Nowhere dense set

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

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

Expanding out the negation of density, it is equivalent that each nonempty open set $U$ contains a nonempty open subset disjoint from $S.${{sfn|Fremlin|2002|loc=3A3F(a)}} It suffices to check either condition on a base for the topology on $X.$ In particular, density nowhere in $\backslash R$ is often described as being dense in no open interval.{{Cite book|last=Oxtoby|first=John C.|title=Measure and Category|publisher=Springer-Verlag|year=1980|isbn=0-387-90508-1|edition=2nd|location=New York|pages=1–2|quote=A set is nowhere dense if it is dense in no interval}}; although note that Oxtoby later gives the interior-of-closure definition on page 40.{{Cite book|last=Natanson|first=Israel P.|url=http://hdl.handle.net/2027/mdp.49015000681685|title=Teoria functsiy veshchestvennoy peremennoy|publisher=Frederick Ungar|year=1955|volume=I (Chapters 1-9)|location=New York|pages=88|hdl=2027/mdp.49015000681685|language=English|translator-last=Boron|translator-first=Leo F.|trans-title=Theory of functions of a real variable|lccn=54-7420}}

The second definition above is equivalent to requiring that the closure, $\backslash operatorname\{cl\}\_X\; S,$ cannot contain any nonempty open set.{{Cite book|last1=Steen|first1=Lynn Arthur|title=Counterexamples in Topology|last2=Seebach Jr.|first2=J. Arthur|publisher=Dover|year=1995|isbn=978-0-486-68735-3|edition=Dover republication of Springer-Verlag 1978|location=New York|pages=7|quote=A subset $A$ of $X$ is said to be nowhere dense in $X$ if no nonempty open set of $X$ is contained in $\backslash overline\{A\}.$}} This is the same as saying that the interior of the closure of $S$ is empty; that is,

$\backslash operatorname\{int\}\_X\; \backslash left(\backslash operatorname\{cl\}\_X\; S\backslash right)\; =\; \backslash varnothing.${{Cite book|last=Gamelin|first=Theodore W.|title=Introduction to Topology|publisher=Dover|year=1999|isbn=0-486-40680-6|edition=2nd|location=Mineola|pages=36–37|via=ProQuest ebook Central}}{{sfn|Rudin|1991|p=41}}

The notion of nowhere dense set is always relative to a given surrounding space. Suppose $A\backslash subseteq\; Y\backslash subseteq\; X,$ where $Y$ has the subspace topology induced from $X.$ The set $A$ may be nowhere dense in $X,$ but not nowhere dense in $Y.$ Notably, a set is always dense in its own subspace topology. So if $A$ is nonempty, it will not be nowhere dense as a subset of itself. However the following results hold:{{sfn|Narici|Beckenstein|2011|loc=Theorem 11.5.4}}{{sfn|Haworth|McCoy|1977|loc=Proposition 1.3}}

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

A set is nowhere dense if and only if its closure is.{{sfn|Bourbaki|1989|loc=ch. IX, section 5.1}}

Every subset of a nowhere dense set is nowhere dense, and a finite union of nowhere dense sets is nowhere dense.{{sfn|Fremlin|2002|loc=3A3F(c)}}{{sfn|Willard|2004|loc=Problem 25A}} 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 $\backslash Q$ is not nowhere dense in $\backslash R.$

The boundary of every open set and of every closed set is closed and nowhere dense.{{sfn|Narici|Beckenstein|2011|loc=Example 11.5.3(e)}}{{sfn|Willard|2004|loc=Problem 4G}} A closed set is nowhere dense if and only if it is equal to its boundary,{{sfn|Narici|Beckenstein|2011|loc=Example 11.5.3(e)}} if and only if it is equal to the boundary of some open set{{sfn|Willard|2004|loc=Problem 4G}} (for example the open set can be taken as the complement of the set). An arbitrary set $A\backslash subseteq\; X$ 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 $A$).

• The set $S=\backslash \{1/n:n=1,2,...\backslash \}$ and its closure $S\backslash cup\backslash \{0\backslash \}$ are nowhere dense in $\backslash R,$ since the closure has empty interior.
• The Cantor set is an uncountable nowhere dense set in $\backslash R.$
• $\backslash R$ viewed as the horizontal axis in the Euclidean plane is nowhere dense in $\backslash R^2.$
• $\backslash Z$ is nowhere dense in $\backslash R$ but the rationals $\backslash Q$ are not (they are dense everywhere).
• $\backslash Z\; \backslash cup\; [(a,\; b)\; \backslash cap\; \backslash Q]$ is {{em|not}} nowhere dense in $\backslash R$: it is dense in the open interval $(a,b),$ and in particular the interior of its closure is $(a,b).$
• The empty set is nowhere dense. In a discrete space, the empty set is the {{em|only}} nowhere dense set.{{sfn|Narici|Beckenstein|2011|loc=Example 11.5.3(a)}}
• 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.{{sfn|Narici|Beckenstein|2011|loc=Example 11.5.3(f)}}

A nowhere dense set is not necessarily negligible in every sense. For example, if $X$ is the unit interval $[0,\; 1],$ 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 $[0,\; 1]$ all dyadic fractions, i.e. fractions of the form $a/2^n$ in lowest terms for positive integers $a,\; n\; \backslash in\; \backslash N,$ and the intervals around them: $\backslash left(a/2^n\; -\; 1/2^\{2n+1\},\; a/2^n\; +\; 1/2^\{2n+1\}\backslash right).$

Since for each $n$ this removes intervals adding up to at most $1/2^\{n+1\},$ the nowhere dense set remaining after all such intervals have been removed has measure of at least $1/2$ (in fact just over $0.535\backslash ldots$ because of overlaps{{cite web| url = http://www.se16.info/hgb/nowhere.htm| title = Some nowhere dense sets with positive measure and a strictly monotonic continuous function with a dense set of points with zero derivative}}) and so in a sense represents the majority of the ambient space $[0,\; 1].$

This set is nowhere dense, as it is closed and has an empty interior: any interval $(a,\; b)$ is not contained in the set since the dyadic fractions in $(a,\; b)$ have been removed.

Generalizing this method, one can construct in the unit interval nowhere dense sets of any measure less than $1,$ 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).{{Cite book|last=Folland|first=G. B.|url=http://hdl.handle.net/2027/mdp.49015000929258|title=Real analysis: modern techniques and their applications|publisher=John Wiley & Sons|year=1984|isbn=0-471-80958-6|location=New York|pages=41|hdl=2027/mdp.49015000929258}}

For another simpler example, if $U$ is any dense open subset of $\backslash R$ having finite Lebesgue measure then $\backslash R\; \backslash setminus\; U$ is necessarily a closed subset of $\backslash R$ having infinite Lebesgue measure that is also nowhere dense in $\backslash R$ (because its topological interior is empty). Such a dense open subset $U$ of finite Lebesgue measure is commonly constructed when proving that the Lebesgue measure of the rational numbers $\backslash Q$ is $0.$ This may be done by choosing any bijection $f\; :\; \backslash N\; \backslash to\; \backslash Q$ (it actually suffices for $f\; :\; \backslash N\; \backslash to\; \backslash Q$ to merely be a surjection) and for every $r\; >\; 0,$ letting

$$U\_r\; ~:=~\; \backslash bigcup\_\{n\; \backslash in\; \backslash N\}\; \backslash left(f(n)\; -\; r/2^n,\; f(n)\; +\; r/2^n\backslash right)\; ~=~\; \backslash bigcup\_\{n\; \backslash in\; \backslash N\}\; f(n)\; +\; \backslash left(-\; r/2^n,\; r/2^n\backslash right)$$

(here, the Minkowski sum notation $f(n)\; +\; \backslash left(-\; r/2^n,\; r/2^n\backslash right)\; :=\; \backslash left(f(n)\; -\; r/2^n,\; f(n)\; +\; r/2^n\backslash right)$ was used to simplify the description of the intervals).

The open subset $U\_r$ is dense in $\backslash R$ because this is true of its subset $\backslash Q$ and its Lebesgue measure is no greater than $\backslash sum\_\{n\; \backslash in\; \backslash N\}\; 2\; r\; /\; 2^n\; =\; 2\; r.$

Taking the union of closed, rather than open, intervals produces the F_{{sigma}}-subset

$$S\_r\; ~:=~\; \backslash bigcup\_\{n\; \backslash in\; \backslash N\}\; f(n)\; +\; \backslash left[-\; r/2^n,\; r/2^n\backslash right]$$

that satisfies $S\_\{r/2\}\; \backslash subseteq\; U\_r\; \backslash subseteq\; S\_r\; \backslash subseteq\; U\_\{2r\}.$ Because $\backslash R\; \backslash setminus\; S\_r$ is a subset of the nowhere dense set $\backslash R\; \backslash setminus\; U\_r,$ it is also nowhere dense in $\backslash R.$

Because $\backslash R$ is a Baire space, the set

$$D\; :=\; \backslash bigcap\_\{m=1\}^\{\backslash infty\}\; U\_\{1/m\}\; =\; \backslash bigcap\_\{m=1\}^\{\backslash infty\}\; S\_\{1/m\}$$

is a dense subset of $\backslash R$ (which means that like its subset $\backslash Q,$ $D$ cannot possibly be nowhere dense in $\backslash R$) with $0$ Lebesgue measure that is also a nonmeager subset of $\backslash R$ (that is, $D$ is of the second category in $\backslash R$), which makes $\backslash R\; \backslash setminus\; D$ a comeager subset of $\backslash R$ whose interior in $\backslash R$ is also empty; however, $\backslash R\; \backslash setminus\; D$ is nowhere dense in $\backslash R$ if and only if its {{em|closure}} in $\backslash R$ has empty interior.

The subset $\backslash Q$ in this example can be replaced by any countable dense subset of $\backslash R$ and furthermore, even the set $\backslash R$ can be replaced by $\backslash R^n$ for any integer $n\; >\; 0.$

• {{annotated link|Baire space}}
• {{annotated link|Smith–Volterra–Cantor set}}
• {{annotated link|Meagre set}}

{{reflist}}

• {{Bourbaki General Topology Part II Chapters 5-10}}
• {{Cite book|last=Fremlin|first=D. H.|title=Measure Theory|publisher=Lulu.com|year=2002|isbn=978-0-9566071-1-9}}
• {{Citation|last1=Haworth|first1=R. C.|last2=McCoy|first2=R. A.|title=Baire Spaces|location=Warszawa|publisher=Instytut Matematyczny Polskiej Akademi Nauk|year=1977|url=http://eudml.org/doc/268479}}
• {{Khaleelulla Counterexamples in Topological Vector Spaces}}
• {{Narici Beckenstein Topological Vector Spaces|edition=2}}
• {{Rudin Walter Functional Analysis|edition=2}}
• {{Schaefer Wolff Topological Vector Spaces|edition=2}}
• {{Trèves François Topological vector spaces, distributions and kernels}}
• {{Willard General Topology}}

• [http://www.se16.info/hgb/nowhere.htm Some nowhere dense sets with positive measure]

Category:General topology

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