Absorbing set

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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

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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 (respectively, the closed 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

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Balanced core and balanced hull

A subset   of   is said to be balanced 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 balanced hull of   while the largest balanced set contained within   denoted by   is called the balanced core of   These sets are given by the formulas   and   (these formulas show that the balanced hull and the balanced core always exist and are unique). 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

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If   and   are subsets of   then   is said to absorb   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  
    • If the scalar field is   then intuitively, "  absorbs  " means that if   is perpetually "scaled up" or "inflated" (referring to   as  ) then eventually (for all positive   sufficiently large), all   will contain   and similarly,   must also eventually contain   for all negative   sufficiently large in magnitude.
    • This definition depends on the underlying scalar field's canonical norm (that is, on the absolute value  ), which thus ties this definition to the usual Euclidean topology on the scalar field. Consequently, the definition of an absorbing set (given below) is also tied to this topology.
  2. There exists a real   such that   for every non-zero[note 1] scalar   satisfying   Or stated more succinctly,   for some  
    • Because this union is equal to   where   is the closed ball with the origin removed, this condition may be restated as:   for some  
    • The non-strict inequality   can be replaced with the strict inequality   which is the next characterization.
  3. There exists a real   such that   for every non-zero[note 1] scalar   satisfying   Or stated more succinctly,   for some  
    • 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[note 1] scalar   such that  

If   (a necessary condition for   to be an absorbing set, or to be a neighborhood of the origin in a topology) 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   (since  ). 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 (since  ) 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[note 1] 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   (according to any defining condition of "absorbs" other than this one).

If   or   then this list can be extended to include:

  1.   absorbs   (according to any defining condition of "absorbs" other than this one).
    • In other words,   may be replaced by   in the characterizations above if   (or trivially, if  ).

A set absorbing a point

A set is said to absorb a point   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 absorbing in   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 bounded if it is absorbed by every neighborhood of the origin. A set is called bornivorous 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 (centered at the origin  ) together with the origin, then   is the one and only non-empty set that   absorbs. Moreover, there does not exist any non-empty subset of   that is absorbed by the unit circle   In contrast, every neighborhood of the origin absorbs every bounded subset of   (and so in particular, absorbs every singleton subset/point).

Absorbing set

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A subset   of a vector space   over a field   is called an absorbing (or absorbent) subset of   and is said to be absorbing in   if it satisfies any of the following equivalent conditions (here ordered so that each condition is an easy consequence of the previous one, starting with the definition):

  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.
    • The only TVS topologies[note 2] on a 1-dimensional vector space are the (non-Hausdorff) trivial topology and the Hausdorff Euclidean topology. Every 1-dimensional vector subspace of   is of the form   for some non-zero   and if this 1-dimensional space   is endowed with the (unique) Hausdorff vector topology, then the map   defined by   is necessarily a TVS-isomorphism (where as usual,   is endowed with its standard Euclidean topology induced by the Euclidean metric).
  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 (i.e. the Euclidean topology).
    • The reason why the Euclidean topology is distinguished in this characterization ultimately stems from the defining requirement on TVS topologies[note 2] that scalar multiplication   be continuous when the scalar field   is given this (Euclidean) topology.
    •  -Neighborhoods are absorbing: This condition gives insight as to why every neighborhood of the origin in every topological vector space (TVS) 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[note 2] 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 (the Euclidean topology).[note 3] Thus   is absorbing.
  8.   contains the origin and for every 1-dimensional vector subspace   of     is absorbing in   (according to any defining condition of "absorbing" other than this one).
    • This characterization shows that the property of being absorbing in   depends only on how   behaves with respect to 1 (or 0) 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   (although it will still be a necessary condition). For this to happen, it suffices for   to be an absorbing set that is also convex, balanced, and closed in   (such a set is called a barrel and it will be a neighborhood of the origin in   because every finite-dimensional Euclidean space, including   is a barrelled space).

If   then to this list can be appended:

  1. The algebraic interior of   contains the origin (that is,  ).

If   is balanced then to this list can be appended:

  1. For every   there exists a scalar   such that  [1] (or equivalently, such that  ).
  2. For every   there exists a scalar   such that  

If   is convex or balanced then to this list can be appended:

  1. For every   there exists a positive real   such that  
    • The proof that a balanced set   satisfying this condition is necessarily absorbing in   follows immediately from condition (10) above and the fact that   for all scalars   (where   is real).
    • The proof that a convex set   satisfying this condition is necessarily absorbing in   is less trivial (but not difficult). A detailed proof is given in this footnote[proof 1] and a summary is given below.
      • Summary of proof: By assumption, for any 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 (  is called an interval since we identify   with   and every non-empty convex subset of   is an interval). 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.  
  2. For every   there exists a positive real   such that  
    • This condition is equivalent to: every   belongs to the set   This happens if and only if   which gives the next characterization.
  3.  
    • It can be shown that for any subset   of     if and only if   for every   where  
  4. For every    

If   (which is necessary for   to be absorbing) then it suffices to check any of the above conditions for all non-zero   rather than all  

Examples and sufficient conditions

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For one set to absorb another

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Let   be a linear map between vector spaces and let   and   be balanced sets. Then   absorbs   if and only if   absorbs  [2]

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 (that is, if and only if   absorbs   for every  ). 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

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The unit ball of any normed vector space (or seminormed vector space) is absorbing. More generally, if   is a topological vector space (TVS) 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 (one or more) 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 (a convex and balanced subset) then   and so in particular, a disk   is always an absorbing subset of  [3] 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 (of the codomain) under a linear operator is again absorbing (in the domain). 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 (also known as an absolutely convex set or a disk) 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   (or over any other set of non-zero scalars having   as a limit point) 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.[4] If this normed space is a Banach space then   is called a Banach disk.

Properties

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Every absorbing set contains the origin. If   is an absorbing disk in a vector space   then there exists an absorbing disk   in   such that  [5]

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 (or equivalently for TVSs, if it is a Baire space) and if   is a closed absorbing subset of   then   necessarily contains a non-empty open subset of   (in other words,  's topological interior will not be empty), which guarantees that   is a neighborhood of the origin in  

Every absorbing set is a total set, meaning that every absorbing subspace is dense.

See also

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Notes

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  1. ^ a b c d The requirement that be scalar   be non-zero cannot be dropped from this characterization.
  2. ^ a b c A topology on a vector space   is called a vector topology or a TVS-topology if its makes vector addition   and scalar multiplication   continuous when the scalar field   is given its usual norm-induced Euclidean topology (that norm being the absolute value  ). Since restrictions of continuous functions are continuous, if   is a vector subspace of a TVS   then  's vector addition   and scalar multiplication   operations will also be continuous. Thus the subspace topology that any vector subspace inherits from a TVS will once again be a vector topology.
  3. ^ If   is a neighborhood of the origin in a TVS   then it would be pathological if there existed any 1-dimensional vector subspace   in which   was not a neighborhood of the origin in at least some TVS topology on   The only TVS topologies on   are the Hausdorff Euclidean topology and the trivial topology, which is a subset of the Euclidean topology. Consequently, this pathology does not occur if and only if   to be a neighborhood of   in the Euclidean topology for all 1-dimensional vector subspaces   which is exactly the condition that   be absorbing in   The fact that all neighborhoods of the origin in all TVSs are necessarily absorbing means that this pathological behavior does not occur.

Proofs

  1. ^ Proof: Let   be a vector space over the field   with   being   or   and endow the field   with its usual normed Euclidean topology. Let   be a convex set such that for every   there exists a positive real   such that   Because   if   then the proof is complete so assume   Clearly, every non-empty convex subset of the real line   is an interval (possibly open, closed, or half-closed; possibly degenerate (that is, a singleton set); possibly bounded or unbounded). Recall that the intersection of convex sets is convex so that for every   the sets   and   are convex, where now the convexity of   (which contains the origin and is contained in the line  ) implies that   is an interval contained in the line   Lemma: If   then the interval   contains an open sub-interval that contains the origin. Proof of lemma: By assumption, since   we can pick some   such that   and (because  ) we can also pick some   such that   where   and   (since  ). Because   is convex and contains the distinct points   and   it contains the convex hull of the points   which (in particular) contains the open sub-interval   where this open sub-interval   contains the origin (to see why, take   which satisfies  ), which proves the lemma.   Now fix   let   Because   was arbitrary, to prove that   is absorbing in   it is necessary and sufficient to show that   is a neighborhood of the origin in   when   is given its usual Hausdorff Euclidean topology, where recall that this topology makes the map   defined by   into a TVS-isomorphism. If   then the fact that the interval   contains an open sub-interval around the origin means exactly that   is a neighborhood of the origin in   which completes the proof. So assume that   Write   so that   and   (naively,   is the " -axis" and   is the " -axis" of  ). The set   is contained in the convex set   so that the convex hull of   is contained in   By the lemma, each of   and   are line segments (intervals) with each segment containing the origin in an open sub-interval; moreover, they clearly intersect at the origin. Pick a real   such that   and   Let   denote the convex hull of   which is contained in the convex hull of   and thus also contained in the convex set   To finish the proof, it suffices to show that   is a neighborhood of   in   Viewed as a subset of the complex plane     is shaped like an open square with its four corners on the positive and negative   and  -axes (that is, in       and  ). So it is readily verified that   contains the open ball   of radius   centered at the origin of   Thus   is a neighborhood of the origin in   as desired.  

Citations

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  1. ^ Narici & Beckenstein 2011, pp. 107–110.
  2. ^ Narici & Beckenstein 2011, pp. 441–457.
  3. ^ Narici & Beckenstein 2011, pp. 67–113.
  4. ^ Narici & Beckenstein 2011, pp. 115–154.
  5. ^ Narici & Beckenstein 2011, pp. 149–153.

References

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