A countable union of nowhere dense sets is called a meagre set. Meagre sets play an important role in the formulation of the Baire category theorem, which is used in the proof of several fundamental results of functional analysis.
Density nowhere can be characterized in different (but equivalent) ways. The simplest definition is the one from density:
A subset of a topological space is said to be dense in another set if the intersection is a dense subsetofisnowhere denseorrareinif is not dense in any nonempty open subset of
Expanding out the negation of density, it is equivalent to require that each nonempty open set contains a nonempty open subset disjoint from [4] It suffices to check either condition on a base for the topology on In particular, density nowhere in is often described as being dense in no open interval.[5][6]
The second definition above is equivalent to requiring that the closure, cannot contain any nonempty open set.[7] This is the same as saying that the interior of the closureof is empty; that is,
The notion of nowhere dense set is always relative to a given surrounding space. Suppose where has the subspace topology induced from The set may be nowhere dense in but not nowhere dense in Notably, a set is always dense in its own subspace topology. So if is nonempty, it will not be nowhere dense as a subset of itself. However the following results hold:[10][11]
If is nowhere dense in then is nowhere dense in
If is open in , then is nowhere dense in if and only if is nowhere dense in
If is dense in , then is nowhere dense in if and only if is nowhere dense in
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 is not nowhere dense in
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 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 exteriorof).
A nowhere dense set is not necessarily negligible in every sense. For example, if is the unit interval 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 all dyadic fractions, i.e. fractions of the form inlowest terms for positive integers and the intervals around them:
Since for each this removes intervals adding up to at most the nowhere dense set remaining after all such intervals have been removed has measure of at least (in fact just over because of overlaps[17]) and so in a sense represents the majority of the ambient space
This set is nowhere dense, as it is closed and has an empty interior: any interval is not contained in the set since the dyadic fractions in have been removed.
Generalizing this method, one can construct in the unit interval nowhere dense sets of any measure less than 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 is any dense open subset of having finite Lebesgue measure then is necessarily a closed subset of having infinite Lebesgue measure that is also nowhere dense in (because its topological interior is empty). Such a dense open subset of finite Lebesgue measure is commonly constructed when proving that the Lebesgue measure of the rational numbers is This may be done by choosing any bijection (it actually suffices for to merely be a surjection) and for every letting
(here, the Minkowski sum notation was used to simplify the description of the intervals).
The open subset is dense in because this is true of its subset and its Lebesgue measure is no greater than
Taking the union of closed, rather than open, intervals produces the F𝜎-subset
that satisfies Because is a subset of the nowhere dense set it is also nowhere dense in
Because is a Baire space, the set
is a dense subset of (which means that like its subset cannot possibly be nowhere dense in ) with Lebesgue measure that is also a nonmeager subsetof (that is, is of the second categoryin), which makes acomeager subsetof whose interior in is also empty; however, is nowhere dense in if and only if its closurein has empty interior.
The subset in this example can be replaced by any countable dense subset of and furthermore, even the set can be replaced by for any integer
Smith–Volterra–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
^Oxtoby, John C. (1980). Measure and Category (2nd ed.). New York: Springer-Verlag. pp. 1–2. ISBN0-387-90508-1. 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.
^Steen, Lynn Arthur; Seebach Jr., J. Arthur (1995). Counterexamples in Topology (Dover republication of Springer-Verlag 1978 ed.). New York: Dover. p. 7. ISBN978-0-486-68735-3. A subset of is said to be nowhere dense in if no nonempty open set of is contained in
^ abGamelin, Theodore W. (1999). Introduction to Topology (2nd ed.). Mineola: Dover. pp. 36–37. ISBN0-486-40680-6 – via ProQuest ebook Central.