Absorbing set

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.
Every neighborhood of the origin in every topological vector space is an absorbing subset.

Definition

Suppose that $X$ is a vector space over the field $\mathbb$ of real numbers $\R$ or complex numbers $\Complex.$

Notation

Products of scalars and vectors

For any $-\infty \leq r \leq R \leq \infty,$ vector $x,$ and subset $A \subseteq X,$ let
$B_r = \ \quad \text \quad B_ = \$
denote the ''open ball'' (respectively, the ''closed ball'') of radius $r$ in $\mathbb$ centered at $0,$ and let
$(r, R) x = \ \quad \text \quad (r, R) A = \.$

Similarly, if $K \subseteq \mathbb$ and $k$ is a scalar then let $K A = \,$ $K x = \,$ $k A = \,$ and $\mathbb x = \ = \operatorname \.$

Balanced core and balanced hull

A subset $S$ of $X$ is said to be '' '' if $a s \in S$ for all $s \in S$ and all scalars $a$ satisfying $, a, \leq 1;$ this condition may be written more succinctly as $B_ S \subseteq S,$ and it holds if and only if $B_ S = S.$

The (respectively, the ) of a set $T,$ denoted by $\operatorname T$ (respectively, by $\operatorname T$), is defined to be the smallest balanced set containing $T$ (respectively, the largest balanced set contained in $T$). The balanced hull and core always exist and are unique.
They are given by the formulas
$\operatorname T ~=~ c \, T = B_ T \qquad \text \qquad \operatorname T ~=~ \begin c \, T & \text 0 \in T \\ \varnothing & \text 0 \not\in T. \\ \end$
A set $T$ is balanced if and only if it is equal to its balanced hull ($T = \operatorname T$) or to its balanced core ($T = \operatorname T$), in which case all three of these sets are equal: $T = \operatorname T = \operatorname T.$

If $c$ is any scalar then
$\operatorname (c \, T) = c \, \operatorname T = , c, \, \operatorname T$
and if $c$ is non-zero or if $0 \in T$ then also
$\operatorname (c \, T) = c \, \operatorname T = , c, \, \operatorname T.$

One set absorbing another

If $S$ and $A$ are subsets of $X,$ then $A$ is said to absorb $S$ if it satisfies any of the following equivalent conditions:
# ''Definition'': There exists a real $r > 0$ such that $S \, \subseteq \, c \, A$ for every scalar $c$ satisfying $, c, \geq r.$ Or stated more succinctly, $S \; \subseteq \; c \, A$ for some $r > 0.$
#* If the scalar field is $\R$ then intuitively, "$A$ absorbs $S$" means that if $A$ is perpetually "scaled up" or "inflated" (referring to $t A$ as $t \to \infty$) then (for all positive $t > 0$ sufficiently large), all $t A$ will contain $S;$ and similarly, $t A$ must also eventually contain $S$ for all negative $t < 0$ sufficiently large in magnitude.
#* This definition depends on the underlying scalar field's canonical norm (that is, on the absolute value $, \cdot,$), 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.
# There exists a real $r > 0$ such that $c \, S \, \subseteq \, A$ for every non-zeroThe requirement that be scalar $c$ be non-zero cannot be dropped from this characterization. scalar $c \neq 0$ satisfying $, c, \leq r.$ Or stated more succinctly, $c \, S \, \subseteq \, A$ for some $r > 0.$
#* Because this union is equal to $\left(B_ \setminus \\right) S,$ where $B_ \setminus \ = \$ is the closed ball with the origin removed, this condition may be restated as: $\left(B_ \setminus \\right) S \, \subseteq \, A$ for some $r > 0.$
#* The non-strict inequality $\,\leq\,$ can be replaced with the strict inequality $\,<\,,$ which is the next characterization.
# There exists a real $r > 0$ such that $c \, S \, \subseteq \, A$ for every non-zero scalar $c \neq 0$ satisfying $, c, < r.$ Or stated more succinctly, $\left(B_r \setminus \\right) S \subseteq \, A$ for some $r > 0.$
#* Here $B_r \setminus \ = \$ is the open ball with the origin removed and $\left(B_r \setminus \\right) S \, = \, c \, S.$

If $A$ is a balanced set then this list can be extended to include:
#

There exists a non-zero scalar $c \neq 0$ such that $S \; \subseteq \, c \, A.$

#* If $0 \in A$ then the requirement $c \neq 0$ may be dropped.
# There exists a non-zero scalar $c \neq 0$ such that $c \, S \, \subseteq \, A.$

If $0 \in A$ (such as when $A$ is a neighborhood of the origin in some topology, as in the definition of "bounded set") then this list can be extended to include:

6. There exists $r > 0$ such that $c \, S \; \subseteq \, A$ for every scalar $c$ satisfying $, c, < r.$ Or stated more succinctly, $B_r \; S \, \subseteq \, A.$


7. There exists $r > 0$ such that $c \, S \; \subseteq \, A$ for every scalar $c$ satisfying $, c, \leq r.$ Or stated more succinctly, $B_ S \, \subseteq \, A.$
   * This set inclusion is equivalent to $B_ S \, \subseteq \, \tfrac A$ (since $B_ = r \, B_$). Because $B_ S \, = \, \operatorname \, S,$ this may be rewritten $\operatorname \, S \, \subseteq \, \tfrac A,$ which gives the next statement.


8. There exists $r > 0$ such that $\operatorname \, S \, \subseteq \, r \, A.$


9. There exists $r > 0$ such that $\operatorname \, S \, \subseteq \, \operatorname (r \, A).$


10. There exists $r > 0$ such that $\;\;\;\;\;\; S \, \subseteq \, \operatorname (r \, A).$
    * The next characterizations follow from those above and the fact that for every scalar $c,$ the balanced hull of $A$ satisfies $\,\operatorname (c \, A) = c \, \operatorname A = , c, \, \operatorname A\,$ and (since $0 \in A$) its balanced core satisfies $\,\operatorname (c \, A) = c \, \operatorname A = , c, \, \operatorname A.$


11. There exists $r > 0$ such that $\;\;\, S \, \subseteq \, r \, \operatorname A.$ In words, a set is absorbed by $A$ if it is contained in some positive scalar multiple of the balanced core of $A.$


12. There exists $r > 0$ such that $r \, S \subseteq \,\;\;\;\; \operatorname A.$


13. There exists a scalar $c$ such that $\operatorname S \, \subseteq \, c \, A.$ In words, $A$ can be scaled to contain the balanced hull of $S.$


14. There exists a scalar $c$ such that $\operatorname S \, \subseteq \, \operatorname (c \, A).$


15. There exists a scalar $c$ such that $\;\;\;\;\;\; S \, \subseteq \, \operatorname (c \, A).$ In words, $A$ can be scaled so that its balanced core contains $S.$


16. There exists a scalar $c$ such that $\;\;\;\;\;\; S \, \subseteq \, c \, \operatorname A.$


17. There exists a non-zero scalar $c \neq 0$ such that $c \, S \, \subseteq \, \operatorname A.$ In words, the balanced core of $A$ contains some non-zero scalar multiple of $S.$

If $0 \not\in S$ or $0 \in A$ this list can be extended to include:
#

$A \cup \$ absorbs $S$ (according to any defining condition of "absorbs" other than this one).

#* In other words, $A$ may be replaced by $A \cup \$ in the characterizations above if $0 \not\in S$ (or trivially, if $0 \in A$).

A set absorbing a point

A set is said to absorb a point $x$ if it absorbs the singleton set $\.$ A set $A$ absorbs the origin if and only if it contains the origin; that is, if and only if $0 \in A.$
As detailed below, a set is said to be if it absorbs every point of $X.$

This notion of one set absorbing another is also used in other definitions:
A subset of a topological vector space $X$ is called if it is absorbed by every neighborhood of the origin.
A set is called if it absorbs every bounded subset.

''Examples''

Every set absorbs the empty set but the empty set does not absorbs any non-empty set. The singleton set $\$ containing the origin is the one and only singleton subset that absorbs itself.

Suppose that $X$ is equal to either $\R^2$ or $\Complex.$ If $A := S^1 \cup \$ is the unit circle (centered at the origin $\mathbf$) together with the origin, then $\$ is the one and only non-empty set that $A$ absorbs. Moreover, there does exist non-empty subset of $X$ that is absorbed by the unit circle $S^1.$ In contrast, every neighborhood of the origin absorbs every bounded subset of $X$ (and so in particular, absorbs every singleton subset/point).

Absorbing set

A subset $A$ of a vector space $X$ over a field $\mathbb$ is called an absorbing (or absorbent) subset of $X$ and is said to be absorbing in $X$ 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'': $A$ absorbs every point of $X;$ that is, for every $x \in X,$ $A$ absorbs $\.$
#*So in particular, $A$ can not be absorbing if $0 \not\in A.$ Every absorbing set must contain the origin.
#$A$ absorbs every finite subset of $X.$
#For every $x \in X,$ there exists a real $r > 0$ such that $x \in c A$ for any scalar $c \in \mathbb$ satisfying $, c, \geq r.$
#For every $x \in X,$ there exists a real $r > 0$ such that $c x \in A$ for any scalar $c \in \mathbb$ satisfying $, c, \leq r.$
#For every $x \in X,$ there exists a real $r > 0$ such that $B_r x \subseteq A.$
#* Here $B_r = \$ is the open ball of radius $r$ in the scalar field centered at the origin and $B_r x = \left\ = \.$
#* The closed ball can be used in place of the open ball.
#For every $x \in X,$ there exists a real $r > 0$ such that $B_r x \subseteq A \cap \mathbb x,$ where $\mathbb x = \operatorname \.$
#* Proof: This follows from the previous condition since $B_r x \subseteq \mathbb x,$ so that $B_r x \subseteq A$ if and only if $B_r x \subseteq A \cap \mathbb x.$
#* Connection to topology: If $\mathbb x$ is given its usual Hausdorff Euclidean topology then the set $B_r x$ is a neighborhood of the origin in $\mathbb x;$ thus, there exists a real $r > 0$ such that $B_r x \subseteq A \cap \mathbb x$ if and only if $A \cap \mathbb x$ is a neighborhood of the origin in $\mathbb x.$
#* Every 1-dimensional vector subspace of $X$ is of the form $\mathbb x = \operatorname \$ for some $x \in X$ and if this 1-dimensional space $\mathbb x$ is endowed with the unique Hausdorff vector topology, then the map $\mathbb \to \mathbb x$ defined by $c \mapsto c x$ is necessarily a TVS-isomorphism (where as usual, $\mathbb$ has the normed Euclidean topology).
#$A$ contains the origin and for every 1-dimensional vector subspace $Y$ of $X,$ $A \cap Y$ is a neighborhood of the origin in $Y$ when $Y$ is given its unique Hausdorff vector topology.
#* The Hausdorff vector topology on a 1-dimensional vector space is necessarily TVS-isomorphic to $\mathbb$ with its usual normed Euclidean topology.
#* Connection to vector/TVS topologies: This condition shows that it is only natural that any neighborhood of $0$ in any topological vector space (TVS) $X$ be absorbing: if $U$ is a neighborhood of the origin in $X$ then it would be pathological if there existed any 1-dimensional vector subspace $Y$ in which $U \cap Y$ was not a neighborhood of the origin in at least TVS topology on $Y.$ The only TVS topologies on $Y$ 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 $U \cap Y$ to be a neighborhood of $0$ in the Euclidean topology for 1-dimensional vector subspaces $Y,$ which is exactly the condition that $U$ be absorbing in $X.$ 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 ultimately stems from the defining requirement on TVS topologies that scalar multiplication $\mathbb \times X \to X$ be continuous when the scalar field $\mathbb$ is given this (Euclidean) topology.
#* This condition is equivalent to: For every $x \in X,$ $A \cap \operatorname \$ is a neighborhood of $0$ in $\operatorname \ = \mathbb x$ when $\operatorname \$ is given its unique Hausdorff TVS topology.
#$A$ contains the origin and for every 1-dimensional vector subspace $Y$ of $X,$ $A \cap Y$ is absorbing in the $Y.$
#* Here "absorbing" means absorbing according to any defining condition other than this one.
#* This shows that the property of being absorbing in $X$ depends on how $A$ behaves with respect to 1 (or 0) dimensional vector subspaces of $X.$ In contrast, if a finite-dimensional vector subspace $Z$ of $X$ has dimension $n > 1$ and is endowed with its unique Hausdorff TVS topology, then $A \cap Z$ being absorbing in $Z$ is no longer sufficient to guarantee that $A \cap Z$ is a neighborhood of the origin in $Z$ (although it will still be a necessary condition). For this to happen, it suffices for $A \cap Z$ to be an absorbing set that is also convex balanced and closed in $Z$ (such a set is called a and it will be a neighborhood of the origin in $Z$ because like every finite-dimensional Euclidean space, $Z$ is a barrelled space).

If $\mathbb = \R$ then to this list can be appended:
#

The algebraic interior of $A$ contains the origin (that is, $0 \in ^A$).

If $A$ is balanced then to this list can be appended:
#

For every $x \in X,$ there exists a scalar $c \neq 0$ such that $x \in c A$ (or equivalently, such that $c x \in A$).

# For every $x \in X,$ there exists a scalar $c$ such that $x \in c A.$

If $A$ is convex ''or'' balanced then to this list can be appended:
#

For every $x \in X,$ there exists a positive real $r > 0$ such that $r x \in A.$
#* The proof that a ''balanced'' set $A$ satisfying this condition is necessarily absorbing in $X$ is almost immediate from the definition of a "balanced set".
#* The proof that a ''convex'' set $A$ satisfying this condition is necessarily absorbing in $X$ is less trivial (but not difficult). A detailed proof is given in this footnote and a summary is given below.

#** Summary of proof: By assumption, for non-zero $0 \neq y \in X,$ it is possible to pick positive real $r > 0$ and $R > 0$ such that $R y \in A$ and $r (- y) \in A$ so that the convex set $A \cap \R y$ contains the open sub-interval $(-r, R) y \,\stackrel\, \,$ which contains the origin ($A \cap \R y$ is called an interval since we identify $\R y$ with $\R$ and every non-empty convex subset of $\R$ is an interval). Give $\mathbb y$ its unique Hausdorff vector topology so it remains to show that $A \cap \mathbb y$ is a neighborhood of the origin in $\mathbb y.$ If $\mathbb = \R$ then we are done, so assume that $\mathbb = \Complex.$ The set $S \,\stackrel\, (A \cap \R y) \,\cup\, (A \cap \R (i y)) \,\subseteq\, A \cap (\Complex y)$ 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 convex hull of $S,$ which is contained in the convex set $A \cap \Complex y,$ clearly contains an open ball around the origin. $\blacksquare$

#For every $x \in X,$ there exists a positive real $r > 0$ such that $x \in r A.$
#* This condition is equivalent to: every $x \in X$ belongs to the set $r A = \ = (0, \infty) A.$ This happens if and only if $X = (0, \infty) A,$ which gives the next characterization.
# $(0, \infty) A = X.$
#*It can be shown that for any subset $T$ of $X,$ $(0, \infty) T = X$ if and only if $T \cap (0, \infty) x \neq \varnothing$ for every $x \in X,$ where $(0, \infty) x \,\stackrel\, \.$
# For every $x \in X,$ $A \cap (0, \infty) x \neq \varnothing.$

If $0 \in A$ (which is necessary for $A$ to be absorbing) then it suffices to check any of the above conditions for all non-zero $x \in X,$ rather than all $x \in X.$

Examples and sufficient conditions

For one set to absorb another

Let $F : X \to Y$ be a linear map between vector spaces and let $B \subseteq X$ and $C \subseteq Y$ be balanced sets. Then $C$ absorbs $F(B)$ if and only if $F^(C)$ absorbs $B.$

If a set $A$ absorbs another set $B$ then any superset of $A$ also absorbs $B.$
A set $A$ absorbs the origin if and only if the origin is an element of $A.$

A set $A$ absorbs a finite union $B_1 \cup \cdots \cup B_n$ if and only it absorbs each set individuality (that is, if and only if $A$ absorbs $B_i$ for every $i = 1, \ldots, n$). In particular, a set $A$ is an absorbing subset of $X$ if and only if it absorbs every finite subset of $X.$

For a set to be absorbing

In a semi normed vector space the unit ball is absorbing.
More generally, if $X$ is a topological vector space (TVS) then any neighborhood of the origin in $X$ is absorbing in $X.$ This fact is one of the primary motivations for even defining the property "absorbing in $X.$"

If $D \neq \varnothing$ is a disk (convex and balanced) then $\operatorname D = n D$ so that in particular, $D$ is an absorbing subset of $\operatorname D.$
Thus if $D$ is a disk in $X,$ then $D$ is absorbing in $X$ if and only if $\operatorname D = X.$
This conclusion is not guaranteed if the set $D \neq \varnothing$ is balanced but not convex; for example, the union $D$ of the $x$ and $y$ axes in $X = \R^2$ is a non-convex balanced set that is not absorbing in $\operatorname D = \R^2.$

Any superset of an absorbing set is absorbing. Thus the union of any 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).
The intersection of a finite family of (one or more) absorbing sets is absorbing. If $A$ absorbing then the same is true of the symmetric set $u A \subseteq A.$

Auxiliary normed spaces

If $W$ is convex and absorbing in $X$ then the symmetric set $D := u W$ will be convex and balanced (also known as an or a ) in addition to being absorbing in $X.$
This guarantees that the Minkowski functional $p_D : X \to \R$ of $D$ will be a seminorm on $X,$ thereby making $\left(X, p_D\right)$ into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples $r D$ as $r$ ranges over $\left\$ (or over any other set of non-zero scalars having $0$ as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If $X$ is a topological vector space and if this convex absorbing subset $W$ is also a bounded subset of $X,$ then the same will be true of the absorbing disk $D := u W,$ in which case $p_D$ will be a norm and $\left(X, p_D\right)$ will form what is known as an auxiliary normed space. If this normed space is a Banach space then $D$ is called a .

Properties

Every absorbing set contains the origin.
If $D$ is an absorbing disk in a vector space $X$ then there exists an absorbing disk $E$ in $X$ such that $E + E \subseteq D.$

If $A$ is an absorbing subset of $X$ then $X = n A$ and more generally, $X = s_n A$ for any sequence of scalars $s_1, s_2, \ldots$ such that $\left, s_n\ \to \infty.$ Consequently, if a topological vector space $X$ is a non-meager subset of itself (or equivalently for TVSs, if it is a Baire space) and if $A$ is a closed absorbing subset of $X$ then $A$ necessarily contains a non-empty open subset of $X$ (in other words, $A$'s topological interior will not be empty), which guarantees that $A - A$ is a neighborhood of the origin in $X.$

See also

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Notes

Proofs

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{{TopologicalVectorSpaces

Functional analysis