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 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 \Complex or the real numbers \R, unless clearly stated otherwise.

Motivation

Normed spaces

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: Thus all Banach spaces and Hilbert spaces are examples of topological vector spaces.

Non-normed 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. 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.

Definition

A topological vector space (TVS) X is a vector space over a topological field \mathbb{K} (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 **' or a **' on X. 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 X to be T1; it then follows that the space is Hausdorff, and even Tychonoff. A topological vector space is said to be 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 \mathbb{K} is commonly denoted or The objects are the topological vector spaces over \mathbb{K} and the morphisms are the continuous \mathbb{K}-linear maps from one object to another. A (abbreviated ), also called a, is a continuous linear map u : X \to Y between topological vector spaces (TVSs) such that the induced map is an open mapping when which is the range or image of u, is given the subspace topology induced by Y. A (abbreviated ), also called a, is an injective topological homomorphism. Equivalently, a TVS-embedding is a linear map that is also a topological embedding. A (abbreviated ), also called a or an, is a bijective linear homeomorphism. Equivalently, it is a surjective TVS embedding 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 \mathcal{N} of subsets of a vector space is called if for every there exists some such that 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 x_0 \in X, the map X \to X defined by 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.

Defining topologies using strings

Let X be**** a vector**** space and let be**** a sequence**** of**** subsets of**** X.**** Each**** set in**** the sequence**** U{*b**ul**l***}**** is**** called**** a **** of**** U{*b**ul**l***}**** and for every index i,**** Ui is**** called**** the i-th**** knot**** of**** U{*b**ul**l***}. The set U_1 is called the beginning of U_{\bull}. The sequence U_{\bull} is/is a: If U is an absorbing disk in a vector space X then the sequence defined by forms a string beginning with U_1 = U. This is called the natural string of U 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. 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 X and if s is a scalar, then by definition: If \mathbb{S} is a collection sequences of subsets of X, then \mathbb{S} is said to be directed (downwards) under inclusion or simply directed downward if \mathbb{S} is not empty and for all there exists some such that and (said differently, if and only if \mathbb{S} is a prefilter with respect to the containment defined above). Notation: Let be the set of all knots of all strings in \mathbb{S}. Defining vector topologies using collections of strings is particularly useful for defining classes of TVSs that are not necessarily locally convex. If \mathbb{S} is the set of all topological strings in a TVS (X, \tau) then A Hausdorff TVS is metrizable if and only if its topology can be induced by a single topological string.

Topological structure

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 X be a topological vector space. Given a subspace the quotient space X / M with the usual quotient topology is a Hausdorff topological vector space if and only if M is closed. This permits the following construction: given a topological vector space X (that is probably not Hausdorff), form the quotient space X / M where M is the closure of {0}. X / M is then a Hausdorff topological vector space that can be studied instead of X.

Invariance of vector topologies

One of the most used properties of vector topologies is that every vector topology is : Scalar multiplication by a non-zero scalar is a TVS-isomorphism. This means that if s \neq 0 then the linear map X \to X defined by is a homeomorphism. Using s = -1 produces the negation map X \to X defined by which is consequently a linear homeomorphism and thus a TVS-isomorphism. If x \in X and any subset then and moreover, if 0 \in 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.

Local notions

A subset E of a vector space X is said to be Every neighborhood of the origin is an absorbing set and contains an open balanced neighborhood of 0 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 the origin. Bounded subsets A subset E of a topological vector space X is bounded if for every neighborhood V of the origin there exists t such that. 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. Also, E is bounded if and only if for every balanced neighborhood V of the origin, there exists t such that 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. 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.

Metrizability

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

Completeness and uniform structure

The canonical uniformity on a TVS (X, \tau) is the unique translation-invariant uniformity that induces the topology \tau on X. Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs into uniform spaces. This allows one to talk 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 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 V of 0, there exists some index n such that whenever i \geq n and j \geq n. 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.

Examples

Finest and coarsest vector topology

Let X be a real or complex vector space. Trivial topology The trivial topology or indiscrete topology 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**** *tau**_f** on*** X,**** called**** the **** on**** X,**** that**** is**** finer than**** every other TVS-topology**** on**** X (that is****,**** any TVS-topology**** on**** X is**** necessarily a subset**** of**** *tau**_f**)*.** Every linear map from into another TVS is necessarily continuous. If X has an uncountable Hamel basis then \tau_f is locally convex and metrizable.

Cartesian products

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 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, becomes a topological vector space whose topology is called The reason for this name is the following: if is a sequence (or more generally, a net) of elements in X and if f \in 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 (that is, 1-dimensional vector subspaces, which are subsets of the form with f \neq 0).

Finite-dimensional spaces

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 \mathbb{K} denote \R or \Complex and endow \mathbb{K} with its usual Hausdorff normed Euclidean topology. Let X be a vector space over \mathbb{K} of finite dimension n := \dim X and so that X is vector space isomorphic to (explicitly, this means that there exists a linear isomorphism between the vector spaces X and ). This finite-dimensional vector space X always has a unique 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 X. X has a unique vector topology if and only if \dim X = 0. If then although X does not have a unique vector topology, it does have a unique vector topology.

Non-vector topologies

Discrete and cofinite topologies If X is a non-trivial vector space (that is, of non-zero dimension) then the discrete topology on X (which is always metrizable) is 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 a TVS topology on X.

Linear maps

A linear operator between two topological vector spaces which is continuous at one point is continuous on the whole domain. Moreover, a linear operator f is continuous if f(X) is bounded (as defined below) for some neighborhood X of the origin. A hyperplane in a topological vector space X is either dense or closed. A linear functional f on a topological vector space X has either dense or closed kernel. Moreover, f is continuous if and only if its kernel is closed.

Types

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."

Dual space

Every topological vector space has a continuous dual space—the set X' of all continuous linear functionals, that is, continuous linear maps from the space into the base field \mathbb{K}. 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 X is a non-normable locally convex space, then the pairing map is never continuous, no matter which vector space topology one chooses on X'. A topological vector space has a non-trivial continuous dual space if and only if it has a proper convex neighborhood of the origin.

Properties

For any of a TVS X, the convex (resp. balanced, disked, closed convex, closed balanced, closed disked') hull of S is the smallest subset of X that has this property and contains S. The closure (respectively, interior, convex hull, balanced hull, disked hull) of a set S is sometimes denoted by (respectively, ). The convex hull of a subset S is equal to the set of all of elements in S, which are finite linear combinations of the form where n \geq 1 is an integer, and sum to 1. 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 U is an open subset of X then S + U is an open set in X and if has non-empty interior then S - S is a neighborhood of the origin. The open convex subsets of a TVS X (not necessarily Hausdorff or locally convex) are exactly those that are of the form for some z \in X and some positive continuous sublinear functional p on X. If K is an absorbing disk in a TVS X and if p := p_K is the Minkowski functional of K then where importantly, it was assumed that K had any topological properties nor that p was continuous (which happens if and only if K is a neighborhood of the origin). Let \tau and \nu be two vector topologies on X. Then if and only if whenever a net in X converges 0 in (X, \nu) then in (X, \tau). Let \mathcal{N} be a neighborhood basis of the origin in X, let and let x \in X. Then if and only if there exists a net in S (indexed by \mathcal{N}) such that in X. 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 X is a TVS that is of the second category in itself (that is, a nonmeager space) then any closed convex absorbing subset of X is a neighborhood of the origin. This is no longer guaranteed if the set is not convex (a counter-example exists even in X = \R^2) or if X is not of the second category in itself. Interior If and S has non-empty interior then and The topological interior of a disk is not empty if and only if this interior contains the origin. More generally, if S is a balanced set with non-empty interior in a TVS X then will necessarily be balanced; consequently, will be balanced if and only if it contains the origin. For this (i.e. ) to be true, it suffices for S to also be convex (in addition to being balanced and having non-empty interior).; The conclusion could be false if S is not also convex; for example, in X := \R^2, the interior of the closed and balanced set is If C is convex and then Explicitly, this means that if C is a convex subset of a TVS X (not necessarily Hausdorff or locally convex), and then the open line segment joining x and y belongs to the interior of C; that is, If is any balanced neighborhood of the origin in X then where B_1 is the set of all scalars a such that |a| < 1. If x belongs to the interior of a convex set and then the half-open line segment and If N is a balanced neighborhood of 0 in X and then by considering intersections of the form N \cap \R x (which are convex symmetric neighborhoods of 0 in the real TVS \R x) it follows that: and furthermore, if then and if then

Non-Hausdorff spaces and the closure of the origin

A topological vector space X is Hausdorff if and only if {0} is a closed subset of X, or equivalently, if and only if Because {0} is a vector subspace of X, the same is true of its closure which is referred to as in X. This vector space satisfies so that in particular, every neighborhood of the origin in X 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 X. In fact, a vector subspace of a TVS is bounded if and only if it is contained in the closure of {0}. Every subset of also carries the trivial topology and so is itself a compact, and thus also complete, subspace (see footnote for a proof). In particular, if X is not Hausdorff then there exist subsets that are both but in X; 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), which is not guaranteed for arbitrary non-Hausdorff topological spaces. For every subset and consequently, if is open or closed in X then (so that this open closed subsets S can be described as a "tube" whose vertical side is the vector space ). For any subset of this TVS X, the following are equivalent: If M is a vector subspace of a TVS X then X / M is Hausdorff if and only if M is closed in X. Moreover, the quotient map is always a closed map onto the (necessarily) Hausdorff TVS. Every vector subspace of X that is an algebraic complement of (that is, a vector subspace H that satisfies and ) is a topological complement of Consequently, if H is an algebraic complement of in X then the addition map defined by is a TVS-isomorphism, where H is necessarily Hausdorff and has the indiscrete topology. Moreover, if C is a Hausdorff completion of H 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 S of a TVS X is totally bounded if and only if is totally bounded, if and only if its image under the canonical quotient map is totally bounded. 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 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. 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 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 M is a vector subspace of X and N is a closed neighborhood of the origin in X such that U \cap N is closed in X then M is closed in X. 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 for examples). If and a is a scalar then where if X is Hausdorff, then equality holds: In particular, every non-zero scalar multiple of a closed set is closed. If and if A is a set of scalars such that neither contain zero then If then is convex. If then and so consequently, if R + S is closed then so is If X is a real TVS and then where the left hand side is independent of the topology on X; moreover, if S is a convex neighborhood of the origin then equality holds. For any subset where \mathcal{N} is any neighborhood basis at the origin for X. However, and it is possible for this containment to be proper (for example, if X = \R and S is the rational numbers). It follows that for every neighborhood U of the origin in X. Closed hulls In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general. If and the closed convex hull of one of the sets S or R is compact then If each have a closed convex hull that is compact (that is, and are compact) then Hulls and compactness In a general TVS, the closed convex hull of a compact set may to be compact. The balanced hull of a compact (respectively, totally bounded) set has that same property. The convex hull of a finite union of compact sets is again compact and convex.

Other properties

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. A vector subspace of a TVS that is closed but not open is nowhere dense. 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. 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 Every nonmeager locally convex TVS is a barrelled space. Important algebraic facts and common misconceptions If then ; if S is convex then equality holds. For an example where equality does hold, let x be non-zero and set also works. A subset C is convex if and only if for all positive real or equivalently, if and only if for all The convex balanced hull of a set is equal to the convex hull of the balanced hull of S; 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 \R^2). If and a is a scalar then If are convex non-empty disjoint sets and then or 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 of F-seminorms. If P(x) is some unary predicate (a true or false statement dependent on x \in X) then for any z \in X, So for example, if P(x) denotes "|x| < 1" then for any z \in X, Similarly, if s \neq 0 is a scalar then The elements x \in X of these sets must range over a vector space (that is, over X) rather than not just a subset or else these equalities are no longer guaranteed; similarly, z must belong to this vector space (that is, z \in X).

Properties preserved by set operators

The following table, the color of each cell indicates whether or not a given property of subsets of X (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 "R \cup 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.

Proofs

Citations