Meagre set explained
In the mathematical field of general topology, a meagre set (also called a meager set or a set of first category) is a subset of a topological space that is small or negligible in a precise sense detailed below. A set that is not meagre is called nonmeagre, or of the second category. See below for definitions of other related terms.
The meagre subsets of a fixed space form a σ-ideal of subsets; that is, any subset of a meagre set is meagre, and the union of countably many meagre sets is meagre.
Meagre sets play an important role in the formulation of the notion of Baire space and of the Baire category theorem, which is used in the proof of several fundamental results of functional analysis.
Definitions
Throughout,
will be a
topological space.
The definition of meagre set uses the notion of a nowhere dense subset of
that is, a subset of
whose
closure has empty
interior. See the corresponding article for more details.
A subset of
is called
a
of
or of the
in
if it is a countable union of
nowhere dense subsets of
. Otherwise, the subset is called
a
of
or of the
in
The qualifier "in
" can be omitted if the ambient space is fixed and understood from context.
A topological space is called (respectively, ) if it is a meagre (respectively, nonmeagre) subset of itself.
A subset
of
is called
in
or
in
if its
complement
is meagre in
. (This use of the prefix "co" is consistent with its use in other terms such as "
cofinite".)A subset is comeagre in
if and only if it is equal to a countable
intersection of sets, each of whose interior is dense in
Remarks on terminology
The notions of nonmeagre and comeagre should not be confused. If the space
is meagre, every subset is both meagre and comeagre, and there are no nonmeagre sets. If the space
is nonmeager, no set is at the same time meagre and comeager, every comeagre set is nonmeagre, and there can be nonmeagre sets that are not comeagre, that is, with nonmeagre complement. See the Examples section below.
As an additional point of terminology, if a subset
of a topological space
is given the
subspace topology induced from
, one can talk about it being a meagre space, namely being a meagre subset of itself (when considered as a topological space in its own right). In this case
can also be called a
meagre subspace of
, meaning a meagre space when given the subspace topology. Importantly, this is not the same as being meagre in the whole space
. (See the Properties and Examples sections below for the relationship between the two.) Similarly, a
nonmeagre subspace will be a set that is nonmeagre in itself, which is not the same as being nonmeagre in the whole space. Be aware however that in the context of
topological vector spaces some authors may use the phrase "meagre/nonmeagre subspace" to mean a vector subspace that is a meagre/nonmeagre set relative to the whole space.
[1] The terms first category and second category were the original ones used by René Baire in his thesis of 1899.[2] The meagre terminology was introduced by Bourbaki in 1948.[3]
Examples
The empty set is always a closed nowhere dense (and thus meagre) subset of every topological space.
In the nonmeagre space
the set
is meagre. The set
is nonmeagre and comeagre.
In the nonmeagre space
the set
is nonmeagre. But it is not comeagre, as its complement
is also nonmeagre.
A countable T1 space without isolated point is meagre. So it is also meagre in any space that contains it as a subspace. For example,
is both a meagre subspace of
(that is, meagre in itself with the subspace topology induced from
) and a meagre subset of
The Cantor set is nowhere dense in
and hence meagre in
But it is nonmeagre in itself, since it is a
complete metric space.
The set
is not nowhere dense in
, but it is meagre in
. It is nonmeagre in itself (since as a subspace it contains an isolated point).
The line
is meagre in the plane
But it is a nonmeagre subspace, that is, it is nonmeagre in itself.
The set
S=(\Q x \Q)\cup(\Reals x \{0\})
is a meagre sub of
even though its meagre subset
is a nonmeagre sub (that is,
is not a meagre topological space). A countable Hausdorff space without
isolated points is meagre, whereas any topological space that contains an isolated point is nonmeagre. Because the
rational numbers are countable, they are meagre as a subset of the reals and as a space—that is, they do not form a
Baire space.
Any topological space that contains an isolated point is nonmeagre (because no set containing the isolated point can be nowhere dense). In particular, every nonempty discrete space is nonmeagre.
There is a subset
of the real numbers
that splits every nonempty open set into two nonmeagre sets. That is, for every nonempty open set
, the sets
and
are both nonmeagre.
In the space
of continuous real-valued functions on
with the topology of
uniform convergence, the set
of continuous real-valued functions on
that have a derivative at some point is meagre.
[4] Since
is a complete metric space, it is nonmeagre. So the complement of
, which consists of the continuous real-valued nowhere differentiable functions on
is comeagre and nonmeagre. In particular that set is not empty. This is one way to show the existence of continuous nowhere differentiable functions.
On an infinite-dimensional Banach, there exists a discontinuous linear functional whose kernel is nonmeagre.[5] Also, under Martin's axiom, on each separable Banach space, there exists a discontinuous linear functional whose kernel is meagre (this statement disproves the Wilansky–Klee conjecture[6]).
Characterizations and sufficient conditions
Every nonempty Baire space is nonmeagre. In particular, by the Baire category theorem every nonempty complete metric space and every nonempty locally compact Hausdorff space is nonmeagre.
Every nonempty Baire space is nonmeagre but there exist nonmeagre spaces that are not Baire spaces. Since complete (pseudo)metric spaces as well as Hausdorff locally compact spaces are Baire spaces, they are also nonmeagre spaces.
Any subset of a meagre set is a meagre set, as is the union of countably many meagre sets. If
is a
homeomorphism then a subset
is meagre if and only if
is meagre.
Every nowhere dense subset is a meagre set. Consequently, any closed subset of
whose interior in
is empty is of the first category of
(that is, it is a meager subset of
).
The states that in any space
the union of any family of open sets of the first category is of the first category.
All subsets and all countable unions of meagre sets are meagre. Thus the meagre subsets of a fixed space form a σ-ideal of subsets, a suitable notion of negligible set. Dually, all supersets and all countable intersections of comeagre sets are comeagre. Every superset of a nonmeagre set is nonmeagre.
Suppose
where
has the
subspace topology induced from
The set
may be meagre in
without being meagre in
However the following results hold:
is meagre in
then
is meagre in
is open in
then
is meagre in
if and only if
is meagre in
is dense in
then
is meagre in
if and only if
is meagre in
And correspondingly for nonmeagre sets:
is nonmeagre in
then
is nonmeagre in
is open in
then
is nonmeagre in
if and only if
is nonmeagre in
is dense in
then
is nonmeagre in
if and only if
is nonmeagre in
In particular, every subset of
that is meagre in itself is meagre in
Every subset of
that is nonmeagre in
is nonmeagre in itself. And for an open set or a dense set in
being meagre in
is equivalent to being meagre in itself, and similarly for the nonmeagre property.
A topological space
is nonmeagre if and only if every countable intersection of dense open sets in
is nonempty.
Properties
A nonmeagre locally convex topological vector space is a barreled space.
Every nowhere dense subset of
is meagre. Consequently, any closed subset with empty interior is meagre. Thus a closed subset of
that is of the second category in
must have non-empty interior in
(because otherwise it would be nowhere dense and thus of the first category).
If
is of the second category in
and if
are subsets of
such that
then at least one
is of the second category in
Meagre subsets and Lebesgue measure
There exist nowhere dense subsets (which are thus meagre subsets) that have positive Lebesgue measure.
A meagre set in
need not have
Lebesgue measure zero, and can even have full measure. For example, in the interval
fat Cantor sets, like the
Smith–Volterra–Cantor set, are closed nowhere dense and they can be constructed with a measure arbitrarily close to
The union of a countable number of such sets with measure approaching
gives a meagre subset of
with measure
[7] Dually, there can be nonmeagre sets with measure zero. The complement of any meagre set of measure
in
(for example the one in the previous paragraph) has measure
and is comeagre in
and hence nonmeagre in
since
is a Baire space.
Here is another example of a nonmeagre set in
with measure
:
where
is a sequence that enumerates the rational numbers.
Relation to Borel hierarchy
Just as a nowhere dense subset need not be closed, but is always contained in a closed nowhere dense subset (viz, its closure), a meagre set need not be an
set (countable union of closed sets), but is always contained in an
set made from nowhere dense sets (by taking the closure of each set).
Dually, just as the complement of a nowhere dense set need not be open, but has a dense interior (contains a dense open set), a comeagre set need not be a
set (countable intersection of open sets), but contains a dense
set formed from dense open sets.
Banach–Mazur game
Meagre sets have a useful alternative characterization in terms of the Banach–Mazur game. Let
be a topological space,
be a family of subsets of
that have nonempty interiors such that every nonempty open set has a subset belonging to
and
be any subset of
Then there is a Banach–Mazur game
In the Banach–Mazur game, two players,
and
alternately choose successively smaller elements of
to produce a sequence
W1\supseteqW2\supseteqW3\supseteq … .
Player
wins if the intersection of this sequence contains a point in
; otherwise, player
wins.
Erdos–Sierpinski duality
Many arguments about meagre sets also apply to null sets, i.e. sets of Lebesgue measure 0. The Erdos–Sierpinski duality theorem states that if the continuum hypothesis holds, there is an involution from reals to reals where the image of a null set of reals is a meagre set, and vice versa.[8] In fact, the image of a set of reals under the map is null if and only if the original set was meagre, and vice versa.[9]
See also
- , for analogs to residual
- , for analogs to meagre
Bibliography
- Book: Oxtoby, John C.. Measure and Category. The Banach Category Theorem. New York. Springer. Second. 1980. 0-387-90508-1. . 62–65.
Notes and References
- Web site: Schaefer . Helmut H. . Topological Vector Spaces . Macmillan . 1966.
- Baire . René . Sur les fonctions de variables réelles . Annali di Mat. Pura ed Appl. . 1899 . 1-123 . 3., page 65
- Oxtoby . J. . Cartesian products of Baire spaces . . 1961 . 49 . 2 . 157–166 . 10.4064/fm-49-2-157-166 . "Following Bourbaki [...], a topological space is called a Baire space if ..."
- Banach, S.. Über die Baire'sche Kategorie gewisser Funktionenmengen. Studia Math.. 3. 1. 1931. 174–179. 10.4064/sm-3-1-174-179. free.
- https://mathoverflow.net/questions/3188/are-proper-linear-subspaces-of-banach-spaces-always-meager
- https://www.ams.org/journals/bull/1966-72-04/S0002-9904-1966-11547-1/S0002-9904-1966-11547-1.pdf
- Web site: Is there a measure zero set which isn't meagre? . MathOverflow.
- Quintanilla. M.. 2206.10754. The real numbers in inner models of set theory. 2022. (p.25)
- S. Saito, The Erdos-Sierpinski Duality Theorem, notes. Accessed 18 January 2023.