Nowhere dense set

Density nowhere can be characterized in different (but equivalent) ways. The simplest definition is the one from density:

A subset ${\displaystyle S}$ of a topological space ${\displaystyle X}$ is said to be dense in another set ${\displaystyle U}$ if the intersection ${\displaystyle S\cap U}$ is a dense subset of ${\displaystyle U.}$ ${\displaystyle S}$ is nowhere dense or rare in ${\displaystyle X}$ if ${\displaystyle S}$ is not dense in any nonempty open subset ${\displaystyle U}$ of ${\displaystyle X.}$

Expanding out the negation of density, it is equivalent to require that each nonempty open set ${\displaystyle U}$ contains a nonempty open subset disjoint from ${\displaystyle S.}$[4] It suffices to check either condition on a base for the topology on ${\displaystyle X.}$ In particular, density nowhere in ${\displaystyle \mathbb {R} }$ is often described as being dense in no open interval.[5][6]

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:[10][11]

• 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.[1]

Every subset of a nowhere dense set is nowhere dense, and a finite union of nowhere dense sets is nowhere dense.[12] 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.[13][2] A closed set is nowhere dense if and only if it is equal to its boundary,[13] if and only if it is equal to the boundary of some open set[2] (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 ${\displaystyle A}$).

• 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.
• ${\displaystyle \mathbb {R} }$ 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.[14]
• 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.[15]

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.

For one 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 ${\displaystyle 0.535\ldots }$ because of overlaps[16]) 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).[17]

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

$${\displaystyle D:=\bigcap _{m=1}^{\infty }U_{1/m}=\bigcap _{m=1}^{\infty }S_{1/m}}$$

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 ${\displaystyle n>0.}$

• Baire space – Concept in topology
• Fat Cantor set – set that is nowhere dense (in particular it contains no intervals), yet has positive measurePages displaying wikidata descriptions as a fallback
• Meagre set – "Small" subset of a topological space

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• Some nowhere dense sets with positive measure