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Topological vector space

From Wikipedia, the free encyclopedia
Vector space with a notion of nearness

Inmathematics, atopological vector space (also called alinear topological space and commonly abbreviatedTVS ort.v.s.) is one of the basic structures investigated infunctional analysis.A topological vector space is avector space that is also atopological space with the property that the vector space operations (vector addition and scalar multiplication) are alsocontinuous functions. Such a topology is called avector topology and every topological vector space has auniform topological structure, allowing a notion ofuniform convergence andcompleteness. Some authors also require that the space is aHausdorff space (although this article does not). One of the most widely studied categories of TVSs arelocally convex topological vector spaces. This article focuses on TVSs that are not necessarily locally convex. Other well-known examples of TVSs includeBanach spaces,Hilbert spaces andSobolev spaces.

Many topological vector spaces are spaces offunctions, orlinear operators acting on topological vector spaces, and the topology is often defined so as to capture a particular notion ofconvergence of sequences of functions.

In this article, thescalar field of a topological vector space will be assumed to be either thecomplex numbersC{\displaystyle \mathbb {C} } or thereal numbersR,{\displaystyle \mathbb {R} ,} unless clearly stated otherwise.

Motivation

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Normed spaces

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Everynormed vector space has a naturaltopological structure: the norm induces ametric and the metric induces a topology.This is a topological vector space because[citation needed]:

  1. The vector addition map⋅+⋅:X×X→X{\displaystyle \cdot \,+\,\cdot \;:X\times X\to X} defined by(x,y)↦x+y{\displaystyle (x,y)\mapsto x+y} is (jointly) continuous with respect to this topology. This follows directly from thetriangle inequality obeyed by the norm.
  2. The scalar multiplication map⋅:K×X→X{\displaystyle \cdot :\mathbb {K} \times X\to X} defined by(s,x)↦s⋅x,{\displaystyle (s,x)\mapsto s\cdot x,} whereK{\displaystyle \mathbb {K} } is the underlying scalar field ofX,{\displaystyle X,} is (jointly) continuous. This follows from the triangle inequality and homogeneity of the norm.

Thus allBanach spaces andHilbert spaces are examples of topological vector spaces.

Non-normed spaces

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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 ofholomorphic functions on an open domain, spaces ofinfinitely differentiable functions, theSchwartz spaces, and spaces oftest functions and the spaces ofdistributions on them.[1] These are all examples ofMontel spaces. An infinite-dimensional Montel space is never normable. The existence of a norm for a given topological vector space is characterized byKolmogorov's normability criterion.

Atopological field is a topological vector space over each of itssubfields.

Definition

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A family of neighborhoods of the origin with the above two properties determines uniquely a topological vector space. The system of neighborhoods of any other point in the vector space is obtained bytranslation.

Atopological vector space (TVS)X{\displaystyle X} is avector space over atopological fieldK{\displaystyle \mathbb {K} } (most often thereal orcomplex numbers with their standard topologies) that is endowed with atopology such that vector addition⋅+⋅:X×X→X{\displaystyle \cdot \,+\,\cdot \;:X\times X\to X} and scalar multiplication⋅:K×X→X{\displaystyle \cdot :\mathbb {K} \times X\to X} arecontinuous functions (where the domains of these functions are endowed withproduct topologies). Such a topology is called avector topology or aTVS topology onX.{\displaystyle X.}

Every topological vector space is also a commutativetopological group under addition.

Hausdorff assumption

Many authors (for example,Walter Rudin), but not this page, require the topology onX{\displaystyle X} to beT1; it then follows that the space isHausdorff, and evenTychonoff. A topological vector space is said to beseparated if it is Hausdorff; importantly, "separated" does not meanseparable. The topological and linear algebraic structures can be tied together even more closely with additional assumptions, the most common of which are listedbelow.

Category and morphisms

Thecategory of topological vector spaces over a given topological fieldK{\displaystyle \mathbb {K} } is commonly denotedTVSK{\displaystyle \mathrm {TVS} _{\mathbb {K} }} orTVectK.{\displaystyle \mathrm {TVect} _{\mathbb {K} }.} Theobjects are the topological vector spaces overK{\displaystyle \mathbb {K} } and themorphisms are thecontinuousK{\displaystyle \mathbb {K} }-linear maps from one object to another.

Atopological vector space homomorphism (abbreviatedTVS homomorphism), also called atopological homomorphism,[2][3] is acontinuouslinear mapu:X→Y{\displaystyle u:X\to Y} between topological vector spaces (TVSs) such that the induced mapu:X→Im⁡u{\displaystyle u:X\to \operatorname {Im} u} is anopen mapping whenIm⁡u:=u(X),{\displaystyle \operatorname {Im} u:=u(X),} which is the range or image ofu,{\displaystyle u,} is given thesubspace topology induced byY.{\displaystyle Y.}

Atopological vector space embedding (abbreviatedTVS embedding), also called atopologicalmonomorphism, is aninjective topological homomorphism. Equivalently, a TVS-embedding is a linear map that is also atopological embedding.[2]

Atopological vector space isomorphism (abbreviatedTVS isomorphism), also called atopological vector isomorphism[4] or anisomorphism in the category of TVSs, is a bijectivelinearhomeomorphism. Equivalently, it is asurjective TVS embedding[2]

Many properties of TVSs that are studied, such aslocal convexity,metrizability,completeness, andnormability, are invariant under TVS isomorphisms.

A necessary condition for a vector topology

A collectionN{\displaystyle {\mathcal {N}}} of subsets of a vector space is calledadditive[5] if for everyN∈N,{\displaystyle N\in {\mathcal {N}},} there exists someU∈N{\displaystyle U\in {\mathcal {N}}} such thatU+U⊆N.{\displaystyle U+U\subseteq N.}

Characterization of continuity of addition at0{\displaystyle 0}[5]—If(X,+){\displaystyle (X,+)} is agroup (as all vector spaces are),τ{\displaystyle \tau } is a topology onX,{\displaystyle X,} andX×X{\displaystyle X\times X} is endowed with theproduct topology, then the addition mapX×X→X{\displaystyle X\times X\to X} (defined by(x,y)↦x+y{\displaystyle (x,y)\mapsto x+y}) is continuous at the origin ofX×X{\displaystyle X\times X} if and only if the set ofneighborhoods of the origin in(X,τ){\displaystyle (X,\tau )} 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

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Since every vector topology is translation invariant (which means that for allx0∈X,{\displaystyle x_{0}\in X,} the mapX→X{\displaystyle X\to X} defined byx↦x0+x{\displaystyle x\mapsto x_{0}+x} is ahomeomorphism), to define a vector topology it suffices to define aneighborhood basis (or subbasis) for it at the origin.

Theorem[6] (Neighborhood filter of the origin)—Suppose thatX{\displaystyle X} is a real or complex vector space. IfB{\displaystyle {\mathcal {B}}} is anon-empty additive collection ofbalanced andabsorbing subsets ofX{\displaystyle X} thenB{\displaystyle {\mathcal {B}}} is aneighborhood base at0{\displaystyle 0} for a vector topology onX.{\displaystyle X.} That is, the assumptions are thatB{\displaystyle {\mathcal {B}}} is afilter base that satisfies the following conditions:

  1. EveryB∈B{\displaystyle B\in {\mathcal {B}}} isbalanced andabsorbing,
  2. B{\displaystyle {\mathcal {B}}} is additive: For everyB∈B{\displaystyle B\in {\mathcal {B}}} there exists aU∈B{\displaystyle U\in {\mathcal {B}}} such thatU+U⊆B,{\displaystyle U+U\subseteq B,}

IfB{\displaystyle {\mathcal {B}}} satisfies the above two conditions but isnot a filter base then it will form a neighborhoodsubbasis at0{\displaystyle 0} (rather than a neighborhood basis) for a vector topology onX.{\displaystyle X.}

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]

Defining topologies using strings

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LetX{\displaystyle X} be a vector space and letU∙=(Ui)i=1∞{\displaystyle U_{\bullet }=\left(U_{i}\right)_{i=1}^{\infty }} be a sequence of subsets ofX.{\displaystyle X.} Each set in the sequenceU∙{\displaystyle U_{\bullet }} is called aknot ofU∙{\displaystyle U_{\bullet }} and for every indexi,{\displaystyle i,}Ui{\displaystyle U_{i}} is called thei{\displaystyle i}-th knot ofU∙.{\displaystyle U_{\bullet }.} The setU1{\displaystyle U_{1}} is called thebeginning ofU∙.{\displaystyle U_{\bullet }.} The sequenceU∙{\displaystyle U_{\bullet }} is/is a:[7][8][9]

IfU{\displaystyle U} is anabsorbingdisk in a vector spaceX{\displaystyle X} then the sequence defined byUi:=21−iU{\displaystyle U_{i}:=2^{1-i}U} forms a string beginning withU1=U.{\displaystyle U_{1}=U.} This is called thenatural string ofU{\displaystyle U}[7] Moreover, if a vector spaceX{\displaystyle X} has countable dimension then every string contains anabsolutely convex string.

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

Theorem (R{\displaystyle \mathbb {R} }-valued function induced by a string)—LetU∙=(Ui)i=0∞{\displaystyle U_{\bullet }=\left(U_{i}\right)_{i=0}^{\infty }} be a collection of subsets of a vector space such that0∈Ui{\displaystyle 0\in U_{i}} andUi+1+Ui+1⊆Ui{\displaystyle U_{i+1}+U_{i+1}\subseteq U_{i}} for alli≥0.{\displaystyle i\geq 0.} For allu∈U0,{\displaystyle u\in U_{0},} letS(u):={n∙=(n1,…,nk) : k≥1,ni≥0 for all i, and u∈Un1+⋯+Unk}.{\displaystyle \mathbb {S} (u):=\left\{n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)~:~k\geq 1,n_{i}\geq 0{\text{ for all }}i,{\text{ and }}u\in U_{n_{1}}+\cdots +U_{n_{k}}\right\}.}

Definef:X→[0,1]{\displaystyle f:X\to [0,1]} byf(x)=1{\displaystyle f(x)=1} ifx∉U0{\displaystyle x\not \in U_{0}} and otherwise letf(x):=inf{2−n1+⋯2−nk : n∙=(n1,…,nk)∈S(x)}.{\displaystyle f(x):=\inf _{}\left\{2^{-n_{1}}+\cdots 2^{-n_{k}}~:~n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)\in \mathbb {S} (x)\right\}.}

Thenf{\displaystyle f} is subadditive (meaningf(x+y)≤f(x)+f(y){\displaystyle f(x+y)\leq f(x)+f(y)} for allx,y∈X{\displaystyle x,y\in X}) andf=0{\displaystyle f=0} on⋂i≥0Ui;{\textstyle \bigcap _{i\geq 0}U_{i};} so in particular,f(0)=0.{\displaystyle f(0)=0.} If allUi{\displaystyle U_{i}} aresymmetric sets thenf(−x)=f(x){\displaystyle f(-x)=f(x)} and if allUi{\displaystyle U_{i}} are balanced thenf(sx)≤f(x){\displaystyle f(sx)\leq f(x)} for all scalarss{\displaystyle s} such that|s|≤1{\displaystyle |s|\leq 1} and allx∈X.{\displaystyle x\in X.} IfX{\displaystyle X} is a topological vector space and if allUi{\displaystyle U_{i}} are neighborhoods of the origin thenf{\displaystyle f} is continuous, where if in additionX{\displaystyle X} is Hausdorff andU∙{\displaystyle U_{\bullet }} forms a basis of balanced neighborhoods of the origin inX{\displaystyle X} thend(x,y):=f(x−y){\displaystyle d(x,y):=f(x-y)} is a metric defining the vector topology onX.{\displaystyle X.}

A proof of the above theorem is given in the article onmetrizable topological vector spaces.

IfU∙=(Ui)i∈N{\displaystyle U_{\bullet }=\left(U_{i}\right)_{i\in \mathbb {N} }} andV∙=(Vi)i∈N{\displaystyle V_{\bullet }=\left(V_{i}\right)_{i\in \mathbb {N} }} are two collections of subsets of a vector spaceX{\displaystyle X} and ifs{\displaystyle s} is a scalar, then by definition:[7]

IfS{\displaystyle \mathbb {S} } is a collection sequences of subsets ofX,{\displaystyle X,} thenS{\displaystyle \mathbb {S} } is said to bedirected (downwards)under inclusion or simplydirected downward ifS{\displaystyle \mathbb {S} } is not empty and for allU∙,V∙∈S,{\displaystyle U_{\bullet },V_{\bullet }\in \mathbb {S} ,} there exists someW∙∈S{\displaystyle W_{\bullet }\in \mathbb {S} } such thatW∙⊆U∙{\displaystyle W_{\bullet }\subseteq U_{\bullet }} andW∙⊆V∙{\displaystyle W_{\bullet }\subseteq V_{\bullet }} (said differently, if and only ifS{\displaystyle \mathbb {S} } is aprefilter with respect to the containment⊆{\displaystyle \,\subseteq \,} defined above).

Notation: LetKnots⁡S:=⋃U∙∈SKnots⁡U∙{\textstyle \operatorname {Knots} \mathbb {S} :=\bigcup _{U_{\bullet }\in \mathbb {S} }\operatorname {Knots} U_{\bullet }} be the set of all knots of all strings inS.{\displaystyle \mathbb {S} .}

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(X,τ){\displaystyle (X,\tau )} is a topological vector space then there exists a setS{\displaystyle \mathbb {S} }[proof 1] of neighborhood strings inX{\displaystyle X} that is directed downward and such that the set of all knots of all strings inS{\displaystyle \mathbb {S} } is aneighborhood basis at the origin for(X,τ).{\displaystyle (X,\tau ).} Such a collection of strings is said to beτ{\displaystyle \tau }fundamental.

Conversely, ifX{\displaystyle X} is a vector space and ifS{\displaystyle \mathbb {S} } is a collection of strings inX{\displaystyle X} that is directed downward, then the setKnots⁡S{\displaystyle \operatorname {Knots} \mathbb {S} } of all knots of all strings inS{\displaystyle \mathbb {S} } forms aneighborhood basis at the origin for a vector topology onX.{\displaystyle X.} In this case, this topology is denoted byτS{\displaystyle \tau _{\mathbb {S} }} and it is called thetopology generated byS.{\displaystyle \mathbb {S} .}

IfS{\displaystyle \mathbb {S} } is the set of all topological strings in a TVS(X,τ){\displaystyle (X,\tau )} thenτS=τ.{\displaystyle \tau _{\mathbb {S} }=\tau .}[7] A Hausdorff TVS ismetrizableif and only if its topology can be induced by a single topological string.[10]

Topological structure

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A vector space is anabelian 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{\displaystyle -1}). Hence, every topological vector space is an abeliantopological group. Every TVS iscompletely regular but a TVS need not benormal.[11]

LetX{\displaystyle X} be a topological vector space. Given asubspaceM⊆X,{\displaystyle M\subseteq X,} the quotient spaceX/M{\displaystyle X/M} with the usualquotient topology is a Hausdorff topological vector space if and only ifM{\displaystyle M} is closed.[note 2] This permits the following construction: given a topological vector spaceX{\displaystyle X} (that is probably not Hausdorff), form the quotient spaceX/M{\displaystyle X/M} whereM{\displaystyle M} is the closure of{0}.{\displaystyle \{0\}.}X/M{\displaystyle X/M} is then a Hausdorff topological vector space that can be studied instead ofX.{\displaystyle X.}

Invariance of vector topologies

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

for allx0∈X,{\displaystyle x_{0}\in X,} the mapX→X{\displaystyle X\to X} defined byx↦x0+x{\displaystyle x\mapsto x_{0}+x} is ahomeomorphism, but ifx0≠0{\displaystyle x_{0}\neq 0} then it is not linear and so not a TVS-isomorphism.

Scalar multiplication by a non-zero scalar is a TVS-isomorphism. This means that ifs≠0{\displaystyle s\neq 0} then the linear mapX→X{\displaystyle X\to X} defined byx↦sx{\displaystyle x\mapsto sx} is a homeomorphism. Usings=−1{\displaystyle s=-1} produces the negation mapX→X{\displaystyle X\to X} defined byx↦−x,{\displaystyle x\mapsto -x,} which is consequently a linear homeomorphism and thus a TVS-isomorphism.

Ifx∈X{\displaystyle x\in X} and any subsetS⊆X,{\displaystyle S\subseteq X,} thenclX⁡(x+S)=x+clX⁡S{\displaystyle \operatorname {cl} _{X}(x+S)=x+\operatorname {cl} _{X}S}[6] and moreover, if0∈S{\displaystyle 0\in S} thenx+S{\displaystyle x+S} is aneighborhood (resp. open neighborhood, closed neighborhood) ofx{\displaystyle x} inX{\displaystyle X} if and only if the same is true ofS{\displaystyle S} at the origin.

Local notions

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A subsetE{\displaystyle E} of a vector spaceX{\displaystyle X} is said to be

Every neighborhood of the origin is anabsorbing set and contains an openbalanced neighborhood of0{\displaystyle 0}[6] so every topological vector space has a local base of absorbing andbalanced sets. The origin even has a neighborhood basis consisting of closed balanced neighborhoods of0;{\displaystyle 0;} if the space islocally convex then it also has a neighborhood basis consisting of closed convex balanced neighborhoods of the origin.

Bounded subsets

A subsetE{\displaystyle E} of a topological vector spaceX{\displaystyle X} isbounded[13] if for every neighborhoodV{\displaystyle V} of the origin there existst{\displaystyle t} such thatE⊆tV{\displaystyle E\subseteq tV}.

The definition of boundedness can be weakened a bit;E{\displaystyle 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.[14] Also,E{\displaystyle E} is bounded if and only if for every balanced neighborhoodV{\displaystyle V} of the origin, there existst{\displaystyle t} such thatE⊆tV.{\displaystyle E\subseteq tV.}

Moreover, whenX{\displaystyle X} is locally convex, the boundedness can be characterized byseminorms: the subsetE{\displaystyle E} is bounded if and only if every continuous seminormp{\displaystyle p} is bounded onE.{\displaystyle E.}[15]

Everytotally bounded set is bounded.[14] IfM{\displaystyle M} is a vector subspace of a TVSX,{\displaystyle X,} then a subset ofM{\displaystyle M} is bounded inM{\displaystyle M} if and only if it is bounded inX.{\displaystyle X.}[14]

Metrizability

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Birkhoff–Kakutani theorem—If(X,τ){\displaystyle (X,\tau )} is a topological vector space then the following four conditions are equivalent:[16][note 3]

  1. The origin{0}{\displaystyle \{0\}} is closed inX{\displaystyle X} and there is acountablebasis of neighborhoods at the origin inX.{\displaystyle X.}
  2. (X,τ){\displaystyle (X,\tau )} ismetrizable (as a topological space).
  3. There is atranslation-invariant metric onX{\displaystyle X} that induces onX{\displaystyle X} the topologyτ,{\displaystyle \tau ,} which is the given topology onX.{\displaystyle X.}
  4. (X,τ){\displaystyle (X,\tau )} is ametrizable topological vector space.[note 4]

By the Birkhoff–Kakutani theorem, it follows that there is anequivalent metric that is translation-invariant.

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

More strongly: a topological vector space is said to benormable 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]

LetK{\displaystyle \mathbb {K} } be a non-discretelocally compact topological field, for example the real or complex numbers. AHausdorff topological vector space overK{\displaystyle \mathbb {K} } is locally compact if and only if it isfinite-dimensional, that is, isomorphic toKn{\displaystyle \mathbb {K} ^{n}} for some natural numbern.{\displaystyle n.}[18]

Completeness and uniform structure

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Main article:Complete topological vector space

Thecanonical uniformity[19] on a TVS(X,τ){\displaystyle (X,\tau )} is the unique translation-invariantuniformity that induces the topologyτ{\displaystyle \tau } onX.{\displaystyle X.}

Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs intouniform spaces. Therefore making sense to related notions such ascompleteness,uniform convergence, Cauchy nets, anduniform continuity, etc., which are always assumed to be with respect to this uniformity (unless indicated other). This implies that every Hausdorff topological vector space isTychonoff.[20] A subset of a TVS iscompact if and only if it is complete andtotally bounded (for Hausdorff TVSs, a set being totally bounded is equivalent to it beingprecompact). 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 arerelatively compact).

With respect to this uniformity, anet (orsequence)x∙=(xi)i∈I{\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}} isCauchy if and only if for every neighborhoodV{\displaystyle V} of0,{\displaystyle 0,} there exists some indexn{\displaystyle n} such thatxi−xj∈V{\displaystyle x_{i}-x_{j}\in V} wheneveri≥n{\displaystyle i\geq n} andj≥n.{\displaystyle j\geq n.}

EveryCauchy sequence is bounded, although Cauchy nets and Cauchy filters may not be bounded. A topological vector space where every Cauchy sequence converges is calledsequentially 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 anopen map. Scalar multiplication isCauchy continuous but in general, it is almost never uniformly continuous. Because of this, every topological vector space can be completed and is thus adenselinear subspace of acomplete topological vector space.

Examples

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Finest and coarsest vector topology

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LetX{\displaystyle X} be a real or complex vector space.

Trivial topology

Thetrivial topology orindiscrete topology{X,∅}{\displaystyle \{X,\varnothing \}} is always a TVS topology on any vector spaceX{\displaystyle X} and it is the coarsest TVS topology possible. An important consequence of this is that the intersection of any collection of TVS topologies onX{\displaystyle X} always contains a TVS topology. Any vector space (including those that are infinite dimensional) endowed with the trivial topology is a compact (and thuslocally compact)completepseudometrizableseminormablelocally convex topological vector space. It isHausdorff if and only ifdim⁡X=0.{\displaystyle \dim X=0.}

Finest vector topology

There exists a TVS topologyτf{\displaystyle \tau _{f}} onX,{\displaystyle X,} called thefinest vector topology onX,{\displaystyle X,} that is finer than every other TVS-topology onX{\displaystyle X} (that is, any TVS-topology onX{\displaystyle X} is necessarily a subset ofτf{\displaystyle \tau _{f}}).[23][24] Every linear map from(X,τf){\displaystyle \left(X,\tau _{f}\right)} into another TVS is necessarily continuous. IfX{\displaystyle X} has an uncountableHamel basis thenτf{\displaystyle \tau _{f}} isnotlocally convex andnotmetrizable.[24]

Cartesian products

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ACartesian product of a family of topological vector spaces, when endowed with theproduct topology, is a topological vector space. Consider for instance the setX{\displaystyle X} of all functionsf:R→R{\displaystyle f:\mathbb {R} \to \mathbb {R} } whereR{\displaystyle \mathbb {R} } carries its usualEuclidean topology. This setX{\displaystyle 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) theCartesian productRR,,{\displaystyle \mathbb {R} ^{\mathbb {R} },,} which carries the naturalproduct topology. With this product topology,X:=RR{\displaystyle X:=\mathbb {R} ^{\mathbb {R} }} becomes a topological vector space whose topology is calledthe topology ofpointwise convergence onR.{\displaystyle \mathbb {R} .} The reason for this name is the following: if(fn)n=1∞{\displaystyle \left(f_{n}\right)_{n=1}^{\infty }} is asequence (or more generally, anet) of elements inX{\displaystyle X} and iff∈X{\displaystyle f\in X} thenfn{\displaystyle f_{n}}converges tof{\displaystyle f} inX{\displaystyle X} if and only if for every real numberx,{\displaystyle x,}fn(x){\displaystyle f_{n}(x)} converges tof(x){\displaystyle f(x)} inR.{\displaystyle \mathbb {R} .} This TVS iscomplete,Hausdorff, andlocally convex but notmetrizable and consequently notnormable; indeed, every neighborhood of the origin in the product topology contains lines (that is, 1-dimensional vector subspaces, which are subsets of the formRf:={rf:r∈R}{\displaystyle \mathbb {R} f:=\{rf:r\in \mathbb {R} \}} withf≠0{\displaystyle f\neq 0}).

Finite-dimensional spaces

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ByF. Riesz's theorem, a Hausdorff topological vector space is finite-dimensional if and only if it islocally compact, which happens if and only if it has a compactneighborhood of the origin.

LetK{\displaystyle \mathbb {K} } denoteR{\displaystyle \mathbb {R} } orC{\displaystyle \mathbb {C} } and endowK{\displaystyle \mathbb {K} } with its usual Hausdorff normedEuclidean topology. LetX{\displaystyle X} be a vector space overK{\displaystyle \mathbb {K} } of finite dimensionn:=dim⁡X{\displaystyle n:=\dim X} and so thatX{\displaystyle X} is vector space isomorphic toKn{\displaystyle \mathbb {K} ^{n}} (explicitly, this means that there exists alinear isomorphism between the vector spacesX{\displaystyle X} andKn{\displaystyle \mathbb {K} ^{n}}). This finite-dimensional vector spaceX{\displaystyle X} always has a uniqueHausdorff vector topology, which makes it TVS-isomorphic toKn,{\displaystyle \mathbb {K} ^{n},} whereKn{\displaystyle \mathbb {K} ^{n}} is endowed with the usual Euclidean topology (which is the same as theproduct topology). This Hausdorff vector topology is also the (unique)finest vector topology onX.{\displaystyle X.}X{\displaystyle X} has a unique vector topology if and only ifdim⁡X=0.{\displaystyle \dim X=0.} Ifdim⁡X≠0{\displaystyle \dim X\neq 0} then althoughX{\displaystyle X} does not have a unique vector topology, it does have a uniqueHausdorff vector topology.

Proof outline

The proof of this dichotomy (i.e. that a vector topology is either trivial or isomorphic toK{\displaystyle \mathbb {K} }) is straightforward so only an outline with the important observations is given. As usual,K{\displaystyle \mathbb {K} } is assumed have the (normed) Euclidean topology. LetBr:={a∈K:|a|<r}{\displaystyle B_{r}:=\{a\in \mathbb {K} :|a|<r\}} for allr>0.{\displaystyle r>0.} LetX{\displaystyle X} be a1{\displaystyle 1}-dimensional vector space overK.{\displaystyle \mathbb {K} .} IfS⊆X{\displaystyle S\subseteq X} andB⊆K{\displaystyle B\subseteq \mathbb {K} } is a ball centered at0{\displaystyle 0} thenB⋅S=X{\displaystyle B\cdot S=X} wheneverS{\displaystyle S} contains an "unbounded sequence", by which it is meant a sequence of the form(aix)i=1∞{\displaystyle \left(a_{i}x\right)_{i=1}^{\infty }} where0≠x∈X{\displaystyle 0\neq x\in X} and(ai)i=1∞⊆K{\displaystyle \left(a_{i}\right)_{i=1}^{\infty }\subseteq \mathbb {K} } is unbounded in normed spaceK{\displaystyle \mathbb {K} } (in the usual sense). Any vector topology onX{\displaystyle X} will be translation invariant and invariant under non-zero scalar multiplication, and for every0≠x∈X,{\displaystyle 0\neq x\in X,} the mapMx:K→X{\displaystyle M_{x}:\mathbb {K} \to X} given byMx(a):=ax{\displaystyle M_{x}(a):=ax} is a continuous linear bijection. BecauseX=Kx{\displaystyle X=\mathbb {K} x} for any suchx,{\displaystyle x,} every subset ofX{\displaystyle X} can be written asFx=Mx(F){\displaystyle Fx=M_{x}(F)} for some unique subsetF⊆K.{\displaystyle F\subseteq \mathbb {K} .} And if this vector topology onX{\displaystyle X} has a neighborhoodW{\displaystyle W} of the origin that is not equal to all ofX,{\displaystyle X,} then the continuity of scalar multiplicationK×X→X{\displaystyle \mathbb {K} \times X\to X} at the origin guarantees the existence of an open ballBr⊆K{\displaystyle B_{r}\subseteq \mathbb {K} } centered at0{\displaystyle 0} and an open neighborhoodS{\displaystyle S} of the origin inX{\displaystyle X} such thatBr⋅S⊆W≠X,{\displaystyle B_{r}\cdot S\subseteq W\neq X,} which implies thatS{\displaystyle S} doesnot contain any "unbounded sequence". This implies that for every0≠x∈X,{\displaystyle 0\neq x\in X,} there exists some positive integern{\displaystyle n} such thatS⊆Bnx.{\displaystyle S\subseteq B_{n}x.} From this, it can be deduced that ifX{\displaystyle X} does not carry the trivial topology and if0≠x∈X,{\displaystyle 0\neq x\in X,} then for any ballB⊆K{\displaystyle B\subseteq \mathbb {K} } center at 0 inK,{\displaystyle \mathbb {K} ,}Mx(B)=Bx{\displaystyle M_{x}(B)=Bx} contains an open neighborhood of the origin inX,{\displaystyle X,} which then proves thatMx{\displaystyle M_{x}} is a linearhomeomorphism.Q.E.D.◼{\displaystyle \blacksquare }

Non-vector topologies

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Discrete and cofinite topologies

IfX{\displaystyle X} is a non-trivial vector space (that is, of non-zero dimension) then thediscrete topology onX{\displaystyle X} (which is alwaysmetrizable) isnot a TVS topology because despite making addition and negation continuous (which makes it into atopological group under addition), it fails to make scalar multiplication continuous. Thecofinite topology onX{\displaystyle X} (where a subset is open if and only if its complement is finite) is alsonot a TVS topology onX.{\displaystyle X.}

Linear maps

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A linear operator between two topological vector spaces which is continuous at one point is continuous on the whole domain. Moreover, a linear operatorf{\displaystyle f} is continuous iff(X){\displaystyle f(X)} is bounded (as defined below) for some neighborhoodX{\displaystyle X} of the origin.

Ahyperplane in a topological vector spaceX{\displaystyle X} is either dense or closed. Alinear functionalf{\displaystyle f} on a topological vector spaceX{\displaystyle X} has either dense or closed kernel. Moreover,f{\displaystyle f} is continuous if and only if itskernel isclosed.

Types

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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: theclosed graph theorem, theopen 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

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Main articles:Algebraic dual space,Continuous dual space, andStrong dual space

Every topological vector space has acontinuous dual space—the setX′{\displaystyle X'} of all continuous linear functionals, that is,continuous linear maps from the space into the base fieldK.{\displaystyle \mathbb {K} .} A topology on the dual can be defined to be the coarsest topology such that the dual pairing each point evaluationX′→K{\displaystyle X'\to \mathbb {K} } is continuous. This turns the dual into a locally convex topological vector space. This topology is called theweak-* topology.[27] This may not be the onlynatural 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 (seeBanach–Alaoglu theorem). Caution: WheneverX{\displaystyle X} is a non-normable locally convex space, then the pairing mapX′×X→K{\displaystyle X'\times X\to \mathbb {K} } is never continuous, no matter which vector space topology one chooses onX′.{\displaystyle 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.[28]

Properties

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See also:Locally convex topological vector space § Properties

For anyS⊆X{\displaystyle S\subseteq X} of a TVSX,{\displaystyle X,} theconvex (resp.balanced,disked, closed convex, closed balanced, closed disked')hull ofS{\displaystyle S} is the smallest subset ofX{\displaystyle X} that has this property and containsS.{\displaystyle S.} The closure (respectively, interior,convex hull, balanced hull, disked hull) of a setS{\displaystyle S} is sometimes denoted byclX⁡S{\displaystyle \operatorname {cl} _{X}S} (respectively,IntX⁡S,{\displaystyle \operatorname {Int} _{X}S,}co⁡S,{\displaystyle \operatorname {co} S,}bal⁡S,{\displaystyle \operatorname {bal} S,}cobal⁡S{\displaystyle \operatorname {cobal} S}).

Theconvex hullco⁡S{\displaystyle \operatorname {co} S} of a subsetS{\displaystyle S} is equal to the set of allconvex combinations of elements inS,{\displaystyle S,} which are finitelinear combinations of the formt1s1+⋯+tnsn{\displaystyle t_{1}s_{1}+\cdots +t_{n}s_{n}} wheren≥1{\displaystyle n\geq 1} is an integer,s1,…,sn∈S{\displaystyle s_{1},\ldots ,s_{n}\in S} andt1,…,tn∈[0,1]{\displaystyle t_{1},\ldots ,t_{n}\in [0,1]} sum to1.{\displaystyle 1.}[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]

Neighborhoods and open sets

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Properties of neighborhoods and open sets

Every TVS isconnected[6] andlocally connected[30] and any connected open subset of a TVS isarcwise connected. IfS⊆X{\displaystyle S\subseteq X} andU{\displaystyle U} is an open subset ofX{\displaystyle X} thenS+U{\displaystyle S+U} is an open set inX{\displaystyle X}[6] and ifS⊆X{\displaystyle S\subseteq X} has non-empty interior thenS−S{\displaystyle S-S} is a neighborhood of the origin.[6]

The open convex subsets of a TVSX{\displaystyle X} (not necessarily Hausdorff or locally convex) are exactly those that are of the formz+{x∈X:p(x)<1} = {x∈X:p(x−z)<1}{\displaystyle z+\{x\in X:p(x)<1\}~=~\{x\in X:p(x-z)<1\}} for somez∈X{\displaystyle z\in X} and some positive continuoussublinear functionalp{\displaystyle p} onX.{\displaystyle X.}[28]

IfK{\displaystyle K} is anabsorbingdisk in a TVSX{\displaystyle X} and ifp:=pK{\displaystyle p:=p_{K}} is theMinkowski functional ofK{\displaystyle K} then[31]IntX⁡K ⊆ {x∈X:p(x)<1} ⊆ K ⊆ {x∈X:p(x)≤1} ⊆ clX⁡K{\displaystyle \operatorname {Int} _{X}K~\subseteq ~\{x\in X:p(x)<1\}~\subseteq ~K~\subseteq ~\{x\in X:p(x)\leq 1\}~\subseteq ~\operatorname {cl} _{X}K} where importantly, it wasnot assumed thatK{\displaystyle K} had any topological properties nor thatp{\displaystyle p} was continuous (which happens if and only ifK{\displaystyle K} is a neighborhood of the origin).

Letτ{\displaystyle \tau } andν{\displaystyle \nu } be two vector topologies onX.{\displaystyle X.} Thenτ⊆ν{\displaystyle \tau \subseteq \nu } if and only if whenever a netx∙=(xi)i∈I{\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}} inX{\displaystyle X} converges0{\displaystyle 0} in(X,ν){\displaystyle (X,\nu )} thenx∙→0{\displaystyle x_{\bullet }\to 0} in(X,τ).{\displaystyle (X,\tau ).}[32]

LetN{\displaystyle {\mathcal {N}}} be a neighborhood basis of the origin inX,{\displaystyle X,} letS⊆X,{\displaystyle S\subseteq X,} and letx∈X.{\displaystyle x\in X.} Thenx∈clX⁡S{\displaystyle x\in \operatorname {cl} _{X}S} if and only if there exists a nets∙=(sN)N∈N{\displaystyle s_{\bullet }=\left(s_{N}\right)_{N\in {\mathcal {N}}}} inS{\displaystyle S} (indexed byN{\displaystyle {\mathcal {N}}}) such thats∙→x{\displaystyle s_{\bullet }\to x} inX.{\displaystyle X.}[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.

IfX{\displaystyle X} is a TVS that is of thesecond category in itself (that is, anonmeager space) then any closed convexabsorbing subset ofX{\displaystyle X} is a neighborhood of the origin.[34] This is no longer guaranteed if the set is not convex (a counter-example exists even inX=R2{\displaystyle X=\mathbb {R} ^{2}}) or ifX{\displaystyle X} is not of the second category in itself.[34]

Interior

IfR,S⊆X{\displaystyle R,S\subseteq X} andS{\displaystyle S} has non-empty interior thenIntX⁡S = IntX⁡(clX⁡S)  and  clX⁡S = clX⁡(IntX⁡S){\displaystyle \operatorname {Int} _{X}S~=~\operatorname {Int} _{X}\left(\operatorname {cl} _{X}S\right)~{\text{ and }}~\operatorname {cl} _{X}S~=~\operatorname {cl} _{X}\left(\operatorname {Int} _{X}S\right)}andIntX⁡(R)+IntX⁡(S) ⊆ R+IntX⁡S⊆IntX⁡(R+S).{\displaystyle \operatorname {Int} _{X}(R)+\operatorname {Int} _{X}(S)~\subseteq ~R+\operatorname {Int} _{X}S\subseteq \operatorname {Int} _{X}(R+S).}

Thetopological interior of adisk is not empty if and only if this interior contains the origin.[35] More generally, ifS{\displaystyle S} is abalanced set with non-empty interiorIntX⁡S≠∅{\displaystyle \operatorname {Int} _{X}S\neq \varnothing } in a TVSX{\displaystyle X} then{0}∪IntX⁡S{\displaystyle \{0\}\cup \operatorname {Int} _{X}S} will necessarily be balanced;[6] consequently,IntX⁡S{\displaystyle \operatorname {Int} _{X}S} will be balanced if and only if it contains the origin.[proof 2] For this (i.e.0∈IntX⁡S{\displaystyle 0\in \operatorname {Int} _{X}S}) to be true, it suffices forS{\displaystyle S} to also be convex (in addition to being balanced and having non-empty interior).;[6] The conclusion0∈IntX⁡S{\displaystyle 0\in \operatorname {Int} _{X}S} could be false ifS{\displaystyle S} is not also convex;[35] for example, inX:=R2,{\displaystyle X:=\mathbb {R} ^{2},} the interior of the closed and balanced setS:={(x,y):xy≥0}{\displaystyle S:=\{(x,y):xy\geq 0\}} is{(x,y):xy>0}.{\displaystyle \{(x,y):xy>0\}.}

IfC{\displaystyle C} is convex and0<t≤1,{\displaystyle 0<t\leq 1,} then[36]tInt⁡C+(1−t)cl⁡C ⊆ Int⁡C.{\displaystyle t\operatorname {Int} C+(1-t)\operatorname {cl} C~\subseteq ~\operatorname {Int} C.} Explicitly, this means that ifC{\displaystyle C} is a convex subset of a TVSX{\displaystyle X} (not necessarily Hausdorff or locally convex),y∈intX⁡C,{\displaystyle y\in \operatorname {int} _{X}C,} andx∈clX⁡C{\displaystyle x\in \operatorname {cl} _{X}C} then the open line segment joiningx{\displaystyle x} andy{\displaystyle y} belongs to the interior ofC;{\displaystyle C;} that is,{tx+(1−t)y:0<t<1}⊆intX⁡C.{\displaystyle \{tx+(1-t)y:0<t<1\}\subseteq \operatorname {int} _{X}C.}[37][38][proof 3]

IfN⊆X{\displaystyle N\subseteq X} is any balanced neighborhood of the origin inX{\displaystyle X} thenIntX⁡N⊆B1N=⋃0<|a|<1aN⊆N{\textstyle \operatorname {Int} _{X}N\subseteq B_{1}N=\bigcup _{0<|a|<1}aN\subseteq N} whereB1{\displaystyle B_{1}} is the set of all scalarsa{\displaystyle a} such that|a|<1.{\displaystyle |a|<1.}

Ifx{\displaystyle x} belongs to the interior of a convex setS⊆X{\displaystyle S\subseteq X} andy∈clX⁡S,{\displaystyle y\in \operatorname {cl} _{X}S,} then the half-open line segment[x,y):={tx+(1−t)y:0<t≤1}⊆IntX⁡ if x≠y{\displaystyle [x,y):=\{tx+(1-t)y:0<t\leq 1\}\subseteq \operatorname {Int} _{X}{\text{ if }}x\neq y} and[37][x,x)=∅ if x=y.{\displaystyle [x,x)=\varnothing {\text{ if }}x=y.} IfN{\displaystyle N} is abalanced neighborhood of0{\displaystyle 0} inX{\displaystyle X} andB1:={a∈K:|a|<1},{\displaystyle B_{1}:=\{a\in \mathbb {K} :|a|<1\},} then by considering intersections of the formN∩Rx{\displaystyle N\cap \mathbb {R} x} (which are convexsymmetric neighborhoods of0{\displaystyle 0} in the real TVSRx{\displaystyle \mathbb {R} x}) it follows that:Int⁡N=[0,1)Int⁡N=(−1,1)N=B1N,{\displaystyle \operatorname {Int} N=[0,1)\operatorname {Int} N=(-1,1)N=B_{1}N,} and furthermore, ifx∈Int⁡N and r:=sup{r>0:[0,r)x⊆N}{\displaystyle x\in \operatorname {Int} N{\text{ and }}r:=\sup\{r>0:[0,r)x\subseteq N\}} thenr>1 and [0,r)x⊆Int⁡N,{\displaystyle r>1{\text{ and }}[0,r)x\subseteq \operatorname {Int} N,} and ifr≠∞{\displaystyle r\neq \infty } thenrx∈cl⁡N∖Int⁡N.{\displaystyle rx\in \operatorname {cl} N\setminus \operatorname {Int} N.}

Non-Hausdorff spaces and the closure of the origin

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A topological vector spaceX{\displaystyle X} is Hausdorff if and only if{0}{\displaystyle \{0\}} is a closed subset ofX,{\displaystyle X,} or equivalently, if and only if{0}=clX⁡{0}.{\displaystyle \{0\}=\operatorname {cl} _{X}\{0\}.} Because{0}{\displaystyle \{0\}} is a vector subspace ofX,{\displaystyle X,} the same is true of its closureclX⁡{0},{\displaystyle \operatorname {cl} _{X}\{0\},} which is referred to asthe closure of the origin inX.{\displaystyle X.} This vector space satisfiesclX⁡{0}=⋂N∈N(0)N{\displaystyle \operatorname {cl} _{X}\{0\}=\bigcap _{N\in {\mathcal {N}}(0)}N} so that in particular, every neighborhood of the origin inX{\displaystyle X} contains the vector spaceclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} as a subset. Thesubspace topology onclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} is always thetrivial topology, which in particular implies that the topological vector spaceclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} acompact space (even if its dimension is non-zero or even infinite) and consequently also abounded subset ofX.{\displaystyle X.} In fact, a vector subspace of a TVS is bounded if and only if it is contained in the closure of{0}.{\displaystyle \{0\}.}[14] Every subset ofclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} 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, ifX{\displaystyle X} is not Hausdorff then there exist subsets that are bothcompact and complete butnot closed inX{\displaystyle X};[39] for instance, this will be true of any non-empty proper subset ofclX⁡{0}.{\displaystyle \operatorname {cl} _{X}\{0\}.}

IfS⊆X{\displaystyle S\subseteq X} is compact, thenclX⁡S=S+clX⁡{0}{\displaystyle \operatorname {cl} _{X}S=S+\operatorname {cl} _{X}\{0\}} and this set is compact. Thus the closure of a compact subset of a TVS is compact (said differently, all compact sets arerelatively compact),[40] which is not guaranteed for arbitrary non-Hausdorfftopological spaces.[note 6]

For every subsetS⊆X,{\displaystyle S\subseteq X,}S+clX⁡{0}⊆clX⁡S{\displaystyle S+\operatorname {cl} _{X}\{0\}\subseteq \operatorname {cl} _{X}S} and consequently, ifS⊆X{\displaystyle S\subseteq X} is open or closed inX{\displaystyle X} thenS+clX⁡{0}=S{\displaystyle S+\operatorname {cl} _{X}\{0\}=S}[proof 5] (so that thisarbitrary openor closed subsetsS{\displaystyle S} can be described as a"tube" whose vertical side is the vector spaceclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}}). For any subsetS⊆X{\displaystyle S\subseteq X} of this TVSX,{\displaystyle X,} the following are equivalent:

IfM{\displaystyle M} is a vector subspace of a TVSX{\displaystyle X} thenX/M{\displaystyle X/M} is Hausdorff if and only ifM{\displaystyle M} is closed inX.{\displaystyle X.} Moreover, thequotient mapq:X→X/clX⁡{0}{\displaystyle q:X\to X/\operatorname {cl} _{X}\{0\}} is always aclosed map onto the (necessarily) Hausdorff TVS.[44]

Every vector subspace ofX{\displaystyle X} that is an algebraic complement ofclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} (that is, a vector subspaceH{\displaystyle H} that satisfies{0}=H∩clX⁡{0}{\displaystyle \{0\}=H\cap \operatorname {cl} _{X}\{0\}} andX=H+clX⁡{0}{\displaystyle X=H+\operatorname {cl} _{X}\{0\}}) is atopological complement ofclX⁡{0}.{\displaystyle \operatorname {cl} _{X}\{0\}.} Consequently, ifH{\displaystyle H} is an algebraic complement ofclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} inX{\displaystyle X} then the addition mapH×clX⁡{0}→X,{\displaystyle H\times \operatorname {cl} _{X}\{0\}\to X,} defined by(h,n)↦h+n{\displaystyle (h,n)\mapsto h+n} is a TVS-isomorphism, whereH{\displaystyle H} is necessarily Hausdorff andclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} has theindiscrete topology.[45] Moreover, ifC{\displaystyle C} is a Hausdorffcompletion ofH{\displaystyle H} thenC×clX⁡{0}{\displaystyle C\times \operatorname {cl} _{X}\{0\}} is a completion ofX≅H×clX⁡{0}.{\displaystyle X\cong H\times \operatorname {cl} _{X}\{0\}.}[41]

Closed and compact sets

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Compact and totally bounded sets

A subset of a TVS is compact if and only if it is complete andtotally bounded.[39] Thus, in acomplete topological vector space, a closed and totally bounded subset is compact.[39] A subsetS{\displaystyle S} of a TVSX{\displaystyle X} istotally bounded if and only ifclX⁡S{\displaystyle \operatorname {cl} _{X}S} is totally bounded,[42][43] if and only if its image under the canonical quotient mapX→X/clX⁡({0}){\displaystyle X\to X/\operatorname {cl} _{X}(\{0\})} 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] IfS{\displaystyle S} is a subset of a TVSX{\displaystyle X} such that every sequence inS{\displaystyle S} has a cluster point inS{\displaystyle S} thenS{\displaystyle S} is totally bounded.[41]

IfK{\displaystyle K} is a compact subset of a TVSX{\displaystyle X} andU{\displaystyle U} is an open subset ofX{\displaystyle X} containingK,{\displaystyle K,} then there exists a neighborhoodN{\displaystyle N} of 0 such thatK+N⊆U.{\displaystyle K+N\subseteq U.}[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 abarrel.

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] IfM{\displaystyle M} is a vector subspace ofX{\displaystyle X} andN{\displaystyle N} is a closed neighborhood of the origin inX{\displaystyle X} such thatU∩N{\displaystyle U\cap N} is closed inX{\displaystyle X} thenM{\displaystyle M} is closed inX.{\displaystyle X.}[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).

IfS⊆X{\displaystyle S\subseteq X} anda{\displaystyle a} is a scalar thenaclX⁡S⊆clX⁡(aS),{\displaystyle a\operatorname {cl} _{X}S\subseteq \operatorname {cl} _{X}(aS),} where ifX{\displaystyle X} is Hausdorff,a≠0, or S=∅{\displaystyle a\neq 0,{\text{ or }}S=\varnothing } then equality holds:clX⁡(aS)=aclX⁡S.{\displaystyle \operatorname {cl} _{X}(aS)=a\operatorname {cl} _{X}S.} In particular, every non-zero scalar multiple of a closed set is closed. IfS⊆X{\displaystyle S\subseteq X} and ifA{\displaystyle A} is a set of scalars such that neithercl⁡S nor cl⁡A{\displaystyle \operatorname {cl} S{\text{ nor }}\operatorname {cl} A} contain zero then[47](cl⁡A)(clX⁡S)=clX⁡(AS).{\displaystyle \left(\operatorname {cl} A\right)\left(\operatorname {cl} _{X}S\right)=\operatorname {cl} _{X}(AS).}

IfS⊆X and S+S⊆2clX⁡S{\displaystyle S\subseteq X{\text{ and }}S+S\subseteq 2\operatorname {cl} _{X}S} thenclX⁡S{\displaystyle \operatorname {cl} _{X}S} is convex.[47]

IfR,S⊆X{\displaystyle R,S\subseteq X} then[6]clX⁡(R)+clX⁡(S) ⊆ clX⁡(R+S)  and  clX⁡[clX⁡(R)+clX⁡(S)] = clX⁡(R+S){\displaystyle \operatorname {cl} _{X}(R)+\operatorname {cl} _{X}(S)~\subseteq ~\operatorname {cl} _{X}(R+S)~{\text{ and }}~\operatorname {cl} _{X}\left[\operatorname {cl} _{X}(R)+\operatorname {cl} _{X}(S)\right]~=~\operatorname {cl} _{X}(R+S)} and so consequently, ifR+S{\displaystyle R+S} is closed then so isclX⁡(R)+clX⁡(S).{\displaystyle \operatorname {cl} _{X}(R)+\operatorname {cl} _{X}(S).}[47]

IfX{\displaystyle X} is a real TVS andS⊆X,{\displaystyle S\subseteq X,} then⋂r>1rS⊆clX⁡S{\displaystyle \bigcap _{r>1}rS\subseteq \operatorname {cl} _{X}S} where the left hand side is independent of the topology onX;{\displaystyle X;} moreover, ifS{\displaystyle S} is a convex neighborhood of the origin then equality holds.

For any subsetS⊆X,{\displaystyle S\subseteq X,}clX⁡S = ⋂N∈N(S+N){\displaystyle \operatorname {cl} _{X}S~=~\bigcap _{N\in {\mathcal {N}}}(S+N)} whereN{\displaystyle {\mathcal {N}}} is any neighborhood basis at the origin forX.{\displaystyle X.}[48] However,clX⁡U ⊇ ⋂{U:S⊆U,U is open in X}{\displaystyle \operatorname {cl} _{X}U~\supseteq ~\bigcap \{U:S\subseteq U,U{\text{ is open in }}X\}} and it is possible for this containment to be proper[49] (for example, ifX=R{\displaystyle X=\mathbb {R} } andS{\displaystyle S} is the rational numbers). It follows thatclX⁡U⊆U+U{\displaystyle \operatorname {cl} _{X}U\subseteq U+U} for every neighborhoodU{\displaystyle U} of the origin inX.{\displaystyle X.}[50]

Closed hulls

In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.[14]

IfR,S⊆X{\displaystyle R,S\subseteq X} and the closed convex hull of one of the setsS{\displaystyle S} orR{\displaystyle R} is compact then[51]clX⁡(co⁡(R+S)) = clX⁡(co⁡R)+clX⁡(co⁡S).{\displaystyle \operatorname {cl} _{X}(\operatorname {co} (R+S))~=~\operatorname {cl} _{X}(\operatorname {co} R)+\operatorname {cl} _{X}(\operatorname {co} S).} IfR,S⊆X{\displaystyle R,S\subseteq X} each have a closed convex hull that is compact (that is,clX⁡(co⁡R){\displaystyle \operatorname {cl} _{X}(\operatorname {co} R)} andclX⁡(co⁡S){\displaystyle \operatorname {cl} _{X}(\operatorname {co} S)} are compact) then[51]clX⁡(co⁡(R∪S)) = co⁡[clX⁡(co⁡R)∪clX⁡(co⁡S)].{\displaystyle \operatorname {cl} _{X}(\operatorname {co} (R\cup S))~=~\operatorname {co} \left[\operatorname {cl} _{X}(\operatorname {co} R)\cup \operatorname {cl} _{X}(\operatorname {co} S)\right].}

Hulls and compactness

In a general TVS, the closed convex hull of a compact set mayfail 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 compactconvex sets is again compact and convex.[6]

Other properties

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Meager, nowhere dense, and Baire

Adisk in a TVS is notnowhere 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 isnowhere dense.[9]

SupposeX{\displaystyle X} is a TVS that does not carry theindiscrete topology. ThenX{\displaystyle X} is aBaire space if and only ifX{\displaystyle X} has no balanced absorbing nowhere dense subset.[9]

A TVSX{\displaystyle X} is a Baire space if and only ifX{\displaystyle X} isnonmeager, which happens if and only if there does not exist anowhere dense setD{\displaystyle D} such thatX=⋃n∈NnD.{\textstyle X=\bigcup _{n\in \mathbb {N} }nD.}[9] Everynonmeager locally convex TVS is abarrelled space.[9]

Important algebraic facts and common misconceptions

IfS⊆X{\displaystyle S\subseteq X} then2S⊆S+S{\displaystyle 2S\subseteq S+S}; ifS{\displaystyle S} is convex then equality holds. For an example where equality doesnot hold, letx{\displaystyle x} be non-zero and setS={−x,x};{\displaystyle S=\{-x,x\};}S={x,2x}{\displaystyle S=\{x,2x\}} also works.

A subsetC{\displaystyle C} is convex if and only if(s+t)C=sC+tC{\displaystyle (s+t)C=sC+tC} for all positive reals>0 and t>0,{\displaystyle s>0{\text{ and }}t>0,}[29] or equivalently, if and only iftC+(1−t)C⊆C{\displaystyle tC+(1-t)C\subseteq C} for all0≤t≤1.{\displaystyle 0\leq t\leq 1.}[52]

Theconvex balanced hull of a setS⊆X{\displaystyle S\subseteq X} is equal to the convex hull of thebalanced hull ofS;{\displaystyle S;} that is, it is equal toco⁡(bal⁡S).{\displaystyle \operatorname {co} (\operatorname {bal} S).} But in general,bal⁡(co⁡S) ⊆ cobal⁡S = co⁡(bal⁡S),{\displaystyle \operatorname {bal} (\operatorname {co} S)~\subseteq ~\operatorname {cobal} S~=~\operatorname {co} (\operatorname {bal} S),} where the inclusion might be strict since thebalanced hull of a convex set need not be convex (counter-examples exist even inR2{\displaystyle \mathbb {R} ^{2}}).

IfR,S⊆X{\displaystyle R,S\subseteq X} anda{\displaystyle a} is a scalar then[6]a(R+S)=aR+aS,  and  co⁡(R+S)=co⁡R+co⁡S,  and  co⁡(aS)=aco⁡S.{\displaystyle a(R+S)=aR+aS,~{\text{ and }}~\operatorname {co} (R+S)=\operatorname {co} R+\operatorname {co} S,~{\text{ and }}~\operatorname {co} (aS)=a\operatorname {co} S.} IfR,S⊆X{\displaystyle R,S\subseteq X} are convex non-empty disjoint sets andx∉R∪S,{\displaystyle x\not \in R\cup S,} thenS∩co⁡(R∪{x})=∅{\displaystyle S\cap \operatorname {co} (R\cup \{x\})=\varnothing } orR∩co⁡(S∪{x})=∅.{\displaystyle R\cap \operatorname {co} (S\cup \{x\})=\varnothing .}

In any non-trivial vector spaceX,{\displaystyle X,} there exist two disjoint non-empty convex subsets whose union isX.{\displaystyle X.}

Other properties

Every TVS topology can be generated by afamily ofF-seminorms.[53]

IfP(x){\displaystyle P(x)} is some unarypredicate (a true or false statement dependent onx∈X{\displaystyle x\in X}) then for anyz∈X,{\displaystyle z\in X,}z+{x∈X:P(x)}={x∈X:P(x−z)}.{\displaystyle z+\{x\in X:P(x)\}=\{x\in X:P(x-z)\}.}[proof 6] So for example, ifP(x){\displaystyle P(x)} denotes "‖x‖<1{\displaystyle \|x\|<1}" then for anyz∈X,{\displaystyle z\in X,}z+{x∈X:‖x‖<1}={x∈X:‖x−z‖<1}.{\displaystyle z+\{x\in X:\|x\|<1\}=\{x\in X:\|x-z\|<1\}.} Similarly, ifs≠0{\displaystyle s\neq 0} is a scalar thens{x∈X:P(x)}={x∈X:P(1sx)}.{\displaystyle s\{x\in X:P(x)\}=\left\{x\in X:P\left({\tfrac {1}{s}}x\right)\right\}.} The elementsx∈X{\displaystyle x\in X} of these sets must range over a vector space (that is, overX{\displaystyle X}) rather than not just a subset or else these equalities are no longer guaranteed; similarly,z{\displaystyle z} must belong to this vector space (that is,z∈X{\displaystyle z\in X}).

Properties preserved by set operators

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  • The balanced hull of a compact (respectively,totally bounded, open) set has that same property.[6]
  • The(Minkowski) sum of two compact (respectively, bounded, balanced, convex) sets has that same property.[6] But the sum of two closed sets neednot be closed.
  • The convex hull of a balanced (resp. open) set is balanced (respectively, open). However, the convex hull of a closed set neednot be closed.[6] And the convex hull of a bounded set neednot be bounded.

The following table, the color of each cell indicates whether or not a given property of subsets ofX{\displaystyle 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∪S{\displaystyle 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.

Properties preserved by set operators
OperationProperty ofR,{\displaystyle R,}S,{\displaystyle S,} and any other subsets ofX{\displaystyle X} that is considered
AbsorbingBalancedConvexSymmetricConvex
Balanced		Vector
subspace		OpenNeighborhood
of 0		ClosedClosed
Balanced		Closed
Convex		Closed
Convex
Balanced		BarrelClosed
Vector
subspace		Totally
bounded		CompactCompact
Convex		Relatively compactCompleteSequentially
Complete		Banach
disk		BoundedBornivorousInfrabornivorousNowhere
dense (inX{\displaystyle X})		MeagerSeparablePseudometrizableOperation
R∪S{\displaystyle R\cup S}YesYesNoYesNoNoYesYesYesYesNoNoNoYesYesNoYesYesYesYesYesYesYesYesYesR∪S{\displaystyle R\cup S}
∪{\displaystyle \cup } of increasing nonempty chainYesYesYesYesYesYesYesYesNoNoNoNoNoNoNoNoNoNoNoNoNoYesYesNoNo∪{\displaystyle \cup } of increasing nonempty chain
Arbitrary unions (of at least 1 set)YesYesNoYesNoNoYesYesNoNoNoNoNoNoNoNoNoNoNoNoNoYesYesNoNoArbitrary unions (of at least 1 set)
R∩S{\displaystyle R\cap S}YesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesR∩S{\displaystyle R\cap S}
∩{\displaystyle \cap } of decreasing nonempty chainNoYesYesYesYesYesNoNoYesYesYesYesNoYesYesYesYesYes∩{\displaystyle \cap } of decreasing nonempty chain
Arbitrary intersections (of at least 1 set)NoYesYesYesYesNoYesNoYesYesYesYesNoYesYesYesYesYesArbitrary intersections (of at least 1 set)
R+S{\displaystyle R+S}YesYesYesYesYesYesYesYesNoNoYesYesYesR+S{\displaystyle R+S}
Scalar multipleNoYesYesYesYesYesNoNoNoNoNoNoNoNoYesYesYesYesYesYesYesNoNoYesYesYesYesScalar multiple
Non-0 scalar multipleYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesNon-0 scalar multiple
Positive scalar multipleYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesPositive scalar multiple
ClosureYesYesYesYesYesYesNoYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesClosure
InteriorNoNoYesYesNoYesYesNoNoNoNoNoNoNoNoYesYesNoInterior
Balanced coreYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesYesBalanced core
Balanced hullYesYesNoYesYesYesYesYesNoYesYesYesYesYesYesNoYesYesYesYesNoNoBalanced hull
Convex hullYesYesYesYesYesYesYesYesNoYesYesYesYesYesYesNoYesYesNoNoConvex hull
Convex balanced hullYesYesYesYesYesYesYesYesYesYesYesYesYesYesNoYesYesNoNoConvex balanced hull
Closed balanced hullYesYesNoYesYesYesNoYesYesYesYesYesYesYesYesYesYesNoNoClosed balanced hull
Closed convex hullYesYesYesYesYesYesNoYesYesYesYesYesYesYesNoYesYesNoNoClosed convex hull
Closed convex balanced hullYesYesYesYesYesYesNoYesYesYesYesYesYesYesNoYesYesNoNoClosed convex balanced hull
Linear spanYesYesYesYesYesYesYesYesYesYesNoNoNoNoYesNoYesYesNoNoLinear span
Pre-image under a continuous linear mapYesYesYesYesYesYesYesYesYesYesYesYesYesYesNoNoNoNoNoNoNoPre-image under a continuous linear map
Image under a continuous linear mapNoYesYesYesYesYesNoNoNoNoNoNoNoNoYesYesYesYesNoYesImage under a continuous linear map
Image under a continuous linear surjectionYesYesYesYesYesYesYesYesYesYesNoYesImage under a continuous linear surjection
Non-empty subset ofR{\displaystyle R}NoNoNoNoNoNoNoNoNoNoNoNoNoNoYesNoNoNoNoYesNoNoYesYesYesNon-empty subset ofR{\displaystyle R}
OperationAbsorbingBalancedConvexSymmetricConvex
Balanced		Vector
subspace		OpenNeighborhood
of 0		ClosedClosed
Balanced		Closed
Convex		Closed
Convex
Balanced		BarrelClosed
Vector
subspace		Totally
bounded		CompactCompact
Convex		Relatively compactCompleteSequentially
Complete		Banach
disk		BoundedBornivorousInfrabornivorousNowhere
dense (inX{\displaystyle X})		MeagerSeparablePseudometrizableOperation

See also

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Notes

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  1. ^The topological properties of course also require thatX{\displaystyle X} be a TVS.
  2. ^In particular,X{\displaystyle X} is Hausdorff if and only if the set{0}{\displaystyle \{0\}} is closed (that is,X{\displaystyle X} is aT1 space).
  3. ^In fact, this is true for topological group, since the proof does not use the scalar multiplications.
  4. ^Also called ametric 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.
  5. ^A series∑i=1∞xi{\textstyle \sum _{i=1}^{\infty }x_{i}} is said toconverge in a TVSX{\displaystyle X} if the sequence of partial sums converges.
  6. ^In general topology, the closure of a compact subset of a non-Hausdorff space may fail to be compact (for example, theparticular point topology on an infinite set). This result shows that this does not happen in non-Hausdorff TVSs.S+clX⁡{0}{\displaystyle S+\operatorname {cl} _{X}\{0\}} is compact because it is the image of the compact setS×clX⁡{0}{\displaystyle S\times \operatorname {cl} _{X}\{0\}} under the continuous addition map⋅+⋅:X×X→X.{\displaystyle \cdot \,+\,\cdot \;:X\times X\to X.} Recall also that the sum of a compact set (that is,S{\displaystyle S}) and a closed set is closed soS+clX⁡{0}{\displaystyle S+\operatorname {cl} _{X}\{0\}} is closed inX.{\displaystyle X.}
  7. ^InR2,{\displaystyle \mathbb {R} ^{2},} the sum of they{\displaystyle y}-axis and the graph ofy=1x,{\displaystyle y={\frac {1}{x}},} which is the complement of they{\displaystyle y}-axis, is open inR2.{\displaystyle \mathbb {R} ^{2}.} InR,{\displaystyle \mathbb {R} ,} theMinkowski sumZ+2Z{\displaystyle \mathbb {Z} +{\sqrt {2}}\mathbb {Z} } is a countable dense subset ofR{\displaystyle \mathbb {R} } so not closed inR.{\displaystyle \mathbb {R} .}

Proofs

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  1. ^This condition is satisfied ifS{\displaystyle \mathbb {S} } denotes the set of all topological strings in(X,τ).{\displaystyle (X,\tau ).}
  2. ^This is because every non-empty balanced set must contain the origin and because0∈IntX⁡S{\displaystyle 0\in \operatorname {Int} _{X}S} if and only ifIntX⁡S={0}∪IntX⁡S.{\displaystyle \operatorname {Int} _{X}S=\{0\}\cup \operatorname {Int} _{X}S.}
  3. ^Fix0<r<1{\displaystyle 0<r<1} so it remains to show thatw0 =def rx+(1−r)y{\displaystyle w_{0}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~rx+(1-r)y} belongs tointX⁡C.{\displaystyle \operatorname {int} _{X}C.} By replacingC,x,y{\displaystyle C,x,y} withC−w0,x−w0,y−w0{\displaystyle C-w_{0},x-w_{0},y-w_{0}} if necessary, we may assume without loss of generality thatrx+(1−r)y=0,{\displaystyle rx+(1-r)y=0,} and so it remains to show thatC{\displaystyle C} is a neighborhood of the origin. Lets =def rr−1<0{\displaystyle s~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\tfrac {r}{r-1}}<0} so thaty=rr−1x=sx.{\displaystyle y={\tfrac {r}{r-1}}x=sx.} Since scalar multiplication bys≠0{\displaystyle s\neq 0} is a linear homeomorphismX→X,{\displaystyle X\to X,}clX⁡(1sC)=1sclX⁡C.{\displaystyle \operatorname {cl} _{X}\left({\tfrac {1}{s}}C\right)={\tfrac {1}{s}}\operatorname {cl} _{X}C.} Sincex∈int⁡C{\displaystyle x\in \operatorname {int} C} andy∈cl⁡C,{\displaystyle y\in \operatorname {cl} C,} it follows thatx=1sy∈cl⁡(1sC)∩int⁡C{\displaystyle x={\tfrac {1}{s}}y\in \operatorname {cl} \left({\tfrac {1}{s}}C\right)\cap \operatorname {int} C} where becauseint⁡C{\displaystyle \operatorname {int} C} is open, there exists somec0∈(1sC)∩int⁡C,{\displaystyle c_{0}\in \left({\tfrac {1}{s}}C\right)\cap \operatorname {int} C,} which satisfiessc0∈C.{\displaystyle sc_{0}\in C.} Defineh:X→X{\displaystyle h:X\to X} byx↦rx+(1−r)sc0=rx−rc0,{\displaystyle x\mapsto rx+(1-r)sc_{0}=rx-rc_{0},} which is a homeomorphism because0<r<1.{\displaystyle 0<r<1.} The seth(int⁡C){\displaystyle h\left(\operatorname {int} C\right)} is thus an open subset ofX{\displaystyle X} that moreover containsh(c0)=rc0−rc0=0.{\textstyle h(c_{0})=rc_{0}-rc_{0}=0.} Ifc∈int⁡C{\displaystyle c\in \operatorname {int} C} thenh(c)=rc+(1−r)sc0∈C{\textstyle h(c)=rc+(1-r)sc_{0}\in C} sinceC{\displaystyle C} is convex,0<r<1,{\displaystyle 0<r<1,} andsc0,c∈C,{\displaystyle sc_{0},c\in C,} which proves thath(int⁡C)⊆C.{\displaystyle h\left(\operatorname {int} C\right)\subseteq C.} Thush(int⁡C){\displaystyle h\left(\operatorname {int} C\right)} is an open subset ofX{\displaystyle X} that contains the origin and is contained inC.{\displaystyle C.} Q.E.D.
  4. ^SinceclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} 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.
  5. ^Ifs∈S{\displaystyle s\in S} thens+clX⁡{0}=clX⁡(s+{0})=clX⁡{s}⊆clX⁡S.{\displaystyle s+\operatorname {cl} _{X}\{0\}=\operatorname {cl} _{X}(s+\{0\})=\operatorname {cl} _{X}\{s\}\subseteq \operatorname {cl} _{X}S.} BecauseS⊆S+clX⁡{0}⊆clX⁡S,{\displaystyle S\subseteq S+\operatorname {cl} _{X}\{0\}\subseteq \operatorname {cl} _{X}S,} ifS{\displaystyle S} is closed then equality holds. Using the fact thatclX⁡{0}{\displaystyle \operatorname {cl} _{X}\{0\}} is a vector space, it is readily verified that the complement inX{\displaystyle X} of any setS{\displaystyle S} satisfying the equalityS+clX⁡{0}=S{\displaystyle S+\operatorname {cl} _{X}\{0\}=S} must also satisfy this equality (whenX∖S{\displaystyle X\setminus S} is substituted forS{\displaystyle S}).
  6. ^z+{x∈X:P(x)}={z+x:x∈X,P(x)}={z+x:x∈X,P((z+x)−z)}{\displaystyle z+\{x\in X:P(x)\}=\{z+x:x\in X,P(x)\}=\{z+x:x\in X,P((z+x)-z)\}} and so usingy=z+x{\displaystyle y=z+x} and the fact thatz+X=X,{\displaystyle z+X=X,} this is equal to{y:y−z∈X,P(y−z)}={y:y∈X,P(y−z)}={y∈X:P(y−z)}.{\displaystyle \{y:y-z\in X,P(y-z)\}=\{y:y\in X,P(y-z)\}=\{y\in X:P(y-z)\}.}Q.E.D.◼{\displaystyle \blacksquare }

Citations

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  1. ^Rudin 1991, p. 4-5 §1.3.
  2. ^abcKöthe 1983, p. 91.
  3. ^Schaefer & Wolff 1999, pp. 74–78.
  4. ^Grothendieck 1973, pp. 34–36.
  5. ^abcWilansky 2013, pp. 40–47.
  6. ^abcdefghijklmnopqrstNarici & Beckenstein 2011, pp. 67–113.
  7. ^abcdeAdasch, Ernst & Keim 1978, pp. 5–9.
  8. ^Schechter 1996, pp. 721–751.
  9. ^abcdefNarici & Beckenstein 2011, pp. 371–423.
  10. ^Adasch, Ernst & Keim 1978, pp. 10–15.
  11. ^Wilansky 2013, p. 53.
  12. ^abcRudin 1991, p. 6 §1.4.
  13. ^Rudin 1991, p. 8.
  14. ^abcdeNarici & Beckenstein 2011, pp. 155–176.
  15. ^Rudin 1991, p. 27-28 Theorem 1.37.
  16. ^Köthe 1983, section 15.11.
  17. ^"Topological vector space",Encyclopedia of Mathematics,EMS Press, 2001 [1994], retrieved26 February 2021
  18. ^Rudin 1991, p. 17 Theorem 1.22.
  19. ^Schaefer & Wolff 1999, pp. 12–19.
  20. ^Schaefer & Wolff 1999, p. 16.
  21. ^abcNarici & Beckenstein 2011, pp. 115–154.
  22. ^Swartz 1992, pp. 27–29.
  23. ^"A quick application of the closed graph theorem".What's new. 2016-04-22. Retrieved2020-10-07.
  24. ^abNarici & Beckenstein 2011, p. 111.
  25. ^abcRudin 1991, p. 9 §1.8.
  26. ^Rudin 1991, p. 27 Theorem 1.36.
  27. ^Rudin 1991, p. 62-68 §3.8-3.14.
  28. ^abNarici & Beckenstein 2011, pp. 177–220.
  29. ^abcRudin 1991, p. 38.
  30. ^Schaefer & Wolff 1999, p. 35.
  31. ^Narici & Beckenstein 2011, p. 119-120.
  32. ^Wilansky 2013, p. 43.
  33. ^Wilansky 2013, p. 42.
  34. ^abRudin 1991, p. 55.
  35. ^abNarici & Beckenstein 2011, p. 108.
  36. ^Jarchow 1981, pp. 101–104.
  37. ^abSchaefer & Wolff 1999, p. 38.
  38. ^Conway 1990, p. 102.
  39. ^abcdefNarici & Beckenstein 2011, pp. 47–66.
  40. ^Narici & Beckenstein 2011, p. 156.
  41. ^abcdeSchaefer & Wolff 1999, pp. 12–35.
  42. ^abSchaefer & Wolff 1999, p. 25.
  43. ^abJarchow 1981, pp. 56–73.
  44. ^Narici & Beckenstein 2011, pp. 107–112.
  45. ^Wilansky 2013, p. 63.
  46. ^abNarici & Beckenstein 2011, pp. 19–45.
  47. ^abcWilansky 2013, pp. 43–44.
  48. ^Narici & Beckenstein 2011, pp. 80.
  49. ^Narici & Beckenstein 2011, pp. 108–109.
  50. ^Jarchow 1981, pp. 30–32.
  51. ^abcNarici & Beckenstein 2011, p. 109.
  52. ^Rudin 1991, p. 6.
  53. ^Swartz 1992, p. 35.

Bibliography

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Further reading

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