Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$

Let $A\subseteq \mathbb{R}$, with interior $A^{\circ }$ and closure $\bar{A}$ as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$.

• $A$ is nowhere dense when the interior of its closure is empty: $(\bar{A}{)}^{\circ }=\mathrm{\varnothing }\mathrm{.}$
• $A$ is meager, or of the first category, when there is a sequence $(A_n{)}_{n\in \mathbb{N}}$ of nowhere dense subsets of $\mathbb{R}$ with $A={\bigcup }_{n\in \mathbb{N}}{A}_n\mathrm{.}$
• $A$ is of the second category when it is not meager.
• $A$ is residual (also comeager) when $\mathbb{R}\setminus A$ is meager.

Why a sequence, and why that is the same as "an at most countable union". Sequences here are indexed by $\mathbb{N}$, which contains $0$. A finite family $A_0,\dots ,A_m$ of nowhere dense sets is turned into a sequence by setting $A_n:=\mathrm{\varnothing }$ for $n>m$, and $\mathrm{\varnothing }$ is nowhere dense because $\bar{\mathrm{\varnothing }}=\mathrm{\varnothing }$ has empty interior; the empty family is handled the same way and gives $A=\mathrm{\varnothing }$. So "a union of an at most countable family of nowhere dense sets" (Finite, countably infinite, countable, uncountable) and the displayed condition define the same class, and the sequence form is used below because it carries an explicit index and needs no case split.

Nowhere dense means exactly that the complement of the closure is dense. For $A\subseteq \mathbb{R}$,

$(\bar{A}{)}^{\circ }=\mathrm{\varnothing }⟺\mathbb{R}\setminus \bar{A}\text{ is dense in }\mathbb{R}\mathrm{.}$

Indeed, by the pointwise description of the interior (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$), $(\bar{A}{)}^{\circ }=\mathrm{\varnothing }$ says that no $x\in \mathbb{R}$ admits a real $\epsilon >0$ with $N_{\epsilon }(x)\subseteq \bar{A}$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$), that is, that every $N_{\epsilon }(x)$ meets $\mathbb{R}\setminus \bar{A}$. By claim 1 of The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points that says precisely that every $x\in \mathbb{R}$ is adherent to $\mathbb{R}\setminus \bar{A}$, that is, $\overline{\mathbb{R}\setminus \stackrel{‾}{A}}=\mathbb{R}$, which is density (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$).

A closed set is nowhere dense exactly when its interior is empty, since a closed set equals its own closure (claim 4 of The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). This is the form in which nowhere density is verified nearly every time below. (The phrase almost everywhere is avoided throughout this pair: it is a measure-theoretic term, and the only measure notion defined here is measure zero.)

Both classes are closed downwards. If $B\subseteq A$ then $\bar{B}\subseteq \bar{A}$ and hence $(\bar{B}{)}^{\circ }\subseteq (\bar{A}{)}^{\circ }$ (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$), so a subset of a nowhere dense set is nowhere dense. If $B\subseteq A={\bigcup }_n{A}_n$ with each $A_n$ nowhere dense, then $B={\bigcup }_n(A_n\cap B)$ and each $A_n\cap B$ is nowhere dense by the previous sentence, so a subset of a meager set is meager.

A union of two meager sets is meager. Let $M={\bigcup }_n{A}_n$ and $M^{\mathrm{\prime }}={\bigcup }_n{B}_n$ with all $A_n$ and all $B_n$ nowhere dense; fixing one witnessing sequence for $M$ and one for $M^{\mathrm{\prime }}$ is two instantiations of an existential statement, not a choice principle. Let $J:\mathbb{N}\times \mathbb{N}\to \mathbb{N}$ be a bijection ($\mathbb{N}\times \mathbb{N}\approx \mathbb{N}$) and define a sequence $(C_j{)}_{j\in \mathbb{N}}$ by

$C_{J(m,n)}:=\{\begin{array}{ll}{A}_n& m=0,\\ B_n& m\mathrm{\ne }0.\end{array}$

This is a total definition because $J$ is a bijection, every $C_j$ is nowhere dense, and ${\bigcup }_j{C}_j=M\cup M^{\mathrm{\prime }}$, since $A_n=C_{J(0,n)}$ and $B_n=C_{J(1,n)}$ and every $C_j$ is one of the $A_n$ or one of the $B_n$.

• The countably infinite version of the last observation is a different statement. To show that ${\bigcup }_i{M}_i$ is meager for a sequence $(M_i)$ of meager sets one must select a witnessing sequence of nowhere dense sets for every $i$ at once, which is an application of countable choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)); the two-set case above avoids it because two selections are two instantiations. Nothing on this page uses the countably infinite version, and every meager set met below is presented together with an explicit witnessing sequence.

• Nowhere dense is strictly stronger than having empty interior. $\mathbb{Q}$ has empty interior, since no neighbourhood consists of rationals alone, yet $\bar{\mathbb{Q}}=\mathbb{R}$ has interior $\mathbb{R}$, so $\mathbb{Q}$ is not nowhere dense. It is nevertheless meager, being a union of singletons; that computation is $\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$.

• First category, second category and residual are not a trichotomy. A set is meager or of the second category, and those two are exhaustive and exclusive by definition. Residual is a separate condition on the complement: a residual set is of the second category once $\mathbb{R}$ is known not to be meager in itself (Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets), but before that theorem nothing rules out a set that is both meager and residual.

• Category is a notion of topological smallness, and it is independent of smallness in the sense of measure. Neither of the two implications between "nowhere dense" and "measure zero" (Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover)) holds, and $\mathbb{R}$ itself splits into a meager set and a set of measure zero; the three items settling this are FALSE: every nowhere dense subset of $\mathbb{R}$ has measure zero, FALSE: every subset of $\mathbb{R}$ of measure zero is nowhere dense and $\mathbb{R}$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions ↗.