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 that is also a topological space with the property that the vector space operations (vector addition and scalar multiplication) are also continuous functions. Such a topology is called a vector topology and every topological vector space has a uniform topological structure, allowing a notion of uniform convergence and completeness. Some authors also require that the space is a Hausdorff space (although this article does not). One of the most widely studied categories of TVSs are locally convex topological vector spaces. This article focuses on TVSs that are not necessarily locally convex. Other well-known examples of TVSs include Banach spaces, Hilbert spaces and Sobolev spaces.
Many topological vector spaces are spaces of 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.
In this article, the scalar field of a topological vector space will be assumed to be either the complex numbers or the real numbers unless clearly stated otherwise.
Every normed vector space has a natural topological structure: the norm induces a metric and the metric induces a topology. This is a topological vector space because^{[citation needed]}:
Thus all Banach spaces and Hilbert spaces are examples of topological vector spaces.
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.^{[1]} 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 field is a topological vector space over each of its subfields.
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.
Hausdorff assumption
Many authors (for example, Walter Rudin), but not this page, require the topology on to be T_{1}; 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 or 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), also called a topological homomorphism,^{[2]}^{[3]} 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
A topological vector space embedding (abbreviated TVS embedding), also called a topological monomorphism, is an injective topological homomorphism. Equivalently, a TVS-embedding is a linear map that is also a topological embedding.^{[2]}
A topological vector space isomorphism (abbreviated TVS isomorphism), also called a topological vector isomorphism^{[4]} or an isomorphism in the category of TVSs, is a bijective linear homeomorphism. Equivalently, it is a surjective TVS embedding^{[2]}
Many properties of TVSs that are studied, such as local convexity, metrizability, completeness, and normability, are invariant under TVS isomorphisms.
A necessary condition for a vector topology
A collection of subsets of a vector space is called additive^{[5]} if for every there exists some such that
Characterization of continuity of addition at ^{[5]} — 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.
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^{[6]} (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.^{[5]}
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:^{[7]}^{[8]}^{[9]}
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 ^{[7]} 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 topological vector spaces.
If and are two collections of subsets of a vector space and if is a scalar, then by definition:^{[7]}
If is a collection sequences of subsets of then is said to be directed (downwards) under inclusion or simply directed downward 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^{[7]} (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
If is the set of all topological strings in a TVS then ^{[7]} A Hausdorff TVS is metrizable if and only if its topology can be induced by a single topological string.^{[10]}
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 ). Hence, every topological vector space is an abelian topological group. Every TVS is completely regular but a TVS need not be normal.^{[11]}
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
One of the most used properties of vector topologies is that every vector topology is translation invariant:
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.
If and any subset then ^{[6]} and moreover, if then is a neighborhood (resp. open neighborhood, closed neighborhood) of in if and only if the same is true of at the origin.
A subset of a vector space is said to be
Every neighborhood of the origin is an absorbing set and contains an open balanced neighborhood of ^{[6]} 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 if the space is locally convex then it also has a neighborhood basis consisting of closed convex balanced neighborhoods of the origin.
Bounded subsets
A subset of a topological vector space is bounded^{[13]} if for every neighborhood of the origin there exists such that .
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.^{[14]} Also, is bounded if and only if for every balanced neighborhood of the origin, 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 ^{[15]}
Every totally bounded set is bounded.^{[14]} If is a vector subspace of a TVS then a subset of is bounded in if and only if it is bounded in ^{[14]}
Birkhoff–Kakutani theorem — If is a topological vector space then the following four conditions are equivalent:^{[16]}^{[note 3]}
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 the origin.^{[17]}
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 ^{[18]}
The canonical uniformity^{[19]} on a TVS is the unique translation-invariant uniformity that induces the topology on
Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs into uniform spaces. This allows one to talk^{[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.^{[20]} 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) 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.
Let be a real or complex vector space.
Trivial topology
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 ).^{[23]}^{[24]} Every linear map from into another TVS is necessarily continuous. If has an uncountable Hamel basis then is not locally convex and not metrizable.^{[24]}
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 ).
By F. Riesz's theorem, a Hausdorff topological vector space is finite-dimensional if and only if it is locally compact, which happens if and only if it has a compact neighborhood of the origin.
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.
The proof of this dichotomy (i.e. that a vector topology is either trivial or isomorphic to ) 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 -dimensional vector space over If and is a ball centered at 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. Q.E.D.
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 in 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."
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.^{[27]} 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.^{[28]}
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 ^{[29]} 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.^{[29]}
Properties of neighborhoods and open sets
Every TVS is connected^{[6]} and locally connected^{[30]} and any connected open subset of a TVS is arcwise connected. If and is an open subset of then is an open set in ^{[6]} and if has non-empty interior then is a neighborhood of the origin.^{[6]}
The open convex subsets of a TVS (not necessarily Hausdorff or locally convex) are exactly those that are of the form for some and some positive continuous sublinear functional on ^{[28]}
If is an absorbing disk in a TVS and if is the Minkowski functional of then^{[31]} where importantly, it was not assumed that had any topological properties nor that was continuous (which happens if and only if is a neighborhood of the origin).
Let and be two vector topologies on Then if and only if whenever a net in converges in then in ^{[32]}
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 ^{[33]} 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.^{[34]} 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.^{[34]}
Interior
If and has non-empty interior then and
The topological interior of a disk is not empty if and only if this interior contains the origin.^{[35]} More generally, if is a balanced set with non-empty interior in a TVS then will necessarily be balanced;^{[6]} consequently, will be balanced if and only if it contains the origin.^{[proof 2]} For this (i.e. ) to be true, it suffices for to also be convex (in addition to being balanced and having non-empty interior).;^{[6]} The conclusion could be false if is not also convex;^{[35]} for example, in the interior of the closed and balanced set is
If is convex and then^{[36]} Explicitly, this means that if is a convex subset of a TVS (not necessarily Hausdorff or locally convex), and then the open line segment joining and belongs to the interior of that is, ^{[37]}^{[38]}^{[proof 3]}
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^{[37]} 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
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 so that in particular, every neighborhood of the origin in contains the vector space as a subset. The subspace topology on is always the trivial topology, which in particular implies that the topological vector space a compact space (even if its dimension is non-zero or even infinite) and consequently also a bounded subset of In fact, a vector subspace of a TVS is bounded if and only if it is contained in the closure of ^{[14]} Every subset of also carries the trivial topology and so is itself a compact, and thus also complete, subspace (see footnote for a proof).^{[proof 4]} In particular, if is not Hausdorff then there exist subsets that are both compact and complete but not closed in ;^{[39]} for instance, this will be true of any non-empty proper subset of
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),^{[40]} which is not guaranteed for arbitrary non-Hausdorff topological spaces.^{[note 6]}
For every subset and consequently, if is open or closed in then ^{[proof 5]} (so that this arbitrary open or closed subsets can be described as a "tube" whose vertical side is the vector space ). For any subset of this TVS the following are equivalent:
If is a vector subspace of a TVS then is Hausdorff if and only if is closed in Moreover, the quotient map is always a closed map onto the (necessarily) Hausdorff TVS.^{[44]}
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.^{[45]} Moreover, if is a Hausdorff completion of then is a completion of ^{[41]}
Compact and totally bounded sets
A subset of a TVS is compact if and only if it is complete and totally bounded.^{[39]} Thus, in a complete topological vector space, a closed and totally bounded subset is compact.^{[39]} A subset of a TVS is totally bounded if and only if is totally bounded,^{[42]}^{[43]} if and only if its image under the canonical quotient map is totally bounded.^{[41]}
Every relatively compact set is totally bounded^{[39]} and the closure of a totally bounded set is totally bounded.^{[39]} The image of a totally bounded set under a uniformly continuous map (such as a continuous linear map for instance) is totally bounded.^{[39]} If is a subset of a TVS such that every sequence in has a cluster point in then is totally bounded.^{[41]}
If is a compact subset of a TVS and is an open subset of containing then there exists a neighborhood of 0 such that ^{[46]}
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.^{[6]} If is a vector subspace of and is a closed neighborhood of the origin in such that is closed in then is closed in ^{[46]} The sum of a compact set and a closed set is closed. However, the sum of two closed subsets may fail to be closed^{[6]} (see this footnote^{[note 7]} for examples).
If and is a scalar then where if is Hausdorff, then equality holds: In particular, every non-zero scalar multiple of a closed set is closed. If and if is a set of scalars such that neither contain zero then^{[47]}
If then is convex.^{[47]}
If then^{[6]} and so consequently, if is closed then so is ^{[47]}
If is a real TVS and then where the left hand side is independent of the topology on moreover, if is a convex neighborhood of the origin then equality holds.
For any subset where is any neighborhood basis at the origin for ^{[48]} However, and it is possible for this containment to be proper^{[49]} (for example, if and is the rational numbers). It follows that for every neighborhood of the origin in ^{[50]}
Closed hulls
In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.^{[14]}
If and the closed convex hull of one of the sets or is compact then^{[51]} If each have a closed convex hull that is compact (that is, and are compact) then^{[51]}
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 (respectively, totally bounded) set has that same property.^{[6]} The convex hull of a finite union of compact convex sets is again compact and convex.^{[6]}
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.^{[9]} A vector subspace of a TVS that is closed but not open is nowhere dense.^{[9]}
Suppose is a TVS that does not carry the indiscrete topology. Then is a Baire space if and only if has no balanced absorbing nowhere dense subset.^{[9]}
A TVS is a Baire space if and only if is nonmeager, which happens if and only if there does not exist a nowhere dense set such that ^{[9]} Every nonmeager locally convex TVS is a barrelled space.^{[9]}
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.
A subset is convex if and only if for all positive real ^{[29]} or equivalently, if and only if for all ^{[52]}
The convex balanced hull of a set is equal to the convex hull of the balanced hull of that is, it is equal to But in general, where the inclusion might be strict since the balanced hull of a convex set need not be convex (counter-examples exist even in ).
If and is a scalar then^{[6]} If are convex non-empty disjoint sets and then or
In any non-trivial vector space there exist two disjoint non-empty convex subsets whose union is
Other properties
Every TVS topology can be generated by a family of F-seminorms.^{[53]}
If is some unary predicate (a true or false statement dependent on ) then for any ^{[proof 6]} 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, ).
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 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Absorbing | Balanced | Convex | Symmetric | Convex Balanced |
Vector subspace |
Open | Neighborhood of 0 |
Closed | Closed Balanced |
Closed Convex |
Closed Convex Balanced |
Barrel | Closed Vector subspace |
Totally bounded |
Compact | Compact Convex |
Relatively compact | Complete | Sequentially Complete |
Banach disk |
Bounded | Bornivorous | Infrabornivorous | Nowhere dense (in ) |
Meager | Separable | Pseudometrizable | Operation | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Closure | Closure | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Interior | Interior | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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 |