The Smith-Volterra-Cantor set is nowhere dense and does not have measure zero

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

The witness is the Smith-Volterra-Cantor set $S$ (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \ge 1$, an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals): the subset of $[0,1]$ obtained by removing, at stage $n$, an open interval of length $4^{-n-1}$ from the middle of each of the $2^{n}$ intervals then present. It is nowhere dense (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$) and no cover of it by intervals has total length below $2^{-1}$, so it is not of measure zero (Measure zero (a countable cover by intervals of total length below every $\varepsilon$) and content zero (a finite such cover)). This item records the witness and says what makes it work; the refutation is carried out in full in FALSE: every nowhere dense subset of $\mathbb{R}$ has measure zero and The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero.

• What makes it work is a change of proportion, not of shape. The middle-thirds construction removes a fixed fraction of each piece and loses total length $1$; this one removes a fixed length $4^{-n-1}$ at stage $n$ and loses only $2^{-1}$ (The intervals removed from the Smith-Volterra-Cantor set have total length $1/2$, so the set cannot be covered by intervals of total length less than $1/2$). Topologically the two sets are indistinguishable at the level of the properties proved here: both are compact, perfect and nowhere dense (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points).

• The complementary witness. For the reverse implication the witness is $\mathbb{Q}$, which is null and not nowhere dense ($\mathbb{Q}$ is dense in $\mathbb{R}$ and has measure zero). The two counterexamples together show the two smallness notions are independent.