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

Statement refuted

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.

Facts & Assumptions

Given: The set $\mathbb{Q}_{\mathbb{R}} \subseteq \mathbb{R}$ of rationals.

[A1]

The refuted claim: every subset of $\mathbb{R}$ of measure zero is nowhere dense.

[L1]

$\mathbb{Q}$ is countably infinite, so $\mathbb{Q}_{\mathbb{R}}$ is at most countable and therefore null ( $\mathbb{Q}$ is countably infinite, Finite, countably infinite, countable, uncountable, The rationals embed densely in the reals, Every at most countable subset of $\mathbb{R}$ has measure zero, Measure zero (a countable cover by intervals of total length below every $\varepsilon$ ) and content zero (a finite such cover)).

[L2]

$\mathbb{Q}_{\mathbb{R}}$ is dense: $\overline{\mathbb{Q}_{\mathbb{R}}} = \mathbb{R}$ (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable).

[L3]

Nowhere dense means the interior of the closure is empty; $\mathbb{R}$ is open, so its interior is itself (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$ , Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ , Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

Counterexample

technique · direct

1.1

$\mathbb{Q}_{\mathbb{R}}$ has measure zero, by [L1].

1.2

$\mathbb{Q}_{\mathbb{R}}$ is not nowhere dense: its closure is $\mathbb{R}$ by [L2] and the interior of $\mathbb{R}$ is $\mathbb{R}$ by [L3], which is not empty.

L2L3

2.1

So $\mathbb{Q}_{\mathbb{R}}$ witnesses the failure of [A1].

step 1.1step 1.2A1∎

Remarks

The cover is explicit and startling. For every $\varepsilon > 0$ the rationals are covered by open intervals of total length exactly $\varepsilon$ ( $\mathbb{Q}$ is covered by open intervals of total length $\varepsilon$ , for every $\varepsilon > 0$ ), 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).

Depends on

Used by

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Sources

Null set (Wikipedia) (standard reference, not scraped)
Dense set (Wikipedia) (standard reference, not scraped)
E. Zakon, Mathematical Analysis, §6.8: Baire Categories (standard reference, not scraped)
MIT 18.125, Homework 2: Measure-zero sets (standard reference, not scraped)

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