Nowhere dense set

Properties

The notion of nowhere dense set is always relative to a given surrounding space. Suppose ${\displaystyle A\subseteq Y\subseteq X,}$ where ${\displaystyle Y}$ has the subspace topology induced from ${\displaystyle X.}$ The set ${\displaystyle A}$ may be nowhere dense in ${\displaystyle X,}$ but not nowhere dense in ${\displaystyle Y.}$ Notably, a set is always dense in its own subspace topology. So if ${\displaystyle A}$ is nonempty, it will not be nowhere dense as a subset of itself. However the following results hold:

• If ${\displaystyle A}$ is nowhere dense in ${\displaystyle Y,}$ then ${\displaystyle A}$ is nowhere dense in ${\displaystyle X.}$
• If ${\displaystyle Y}$ is open in ${\displaystyle X}$, then ${\displaystyle A}$ is nowhere dense in ${\displaystyle Y}$ if and only if ${\displaystyle A}$ is nowhere dense in ${\displaystyle X.}$
• If ${\displaystyle Y}$ is dense in ${\displaystyle X}$, then ${\displaystyle A}$ is nowhere dense in ${\displaystyle Y}$ if and only if ${\displaystyle A}$ is nowhere dense in ${\displaystyle X.}$

A set is nowhere dense if and only if its closure is.

Every subset of a nowhere dense set is nowhere dense, and a finite union of nowhere dense sets is nowhere dense. Thus the nowhere dense sets form an ideal of sets, a suitable notion of negligible set. In general they do not form a 𝜎-ideal, as meager sets, which are the countable unions of nowhere dense sets, need not be nowhere dense. For example, the set ${\displaystyle \mathbb{Q}}$ is not nowhere dense in ${\displaystyle \mathbb{R}.}$

The boundary of every open set and of every closed set is closed and nowhere dense. A closed set is nowhere dense if and only if it is equal to its boundary, if and only if it is equal to the boundary of some open set (for example the open set can be taken as the complement of the set). An arbitrary set ${\displaystyle A\subseteq X}$ is nowhere dense if and only if it is a subset of the boundary of some open set (for example the open set can be taken as the exterior of ${\displaystyle A}$).

Examples

• The set ${\displaystyle S=\{1/n:n=1,2,...\}}$ and its closure ${\displaystyle S\cup \{0\}}$ are nowhere dense in ${\displaystyle \mathbb{R},}$ since the closure has empty interior.
• The Cantor set is an uncountable nowhere dense set in ${\displaystyle \mathbb{R}.}$
• ${\displaystyle \mathbb{R}}$ viewed as the horizontal axis in the Euclidean plane is nowhere dense in ${\displaystyle {\mathbb{R}}^2.}$
• ${\displaystyle \mathbb{Z}}$ is nowhere dense in ${\displaystyle \mathbb{R}}$ but the rationals ${\displaystyle \mathbb{Q}}$ are not (they are dense everywhere).
• ${\displaystyle \mathbb{Z}\cup [(a,b)\cap \mathbb{Q}]}$ is not nowhere dense in ${\displaystyle \mathbb{R}}$: it is dense in the open interval ${\displaystyle (a,b),}$ and in particular the interior of its closure is ${\displaystyle (a,b).}$
• The empty set is nowhere dense. In a discrete space, the empty set is the only nowhere dense set.
• In a T₁ space, any singleton set that is not an isolated point is nowhere dense.
• A vector subspace of a topological vector space is either dense or nowhere dense.

Nowhere dense sets with positive measure

A nowhere dense set is not necessarily negligible in every sense. For example, if ${\displaystyle X}$ is the unit interval ${\displaystyle [0,1],}$ not only is it possible to have a dense set of Lebesgue measure zero (such as the set of rationals), but it is also possible to have a nowhere dense set with positive measure. One such example is the Smith–Volterra–Cantor set.

For another example (a variant of the Cantor set), remove from ${\displaystyle [0,1]}$ all dyadic fractions, i.e. fractions of the form ${\displaystyle a/2^n}$ in lowest terms for positive integers ${\displaystyle a,n\in \mathbb{N},}$ and the intervals around them: ${\displaystyle \left(a/2^n-1/2^{2n+1},a/2^n+1/2^{2n+1}\right).}$ Since for each ${\displaystyle n}$ this removes intervals adding up to at most ${\displaystyle 1/2^{n+1},}$ the nowhere dense set remaining after all such intervals have been removed has measure of at least ${\displaystyle 1/2}$ (in fact just over ${\displaystyle 0.535\dots }$ because of overlaps) and so in a sense represents the majority of the ambient space ${\displaystyle [0,1].}$ This set is nowhere dense, as it is closed and has an empty interior: any interval ${\displaystyle (a,b)}$ is not contained in the set since the dyadic fractions in ${\displaystyle (a,b)}$ have been removed.

Generalizing this method, one can construct in the unit interval nowhere dense sets of any measure less than ${\displaystyle 1,}$ although the measure cannot be exactly 1 (because otherwise the complement of its closure would be a nonempty open set with measure zero, which is impossible).

For another simpler example, if ${\displaystyle U}$ is any dense open subset of ${\displaystyle \mathbb{R}}$ having finite Lebesgue measure then ${\displaystyle \mathbb{R}\setminus U}$ is necessarily a closed subset of ${\displaystyle \mathbb{R}}$ having infinite Lebesgue measure that is also nowhere dense in ${\displaystyle \mathbb{R}}$ (because its topological interior is empty). Such a dense open subset ${\displaystyle U}$ of finite Lebesgue measure is commonly constructed when proving that the Lebesgue measure of the rational numbers ${\displaystyle \mathbb{Q}}$ is ${\displaystyle 0.}$ This may be done by choosing any bijection ${\displaystyle f:\mathbb{N}\to \mathbb{Q}}$ (it actually suffices for ${\displaystyle f:\mathbb{N}\to \mathbb{Q}}$ to merely be a surjection) and for every ${\displaystyle r>0,}$ letting $${\displaystyle U_r:=\bigcup _{n\in \mathbb{N}}\left(f(n)-r/2^n,f(n)+r/2^n\right)=\bigcup _{n\in \mathbb{N}}f(n)+\left(-r/2^n,r/2^n\right)}$$ (here, the Minkowski sum notation ${\displaystyle f(n)+\left(-r/2^n,r/2^n\right):=\left(f(n)-r/2^n,f(n)+r/2^n\right)}$ was used to simplify the description of the intervals). The open subset ${\displaystyle U_r}$ is dense in ${\displaystyle \mathbb{R}}$ because this is true of its subset ${\displaystyle \mathbb{Q}}$ and its Lebesgue measure is no greater than ${\displaystyle \sum _{n\in \mathbb{N}}2r/2^n=2r.}$ Taking the union of closed, rather than open, intervals produces the F_𝜎-subset $${\displaystyle S_r:=\bigcup _{n\in \mathbb{N}}f(n)+\left[-r/2^n,r/2^n\right]}$$ that satisfies ${\displaystyle S_{r/2}\subseteq U_r\subseteq S_r\subseteq U_{2r}.}$ Because ${\displaystyle \mathbb{R}\setminus S_r}$ is a subset of the nowhere dense set ${\displaystyle \mathbb{R}\setminus U_r,}$ it is also nowhere dense in ${\displaystyle \mathbb{R}.}$ Because ${\displaystyle \mathbb{R}}$ is a Baire space, the set $${\displaystyle D:=\bigcap _{m=1}^{\mathrm{\infty }}{U}_{1/m}=\bigcap _{m=1}^{\mathrm{\infty }}{S}_{1/m}}$$ is a dense subset of ${\displaystyle \mathbb{R}}$ (which means that like its subset ${\displaystyle \mathbb{Q},}$ ${\displaystyle D}$ cannot possibly be nowhere dense in ${\displaystyle \mathbb{R}}$) with ${\displaystyle 0}$ Lebesgue measure that is also a nonmeager subset of ${\displaystyle \mathbb{R}}$ (that is, ${\displaystyle D}$ is of the second category in ${\displaystyle \mathbb{R}}$), which makes ${\displaystyle \mathbb{R}\setminus D}$ a comeager subset of ${\displaystyle \mathbb{R}}$ whose interior in ${\displaystyle \mathbb{R}}$ is also empty; however, ${\displaystyle \mathbb{R}\setminus D}$ is nowhere dense in ${\displaystyle \mathbb{R}}$ if and only if its closure in ${\displaystyle \mathbb{R}}$ has empty interior. The subset ${\displaystyle \mathbb{Q}}$ in this example can be replaced by any countable dense subset of ${\displaystyle \mathbb{R}}$ and furthermore, even the set ${\displaystyle \mathbb{R}}$ can be replaced by ${\displaystyle {\mathbb{R}}^n}$ for any integer ${\displaystyle n>0.}$