}}_{\mathbb{R}}$ and $X$ partition $\mathbb{R}$ and both are dense, so density says nothing about size: one of them is countable and the other is not (The irrationals are uncountable). What density does say is that neither has interior: a set whose complement is dense has empty interior, which is the computation carried out for $\mathbb{Q}$ in $\mathbb{Q}$ has closure $\mathbb{R}$, empty interior, and boundary $\mathbb{R}$ ↗.

• Claim 3 is a statement about open sets, not about intervals. It follows from the uncountability of intervals (Every nondegenerate interval of $\mathbb{R}$ is uncountable) only because openness supplies an interval inside the set at each of its points. A nonempty set with empty interior can perfectly well be countable, as ${\mathbb{Q}}_{\mathbb{R}}$ shows.

• An explicit irrational is not produced here. Step 2.1 is a counting argument and exhibits nothing. The library does exhibit one separately, $\sqrt{2}$ (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), and an explicit irrational in a given interval $(a,b)$ can be built from it as $q_1+(q_2-q_1)\mathrm{/}\sqrt{2}$ for suitable rationals $q_1<q_2$ in the interval; that route is longer and is not the one taken above.