Nowhere dense set

Density nowhere can be characterized in three 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 ${\displaystyle S.}$[2] It suffices to check either condition on a base for the topology on ${\displaystyle X,}$ and density nowhere in ${\displaystyle \mathbb {R} }$ is often described as being dense in no open interval.[3][4]

• A set is nowhere dense if and only if its closure is.[1] Thus a nowhere dense set need not be closed (for instance, the set ${\displaystyle \left\{{\frac {1}{n}}:n\in \mathbb {N} \right\}}$ is nowhere dense in the reals), but is then properly contained in a nowhere dense closed set.
• Suppose ${\displaystyle A\subseteq B\subseteq X.}$
  • If ${\displaystyle A}$ is nowhere dense in ${\displaystyle B}$ then ${\displaystyle A}$ is nowhere dense in ${\displaystyle X.}$
  • If ${\displaystyle A}$ is nowhere dense in ${\displaystyle X}$ and ${\displaystyle B}$ is an open subset of ${\displaystyle X}$ then ${\displaystyle A}$ is nowhere dense in ${\displaystyle B.}$[1]
• Every subset of a nowhere dense set is nowhere dense.[9]
• The union of finitely many nowhere dense sets is nowhere dense.[9]

Thus the nowhere dense sets form an ideal of sets, a suitable notion of negligible set.

The union of countably many nowhere dense sets, however, need not be nowhere dense. (Thus, the nowhere dense sets do not, in general, form a 𝜎-ideal.) Instead, such a union is called a meagre set or a set of first category.

• ${\displaystyle S:=\left\{{\frac {1}{n}}:n\in \mathbb {N} \right\}}$ is nowhere dense in ${\displaystyle \mathbb {R} }$: although the points get arbitrarily close to ${\displaystyle 0,}$ the closure of the set is ${\displaystyle S\cup \{0\},}$ which has empty interior (and is thus also nowhere dense in ${\displaystyle \mathbb {R} }$).[1]
• ${\displaystyle \mathbb {R} }$ is nowhere dense in ${\displaystyle \mathbb {R} ^{2}.}$[1]
• ${\displaystyle \mathbb {Z} }$ is nowhere dense in ${\displaystyle \mathbb {R} }$ but the rationals ${\displaystyle \mathbb {Q} }$ are not (they are dense everywhere).[1]
• ${\displaystyle \mathbb {Z} \cup [(a,b)\cap \mathbb {Q} ]}$ is not nowhere dense in ${\displaystyle \mathbb {R} }$: it is dense in the 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 such subset.[1]
• In a T₁ space, any singleton set that is not an isolated point is nowhere dense.
• The boundary of every open set and of every closed set is nowhere dense.[1]
• A vector subspace of a topological vector space is either dense or nowhere dense.[1]

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[10]) 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).[11]

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
• Meagre set – A notion in general topology of a "small" or "negligible" subset of a topological space

• Fremlin, D. H. (2002). Measure Theory. Lulu.com. 3A3F(a). ISBN 978-0-9566071-1-9.
• Khaleelulla, S. M. (1982). Counterexamples in Topological Vector Spaces. Lecture Notes in Mathematics. Vol. 936. Berlin, Heidelberg, New York: Springer-Verlag. ISBN 978-3-540-11565-6. OCLC 8588370.
• Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
• Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277.
• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.

• Some nowhere dense sets with positive measure