Topological vector space

Since every vector topology is translation invariant (i.e. for all x₀ ∈ X, the map X → X defined by x ↦ x₀ + x is a homeomorphism), to define a vector topology it suffices to define a neighborhood basis (or subbasis) for it at the origin.

In general, the set of all balanced and absorbing subsets of a vector space does not satisfy the conditions of this theorem and does not form a neighborhood basis at the origin for any vector topology.[3]

Let X be a vector space and let U_• = (U_i)^∞
_i=1 be a sequence of subsets of X. Each set in the sequence U_• is called a knot of U_• and for every index i, U_i is called the i^th knot of U_•. The setU₁ is called the beginning of U_•. The sequence U_• is/is a:[5][6][7]

• Summative if U_i+1 + U_i+1  ⊆  U_i for every index i.
• Balanced (resp. absorbing, closed,[note 1] convex, open, symmetric, barrelled, absolutely convex/disked, etc.) if this is true of every U_i.
• String if U_• is summative, absorbing, and balanced.
• Topological string or a neighborhood string in a TVS X if U_• is a string and each of its knows is a neighborhood of the origin in X.

If U is an absorbing disk in a vector space X then the sequence defined by U_i ≝ 2^{1 − i} U forms a string beginning with U₁ = U. This is called the natural string of U[5] Moreover, if a vector space X has countable dimension then every string contains an absolutely convex string.

Summative sequences of sets have the particularly nice property that they define non-negative continuous real-valued subadditive functions. These functions can then be used to prove many of the basic properties of topological vector spaces.

If U_• = (U_i)_{i ∈ ℕ} and V_• = (V_i)_{i ∈ ℕ} are two collections of subsets of a vector space X and if s is a scalar, then by definition:[5]

• V_• contains U_•:  U_•  ⊆  V_• if and only if U_i  ⊆  V_i for every index i.
• Set of knots:  Knots (U_•)  ≝  { U_i  :  i ∈ ℕ }.
• Kernel:  ker U_•  ≝  ∩i ∈ ℕ U_i.
• Scalar multiple:  s U_•  ≝  (s U_i)_{i ∈ ℕ}.
• Sum:  U_• + V_•  ≝  (U_i + V_i)_{i ∈ ℕ}.
• Intersection:  U_• ∩ V_•  ≝  (U_i ∩ V_i)_{i ∈ ℕ}.

If 𝕊 is a collection sequences of subsets of X, then 𝕊 is said to be directed (downwards) under inclusion or simply directed if 𝕊 is not empty and for all U_•, V_• ∈ 𝕊 there exists some W_• ∈ 𝕊 such that W_• ⊆ U_• and W_• ⊆ V_• (said differently, if and only if 𝕊 is a prefilter with respect to the containment ⊆ defined above).

Notation: Let Knots (𝕊)  ≝  ∪U_• ∈ 𝕊 Knots (U_•) be the set of all knots of all strings in 𝕊.

Defining vector topologies using collections of strings is particularly useful for defining classes of TVSs that are not necessarily locally convex.

If 𝕊 is the set of all topological strings in a TVS (X, 𝜏) then 𝜏_𝕊 = 𝜏.[5] A Hausdorff TVS is metrizable if and only if its topology can be induced by a single topological string.[8]

One of the most used properties of vector topologies is that every vector topology is translation invariant:

for all x₀ ∈ X, the map X → X defined by x ↦ x₀ + x is a homeomorphism, but if x₀ ≠ 0 then it is not linear and so not a TVS-isomorphism.

Scalar multiplication by a non-zero scalar is a TVS-isomorphism. This means that if s ≠ 0 then the linear map X → X defined by x ↦ s x is a homeomorphism. Using s = −1 produces the negation map X → X defined by x ↦ −x, which is consequently a linear homeomorphism and thus a TVS-isomorphism.

If x ∈ X and any subset S ⊆ X, then  cl(x + S) = x + cl(S)[4] and moreover, if 0 ∈ S then x + S is a neighborhood (resp. open neighborhood, closed neighborhood) of x in X if and only if the same is true of S at the origin.

A subset E of a vector space X is said to be

• absorbing (in X): if for every x ∈ X, there exists a real r > 0 such that c x ∈ E for any scalar c satisfying |c| ≤ r.
• balanced or circled: if tE ⊆ E for every scalar |t| ≤ 1.
• convex: if tE + (1−t)E ⊆ E for every real 0 ≤ t ≤ 1.
• a disk or absolutely convex: if E is convex and balanced.
• symmetric: if −E ⊆ E, or equivalently, if −E = E.

Every neighborhood of 0 is an absorbing set and contains an open balanced neighborhood of 0[4] so every topological vector space has a local base of absorbing and balanced sets. The origin even has a neighborhood basis consisting of closed balanced neighborhoods of 0; if the space is locally convex then it also has a neighborhood basis consisting of closed convex balanced neighborhoods of 0.

Bounded subsets

A subset E of a topological vector space X is bounded[10] if for every neighborhood V of 0, then E ⊆ tV when t is sufficiently large.

The definition of boundedness can be weakened a bit; E is bounded if and only if every countable subset of it is bounded. A set is bounded if and only if each of its subsequences is a bounded set.[11] Also, E is bounded if and only if for every balanced neighborhood V of 0, there exists t such that E ⊆ tV. Moreover, when X is locally convex, the boundedness can be characterized by seminorms: the subset E is bounded if and only if every continuous seminorm p is bounded on E.

Every totally bounded set is bounded.[11] If M is a vector subspace of a TVS X, then a subset of M is bounded in M if and only if it is bounded in X.[11]

A TVS is pseudometrizable if and only if it has a countable neighborhood basis at the origin, or equivalent, if and only if its topology is generated by an F-seminorm. A TVS is metrizable if and only if it is Hausdorff and pseudometrizable.

More strongly: a topological vector space is said to be normable if its topology can be induced by a norm. A topological vector space is normable if and only if it is Hausdorff and has a convex bounded neighborhood of 0.[13]

Let 𝕂 be a non-discrete locally compact topological field, for example the real or complex numbers. A Hausdorff topological vector space over 𝕂 is locally compact if and only if it is finite-dimensional, that is, isomorphic to 𝕂ⁿ for some natural number n.

The canonical uniformity[14] on a TVS (X, τ) is the unique translation-invariant uniformity that induces the topology τ on X.

Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs into uniform spaces. This allows one to about related notions such as completeness, uniform convergence, Cauchy nets, and uniform continuity. etc., which are always assumed to be with respect to this uniformity (unless indicated other). This implies that every Hausdorff topological vector space is Tychonoff.[15] A subset of a TVS is compact if and only if it is complete and totally bounded (for Hausdorff TVSs, a set being totally bounded is equivalent to it being precompact). But if the TVS is not Hausdorff then there exist compact subsets that are not closed. However, the closure of a compact subset of a non-Hausdorff TVS is again compact (so compact subsets are relatively compact).

With respect to this uniformity, a net (or sequence) x_• = (x_i)_{i ∈ I} is Cauchy if and only if for every neighborhood V of 0, there exists some index i such that x_m − x_n ∈ V whenever j ≥ i and k ≥ i.

Every Cauchy sequence is bounded, although Cauchy nets and Cauchy filters may not be bounded. A topological vector space where every Cauchy sequence converges is called sequentially complete; in general, it may not be complete (in the sense that all Cauchy filters converge).

The vector space operation of addition is uniformly continuous and an open map. Scalar multiplication is Cauchy continuous but in general, it is almost never uniformly continuous. Because of this, every topological vector space can be completed and is thus a dense linear subspace of a complete topological vector space.

• Every TVS has a completion and every Hausdorff TVS has a Hausdorff completion.[4] Every TVS (even those that are Hausdorff and/or complete) has infinitely many non-isomorphic non-Hausdorff completions.
• A compact subset of a TVS (not necessarily Hausdorff) is complete.[16] A complete subset of a Hausdorff TVS is closed.[16]
• If C is a complete subset of a TVS then any subset of C that is closed in C is complete.[16]
• A Cauchy sequence in a Hausdorff TVS X is not necessarily relatively compact (i.e. its closure in X is not necessarily compact).
• If a Cauchy filter in a TVS has an accumulation point x then it converges to x.
• If a series ∑^∞
  _i=1 x_i converges[note 5] in a TVS X then x_i → 0 in X.[17]

Let X be a real or complex vector space.

Trivial topology

The trivial topology or indiscrete topology { X, ∅ } is always a TVS topology on any vector space X and it is the coarsest TVS topology possible. An important consequence of this is that the intersection of any collection of TVS topologies on X always contains a TVS topology. Any vector space (including those that are infinite dimensional) endowed with the trivial topology is a compact (and thus locally compact) complete pseudometrizable seminormable locally convex topological vector space. It is Hausdorff if and only if dim X = 0.

Finest vector topology

There exists a TVS topology τ_f on X that is finer than every other TVS-topology on X (that is, any TVS-topology on X is necessarily a subset of 𝜏_f).[18][19] Every linear map from (X, τ_f) into another TVS is necessarily continuous. If X has an uncountable Hamel basis then 𝜏_f is not locally convex and not metrizable.[19]

A Cartesian product of a family of topological vector spaces, when endowed with the product topology, is a topological vector space. Consider for instance the set X of all functions f : ℝ → ℝ where ℝ carries its usual Euclidean topology. This set X is a real vector space (where addition and scalar multiplication are defined pointwise, as usual) that can be identified with (and indeed, is often defined to be) the Cartesian product ℝ^ℝ, which carries the natural product topology. With this product topology, X ≝ ℝ^ℝ becomes a topological vector space whose topology is called the topology of pointwise convergence on ℝ. The reason for this name is the following: if (f_n) is a sequence (or more generally, a net) of elements in X and if f ∈ X then f_n converges to f in X if and only if for every real number x, f_n(x) converges to f(x) in ℝ. This TVS is complete, Hausdorff, and locally convex but not metrizable and consequently not normable; indeed, every neighborhood of the origin in the product topology contains lines (i.e. 1-dimensional vector subspaces, which are subsets of the form ℝ f ≝ { r f : r ∈ ℝ } with f ≠ 0).

Let 𝔽 denote ℝ or ℂ and endow 𝔽 with its usual Hausdorff normed Euclidean topology. Let X be a vector space over 𝔽 of finite dimension n = dim X and so that X is vector space isomorphic to 𝔽ⁿ (explicitly, this means that there exists a linear isomorphism between the vector spaces X and 𝔽ⁿ). This finite-dimensional vector space X always has a unique Hausdorff vector topology, which makes it TVS-isomorphic to 𝔽ⁿ, where 𝔽ⁿ is endowed with the usual Euclidean topology (which is the same as the product topology). This Hausdorff vector topology is also the (unique) finest vector topology on X. X has a unique vector topology if and only if dim X = 0. If dim X ≠ 0 then although X does not have a unique vector topology, it does have a unique Hausdorff vector topology.

• If dim X = 0 then X = { 0 } has exactly one vector topology: the trivial topology, which in this case (and only in this case) is Hausdorff. The trivial topology on a vector space is Hausdorff if and only if the vector space has dimension 0.
• If dim X = 1 then X has two vector topologies: the usual Euclidean topology and the (non-Hausdorff) trivial topology.
  • Since the field 𝔽 is itself a 1-dimensional topological vector space over 𝔽 and since it plays an important role in the definition of topological vector spaces, this dichotomy plays an important role in the definition of an absorbing set and has consequences that reverberate throughout functional analysis.

  Proof outline

  The proof of this dichotomy is straightforward so only an outline with the important observations is given. As usual, 𝔽 is assumed have the (normed) Euclidean topology. Let X be a 1-dimensional vector space over 𝔽. Observe that if B ⊆ 𝔽 is a ball centered at 0 and if S ⊆ X is a subset containing an "unbounded sequence" then B ⋅ S = X, where an "unbounded sequence" means a sequence of the form (si x)∞ i=1 where 0 ≠ x ∈ X and (si)∞ i=1 ⊆ 𝔽 is unbounded in normed space 𝔽. Any vector topology on X will be translation invariant and invariant under non-zero scalar multiplication, and for every 0 ≠ x ∈ X, the map Mx : 𝔽 → X given by Mx (s) ≝ s x is a continuous linear bijection. In particular, for any such x, X = 𝔽 x so every subset of X can be written as F x = Mx(F) for some unique subset F ⊆ 𝔽. And if this vector topology on X has a neighborhood of 0 that is properly contained in X, then the continuity of scalar multiplication 𝔽 × X → X at the origin forces the existence of an open neighborhood of the origin in X that doesn't contain any "unbounded sequence". From this, one deduces that if X doesn't carry the trivial topology and if 0 ≠ x ∈ X, then for any ball B ⊆ 𝔽 center at 0 in 𝔽, Mx (B) = B x contains an open neighborhood of the origin in X so that Mx is thus a linear homeomorphism. ∎

• If dim X = n ≥ 2 then X has infinitely many distinct vector topologies:
  • Some of these topologies are now described: Every linear functional f on X, which is vector space isomorphic to 𝔽ⁿ, induces a seminorm |f| : X → ℝ defined by |f|(x) = |f (x)| where ker f = ker |f|. Every seminorm induces a (pseudometrizable locally convex) vector topology on X and seminorms with distinct kernels induce distinct topologies so that in particular, seminorms on X that are induced by linear functionals with distinct kernel will induces distinct vector topologies on X.
  • However, while there are infinitely many vector topologies on X when dim X ≥ 2, there are, up to TVS-isomorphism only 1 + dim X vector topologies on X. For instance, if n ≝ dim X = 2 then the vector topologies on X consist of the trivial topology, the Hausdorff Euclidean topology, and then the infinitely many remaining non-trivial non-Euclidean vector topologies on X are all TVS-isomorphic to one another.

Discrete and cofinite topologies

If X is a non-trivial vector space (i.e. of non-zero dimension) then the discrete topology on X (which is always metrizable) is not a TVS topology because despite making addition and negation continuous (which makes it into a topological group under addition), it fails to make scalar multiplication continuous. The cofinite topology on X (where a subset is open if and only if its complement is finite) is also not a TVS topology on X.

Properties of neighborhoods and open sets

• The open convex subsets of a TVS X (not necessarily Hausdorff or locally convex) are exactly those that are of the form z + { x ∈ X : p(x) < 1 } = { x ∈ X : p(x − z) < 1 } for some z ∈ X and some positive continuous sublinear functional p on X.[20]
• If S ⊆ X and U is an open subset of X then S + U is an open set in X.[4]
• If S ⊆ X has non-empty interior then S − S is a neighborhood of 0.[4]
• If K is an absorbing disk in a TVS X and if p ≝ p_K is the Minkowski functional of K then[21]

  Int K   ⊆   { x ∈ X : p(x) < 1 }   ⊆   K   ⊆   { x ∈ X : p(x) ≤ 1 }   ⊆   cl K

  • It was not assumed that K had any topological properties nor that p was continuous (which happens if and only if K is a neighborhood of 0).
• Every TVS is connected[4] and locally connected.[22] Any connected open subset of a TVS is arcwise connected.
• Let 𝜏 and 𝜐 be two vector topologies on X. Then 𝜏 ⊆ 𝜐 if and only if whenever a net x_• = (x_i)_{i ∈ I} in X converges 0 in (X, 𝜐) then x_• → 0 in (X, 𝜏).[23]
• Let 𝒩 be a neighborhood basis of the origin in X, let S ⊆ X, and let x ∈ X. Then x ∈ cl S if and only if there exists a net s_• = (s_N)_{N ∈ 𝒩} in S (indexed by 𝒩) such that s_• → x in X.[24][note 6]

Interior

• If S has non-empty interior then Int S = Int(cl S) and cl S = cl(Int S).
• If R, S ⊆ X and S has non-empty interior then Int(R) + Int(S) ⊆ R + Int(S) ⊆ Int(R+S).
• If S is a disk in X that has non-empty interior then 0 belongs to the interior of S.[25]
  • However, a closed balanced subset of X with non-empty interior may fail to contain 0 in its interior.[25]
• If S is a balanced subset of X with non-empty interior then { 0 } ∪ Int S is balanced; in particular, if the interior of a balanced set contains the origin then Int S is balanced.[4][note 7]
• If x belongs to the interior of a convex set S ⊆ X and y ∈ cl_X S, then the half-open line segment [x, y) ≝ {tx + (1 − t)y : 0 < t ≤ 1} ⊆ Int S.[26] If N is a balanced neighborhood of 0 in X then by considering intersections of the form N ∩ ℝ x (which are convex symmetric neighborhoods of 0 in the real TVS ℝ x) it follows that:
  • Int N  =  [0, 1) Int N  =  [0, 1) N  =  (−1, 1) N  =  B₁ N, where B₁ ≝ { a ∈ 𝕂 : |a| < 1 .
  • if x ∈ Int N and r ≝ sup { r > 0 : [0, r) x ⊆ Int N} then r > 1, [0, r) x ⊆ Int N, and if r ≠ ∞ then r x ∈ cl N ∖ Int N.
• If C is convex and 0 < t ≤ 1, then t Int C + (1 − t) cl C ⊆ Int C.[27]

• X is Hausdorff if and only if { 0 } is closed in X.
• cl_X { 0 } = ∩N ∈ 𝒩(0) N so every neighborhood of the origin contains the closure of { 0 .
• cl_X { 0 } is a vector subspace of X and its subspace topology is the trivial topology (which makes cl_X { 0 } compact).
• Every subset of cl_X { 0 } is compact and thus complete (see footnote for a proof).[proof 2] In particular, if X is not Hausdorff then there exist compact complete subsets that are not closed.[28]
• S + cl_X { 0 }  ⊆  cl_X S for every subset S ⊆ X.[proof 3]
  • So if S ⊆ X is open or closed in X then S + cl_X { 0 }  =  S (so S is a "tube" with vertical side cl_X { 0  ).
  • The quotient map q : X → X / cl_X { 0 } is a closed map onto a Hausdorff TVS.[29]
• A subset S of a TVS X is totally bounded if and only if S + cl { 0 } is totally bounded,[30] if and only if cl_X S is totally bounded,[31][32] if and only if its image under the canonical quotient map X → X / cl_X ({ 0 }) is totally bounded.[30]
• If S ⊆ X is compact, then cl_X S  = S + cl_X { 0 } and this set is compact. Thus the closure of a compact set is compact[note 8] (i.e. all compact sets are relatively compact).[33]
• A vector subspace of a TVS is bounded if and only if it is contained in the closure of { 0  .[11]
• If M is a vector subspace of a TVS X then X / M is Hausdorff if and only if M is closed in X.
• Every vector subspace of X that is an algebraic complement of cl { 0 } is a topological complement of cl { 0  . Thus if H is an algebraic complement of cl_X { 0 } in X then the addition map H × cl_X { 0 } → X, defined by (h, n) ↦ h + n is a TVS-isomorphism, where H is Hausdorff and cl { 0 } has the indiscrete topology.[34] Moreover, if C is a Hausdorff completion of H then C × cl_X { 0 } is a completion of X ≅ H × cl_X { 0 }.[30]

Compact and totally bounded sets

• A subset of a TVS is compact if and only if it is complete and totally bounded.[28]
  • Thus, in a complete TVS, a closed and totally bounded subset is compact.[28]
• A subset S of a TVS X is totally bounded if and only if cl_X S is totally bounded,[31][32] if and only if its image under the canonical quotient map X → X / cl_X ({ 0 }) is totally bounded.[30]
• Every relatively compact set is totally bounded.[28] The closure of a totally bounded set is totally bounded.[28]
• The image of a totally bounded set under a uniformly continuous map (e.g. a continuous linear map) is totally bounded.[28]
• If K is a compact subset of a TVS X and U is an open subset of X containing K, then there exists a neighborhood N of 0 such that K + N ⊆ U.[35]
• If S is a subset of a TVS X such that every sequence in S has a cluster point in S then S is totally bounded.[30]

Closure and closed set

• If S ⊆ X and a is a scalar then acl(S) ⊆ cl(aS); if X is Hausdorff, a ≠ 0, or S = ∅ then equality holds: cl(aS) = acl(S).
  • In particular, every non-zero scalar multiple of a closed set is closed.
• If S ⊆ X and S + S ⊆ 2 cl S then cl S is convex.[36]
• If R, S ⊆ X then cl(R) + cl(S) ⊆ cl(R+S) and cl[cl(R) + cl(S)] = cl(R+S).[4] Thus if R + S is closed then so is cl(R) + cl(S).[36]
• If S ⊆ X and if R is a set of scalars such that neither cl S nor cl R contain zero then (cl R) (cl S) = cl(RS).[36]
• The closure of a vector subspace of a TVS is a vector subspace.
• If S ⊆ X then cl S = ∩N ∈ 𝒩 (S + N) where 𝒩 is any neighborhood basis at the origin for X.[37]
  • However, cl S  ⊇  ∩ { U : S ⊆ U, U open in X } and it's possible for this containment to be proper[38] (e.g. if X = ℝ and S is the rational numbers).
  • It follows that cl U ⊆ U + U for every neighborhood U of the origin in X.[39]
• If X is a real TVS and S ⊆ X, then ∩r > 1 rS ⊆ cl S (observe that the left hand side is independent of the topology on X); if S is a convex neighborhood of the origin then equality holds.
• The sum of a compact set and a closed set is closed. However, the sum of two closed subsets may fail to be closed[4] (see this footnote[note 9] for examples).
• If M is a vector subspace of X and N is a closed neighborhood of 0 in X such that U ∩ N is closed in X then M is closed in X.[35]
• Every finite dimensional vector subspace of a Hausdorff TVS is closed. The sum of a closed vector subspace and a finite-dimensional vector subspace is closed.[4]

Closed hulls

• In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.[11]
• The closed convex hull of a set is equal to the closure of the convex hull of that set (i.e. to cl(co(S))).[4]
• The closed balanced hull of a set is equal to the closure of the balanced hull of that set (i.e. to cl(bal(S))).[4]
• The closed disked hull of a set is equal to the closure of the disked hull of that set (i.e. to cl(cobal(S))).[4]
• If R, S ⊆ X and the closed convex hull of one of the sets S or R is compact then cl(co(R)) + cl(co(S)) = cl(co(R+S)).[4]
• If R, S ⊆ X each have a closed convex hull that is compact (i.e. cl(co(R)) and cl(co(S)) are compact) then cl(co(R ∪ S))  =  co[cl(co(R)) ∪ cl(co(S))].

Hulls and compactness

• In a general TVS, the closed convex hull of a compact set may fail to be compact.
• The balanced hull of a compact (resp. totally bounded) set has that same property.[4]
• The convex hull of a finite union of compact convex sets is again compact and convex.[4]

Meager, nowhere dense, and Baire

• A disk in a TVS is not nowhere dense if and only if its closure is a neighborhood of the origin.[7]
• A vector subspace of a TVS that is closed but not open is nowhere dense.[7]
• Suppose X is a TVS that does not carry the indiscrete topology. Then X is a Baire space if and only if X has no balanced absorbing nowhere dense subset.[7]
• A TVS X is a Baire space if and only if X is nonmeager, which happens if and only if there does not exist a nowhere dense set D such that X = ∪n ∈ ℕ nD.[7]
  • Every nonmeager locally convex TVS is a barrelled space.[7]

Important algebraic facts and common misconceptions

• If S ⊆ X then 2S ⊆ S + S; if S is convex then equality holds.
  • For an example where equality does not hold, let x be non-zero and set S = { −x, x }; S = { x, 2x } also works.
• A subset C is convex if and only if (s + t)C = sC + tC for all positive real s and t.[40]
• The disked hull of a set S ⊆ X is equal to the convex hull of the balanced hull of S (i.e. to co(bal(S))).
  • However, in general co(bal(S)) ≠ bal(co(S)).
• If R, S ⊆ X and a is a scalar then co(R + S) = co R + co S and co(aS) = aco S.[4]
• If R, S ⊆ X are convex non-empty disjoint sets and x ∉ R ∪ S, then S ∩ co(R ∪ { x }) = ∅ or R ∩ co(S ∪ { x }) = ∅.
• In any non-trivial vector space X, there exist two disjoint non-empty convex subsets whose union is X.

Other properties

• Every TVS topology can be generated by a family of F-seminorms.[41]

• The balanced hull of a compact (resp. totally bounded, open) set has that same property.[4]
• The (Minkowski) sum of two compact (resp. bounded, balanced, convex) sets has that same property.[4] But the sum of two closed sets need not be closed.
• The convex hull of a balanced (resp. open) set is balanced (resp. open). However, the convex hull of a closed set need not be closed.[4] And the convex hull of a bounded set need not be bounded.

The following table, the color of each cell indicates whether or not a given property of subsets of X (indicated by the column name e.g. "convex") is preserved under the set operator (indicated by the row's name e.g. "closure"). If in every TVS, a property is preserved under the indicated set operator then that cell will be colored green; otherwise, it will be colored red.

So for instance, since the union of two absorbing sets is again absorbing, the cell in row "R∪S" and column "Absorbing" is colored green. But since the arbitrary intersection of absorbing sets need not be absorbing, the cell in row "Arbitrary intersections (of at least 1 set)" and column "Absorbing" is colored red. If a cell is not colored then that information has yet to be filled in.