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 to include any element of the vector space. Alternative terms are radial or absorbent set.

Definition

Suppose that X is a vector space over the field 𝕂 of real ℝ or complex numbers ℂ.

Notation
Products of scalars and vectors

Notation: Let A be a subset of X, x ∈ X, K ⊆ 𝕂 a set of scalars, k0 ∈ 𝕂 a scalar, and -∞ ≤ t ≤ T ≤ ∞. Define:

  • K A := { c a : k ∈ K, a ∈ A }
    • k0 A := { k0 a : a ∈ A }.
    • K x := { k x : k ∈ K }.
  • 𝕂 x := { k x : k ∈ 𝕂 } = span { x }
  • ℝ x := { r x : r ∈ ℝ }
  • (t, T) x := { r x : t < r < T } and (t, T) A := { r a : t < r < T, a ∈ A }
Balls in 𝕂

Notation: If r > 0 then denote the open ball of radius r centered at the origin in 𝕂 (where 𝕂 is ℝ or ℂ) by

Br := B𝕂
r := { c ∈ 𝕂 : |c| < r }

and denote the closed ball of radius r > 0 centered at the origin in 𝕂 by

Br := B𝕂
r := { c ∈ 𝕂 : |c| ≤ r }.

One set absorbing another

Definition: If S and A are subsets of X, we say that A absorbs S if it satisfies any of the following equivalent conditions:

  1. There exists a real r > 0 such that S ⊆ c A for any scalar c satisfying |c| ≥ r;
  2. There exists a real r > 0 such that c S ⊆ A for any scalar c ≠ 0 satisfying |c| ≤ r;
    • If it is known that 0 ∈ A then we may remove the restriction c ≠ 0.

If A is balanced then we can add to this list:

  1. There exists a scalar c ≠ 0 such that S ⊆ c A;
  2. There exists a scalar c ≠ 0 such that c S ⊆ A.

Absorbing set

The conditions are by and large ordered so that each condition is an easy consequence of the previous condition.

Definition: A subset A of a vector space X over a field 𝕂 is called absorbing or absorbent in X if it satisfies any of the following equivalent conditions:

  1. For every x ∈ X, A absorbs { x }.
  2. For every x ∈ X, there exists a real r > 0 such that x ∈ c A for any scalar c ∈ 𝕂 satisfying |c| ≥ r.
  3. For every x ∈ X, there exists a real r > 0 such that c x ∈ A for any scalar c ∈ 𝕂 satisfying |c| ≤ r.
  4. For every x ∈ X, there exists a real r > 0 such that Br x ⊆ A.
    • Here, Br := B𝕂
      r       := { c ∈ 𝕂 : |c| < r       } is the open ball of radius r in 𝕂 centered at the origin and Br x := { c x : c ∈ B𝕂
      r       } = { c x : c ∈ 𝕂 and |c| < r       }.
    • The closed ball can be used in place of the open ball.
  5. For every x ∈ X, there exists a real r > 0 such that Br x ⊆ A ∩ 𝕂 x, where 𝕂 x = span { x }.
    • Proof: This follows from the previous condition since Br x ⊆ 𝕂 x, so that Br x ⊆ A if and only if Br x ⊆ A ∩ 𝕂 x.
    • Connection to topology: Note that if 𝕂 x is given its usual Hausdorff Euclidean topology then the set Br x is a neighborhood of the origin in 𝕂 x; thus, there exists a real r > 0 such that Br x ⊆ A ∩ 𝕂 x if and only if A ∩ 𝕂 x is a neighborhood of the origin in 𝕂 x.
    • Note that every 1-dimensional vector subspace of X is of the form 𝕂 x = span { x } for some non-zero x ∈ X and that if the 1-dimensional space 𝕂 x is endowed with the unique Hausdorff vector topology, then the map 𝕂 → 𝕂 x given by c ↦ c x is a TVS-isomorphism (where as usual, 𝕂 has the normed Euclidean topology).
  6. A contains the origin and for every 1-dimensional vector subspace Y of X, A ∩ 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 𝕂 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) X be absorbing: if U is a neighborhood of the origin in X then it would be pathological if there was any 1-dimensional vector subspace Y in which U ∩ Y wasn't a neighborhood of the origin in at least some TVS topology on Y. But the only TVS topologies on Y are the Hausdorff Euclidean topology and the trivial topology, which is contained in the Euclidean topology. Consequently, it is natural to expect for U ∩ Y to be a neighborhood of 0 in the Euclidean topology for all 1-dimensional vector subspaces Y, which is exactly the condition that U be absorbing in X. Thus, it is not surprising that all neighborhoods of the origin in all TVSs are necessarily absorbing. The reason why the Euclidean topology is distinguished is due 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 x ∈ X, A ∩ span { x } is a neighborhood of 0 in span { x } = 𝕂 x when span { x } is given its unique Hausdorff TVS topology.
  7. A contains the origin and for every 1-dimensional vector subspace Y of X, A ∩ 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 only on how A behaves with respect to 1 (or 0) dimensional vector subspaces of X.

If 𝕂 = ℝ then we can add to this list:

  1. The algebraic interior of A contains the origin (i.e. 0 ∈ iA).

If A is balanced then we can add to this list:

  1. For every x ∈ X, there exists a scalar c ≠ 0 such that x ∈ c A.[1]

If A is convex or balanced then we can add to this list:

  1. For every x ∈ X, there exists a positive real r > 0 such that r x ∈ 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[2] and a summary is given below.
      • Summary of proof: Observe that for any 0 ≠ y ∈ X, by assumption we may pick positive real r > 0 and R > 0 such that R y ∈ A and r (-y) ∈ A so that the convex set A ∩ ℝ y contains the open sub-interval (- r, R) y := { t y : - r < t < R, t ∈ ℝ }, which contains the origin (we also call A ∩ ℝ y an interval since any non-empty convex subset of ℝ is an interval). Give 𝕂 y its unique Hausdorff vector topology and we must show that A ∩ 𝕂 y is a neighborhood of the origin in 𝕂 y. If 𝕂 = ℝ then we're done while if 𝕂 = ℂ then note that the set S := (A ∩ ℝ y) ∪ (A ∩ ℝ (i y)) ⊆ A ∩ (ℂ y) is a union of the two intervals, each of which contains an open sub-interval that contains the origin, and that the intersection of these two intervals is precisely the origin. So the convex hull of S, which is contained in the convex set A ∩ (ℂ y), clearly contains the origin in its interior.
  2. For every x ∈ X, there exists a positive real r > 0 such that x ∈ r A.
    • This condition is equivalent to every x ∈ X belonging to the set (0, ∞) A := { r a : 0 < r < ∞, a ∈ A } = ∪0 < r < ∞ r A, which happens if and only if X = (0, ∞) A. One may also show that for any subset T of X, (0, ∞) T = X if and only if T ∩ (0, ∞) x ≠ ∅ for every x ∈ X.
  3. (0, ∞) A = X.
  4. For every x ∈ X, A ∩ (0, ∞) x ≠ ∅, where (0, ∞) x := { r x : 0 < r < ∞ }.

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

Examples and sufficient conditions

For one set to absorb another
  • Let F : X → Y be a linear map between vector spaces and let B ⊆ X and C ⊆ Y be balanced sets. Then C absorbs F(B) if and only if F -1(C) absorbs B.[3]

For a set to be absorbing
  • 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."
  • In a semi normed vector space the unit ball is absorbing.
  • If D ≠ ∅ is a disk in X then span D = ∪∞
    n=1     nD     so that in particular, D is an absorbing subset of span D.[4]
    • Thus if D is a disk in X, then X is absorbing in X if and only if span D = X.
  • The intersection of finite but nonempty family of absorbing sets is absorbing.
  • The union of a nonempty arbitrary family of absorbing sets is absorbing.
  • The image of an absorbing set under a surjective linear operator is again absorbing.
  • The inverse image of an absorbing set (in the codomain) under a linear operator is again absorbing (in the domain).

Properties
See also: Topological vector space § 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 ⊆ D.[5]

See also

References
  1. Narici 2011, pp. 107-110.
  2. Proof: Let X be a vector space over the field 𝕂, with 𝕂 being ℝ or ℂ, and endow the field 𝕂 with its usual normed Euclidean topology. Let A be a convex set such that for every z ∈ X, there exists a positive real r > 0 such that r z ∈ A. Since 0 ∈ A, if dim X = 0 then we're done so assume dim X ≠ 0. Clearly, every non-empty convex subset of the real line ℝ is an interval (possibly open, closed, or half-closed, and possibly bounded or unbounded, and possibly even degenerate (i.e. a single point)). Recall that the intersection of convex sets is convex so that for every 0 ≠ y ∈ X, the sets A ∩ 𝕂 y and A ∩ ℝ y are convex, where now the convexity of A ∩ ℝ y (which contains the origin and is contained in the line ℝ y) implies that A ∩ ℝ y is an interval contained in the line ℝ y := { r y : - ∞ < r < ∞ }. Lemma: We will now prove that if 0 ≠ y ∈ X then the interval A ∩ ℝ y contains an open sub-interval that contains the origin. By assumption, since y ∈ X we can pick some R > 0 such that R y ∈ A and (since - y ∈ X) we can pick some r > 0 such that r (- y) ∈ A, where r (- y) = (- r) y and - r y ≠ R y (since y ≠ 0). Since A ∩ ℝ y is convex and contains the distinct points - r y and R y, it contains the convex hull of the points { - r y, R y }, which (in particular) contains the open sub-interval (- r, R) y := { t y : - r < t < R, t ∈ ℝ }, where this open sub-interval (- r, R) y contains the origin (since take t = 0, and note that - r < t = 0 < R ), which proves the lemma. ∎ Now fix 0 ≠ x ∈ X, let Y := span { x } = 𝕂 x, and note that (since 0 ≠ x ∈ X was arbitrary), to prove that A is absorbing in X it is necessary and sufficient to show that A ∩ Y is a neighborhood of the origin in Y when Y is given its usual Hausdorff Euclidean topology, where recall that this makes the map 𝕂 → 𝕂 x given by c ↦ c x into a TVS-isomorphism. If 𝕂 = ℝ then the fact that the interval A ∩ Y = A ∩ ℝ x contains an open sub-interval around the origin implies that A ∩ Y is a neighborhood of the origin in Y = ℝ x so we're done. So assume that 𝕂 = ℂ. Note that i x ∈ Y = ℂ x, where i = √-1, and that Y = ℂ x = (ℝ x) + (ℝ (i x)) (so naively, ℝ x is the "x-axis" and ℝ (i x) is the "y-axis" of ℂ x so that (0, ∞) x (resp. (0, ∞) (i x)) is the strictly positive x-axis (resp. y-axis) while (-∞, 0) x (resp. (-∞, 0) (i x)) is the strictly negative x-axis (resp. y-axis)). Notice that the set S := (A ∩ ℝ x) ∪ (A ∩ ℝ (i x)), which is contained in A ∩ Y, is a union of two line segments (i.e. intervals) intersecting at the origin, with each segment containing the origin in an open sub-interval. So the convex hull of S, which is contained in the convex set A ∩ Y, clearly contains a convex quadrilateral having the origin in its interior (e.g. pick each of its four vertices/corners from one of the non-empty pairwise disjoint sets A ∩ (-∞, 0) x, A ∩ (0, ∞) x, A ∩ (-∞, 0) (i x), and A ∩ (0, ∞) (i x) with each of these four subsets of S (none of which contain the origin) containing exactly one vertex). This shows that A ∩ Y is a neighborhood of the origin in Y = ℂ x, as desired. ∎
  3. Narici 2011, pp. 441-457.
  4. Narici 2011, pp. 67-113.
  5. Narici 2011, pp. 149-153.
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