$\mathbb{Q}$ is dense in $\mathbb{R}$ and has measure zero

Refuted claim: every subset of $\mathbb{R}$ of measure zero is nowhere dense (FALSE: every subset of $\mathbb{R}$ of measure zero is nowhere dense).

The witness is ${\mathbb{Q}}_{\mathbb{R}}$, the set of rationals inside $\mathbb{R}$ (The rationals embed densely in the reals). It is at most countable, hence null (Every at most countable subset of $\mathbb{R}$ has measure zero), and it is dense, so its closure is $\mathbb{R}$ and the interior of that closure is $\mathbb{R}$, as far from empty as possible. The refutation is carried out in full in FALSE: every subset of $\mathbb{R}$ of measure zero is nowhere dense; this item records the witness and the explicit cover.

• The cover is explicit and startling. For every $\epsilon >0$ the rationals are covered by open intervals of total length exactly $\epsilon$ ($\mathbb{Q}$ is covered by open intervals of total length $\epsilon$, for every $\epsilon >0$), although their union is dense because it contains ${\mathbb{Q}}_{\mathbb{R}}$.

• $\mathbb{Q}$ is small in the other sense too, one level up. It is meager, being a countable union of singletons ($\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$); what fails is only nowhere density itself. So the counterexample separates "nowhere dense" from "meager", not category from measure.

• The complementary witness is the Smith-Volterra-Cantor set, nowhere dense and not null (The Smith-Volterra-Cantor set is nowhere dense and does not have measure zero).