Absorbing set


In functional analysis and related areas of mathematics an absorbing set in a vector space is a set which can be "inflated" or "scaled up" to eventually always include any given point of the vector space.
Alternative terms are radial or absorbent set.
Every neighborhood of the origin in every topological vector space is an absorbing subset.

Definition

Notation for scalars
Suppose that is a vector space over the field of real numbers or complex numbers and for any let
denote the open ball of radius in centered at
Define the product of a set of scalars with a set of vectors as and define the product of with a single vector as

Preliminaries

Balanced core and [|balanced hull]
A subset of is said to be [Balanced set|] if for all and all scalars satisfying this condition may be written more succinctly as and it holds if and only if
Given a set the smallest balanced set containing denoted by is called the of while the largest balanced set contained within denoted by is called the of
These sets are given by the formulas
and
.
A set is balanced if and only if it is equal to its balanced hull or to its [|balanced core], in which case all three of these sets are equal:
If is any scalar then
while if is non-zero or if then also

One set absorbing another

If and are subsets of then is said to ' if it satisfies any of the following equivalent conditions:
  1. Definition: There exists a real such that for every scalar satisfying Or stated more succinctly, for some
  2. * If the scalar field is then intuitively, " [|absorbs] " means that if is perpetually "scaled up" or "inflated" then, all will contain and similarly, must also eventually contain for all negative sufficiently large in magnitude.
  3. * This definition depends on the underlying scalar field's canonical norm, which thus ties this definition to the usual Euclidean topology on the scalar field. Consequently, the definition of an absorbing set is also tied to this topology.
  4. There exists a real such that for every non-zero scalar satisfying Or stated more succinctly, for some
  5. * Because this union is equal to where is the closed ball with the origin removed, this condition may be restated as: for some
  6. * The non-strict inequality can be replaced with the strict inequality which is the next characterization.
  7. There exists a real such that for every non-zero scalar satisfying Or stated more succinctly, for some
  8. * Here is the open ball with the origin removed and
If is a balanced set then this list can be extended to include:

  1. There exists a non-zero scalar such that
    • If then the requirement may be dropped.
  2. There exists a non-zero scalar such that
If then this list can be extended to include:

  1. There exists such that for every scalar satisfying Or stated more succinctly,
  2. There exists such that for every scalar satisfying Or stated more succinctly,
    • The inclusion is equivalent to . Because this may be rewritten which gives the next statement.
  3. There exists such that
  4. There exists such that
  5. There exists such that
    • The next characterizations follow from those above and the fact that for every scalar the balanced hull of satisfies and its balanced core satisfies
  6. There exists such that In words, a set is absorbed by if it is contained in some positive scalar multiple of the balanced core of
  7. There exists such that
  8. There exists a non-zero scalar such that In words, the balanced core of contains some non-zero scalar multiple of
  9. There exists a scalar such that In words, can be scaled to contain the balanced hull of
  10. There exists a scalar such that
  11. There exists a scalar such that In words, can be scaled so that its balanced core contains
  12. There exists a scalar such that
  13. There exists a scalar such that In words, the balanced core of can be scaled to contain the balanced hull of
  14. The balanced core of absorbs the balanced hull .
If or then this list can be extended to include:
  1. absorbs .
    • In other words, may be replaced by in the characterizations above if .
A set absorbing a point
A set is said to ' if it absorbs the singleton set A set absorbs the origin if and only if it contains the origin; that is, if and only if
As detailed below, a set is said to be if it absorbs every point of
This notion of one set absorbing another is also used in other definitions:
A subset of a topological vector space is called if it is absorbed by every neighborhood of the origin.
A set is called if it absorbs every bounded subset.
First examples
Every set absorbs the empty set but the empty set does not absorb any non-empty set. The singleton set containing the origin is the one and only singleton subset that absorbs itself.
Suppose that is equal to either or If is the unit circle together with the origin, then is the one and only non-empty set that absorbs. Moreover, there does exist non-empty subset of that is absorbed by the unit circle In contrast, every neighborhood of the origin absorbs every bounded subset of .

Absorbing set

A subset of a vector space over a field is called an of and is said to be if it satisfies any of the following equivalent conditions :

  1. Definition: absorbs every point of that is, for every absorbs
    • So in particular, can not be absorbing if Every absorbing set must contain the origin.
  2. absorbs every finite subset of
  3. For every there exists a real such that for any scalar satisfying
  4. For every there exists a real such that for any scalar satisfying
  5. For every there exists a real such that
    • Here is the open ball of radius in the scalar field centered at the origin and
    • The closed ball can be used in place of the open ball.
    • Because the inclusion holds if and only if This proves the next statement.
  6. For every there exists a real such that where Connection to topology: If is given its usual Hausdorff Euclidean topology then the set is a neighborhood of the origin in thus, there exists a real such that if and only if is a neighborhood of the origin in Consequently, satisfies this condition if and only if for every is a neighborhood of in when is given the Euclidean topology. This gives the next characterization.
  7. contains the origin and for every 1-dimensional vector subspace of is a neighborhood of the origin in when is given its unique Hausdorff vector topology.

    • The reason why the Euclidean topology is distinguished in this characterization ultimately stems from the defining requirement on TVS topologies that scalar multiplication be continuous when the scalar field is given this topology.
    • -Neighborhoods are absorbing: This condition gives insight as to why every neighborhood of the origin in every topological vector space is necessarily absorbing: If is a neighborhood of the origin in a TVS then for every 1-dimensional vector subspace is a neighborhood of the origin in when is endowed with the subspace topology induced on it by This subspace topology is always a vector topology and because is 1-dimensional, the only vector topologies on it are the Hausdorff Euclidean topology and the trivial topology, which is a subset of the Euclidean topology.
      So regardless of which of these vector topologies is on the set will be a neighborhood of the origin in with respect to [|its unique Hausdorff vector topology].
      Thus is absorbing.
  8. contains the origin and for every 1-dimensional vector subspace of is absorbing in .
    • This characterization shows that the property of being absorbing in depends on how behaves with respect to 1 dimensional vector subspaces of In contrast, if a finite-dimensional vector subspace of has dimension and is endowed with its unique Hausdorff TVS topology, then being absorbing in is no longer sufficient to guarantee that is a neighborhood of the origin in . For this to happen, it suffices for to be an absorbing set that is also convex, balanced, and closed in .
If then to this list can be appended:
  2. The algebraic interior of contains the origin.
If is balanced then to this list can be appended:
  2. For every there exists a scalar such that .
  3. For every there exists a scalar such that
If is convex or balanced then to this list can be appended:
  2. For every there exists a positive real such that
  3. * The proof that a balanced set satisfying this condition is necessarily absorbing in follows immediately from condition above and the fact that for all scalars .
  4. * The proof that a convex set satisfying this condition is necessarily absorbing in is less trivial. A detailed proof is given in this footnote and a summary is given below.
  5. ** Summary of proof: By assumption, for non-zero it is possible to pick positive real and such that and so that the convex set contains the open sub-interval which contains the origin. Give its unique Hausdorff vector topology so it remains to show that is a neighborhood of the origin in If then we are done, so assume that The set is a union of two intervals, each of which contains an open sub-interval that contains the origin; moreover, the intersection of these two intervals is precisely the origin. So the quadrilateral-shaped convex hull of which is contained in the convex set clearly contains an open ball around the origin.
  6. For every there exists a positive real such that
  7. * This condition is equivalent to: every belongs to the set This happens if and only if which gives the next characterization.
  9. *It can be shown that for any subset of if and only if for every where
  10. For every
If then it suffices to check any of the above conditions for all non-zero rather than all

Examples and sufficient conditions

For one set to absorb another

Let be a linear map between vector spaces and let and be balanced sets. Then absorbs if and only if absorbs
If a set absorbs another set then any superset of also absorbs
A set absorbs the origin if and only if the origin is an element of
A set absorbs a finite union of sets if and only it absorbs each set individuality. In particular, a set is an absorbing subset of if and only if it absorbs every finite subset of

For a set to be absorbing

The unit ball of any normed vector space is absorbing.
More generally, if is a topological vector space then any neighborhood of the origin in is absorbing in This fact is one of the primary motivations for defining the property "absorbing in "
Every superset of an absorbing set is absorbing. Consequently, the union of any family of absorbing sets is absorbing. The intersection of finitely many absorbing subsets is once again an absorbing subset. However, the open balls of radius are all absorbing in although their intersection is not absorbing.
If is a disk then and so in particular, a disk is always an absorbing subset of
Thus if is a disk in then is absorbing in if and only if
This conclusion is not guaranteed if the set is balanced but not convex; for example, the union of the and axes in is a non-convex balanced set that is not absorbing in
The image of an absorbing set under a surjective linear operator is again absorbing. The inverse image of an absorbing subset under a linear operator is again absorbing.
If absorbing then the same is true of the symmetric set
Auxiliary normed spaces
If is convex and absorbing in then the symmetric set will be convex and balanced in addition to being absorbing in
This guarantees that the Minkowski functional of will be a seminorm on thereby making into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples as ranges over forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If is a topological vector space and if this convex absorbing subset is also a bounded subset of then all this will also be true of the absorbing disk if in addition does not contain any non-trivial vector subspace then will be a norm and will form what is known as an auxiliary normed space. If this normed space is a Banach space then is called a.

Properties

Every absorbing set contains the origin.
If is an absorbing disk in a vector space then there exists an absorbing disk in such that
If is an absorbing subset of then and more generally, for any sequence of scalars such that Consequently, if a topological vector space is a non-meager subset of itself and if is a closed absorbing subset of then necessarily contains a non-empty open subset of, which guarantees that is a neighborhood of the origin in
Every absorbing set is a total set, meaning that every absorbing subspace is dense.