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 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 second definition above is equivalent to requiring that the closure, ${\displaystyle \operatorname {cl} _{X}S,}$  cannot contain any nonempty open set.^[7] This is the same as saying that the interior of the closure of ${\displaystyle S}$  is empty; that is,

${\displaystyle \operatorname {int} _{X}\left(\operatorname {cl} _{X}S\right)=\varnothing .}$ ^[8][9]

Alternatively, the complement of the closure ${\displaystyle X\setminus \left(\operatorname {cl} _{X}S\right)}$  must be a dense subset of ${\displaystyle X;}$ ^[4][8] in other words, the exterior of ${\displaystyle S}$  is dense in ${\displaystyle X.}$ 

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][13] 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.^[14][2] A closed set is nowhere dense if and only if it is equal to its boundary,^[14] 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.
• 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.^[15]
• 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.^[16]

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 ${\displaystyle 0.535\ldots }$  because of overlaps^[17]) 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).^[18]

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
• Smith–Volterra–Cantor set – Set of real numbers in mathematics
• Meagre set – "Small" subset of a topological space

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• Some nowhere dense sets with positive measure