Topological vector space
In mathematics, a topological vector space (also called a linear topological space and commonly abbreviated TVS or t.v.s.) is one of the basic structures investigated in functional analysis. A topological vector space is a vector space which is also a topological space, and where the vector space operations are continuous functions. This implies that the space has a uniform topological structure, allowing a notion of uniform convergence. Some authors also require that the space is a Hausdorff space.
The elements of topological vector spaces are typically functions or linear operators acting on topological vector spaces, and the topology is often defined so as to capture a particular notion of convergence of sequences of functions.
- The vector addition is jointly continuous with respect to this topology. This follows directly from the triangle inequality obeyed by the norm.
- The scalar multiplication where is the underlying scalar field of is jointly continuous. This follows from the triangle inequality and homogeneity of the norm.
There are topological vector spaces whose topology is not induced by a norm, but are still of interest in analysis. Examples of such spaces are spaces of holomorphic functions on an open domain, spaces of infinitely differentiable functions, the Schwartz spaces, and spaces of test functions and the spaces of distributions on them. These are all examples of Montel spaces. An infinite-dimensional Montel space is never normable. The existence of a norm for a given topological vector space is characterized by Kolmogorov's normability criterion.
A topological vector space (TVS) is a vector space over a topological field (most often the real or complex numbers with their standard topologies) that is endowed with a topology such that vector addition and scalar multiplication are continuous functions (where the domains of these functions are endowed with product topologies). Such a topology is called a vector topology or a TVS topology on
Every topological vector space is also a commutative topological group under addition.
Many authors (for example, Walter Rudin), but not this page, require the topology on to be T1; it then follows that the space is Hausdorff, and even Tychonoff. A topological vector space is said to be separated if it is Hausdorff; importantly, "separated" does not mean separable. The topological and linear algebraic structures can be tied together even more closely with additional assumptions, the most common of which are listed below.
Category and morphisms
The category of topological vector spaces over a given topological field is commonly denoted TVS or TVect. The objects are the topological vector spaces over and the morphisms are the continuous -linear maps from one object to another.
A topological vector space homomorphism (abbreviated TVS homomorphism) or topological homomorphism is a continuous linear map between topological vector spaces (TVSs) such that the induced map is an open mapping when which is the range or image of is given the subspace topology induced by Y.
A topological vector space embedding (abbreviated TVS embedding) or a topological monomorphism is an injective topological homomorphism. Equivalently, a TVS-embedding is a linear map that is also a topological embedding.
A topological vector space isomorphism (abbreviated TVS isomorphism), also called a topological vector isomorphism  or an isomorphism in the category of TVSs, is a bijective linear homeomorphism. Equivalently, it is a surjective TVS embedding
A necessary condition for a vector topology
A collection of subsets of a vector space is called additive if for every there exists some such that
Characterization of continuity of addition at  — If is a group (as all vector spaces are), is a topology on and is endowed with the product topology, then the addition map (defined by ) is continuous at the origin of if and only if the set of neighborhoods of the origin in is additive. This statement remains true if the word "neighborhood" is replaced by "open neighborhood."
All of the above conditions are consequently a necessity for a topology to form a vector topology.
Defining topologies using neighborhoods of the origin
Since every vector topology is translation invariant (which means that for all the map defined by is a homeomorphism), to define a vector topology it suffices to define a neighborhood basis (or subbasis) for it at the origin.
Theorem (Neighborhood filter of the origin) — Suppose that is a real or complex vector space. If is a non-empty additive collection of balanced and absorbing subsets of then is a neighborhood base at for a vector topology on That is, the assumptions are that is a filter base that satisfies the following conditions:
If satisfies the above two conditions but is not a filter base then it will form a neighborhood subbasis at (rather than a neighborhood basis) for a vector topology on
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.
Defining topologies using strings
Let be a vector space and let be a sequence of subsets of Each set in the sequence is called a knot of and for every index is called the th knot of The set is called the beginning of The sequence is/is a:
- Summative if for every index
- Balanced (resp. absorbing, closed,[note 1] convex, open, symmetric, barrelled, absolutely convex/disked, etc.) if this is true of every
- String if is summative, absorbing, and balanced.
- Topological string or a neighborhood string in a TVS if is a string and each of its knots is a neighborhood of the origin in
If is an absorbing disk in a vector space then the sequence defined by forms a string beginning with This is called the natural string of  Moreover, if a vector space 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.
Theorem (-valued function induced by a string) — Let be a collection of subsets of a vector space such that and for all For all let
Define by if and otherwise let
Then is subadditive (meaning for all ) and on so in particular If all are symmetric sets then and if all are balanced then for all scalars such that and all If is a topological vector space and if all are neighborhoods of the origin then is continuous, where if in addition is Hausdorff and forms a basis of balanced neighborhoods of the origin in then is a metric defining the vector topology on
A proof of the above theorem is given in the article on metrizable TVSs.
If and are two collections of subsets of a vector space and if is a scalar, then by definition:
- contains : if and only if for every index
- Set of knots:
- Scalar multiple:
If is a collection sequences of subsets of then is said to be directed (downwards) under inclusion or simply directed if is not empty and for all there exists some such that and (said differently, if and only if is a prefilter with respect to the containment defined above).
Notation: Let be the set of all knots of all strings in
Defining vector topologies using collections of strings is particularly useful for defining classes of TVSs that are not necessarily locally convex.
Theorem (Topology induced by strings) — If is a topological vector space then there exists a set [proof 1] of neighborhood strings in that is directed downward and such that the set of all knots of all strings in is a neighborhood basis at the origin for Such a collection of strings is said to be fundamental.
Conversely, if is a vector space and if is a collection of strings in that is directed downward, then the set of all knots of all strings in forms a neighborhood basis at the origin for a vector topology on In this case, this topology is denoted by and it is called the topology generated by .
A vector space is an abelian group with respect to the operation of addition, and in a topological vector space the inverse operation is always continuous (since it is the same as multiplication by −1). Hence, every topological vector space is an abelian topological group. Every TVS is completely regular but a TVS need not be normal.
Let be a topological vector space. Given a subspace the quotient space with the usual quotient topology is a Hausdorff topological vector space if and only if is closed.[note 2] This permits the following construction: given a topological vector space (that is probably not Hausdorff), form the quotient space where is the closure of is then a Hausdorff topological vector space that can be studied instead of
Invariance of vector topologies
One of the most used properties of vector topologies is that every vector topology is translation invariant:
- for all the map defined by is a homeomorphism, but if 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 then the linear map defined by is a homeomorphism. Using produces the negation map defined by which is consequently a linear homeomorphism and thus a TVS-isomorphism.
A subset of a vector space is said to be
- absorbing (in ): if for every there exists a real such that for any scalar satisfying
- balanced or circled: if for every scalar
- convex: if for every real
- a disk or absolutely convex: if is convex and balanced.
- symmetric: if or equivalently, if
Every neighborhood of 0 is an absorbing set and contains an open balanced neighborhood of  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.
The definition of boundedness can be weakened a bit; 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. Also, is bounded if and only if for every balanced neighborhood of 0, there exists such that Moreover, when is locally convex, the boundedness can be characterized by seminorms: the subset is bounded if and only if every continuous seminorm is bounded on
- The origin is closed in and there is a countable basis of neighborhoods for 0 in
- is metrizable (as a topological space).
- There is a translation-invariant metric on that induces on the topology which is the given topology on
- is a metrizable topological vector space.[note 4]
By the Birkhoff–Kakutani theorem, it follows that there is an equivalent metric that is translation-invariant.
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 
Let be a non-discrete locally compact topological field, for example the real or complex numbers. A Hausdorff topological vector space over is locally compact if and only if it is finite-dimensional, that is, isomorphic to for some natural number
Completeness and uniform structure
Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs into uniform spaces. This allows one to[clarification needed] 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. A subspace 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) is Cauchy if and only if for every neighborhood of there exists some index such that whenever and
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. 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. A complete subset of a Hausdorff TVS is closed.
- If is a complete subset of a TVS then any subset of that is closed in is complete.
- A Cauchy sequence in a Hausdorff TVS is not necessarily relatively compact (that is, its closure in is not necessarily compact).
- If a Cauchy filter in a TVS has an accumulation point then it converges to
- If a series converges[note 5] in a TVS then in 
Finest and coarsest vector topology
Let be a real or complex vector space.
The trivial topology or indiscrete topology is always a TVS topology on any vector space and it is the coarsest TVS topology possible. An important consequence of this is that the intersection of any collection of TVS topologies on 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
Finest vector topology
There exists a TVS topology on called the finest vector topology on that is finer than every other TVS-topology on (that is, any TVS-topology on is necessarily a subset of ). Every linear map from into another TVS is necessarily continuous. If has an uncountable Hamel basis then is not locally convex and not metrizable.
Product vector spaces
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 of all functions where carries its usual Euclidean topology. This set 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 which carries the natural product topology. With this product topology, becomes a topological vector space whose topology is called the topology of pointwise convergence on . The reason for this name is the following: if is a sequence (or more generally, a net) of elements in and if then converges to in if and only if for every real number converges to in 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 (that is, 1-dimensional vector subspaces, which are subsets of the form with ).
Let denote or and endow with its usual Hausdorff normed Euclidean topology. Let be a vector space over of finite dimension and so that is vector space isomorphic to (explicitly, this means that there exists a linear isomorphism between the vector spaces and ). This finite-dimensional vector space always has a unique Hausdorff vector topology, which makes it TVS-isomorphic to , where 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 has a unique vector topology if and only if If then although does not have a unique vector topology, it does have a unique Hausdorff vector topology.
- If then 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
- If then has two vector topologies: the usual Euclidean topology and the (non-Hausdorff) trivial topology.
- Since the field is itself a 1-dimensional topological vector space over 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.
The proof of this dichotomy is straightforward so only an outline with the important observations is given. As usual, is assumed have the (normed) Euclidean topology. Let for all . Let be a 1-dimensional vector space over . If and is a ball centered at 0 then whenever contains an "unbounded sequence", by which it is meant a sequence of the form where and is unbounded in normed space (in the usual sense). Any vector topology on will be translation invariant and invariant under non-zero scalar multiplication, and for every , the map given by is a continuous linear bijection. Because for any such , every subset of can be written as for some unique subset . And if this vector topology on has a neighborhood of the origin that is not equal to all of , then the continuity of scalar multiplication at the origin guarantees the existence of an open ball centered at and an open neighborhood of the origin in such that , which implies that does not contain any "unbounded sequence". This implies that for every , there exists some positive integer such that . From this, it can be deduced that if does not carry the trivial topology and if , then for any ball center at 0 in , contains an open neighborhood of the origin in , which then proves that is a linear homeomorphism.
- If then has infinitely many distinct vector topologies:
- Some of these topologies are now described: Every linear functional on , which is vector space isomorphic to , induces a seminorm defined by where . Every seminorm induces a (pseudometrizable locally convex) vector topology on and seminorms with distinct kernels induce distinct topologies so that in particular, seminorms on that are induced by linear functionals with distinct kernel will induces distinct vector topologies on .
- However, while there are infinitely many vector topologies on when , there are, up to TVS-isomorphism only vector topologies on . For instance, if then the vector topologies on consist of the trivial topology, the Hausdorff Euclidean topology, and then the infinitely many remaining non-trivial non-Euclidean vector topologies on are all TVS-isomorphic to one another.
Discrete and cofinite topologies
If is a non-trivial vector space (that is, of non-zero dimension) then the discrete topology on (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 (where a subset is open if and only if its complement is finite) is also not a TVS topology on
A linear operator between two topological vector spaces which is continuous at one point is continuous on the whole domain. Moreover, a linear operator is continuous if is bounded (as defined below) for some neighborhood of the origin.
A hyperplane on a topological vector space is either dense or closed. A linear functional on a topological vector space has either dense or closed kernel. Moreover, is continuous if and only if its kernel is closed.
Depending on the application additional constraints are usually enforced on the topological structure of the space. In fact, several principal results in functional analysis fail to hold in general for topological vector spaces: the closed graph theorem, the open mapping theorem, and the fact that the dual space of the space separates points in the space.
Below are some common topological vector spaces, roughly in order of increasing "niceness."
- F-spaces are complete topological vector spaces with a translation-invariant metric. These include spaces for all
- Locally convex topological vector spaces: here each point has a local base consisting of convex sets. By a technique known as Minkowski functionals it can be shown that a space is locally convex if and only if its topology can be defined by a family of seminorms. Local convexity is the minimum requirement for "geometrical" arguments like the Hahn–Banach theorem. The spaces are locally convex (in fact, Banach spaces) for all but not for
- Barrelled spaces: locally convex spaces where the Banach–Steinhaus theorem holds.
- Bornological space: a locally convex space where the continuous linear operators to any locally convex space are exactly the bounded linear operators.
- Stereotype space: a locally convex space satisfying a variant of reflexivity condition, where the dual space is endowed with the topology of uniform convergence on totally bounded sets.
- Montel space: a barrelled space where every closed and bounded set is compact
- Fréchet spaces: these are complete locally convex spaces where the topology comes from a translation-invariant metric, or equivalently: from a countable family of seminorms. Many interesting spaces of functions fall into this class -- is a Fréchet space under the seminorms . A locally convex F-space is a Fréchet space.
- LF-spaces are limits of Fréchet spaces. ILH spaces are inverse limits of Hilbert spaces.
- Nuclear spaces: these are locally convex spaces with the property that every bounded map from the nuclear space to an arbitrary Banach space is a nuclear operator.
- Normed spaces and seminormed spaces: locally convex spaces where the topology can be described by a single norm or seminorm. In normed spaces a linear operator is continuous if and only if it is bounded.
- Banach spaces: Complete normed vector spaces. Most of functional analysis is formulated for Banach spaces. This class includes the spaces with , the space of functions of bounded variation, and certain spaces of measures.
- Reflexive Banach spaces: Banach spaces naturally isomorphic to their double dual (see below), which ensures that some geometrical arguments can be carried out. An important example which is not reflexive is , whose dual is but is strictly contained in the dual of
- Hilbert spaces: these have an inner product; even though these spaces may be infinite-dimensional, most geometrical reasoning familiar from finite dimensions can be carried out in them. These include spaces, the Sobolev spaces , and Hardy spaces.
- Euclidean spaces: or with the topology induced by the standard inner product. As pointed out in the preceding section, for a given finite there is only one -dimensional topological vector space, up to isomorphism. It follows from this that any finite-dimensional subspace of a TVS is closed. A characterization of finite dimensionality is that a Hausdorff TVS is locally compact if and only if it is finite-dimensional (therefore isomorphic to some Euclidean space).
Every topological vector space has a continuous dual space—the set of all continuous linear functionals, that is, continuous linear maps from the space into the base field A topology on the dual can be defined to be the coarsest topology such that the dual pairing each point evaluation is continuous. This turns the dual into a locally convex topological vector space. This topology is called the weak-* topology. This may not be the only natural topology on the dual space; for instance, the dual of a normed space has a natural norm defined on it. However, it is very important in applications because of its compactness properties (see Banach–Alaoglu theorem). Caution: Whenever is a non-normable locally convex space, then the pairing map is never continuous, no matter which vector space topology one chooses on A topological vector space has a non-trivial continuous dual space if and only if it has a proper convex neighborhood of the origin.
For any of a TVS the convex (resp. balanced, disked, closed convex, closed balanced, closed disked') hull of is the smallest subset of that has this property and contains The closure (respectively, interior, convex hull, balanced hull, disked hull) of a set is sometimes denoted by (respectively, ).
The convex hull of a subset is equal to the set of all convex combinations of elements in which are finite linear combinations of the form where is an integer, and sum to  The intersection of any family of convex sets is convex and the convex hull of a subset is equal to the intersection of all convex sets that contain it.
Neighborhoods and open sets
Properties of neighborhoods and open sets
Every TVS is connected and locally connected and any connected open subset of a TVS is arcwise connected. If and is an open subset of then is an open set in  and if has non-empty interior then is a neighborhood of the origin.
The open convex subsets of a TVS (not necessarily Hausdorff or locally convex) are exactly those that are of the form
Let and be two vector topologies on Then if and only if whenever a net in converges in then in 
Let be a neighborhood basis of the origin in let and let Then if and only if there exists a net in (indexed by ) such that in  This shows, in particular, that it will often suffice to consider nets indexed by a neighborhood basis of the origin rather than nets on arbitrary directed sets.
If is a TVS that is of the second category in itself (that is, a nonmeager space) then any closed convex absorbing subset of is a neighborhood of the origin. This is no longer guaranteed if the set is not convex (a counter-example exists even in ) or if is not of the second category in itself.
If and has non-empty interior then
If is a disk in that has non-empty interior then the origin belongs to the interior of  However, a closed balanced subset of with non-empty interior may fail to contain the origin in its interior.
If is convex and then
If is any balanced neighborhood of the origin in then where is the set of all scalars such that
If belongs to the interior of a convex set and then the half-open line segment and  If is a balanced neighborhood of in and then by considering intersections of the form (which are convex symmetric neighborhoods of in the real TVS ) it follows that: and furthermore, if then and if then
Non-Hausdorff spaces and the closure of the origin
A topological vector space is Hausdorff if and only if is a closed subset of or equivalently, if and only if Because is a vector subspace of the same is true of its closure which is referred to as the closure of the origin in This vector space satisfies
If is compact, then and this set is compact. Thus the closure of a compact subset of a TVS is compact (said differently, all compact sets are relatively compact), which is not guaranteed for arbitrary non-Hausdorff topological spaces.[note 7]
For every subset
- is totally bounded.
- is totally bounded.
- is totally bounded.
- The image if under the canonical quotient map is totally bounded.
Every vector subspace of that is an algebraic complement of (that is, a vector subspace that satisfies and ) is a topological complement of Consequently, if is an algebraic complement of in then the addition map defined by is a TVS-isomorphism, where is necessarily Hausdorff and has the indiscrete topology. Moreover, if is a Hausdorff completion of then is a completion of 
Closed and compact sets
Compact and totally bounded sets
A subset of a TVS is compact if and only if it is complete and totally bounded. Thus, in a complete topological vector space, a closed and totally bounded subset is compact. A subset of a TVS is totally bounded if and only if is totally bounded, if and only if its image under the canonical quotient map
Every relatively compact set is totally bounded and the closure of a totally bounded set is totally bounded. The image of a totally bounded set under a uniformly continuous map (such as a continuous linear map for instance) is totally bounded. If is a subset of a TVS such that every sequence in has a cluster point in then is totally bounded.
If is a compact subset of a TVS and is an open subset of containing then there exists a neighborhood of 0 such that 
Closure and closed set
The closure of any convex (respectively, any balanced, any absorbing) subset of any TVS has this same property. In particular, the closure of any convex, balanced, and absorbing subset is a barrel.
The closure of a vector subspace of a TVS is a vector subspace. 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. If is a vector subspace of and is a closed neighborhood of the origin in such that is closed in then is closed in  The sum of a compact set and a closed set is closed. However, the sum of two closed subsets may fail to be closed (see this footnote[note 8] for examples).
If and is a scalar then
If then is convex.
If is a real TVS and then
For any subset
In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.
- The closed convex hull of a set is equal to the closure of the convex hull of that set; that is, equal to 
- The closed balanced hull of a set is equal to the closure of the balanced hull of that set; that is, equal to 
- The closed disked hull of a set is equal to the closure of the disked hull of that set; that is, equal to 
If and the closed convex hull of one of the sets or is compact then
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. The convex hull of a finite union of compact convex sets is again compact and convex.
Meager, nowhere dense, and Baire
Important algebraic facts and common misconceptions
If then ; if is convex then equality holds. For an example where equality does not hold, let be non-zero and set also works.
If and is a scalar then
In any non-trivial vector space there exist two disjoint non-empty convex subsets whose union is
If is some unary predicate (a true or false statement dependent on ) then for any [proof 4] So for example, if denotes "" then for any Similarly, if is a scalar then The elements of these sets must range over a vector space (that is, over ) rather than not just a subset or else these equalities are no longer guaranteed; similarly, must belong to this vector space (that is, ).
Properties preserved by set operators
- The balanced hull of a compact (respectively, totally bounded, open) set has that same property.
- The (Minkowski) sum of two compact (respectively, bounded, balanced, convex) sets has that same property. But the sum of two closed sets need not be closed.
- The convex hull of a balanced (resp. open) set is balanced (respectively, open). However, the convex hull of a closed set need not be closed. 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 (indicated by the column name, "convex" for instance) is preserved under the set operator (indicated by the row's name, "closure" for instance). 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 "" 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.
|Operation||Property of and any other subsets of that is considered|
dense (in )
|of increasing nonempty chain||of increasing nonempty chain|
|Arbitrary unions (of at least 1 set)||Arbitrary unions (of at least 1 set)|
|of decreasing nonempty chain||of decreasing nonempty chain|
|Arbitrary intersections (of at least 1 set)||Arbitrary intersections (of at least 1 set)|
|Scalar multiple||Scalar multiple|
|Non-0 scalar multiple||Non-0 scalar multiple|
|Positive scalar multiple||Positive scalar multiple|
|Balanced core||Balanced core|
|Balanced hull||Balanced hull|
|Convex hull||Convex hull|
|Convex balanced hull||Convex balanced hull|
|Closed balanced hull||Closed balanced hull|
|Closed convex hull||Closed convex hull|
|Closed convex balanced hull||Closed convex balanced hull|
|Linear span||Linear span|
|Pre-image under a continuous linear map||Pre-image under a continuous linear map|
|Image under a continuous linear map||Image under a continuous linear map|
|Image under a continuous linear surjection||Image under a continuous linear surjection|
|Non-empty subset of||Non-empty subset of|
dense (in )
- Banach space – Normed vector space that is complete
- Complete field
- Hilbert space – Generalization of Euclidean space allowing infinite dimensions
- Normed space
- Locally compact field
- Locally compact group
- Locally compact quantum group
- Locally convex topological vector space – A vector space with a topology defined by convex open sets
- Ordered topological vector space
- Topological abelian group
- Topological field
- Topological group – Group that is a topological space with continuous group action
- Topological module
- Topological ring
- Topological semigroup
- Topological vector lattice
- The topological properties of course also require that be a TVS.
- In particular, is Hausdorff if and only if the set is closed (that is, is a T1 space).
- In fact, this is true for topological group, since the proof does not use the scalar multiplications.
- Also called a metric linear space, which means that it is a real or complex vector space together with a translation-invariant metric for which addition and scalar multiplication are continuous.
- A series is said to converge in a TVS if the sequence of partial sums converges.
- If the interior of a balanced set is non-empty but does not contain the origin (such sets exists even in and ) then the interior of this set can not be a balanced set.
- In general topology, the closure of a compact subset of a non-Hausdorff space may fail to be compact (for example, the particular point topology on an infinite set). This result shows that this does not happen in non-Hausdorff TVSs. is compact because it is the image of the compact set under the continuous addition map Recall also that the sum of a compact set (i.e. ) and a closed set is closed so is closed in
- In the sum of the -axis and the graph of which is the complement of the -axis, is open in In the Minkowski sum is a countable dense subset of so not closed in
- This condition is satisfied if denotes the set of all topological strings in
- Since has the trivial topology, so does each of its subsets, which makes them all compact. It is known that a subset of any uniform space is compact if and only if it is complete and totally bounded.
- If then Because if is closed then equality holds. Using the fact that is a vector space, it is readily verified that the complement in of any set satisfying the equality must also satisfy this equality (when is substituted for ).
- and so using and the fact that this is equal to