In functional analysis and related areas of mathematics an absorbing set in a vector space is a set S 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.
Definition
Suppose that is a vector space over the field of real numbers or complex numbers
Notation
 Products of scalars and vectors
For any let
For any subsets and and any and define
One set absorbing another
If and are subsets of then is said to absorb if it satisfies any of the following equivalent conditions:
 Definition: There exists a real for every scalar satisfying
 If the scalar field is then intuitively, " absorbs " means that if is perpetually "scaled up" or "inflated" (referring to with ) then eventually, all will contain (for sufficiently large); similarly, all will also eventually contain (for sufficiently large).
 This definition depends on the canonical norm on the underlying scalar field, which ties this definition to the usual Euclidean topology on the scalar field.
 There exists a real for every scalar satisfying
 If it is known that then the restriction may be removed, giving the characterization: There exists a real for every scalar satisfying
 There exists a real
 The closed ball (with the origin removed) can be used in place of the open ball, giving the next characterization.
 There exists a real
If is a balanced set then to this list can be appended:
 There exists a scalar
 There exists a scalar
A set is said to absorb a point if and only 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
Absorbing set
A subset of a vector space over a field is called absorbing or absorbent 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):
 Definition: For every absorbs
 So in particular, can not be absorbing if
 For every there exists a real for any scalar satisfying
 For every there exists a real for any scalar satisfying
 For every there exists a real
 Here is the open ball of radius in centered at the origin and
 The closed ball can be used in place of the open ball.
 For every there exists a real where
 Proof: This follows from the previous condition since so that if and only if
 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 if and only if is a neighborhood of the origin in
 Every 1dimensional vector subspace of is of the form for some nonzero and if this 1dimensional space is endowed with the unique Hausdorff vector topology, then the map defined by is necessarily a TVSisomorphism (where as usual, has the normed Euclidean topology).
 contains the origin and for every 1dimensional vector subspace of is a neighborhood of the origin in when is given its unique Hausdorff vector topology.
 The Hausdorff vector topology on a 1dimensional vector space is necessarily TVSisomorphic to with its usual normed Euclidean topology.
 Intuition: This condition shows that it is only natural that any neighborhood of 0 in any topological vector space (TVS) be absorbing: if is a neighborhood of the origin in then it would be pathological if there existed any 1dimensional 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, it is natural to expect for to be a neighborhood of in the Euclidean topology for all 1dimensional 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. The reason why the Euclidean topology is distinguished is ultimately due to the defining requirement on TVS topologies that scalar multiplication be continuous when the scalar field is given the Euclidean topology.
 This condition is equivalent to: For every is a neighborhood of in when is given its unique Hausdorff TVS topology.
 contains the origin and for every 1dimensional vector subspace of is absorbing in the
 Here "absorbing" means absorbing according to any defining condition other than this one.
 This 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 finitedimensional vector subspace of has dimension then being absorbing in is no longer sufficient to guarantee that is a neighborhood of the origin in when is endowed with its unique Hausdorff TVS topology (although it will still be a necessary condition). For this to happen, it suffices for to be a barrel in this Hausdorff TVS (because every finitedimensional Euclidean space is a barrelled space).
If then to this list can be appended:
 The algebraic interior of contains the origin (that is, ).
If is balanced then to this list can be appended:
 For every there exists a scalar ^{[1]}
If is convex or balanced then to this list can be appended:
 For every there exists a positive real
 The proof that a balanced set satisfying this condition is necessarily absorbing in is almost immediate from the definition of a "balanced set".
 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 nonzero it is possible to pick positive real and such that and so that the convex set contains the open subinterval which contains the origin (the set is also called an interval because every nonempty 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 subinterval that contains the origin; moreover, the intersection of these two intervals is precisely the origin. So the convex hull of which is contained in the convex set clearly contains an open ball around the origin.
 For every there exists a positive real
 This condition is equivalent to: every belongs to the set This happens if and only if which gives the next characterization.

 It can be shown that for any subset of if and only if
 For every where
If (which is necessary for to be absorbing) then it suffices to check any of the above conditions for all nonzero 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 ^{[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
For a set to be absorbing
In a semi normed vector space the unit ball 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 even defining the property "absorbing in "
If is a disk in then so that in particular, is an absorbing subset of ^{[3]} Thus if is a disk in then is absorbing in if and only if
Any superset of an absorbing set is absorbing. Thus the union of any family of (one or more) absorbing sets is absorbing. The intersection of a finite family of (one or more) absorbing sets is absorbing.
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).
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 ^{[4]}
See also
 Algebraic interior – Generalization of topological interior
 Absolutely convex set
 Balanced set – Construct in functional analysis
 Bornivorous set – A set that can absorb any bounded subset
 Bounded set (topological vector space)
 Convex set – In geometry, set that intersects every line into a single line segment
 Locally convex topological vector space – A vector space with a topology defined by convex open sets
 Radial set
 Star domain
 Symmetric set
 Topological vector space – Vector space with a notion of nearness
Notes
 ^ 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 nonempty convex subset of the real line is an interval (possibly open, closed, or halfclosed, and possibly bounded or unbounded, and possibly even degenerate (that is, a singleton set)). 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: We will now prove that if then the interval contains an open subinterval that contains the origin. 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 subinterval where this open subinterval 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 TVSisomorphism. If then the fact that the interval contains an open subinterval around the origin implies that is a neighborhood of the origin in so we're done. So assume that Write so that and that (so naively, is the "axis" and is the "axis" of so that the set (resp. ) is the strictly positive axis (resp. axis) while (resp. ) is the strictly negative axis (resp. axis)). 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 (i.e. intervals) with each segment containing the origin in an open subinterval; 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 So 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 corners on the positive and negative and axes. So it is readily verified that contains the open ball of radius centered at the origin of This shows that is a neighborhood of the origin in as desired.
Citations
 ^ Narici & Beckenstein 2011, pp. 107–110.
 ^ Narici & Beckenstein 2011, pp. 441–457.
 ^ Narici & Beckenstein 2011, pp. 67–113.
 ^ Narici & Beckenstein 2011, pp. 149–153.
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