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Metrizable topological vector space

From Wikipedia, the free encyclopedia
Topological vector space whose topology can be defined by a metric

Infunctional analysis and related areas ofmathematics, ametrizable (resp.pseudometrizable)topological vector space (TVS) is a TVS whose topology is induced by a metric (resp.pseudometric). AnLM-space is aninductive limit of a sequence oflocally convex metrizable TVS.

Pseudometrics and metrics

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Apseudometric on a setX{\displaystyle X} is a mapd:X×XR{\displaystyle d:X\times X\rightarrow \mathbb {R} } satisfying the following properties:

  1. d(x,x)=0 for all xX{\displaystyle d(x,x)=0{\text{ for all }}x\in X};
  2. Symmetry:d(x,y)=d(y,x) for all x,yX{\displaystyle d(x,y)=d(y,x){\text{ for all }}x,y\in X};
  3. Subadditivity:d(x,z)d(x,y)+d(y,z) for all x,y,zX.{\displaystyle d(x,z)\leq d(x,y)+d(y,z){\text{ for all }}x,y,z\in X.}

A pseudometric is called ametric if it satisfies:

  1. Identity of indiscernibles: for allx,yX,{\displaystyle x,y\in X,} ifd(x,y)=0{\displaystyle d(x,y)=0} thenx=y.{\displaystyle x=y.}

Ultrapseudometric

A pseudometricd{\displaystyle d} onX{\displaystyle X} is called aultrapseudometric or astrong pseudometric if it satisfies:

  1. Strong/Ultrametric triangle inequality:d(x,z)max{d(x,y),d(y,z)} for all x,y,zX.{\displaystyle d(x,z)\leq \max\{d(x,y),d(y,z)\}{\text{ for all }}x,y,z\in X.}

Pseudometric space

Apseudometric space is a pair(X,d){\displaystyle (X,d)} consisting of a setX{\displaystyle X} and a pseudometricd{\displaystyle d} onX{\displaystyle X} such thatX{\displaystyle X}'s topology is identical to the topology onX{\displaystyle X} induced byd.{\displaystyle d.} We call a pseudometric space(X,d){\displaystyle (X,d)} ametric space (resp.ultrapseudometric space) whend{\displaystyle d} is a metric (resp. ultrapseudometric).

Topology induced by a pseudometric

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Ifd{\displaystyle d} is a pseudometric on a setX{\displaystyle X} then collection ofopen balls:Br(z):={xX:d(x,z)<r}{\displaystyle B_{r}(z):=\{x\in X:d(x,z)<r\}} asz{\displaystyle z} ranges overX{\displaystyle X} andr>0{\displaystyle r>0} ranges over the positive real numbers,forms a basis for a topology onX{\displaystyle X} that is called thed{\displaystyle d}-topology or thepseudometric topology onX{\displaystyle X} induced byd.{\displaystyle d.}

Convention: If(X,d){\displaystyle (X,d)} is a pseudometric space andX{\displaystyle X} is treated as atopological space, then unless indicated otherwise, it should be assumed thatX{\displaystyle X} is endowed with the topology induced byd.{\displaystyle d.}

Pseudometrizable space

A topological space(X,τ){\displaystyle (X,\tau )} is calledpseudometrizable (resp.metrizable,ultrapseudometrizable) if there exists a pseudometric (resp. metric, ultrapseudometric)d{\displaystyle d} onX{\displaystyle X} such thatτ{\displaystyle \tau } is equal to the topology induced byd.{\displaystyle d.}[1]

Pseudometrics and values on topological groups

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An additivetopological group is an additive group endowed with a topology, called agroup topology, under which addition and negation become continuous operators.

A topologyτ{\displaystyle \tau } on a real or complex vector spaceX{\displaystyle X} is called avector topology or aTVS topology if it makes the operations of vector addition andscalar multiplication continuous (that is, if it makesX{\displaystyle X} into atopological vector space).

Everytopological vector space (TVS)X{\displaystyle X} is an additive commutative topological group but not all group topologies onX{\displaystyle X} are vector topologies. This is because despite it making addition and negation continuous, a group topology on a vector spaceX{\displaystyle X} may fail to make scalar multiplication continuous. For instance, thediscrete topology on any non-trivial vector space makes addition and negation continuous but do not make scalar multiplication continuous.

Translation invariant pseudometrics

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IfX{\displaystyle X} is an additive group then we say that a pseudometricd{\displaystyle d} onX{\displaystyle X} istranslation invariant or justinvariant if it satisfies any of the following equivalent conditions:

  1. Translation invariance:d(x+z,y+z)=d(x,y) for all x,y,zX{\displaystyle d(x+z,y+z)=d(x,y){\text{ for all }}x,y,z\in X};
  2. d(x,y)=d(xy,0) for all x,yX.{\displaystyle d(x,y)=d(x-y,0){\text{ for all }}x,y\in X.}

Value/G-seminorm

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IfX{\displaystyle X} is atopological group the avalue orG-seminorm onX{\displaystyle X} (theG stands for Group) is a real-valued mapp:XR{\displaystyle p:X\rightarrow \mathbb {R} } with the following properties:[2]

  1. Non-negative:p0.{\displaystyle p\geq 0.}
  2. Subadditive:p(x+y)p(x)+p(y) for all x,yX{\displaystyle p(x+y)\leq p(x)+p(y){\text{ for all }}x,y\in X};
  3. p(0)=0..{\displaystyle p(0)=0..}
  4. Symmetric:p(x)=p(x) for all xX.{\displaystyle p(-x)=p(x){\text{ for all }}x\in X.}

where we call a G-seminorm aG-norm if it satisfies the additional condition:

  1. Total/Positive definite: Ifp(x)=0{\displaystyle p(x)=0} thenx=0.{\displaystyle x=0.}

Properties of values

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Ifp{\displaystyle p} is a value on a vector spaceX{\displaystyle X} then:

Equivalence on topological groups

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Theorem[2]Suppose thatX{\displaystyle X} is an additive commutative group. Ifd{\displaystyle d} is a translation invariant pseudometric onX{\displaystyle X} then the mapp(x):=d(x,0){\displaystyle p(x):=d(x,0)} is a value onX{\displaystyle X} calledthe value associated withd{\displaystyle d}, and moreover,d{\displaystyle d} generates a group topology onX{\displaystyle X} (i.e. thed{\displaystyle d}-topology onX{\displaystyle X} makesX{\displaystyle X} into a topological group).Conversely, ifp{\displaystyle p} is a value onX{\displaystyle X} then the mapd(x,y):=p(xy){\displaystyle d(x,y):=p(x-y)} is a translation-invariant pseudometric onX{\displaystyle X} and the value associated withd{\displaystyle d} is justp.{\displaystyle p.}

Pseudometrizable topological groups

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Theorem[2]If(X,τ){\displaystyle (X,\tau )} is an additive commutativetopological group then the following are equivalent:

  1. τ{\displaystyle \tau } is induced by a pseudometric; (i.e.(X,τ){\displaystyle (X,\tau )} is pseudometrizable);
  2. τ{\displaystyle \tau } is induced by a translation-invariant pseudometric;
  3. the identity element in(X,τ){\displaystyle (X,\tau )} has a countable neighborhood basis.

If(X,τ){\displaystyle (X,\tau )} is Hausdorff then the word "pseudometric" in the above statement may be replaced by the word "metric." A commutative topological group is metrizable if and only if it is Hausdorff and pseudometrizable.

An invariant pseudometric that doesn't induce a vector topology

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LetX{\displaystyle X} be a non-trivial (i.e.X{0}{\displaystyle X\neq \{0\}}) real or complex vector space and letd{\displaystyle d} be the translation-invarianttrivial metric onX{\displaystyle X} defined byd(x,x)=0{\displaystyle d(x,x)=0} andd(x,y)=1 for all x,yX{\displaystyle d(x,y)=1{\text{ for all }}x,y\in X} such thatxy.{\displaystyle x\neq y.} The topologyτ{\displaystyle \tau } thatd{\displaystyle d} induces onX{\displaystyle X} is thediscrete topology, which makes(X,τ){\displaystyle (X,\tau )} into a commutative topological group under addition but doesnot form a vector topology onX{\displaystyle X} because(X,τ){\displaystyle (X,\tau )} isdisconnected but every vector topology is connected. What fails is that scalar multiplication isn't continuous on(X,τ).{\displaystyle (X,\tau ).}

This example shows that a translation-invariant (pseudo)metric isnot enough to guarantee a vector topology, which leads us to define paranorms andF-seminorms.

Additive sequences

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

Continuity of addition at 0If(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×XX{\displaystyle X\times X\to X} (i.e. the map(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."[5]

All of the above conditions are consequently a necessary for a topology to form a vector topology. Additive 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 and also show that a Hausdorff TVS with a countable basis of neighborhoods is metrizable. The following theorem is true more generally for commutative additivetopological groups.

TheoremLetU=(Ui)i=0{\displaystyle U_{\bullet }=\left(U_{i}\right)_{i=0}^{\infty }} be a collection of subsets of a vector space such that0Ui{\displaystyle 0\in U_{i}} andUi+1+Ui+1Ui{\displaystyle U_{i+1}+U_{i+1}\subseteq U_{i}} for alli0.{\displaystyle i\geq 0.} For alluU0,{\displaystyle u\in U_{0},} letS(u):={n=(n1,,nk) : k1,ni0 for all i, and uUn1++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} ifxU0{\displaystyle x\not \in U_{0}} and otherwise letf(x):=inf{2n1+2nk : 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} issubadditive (meaningf(x+y)f(x)+f(y) for all x,yX{\displaystyle f(x+y)\leq f(x)+f(y){\text{ for all }}x,y\in X}) andf=0{\displaystyle f=0} oni0Ui,{\displaystyle \bigcap _{i\geq 0}U_{i},} so in particularf(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 allxX.{\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(xy){\displaystyle d(x,y):=f(x-y)} is a metric defining the vector topology onX.{\displaystyle X.}

Proof

Assume thatn=(n1,,nk){\displaystyle n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)} always denotes a finite sequence of non-negative integers and use the notation:2n:=2n1++2nk and Un:=Un1++Unk.{\displaystyle \sum 2^{-n_{\bullet }}:=2^{-n_{1}}+\cdots +2^{-n_{k}}\quad {\text{ and }}\quad \sum U_{n_{\bullet }}:=U_{n_{1}}+\cdots +U_{n_{k}}.}

For any integersn0{\displaystyle n\geq 0} andd>2,{\displaystyle d>2,}UnUn+1+Un+1Un+1+Un+2+Un+2Un+1+Un+2++Un+d+Un+d+1+Un+d+1.{\displaystyle U_{n}\supseteq U_{n+1}+U_{n+1}\supseteq U_{n+1}+U_{n+2}+U_{n+2}\supseteq U_{n+1}+U_{n+2}+\cdots +U_{n+d}+U_{n+d+1}+U_{n+d+1}.}

From this it follows that ifn=(n1,,nk){\displaystyle n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)} consists of distinct positive integers thenUnU1+min(n).{\displaystyle \sum U_{n_{\bullet }}\subseteq U_{-1+\min \left(n_{\bullet }\right)}.}

It will now be shown by induction onk{\displaystyle k} that ifn=(n1,,nk){\displaystyle n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)} consists of non-negative integers such that2n2M{\displaystyle \sum 2^{-n_{\bullet }}\leq 2^{-M}} for some integerM0{\displaystyle M\geq 0} thenUnUM.{\displaystyle \sum U_{n_{\bullet }}\subseteq U_{M}.} This is clearly true fork=1{\displaystyle k=1} andk=2{\displaystyle k=2} so assume thatk>2,{\displaystyle k>2,} which implies that allni{\displaystyle n_{i}} are positive. If allni{\displaystyle n_{i}} are distinct then this step is done, and otherwise pick distinct indicesi<j{\displaystyle i<j} such thatni=nj{\displaystyle n_{i}=n_{j}} and constructm=(m1,,mk1){\displaystyle m_{\bullet }=\left(m_{1},\ldots ,m_{k-1}\right)} fromn{\displaystyle n_{\bullet }} by replacing eachni{\displaystyle n_{i}} withni1{\displaystyle n_{i}-1} and deleting thejth{\displaystyle j^{\text{th}}} element ofn{\displaystyle n_{\bullet }} (all other elements ofn{\displaystyle n_{\bullet }} are transferred tom{\displaystyle m_{\bullet }} unchanged). Observe that2n=2m{\displaystyle \sum 2^{-n_{\bullet }}=\sum 2^{-m_{\bullet }}} andUnUm{\displaystyle \sum U_{n_{\bullet }}\subseteq \sum U_{m_{\bullet }}} (becauseUni+UnjUni1{\displaystyle U_{n_{i}}+U_{n_{j}}\subseteq U_{n_{i}-1}}) so by appealing to the inductive hypothesis we conclude thatUnUmUM,{\displaystyle \sum U_{n_{\bullet }}\subseteq \sum U_{m_{\bullet }}\subseteq U_{M},} as desired.

It is clear thatf(0)=0{\displaystyle f(0)=0} and that0f1{\displaystyle 0\leq f\leq 1} so to prove thatf{\displaystyle f} is subadditive, it suffices to prove thatf(x+y)f(x)+f(y){\displaystyle f(x+y)\leq f(x)+f(y)} whenx,yX{\displaystyle x,y\in X} are such thatf(x)+f(y)<1,{\displaystyle f(x)+f(y)<1,} which implies thatx,yU0.{\displaystyle x,y\in U_{0}.} This is an exercise. If allUi{\displaystyle U_{i}} are symmetric thenxUn{\displaystyle x\in \sum U_{n_{\bullet }}} if and only ifxUn{\displaystyle -x\in \sum U_{n_{\bullet }}} from which it follows thatf(x)f(x){\displaystyle f(-x)\leq f(x)} andf(x)f(x).{\displaystyle f(-x)\geq f(x).} If allUi{\displaystyle U_{i}} are balanced then the inequalityf(sx)f(x){\displaystyle f(sx)\leq f(x)} for all unit scalarss{\displaystyle s} such that|s|1{\displaystyle |s|\leq 1} is proved similarly. Becausef{\displaystyle f} is a nonnegative subadditive function satisfyingf(0)=0,{\displaystyle f(0)=0,} as described in the article onsublinear functionals,f{\displaystyle f} is uniformly continuous onX{\displaystyle X} if and only iff{\displaystyle f} is continuous at the origin. If allUi{\displaystyle U_{i}} are neighborhoods of the origin then for any realr>0,{\displaystyle r>0,} pick an integerM>1{\displaystyle M>1} such that2M<r{\displaystyle 2^{-M}<r} so thatxUM{\displaystyle x\in U_{M}} impliesf(x)2M<r.{\displaystyle f(x)\leq 2^{-M}<r.} If the set of allUi{\displaystyle U_{i}} form basis of balanced neighborhoods of the origin then it may be shown that for anyn>1,{\displaystyle n>1,} there exists some0<r2n{\displaystyle 0<r\leq 2^{-n}} such thatf(x)<r{\displaystyle f(x)<r} impliesxUn.{\displaystyle x\in U_{n}.}{\displaystyle \blacksquare }

Paranorms

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IfX{\displaystyle X} is a vector space over the real or complex numbers then aparanorm onX{\displaystyle X} is a G-seminorm (defined above)p:XR{\displaystyle p:X\rightarrow \mathbb {R} } onX{\displaystyle X} that satisfies any of the following additional conditions, each of which begins with "for all sequencesx=(xi)i=1{\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }} inX{\displaystyle X} and all convergent sequences of scalarss=(si)i=1{\displaystyle s_{\bullet }=\left(s_{i}\right)_{i=1}^{\infty }}":[6]

  1. Continuity of multiplication: ifs{\displaystyle s} is a scalar andxX{\displaystyle x\in X} are such thatp(xix)0{\displaystyle p\left(x_{i}-x\right)\to 0} andss,{\displaystyle s_{\bullet }\to s,} thenp(sixisx)0.{\displaystyle p\left(s_{i}x_{i}-sx\right)\to 0.}
  2. Both of the conditions:
  3. Both of the conditions:
  4. Separate continuity:[7]

A paranorm is calledtotal if in addition it satisfies:

Properties of paranorms

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Ifp{\displaystyle p} is a paranorm on a vector spaceX{\displaystyle X} then the mapd:X×XR{\displaystyle d:X\times X\rightarrow \mathbb {R} } defined byd(x,y):=p(xy){\displaystyle d(x,y):=p(x-y)} is a translation-invariant pseudometric onX{\displaystyle X} that defines avector topology onX.{\displaystyle X.}[8]

Ifp{\displaystyle p} is a paranorm on a vector spaceX{\displaystyle X} then:

Examples of paranorms

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F-seminorms

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IfX{\displaystyle X} is a vector space over the real or complex numbers then anF-seminorm onX{\displaystyle X} (theF{\displaystyle F} stands forFréchet) is a real-valued mapp:XR{\displaystyle p:X\to \mathbb {R} } with the following four properties:[11]

  1. Non-negative:p0.{\displaystyle p\geq 0.}
  2. Subadditive:p(x+y)p(x)+p(y){\displaystyle p(x+y)\leq p(x)+p(y)} for allx,yX{\displaystyle x,y\in X}
  3. Balanced:p(ax)p(x){\displaystyle p(ax)\leq p(x)} forxX{\displaystyle x\in X} all scalarsa{\displaystyle a} satisfying|a|1;{\displaystyle |a|\leq 1;}
  4. For everyxX,{\displaystyle x\in X,}p(1nx)0{\displaystyle p\left({\tfrac {1}{n}}x\right)\to 0} asn{\displaystyle n\to \infty }

AnF-seminorm is called anF-norm if in addition it satisfies:

  1. Total/Positive definite:p(x)=0{\displaystyle p(x)=0} impliesx=0.{\displaystyle x=0.}

AnF-seminorm is calledmonotone if it satisfies:

  1. Monotone:p(rx)<p(sx){\displaystyle p(rx)<p(sx)} for all non-zeroxX{\displaystyle x\in X} and all reals{\displaystyle s} andt{\displaystyle t} such thats<t.{\displaystyle s<t.}[12]

F-seminormed spaces

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AnF-seminormed space (resp.F-normed space)[12] is a pair(X,p){\displaystyle (X,p)} consisting of a vector spaceX{\displaystyle X} and anF-seminorm (resp.F-norm)p{\displaystyle p} onX.{\displaystyle X.}

If(X,p){\displaystyle (X,p)} and(Z,q){\displaystyle (Z,q)} areF-seminormed spaces then a mapf:XZ{\displaystyle f:X\to Z} is called anisometric embedding[12] ifq(f(x)f(y))=p(x,y) for all x,yX.{\displaystyle q(f(x)-f(y))=p(x,y){\text{ for all }}x,y\in X.}

Every isometric embedding of oneF-seminormed space into another is atopological embedding, but the converse is not true in general.[12]

Examples ofF-seminorms

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Properties ofF-seminorms

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EveryF-seminorm is a paranorm and every paranorm is equivalent to someF-seminorm.[7] EveryF-seminorm on a vector spaceX{\displaystyle X} is a value onX.{\displaystyle X.} In particular,p(x)=0,{\displaystyle p(x)=0,} andp(x)=p(x){\displaystyle p(x)=p(-x)} for allxX.{\displaystyle x\in X.}

Topology induced by a singleF-seminorm

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Theorem[11]Letp{\displaystyle p} be anF-seminorm on a vector spaceX.{\displaystyle X.} Then the mapd:X×XR{\displaystyle d:X\times X\to \mathbb {R} } defined byd(x,y):=p(xy){\displaystyle d(x,y):=p(x-y)} is a translation invariant pseudometric onX{\displaystyle X} that defines a vector topologyτ{\displaystyle \tau } onX.{\displaystyle X.} Ifp{\displaystyle p} is anF-norm thend{\displaystyle d} is a metric. WhenX{\displaystyle X} is endowed with this topology thenp{\displaystyle p} is a continuous map onX.{\displaystyle X.}

The balanced sets{xX : p(x)r},{\displaystyle \{x\in X~:~p(x)\leq r\},} asr{\displaystyle r} ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of closed set. Similarly, the balanced sets{xX : p(x)<r},{\displaystyle \{x\in X~:~p(x)<r\},} asr{\displaystyle r} ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of open sets.

Topology induced by a family ofF-seminorms

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Suppose thatL{\displaystyle {\mathcal {L}}} is a non-empty collection ofF-seminorms on a vector spaceX{\displaystyle X} and for any finite subsetFL{\displaystyle {\mathcal {F}}\subseteq {\mathcal {L}}} and anyr>0,{\displaystyle r>0,} letUF,r:=pF{xX:p(x)<r}.{\displaystyle U_{{\mathcal {F}},r}:=\bigcap _{p\in {\mathcal {F}}}\{x\in X:p(x)<r\}.}

The set{UF,r : r>0,FL,F finite }{\displaystyle \left\{U_{{\mathcal {F}},r}~:~r>0,{\mathcal {F}}\subseteq {\mathcal {L}},{\mathcal {F}}{\text{ finite }}\right\}} forms a filter base onX{\displaystyle X} that also forms a neighborhood basis at the origin for a vector topology onX{\displaystyle X} denoted byτL.{\displaystyle \tau _{\mathcal {L}}.}[12] EachUF,r{\displaystyle U_{{\mathcal {F}},r}} is abalanced andabsorbing subset ofX.{\displaystyle X.}[12] These sets satisfy[12]UF,r/2+UF,r/2UF,r.{\displaystyle U_{{\mathcal {F}},r/2}+U_{{\mathcal {F}},r/2}\subseteq U_{{\mathcal {F}},r}.}

Fréchet combination

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Suppose thatp=(pi)i=1{\displaystyle p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty }} is a family of non-negative subadditive functions on a vector spaceX.{\displaystyle X.}

TheFréchet combination[8] ofp{\displaystyle p_{\bullet }} is defined to be the real-valued mapp(x):=i=1pi(x)2i[1+pi(x)].{\displaystyle p(x):=\sum _{i=1}^{\infty }{\frac {p_{i}(x)}{2^{i}\left[1+p_{i}(x)\right]}}.}

As anF-seminorm

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Assume thatp=(pi)i=1{\displaystyle p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty }} is an increasing sequence of seminorms onX{\displaystyle X} and letp{\displaystyle p} be the Fréchet combination ofp.{\displaystyle p_{\bullet }.} Thenp{\displaystyle p} is anF-seminorm onX{\displaystyle X} that induces the same locally convex topology as the familyp{\displaystyle p_{\bullet }} of seminorms.[13]

Sincep=(pi)i=1{\displaystyle p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty }} is increasing, a basis of open neighborhoods of the origin consists of all sets of the form{xX : pi(x)<r}{\displaystyle \left\{x\in X~:~p_{i}(x)<r\right\}} asi{\displaystyle i} ranges over all positive integers andr>0{\displaystyle r>0} ranges over all positive real numbers.

Thetranslation invariantpseudometric onX{\displaystyle X} induced by thisF-seminormp{\displaystyle p} isd(x,y)=i=112ipi(xy)1+pi(xy).{\displaystyle d(x,y)=\sum _{i=1}^{\infty }{\frac {1}{2^{i}}}{\frac {p_{i}(x-y)}{1+p_{i}(x-y)}}.}

This metric was discovered byFréchet in his 1906 thesis for the spaces of real and complex sequences with pointwise operations.[14]

As a paranorm

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If eachpi{\displaystyle p_{i}} is a paranorm then so isp{\displaystyle p} and moreover,p{\displaystyle p} induces the same topology onX{\displaystyle X} as the familyp{\displaystyle p_{\bullet }} of paranorms.[8] This is also true of the following paranorms onX{\displaystyle X}:

Generalization

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The Fréchet combination can be generalized by use of a bounded remetrization function.

Abounded remetrization function[15] is a continuous non-negative non-decreasing mapR:[0,)[0,){\displaystyle R:[0,\infty )\to [0,\infty )} that has a bounded range, issubadditive (meaning thatR(s+t)R(s)+R(t){\displaystyle R(s+t)\leq R(s)+R(t)} for alls,t0{\displaystyle s,t\geq 0}), and satisfiesR(s)=0{\displaystyle R(s)=0} if and only ifs=0.{\displaystyle s=0.}

Examples of bounded remetrization functions includearctant,{\displaystyle \arctan t,}tanht,{\displaystyle \tanh t,}tmin{t,1},{\displaystyle t\mapsto \min\{t,1\},} andtt1+t.{\displaystyle t\mapsto {\frac {t}{1+t}}.}[15] Ifd{\displaystyle d} is a pseudometric (respectively, metric) onX{\displaystyle X} andR{\displaystyle R} is a bounded remetrization function thenRd{\displaystyle R\circ d} is a bounded pseudometric (respectively, bounded metric) onX{\displaystyle X} that is uniformly equivalent tod.{\displaystyle d.}[15]

Suppose thatp=(pi)i=1{\displaystyle p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty }} is a family of non-negativeF-seminorm on a vector spaceX,{\displaystyle X,}R{\displaystyle R} is a bounded remetrization function, andr=(ri)i=1{\displaystyle r_{\bullet }=\left(r_{i}\right)_{i=1}^{\infty }} is a sequence of positive real numbers whose sum is finite. Thenp(x):=i=1riR(pi(x)){\displaystyle p(x):=\sum _{i=1}^{\infty }r_{i}R\left(p_{i}(x)\right)}defines a boundedF-seminorm that is uniformly equivalent to thep.{\displaystyle p_{\bullet }.}[16] It has the property that for any netx=(xa)aA{\displaystyle x_{\bullet }=\left(x_{a}\right)_{a\in A}} inX,{\displaystyle X,}p(x)0{\displaystyle p\left(x_{\bullet }\right)\to 0} if and only ifpi(x)0{\displaystyle p_{i}\left(x_{\bullet }\right)\to 0} for alli.{\displaystyle i.}[16]p{\displaystyle p} is anF-norm if and only if thep{\displaystyle p_{\bullet }} separate points onX.{\displaystyle X.}[16]

Characterizations

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Of (pseudo)metrics induced by (semi)norms

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A pseudometric (resp. metric)d{\displaystyle d} is induced by a seminorm (resp. norm) on a vector spaceX{\displaystyle X} if and only ifd{\displaystyle d} is translation invariant andabsolutely homogeneous, which means that for all scalarss{\displaystyle s} and allx,yX,{\displaystyle x,y\in X,} in which case the function defined byp(x):=d(x,0){\displaystyle p(x):=d(x,0)} is a seminorm (resp. norm) and the pseudometric (resp. metric) induced byp{\displaystyle p} is equal tod.{\displaystyle d.}

Of pseudometrizable TVS

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If(X,τ){\displaystyle (X,\tau )} is atopological vector space (TVS) (where note in particular thatτ{\displaystyle \tau } is assumed to be a vector topology) then the following are equivalent:[11]

  1. X{\displaystyle X} is pseudometrizable (i.e. the vector topologyτ{\displaystyle \tau } is induced by a pseudometric onX{\displaystyle X}).
  2. X{\displaystyle X} has a countable neighborhood base at the origin.
  3. The topology onX{\displaystyle X} is induced by a translation-invariant pseudometric onX.{\displaystyle X.}
  4. The topology onX{\displaystyle X} is induced by anF-seminorm.
  5. The topology onX{\displaystyle X} is induced by a paranorm.

Of metrizable TVS

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If(X,τ){\displaystyle (X,\tau )} is a TVS then the following are equivalent:

  1. X{\displaystyle X} is metrizable.
  2. X{\displaystyle X} isHausdorff and pseudometrizable.
  3. X{\displaystyle X} is Hausdorff and has a countable neighborhood base at the origin.[11][12]
  4. The topology onX{\displaystyle X} is induced by a translation-invariant metric onX.{\displaystyle X.}[11]
  5. The topology onX{\displaystyle X} is induced by anF-norm.[11][12]
  6. The topology onX{\displaystyle X} is induced by a monotoneF-norm.[12]
  7. The topology onX{\displaystyle X} is induced by a total paranorm.

Birkhoff–Kakutani theoremIf(X,τ){\displaystyle (X,\tau )} is a topological vector space then the following three conditions are equivalent:[17][note 1]

  1. The origin{0}{\displaystyle \{0\}} is closed inX,{\displaystyle X,} and there is acountablebasis of neighborhoods for0{\displaystyle 0} 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.}

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

Of locally convex pseudometrizable TVS

[edit]

If(X,τ){\displaystyle (X,\tau )} is TVS then the following are equivalent:[13]

  1. X{\displaystyle X} islocally convex and pseudometrizable.
  2. X{\displaystyle X} has a countable neighborhood base at the origin consisting of convex sets.
  3. The topology ofX{\displaystyle X} is induced by a countable family of (continuous) seminorms.
  4. The topology ofX{\displaystyle X} is induced by a countable increasing sequence of (continuous) seminorms(pi)i=1{\displaystyle \left(p_{i}\right)_{i=1}^{\infty }} (increasing means that for alli,{\displaystyle i,}pipi+1.{\displaystyle p_{i}\geq p_{i+1}.}
  5. The topology ofX{\displaystyle X} is induced by anF-seminorm of the form:p(x)=n=12narctanpn(x){\displaystyle p(x)=\sum _{n=1}^{\infty }2^{-n}\operatorname {arctan} p_{n}(x)}where(pi)i=1{\displaystyle \left(p_{i}\right)_{i=1}^{\infty }} are (continuous) seminorms onX.{\displaystyle X.}[18]

Quotients

[edit]

LetM{\displaystyle M} be a vector subspace of a topological vector space(X,τ).{\displaystyle (X,\tau ).}

Examples and sufficient conditions

[edit]

IfX{\displaystyle X} is Hausdorff locally convex TVS thenX{\displaystyle X} with thestrong topology,(X,b(X,X)),{\displaystyle \left(X,b\left(X,X^{\prime }\right)\right),} is metrizable if and only if there exists a countable setB{\displaystyle {\mathcal {B}}} of bounded subsets ofX{\displaystyle X} such that every bounded subset ofX{\displaystyle X} is contained in some element ofB.{\displaystyle {\mathcal {B}}.}[22]

Thestrong dual spaceXb{\displaystyle X_{b}^{\prime }} of a metrizable locally convex space (such as aFréchet space[23])X{\displaystyle X} is aDF-space.[24] The strong dual of a DF-space is aFréchet space.[25] The strong dual of areflexive Fréchet space is abornological space.[24] The strong bidual (that is, thestrong dual space of the strong dual space) of a metrizable locally convex space is a Fréchet space.[26]IfX{\displaystyle X} is a metrizable locally convex space then its strong dualXb{\displaystyle X_{b}^{\prime }} has one of the following properties, if and only if it has all of these properties: (1)bornological, (2)infrabarreled, (3)barreled.[26]

Normability

[edit]

A topological vector space isseminormable if and only if it has aconvex bounded neighborhood of the origin. Moreover, a TVS isnormable if and only if it isHausdorff and seminormable.[14] Every metrizable TVS on a finite-dimensional vector space is a normablelocally convexcomplete TVS, beingTVS-isomorphic toEuclidean space. Consequently, any metrizable TVS that isnot normable must be infinite dimensional.

IfM{\displaystyle M} is a metrizablelocally convex TVS that possess acountable fundamental system of bounded sets, thenM{\displaystyle M} is normable.[27]

IfX{\displaystyle X} is a Hausdorfflocally convex space then the following are equivalent:

  1. X{\displaystyle X} isnormable.
  2. X{\displaystyle X} has a(von Neumann) bounded neighborhood of the origin.
  3. thestrong dual spaceXb{\displaystyle X_{b}^{\prime }} ofX{\displaystyle X} is normable.[28]

and if this locally convex spaceX{\displaystyle X} is also metrizable, then the following may be appended to this list:

  1. the strong dual space ofX{\displaystyle X} is metrizable.[28]
  2. the strong dual space ofX{\displaystyle X} is aFréchet–Urysohn locally convex space.[23]

In particular, if a metrizable locally convex spaceX{\displaystyle X} (such as aFréchet space) isnot normable then itsstrong dual spaceXb{\displaystyle X_{b}^{\prime }} is not aFréchet–Urysohn space and consequently, thiscomplete Hausdorff locally convex spaceXb{\displaystyle X_{b}^{\prime }} is also neither metrizable nor normable.

Another consequence of this is that ifX{\displaystyle X} is areflexivelocally convex TVS whose strong dualXb{\displaystyle X_{b}^{\prime }} is metrizable thenXb{\displaystyle X_{b}^{\prime }} is necessarily a reflexive Fréchet space,X{\displaystyle X} is aDF-space, bothX{\displaystyle X} andXb{\displaystyle X_{b}^{\prime }} are necessarilycomplete Hausdorffultrabornologicaldistinguishedwebbed spaces, and moreover,Xb{\displaystyle X_{b}^{\prime }} is normable if and only ifX{\displaystyle X} is normable if and only ifX{\displaystyle X} is Fréchet–Urysohn if and only ifX{\displaystyle X} is metrizable. In particular, such a spaceX{\displaystyle X} is either aBanach space or else it is not even a Fréchet–Urysohn space.

Metrically bounded sets and bounded sets

[edit]

Suppose that(X,d){\displaystyle (X,d)} is a pseudometric space andBX.{\displaystyle B\subseteq X.} The setB{\displaystyle B} ismetrically bounded ord{\displaystyle d}-bounded if there exists a real numberR>0{\displaystyle R>0} such thatd(x,y)R{\displaystyle d(x,y)\leq R} for allx,yB{\displaystyle x,y\in B}; the smallest suchR{\displaystyle R} is then called thediameter ord{\displaystyle d}-diameter ofB.{\displaystyle B.}[14] IfB{\displaystyle B} isbounded in a pseudometrizable TVSX{\displaystyle X} then it is metrically bounded; the converse is in general false but it is true forlocally convex metrizable TVSs.[14]

Properties of pseudometrizable TVS

[edit]

Theorem[29]All infinite-dimensionalseparable complete metrizable TVS arehomeomorphic.

Completeness

[edit]
Main article:Complete topological vector space

Everytopological vector space (and more generally, atopological group) has a canonicaluniform structure, induced by its topology, which allows the notions of completeness anduniform continuity to be applied to it. IfX{\displaystyle X} is a metrizable TVS andd{\displaystyle d} is a metric that definesX{\displaystyle X}'s topology, then its possible thatX{\displaystyle X} is complete as a TVS (i.e. relative to its uniformity) but the metricd{\displaystyle d} isnot acomplete metric (such metrics exist even forX=R{\displaystyle X=\mathbb {R} }). Thus, ifX{\displaystyle X} is a TVS whose topology is induced by a pseudometricd,{\displaystyle d,} then the notion of completeness ofX{\displaystyle X} (as a TVS) and the notion of completeness of the pseudometric space(X,d){\displaystyle (X,d)} are not always equivalent. The next theorem gives a condition for when they are equivalent:

TheoremIfX{\displaystyle X} is a pseudometrizable TVS whose topology is induced by atranslation invariant pseudometricd,{\displaystyle d,} thend{\displaystyle d} is a complete pseudometric onX{\displaystyle X} if and only ifX{\displaystyle X} is complete as a TVS.[36]

Theorem[37][38] (Klee)Letd{\displaystyle d} beany[note 2] metric on a vector spaceX{\displaystyle X} such that the topologyτ{\displaystyle \tau } induced byd{\displaystyle d} onX{\displaystyle X} makes(X,τ){\displaystyle (X,\tau )} into a topological vector space. If(X,d){\displaystyle (X,d)} is a complete metric space then(X,τ){\displaystyle (X,\tau )} is a complete-TVS.

TheoremIfX{\displaystyle X} is a TVS whose topology is induced by a paranormp,{\displaystyle p,} thenX{\displaystyle X} is complete if and only if for every sequence(xi)i=1{\displaystyle \left(x_{i}\right)_{i=1}^{\infty }} inX,{\displaystyle X,} ifi=1p(xi)<{\displaystyle \sum _{i=1}^{\infty }p\left(x_{i}\right)<\infty } theni=1xi{\displaystyle \sum _{i=1}^{\infty }x_{i}} converges inX.{\displaystyle X.}[39]

IfM{\displaystyle M} is a closed vector subspace of a complete pseudometrizable TVSX,{\displaystyle X,} then the quotient spaceX/M{\displaystyle X/M} is complete.[40] IfM{\displaystyle M} is acomplete vector subspace of a metrizable TVSX{\displaystyle X} and if the quotient spaceX/M{\displaystyle X/M} is complete then so isX.{\displaystyle X.}[40] IfX{\displaystyle X} is not complete thenM:=X,{\displaystyle M:=X,} but not complete, vector subspace ofX.{\displaystyle X.}

ABaireseparabletopological group is metrizable if and only if it is cosmic.[23]

Subsets and subsequences

[edit]

Banach-Saks theorem[45]If(xn)n=1{\displaystyle \left(x_{n}\right)_{n=1}^{\infty }} is a sequence in alocally convex metrizable TVS(X,τ){\displaystyle (X,\tau )} that convergesweakly to somexX,{\displaystyle x\in X,} then there exists a sequencey=(yi)i=1{\displaystyle y_{\bullet }=\left(y_{i}\right)_{i=1}^{\infty }} inX{\displaystyle X} such thatyx{\displaystyle y_{\bullet }\to x} in(X,τ){\displaystyle (X,\tau )} and eachyi{\displaystyle y_{i}} is a convex combination of finitely manyxn.{\displaystyle x_{n}.}

Mackey's countability condition[14]Suppose thatX{\displaystyle X} is a locally convex metrizable TVS and that(Bi)i=1{\displaystyle \left(B_{i}\right)_{i=1}^{\infty }} is a countable sequence of bounded subsets ofX.{\displaystyle X.} Then there exists a bounded subsetB{\displaystyle B} ofX{\displaystyle X} and a sequence(ri)i=1{\displaystyle \left(r_{i}\right)_{i=1}^{\infty }} of positive real numbers such thatBiriB{\displaystyle B_{i}\subseteq r_{i}B} for alli.{\displaystyle i.}

Generalized series

As describedin this article's section on generalized series, for anyI{\displaystyle I}-indexed family family(ri)iI{\displaystyle \left(r_{i}\right)_{i\in I}} of vectors from a TVSX,{\displaystyle X,} it is possible to define their sumiIri{\displaystyle \textstyle \sum \limits _{i\in I}r_{i}} as the limit of thenet of finite partial sumsFFiniteSubsets(I)iFri{\displaystyle F\in \operatorname {FiniteSubsets} (I)\mapsto \textstyle \sum \limits _{i\in F}r_{i}} where the domainFiniteSubsets(I){\displaystyle \operatorname {FiniteSubsets} (I)} isdirected by.{\displaystyle \,\subseteq .\,} IfI=N{\displaystyle I=\mathbb {N} } andX=R,{\displaystyle X=\mathbb {R} ,} for instance, then the generalized seriesiNri{\displaystyle \textstyle \sum \limits _{i\in \mathbb {N} }r_{i}} converges if and only ifi=1ri{\displaystyle \textstyle \sum \limits _{i=1}^{\infty }r_{i}}converges unconditionally in the usual sense (which for real numbers,is equivalent toabsolute convergence). If a generalized seriesiIri{\displaystyle \textstyle \sum \limits _{i\in I}r_{i}} converges in a metrizable TVS, then the set{iI:ri0}{\displaystyle \left\{i\in I:r_{i}\neq 0\right\}} is necessarilycountable (that is, either finite orcountably infinite);[proof 1] in other words, all but at most countably manyri{\displaystyle r_{i}} will be zero and so this generalized seriesiIri = ri0iIri{\displaystyle \textstyle \sum \limits _{i\in I}r_{i}~=~\textstyle \sum \limits _{\stackrel {i\in I}{r_{i}\neq 0}}r_{i}} is actually a sum of at most countably many non-zero terms.

Linear maps

[edit]

IfX{\displaystyle X} is a pseudometrizable TVS andA{\displaystyle A} maps bounded subsets ofX{\displaystyle X} to bounded subsets ofY,{\displaystyle Y,} thenA{\displaystyle A} is continuous.[14] Discontinuous linear functionals exist on any infinite-dimensional pseudometrizable TVS.[46] Thus, a pseudometrizable TVS is finite-dimensional if and only if its continuous dual space is equal to itsalgebraic dual space.[46]

IfF:XY{\displaystyle F:X\to Y} is a linear map between TVSs andX{\displaystyle X} is metrizable then the following are equivalent:

  1. F{\displaystyle F} is continuous;
  2. F{\displaystyle F} is a (locally) bounded map (that is,F{\displaystyle F} maps(von Neumann) bounded subsets ofX{\displaystyle X} to bounded subsets ofY{\displaystyle Y});[12]
  3. F{\displaystyle F} issequentially continuous;[12]
  4. the image underF{\displaystyle F} of every null sequence inX{\displaystyle X} is a bounded set[12] where by definition, anull sequence is a sequence that converges to the origin.
  5. F{\displaystyle F} maps null sequences to null sequences;

Open and almost open maps

Theorem: IfX{\displaystyle X} is a complete pseudometrizable TVS,Y{\displaystyle Y} is a Hausdorff TVS, andT:XY{\displaystyle T:X\to Y} is a closed andalmost open linear surjection, thenT{\displaystyle T} is an open map.[47]
Theorem: IfT:XY{\displaystyle T:X\to Y} is a surjective linear operator from alocally convex spaceX{\displaystyle X} onto abarrelled spaceY{\displaystyle Y} (e.g. every complete pseudometrizable space is barrelled) thenT{\displaystyle T} isalmost open.[47]
Theorem: IfT:XY{\displaystyle T:X\to Y} is a surjective linear operator from a TVSX{\displaystyle X} onto aBaire spaceY{\displaystyle Y} thenT{\displaystyle T} is almost open.[47]
Theorem: SupposeT:XY{\displaystyle T:X\to Y} is a continuous linear operator from a complete pseudometrizable TVSX{\displaystyle X} into a Hausdorff TVSY.{\displaystyle Y.} If the image ofT{\displaystyle T} is non-meager inY{\displaystyle Y} thenT:XY{\displaystyle T:X\to Y} is a surjective open map andY{\displaystyle Y} is acomplete metrizable space.[47]

Hahn-Banach extension property

[edit]
Main article:Hahn-Banach theorem

A vector subspaceM{\displaystyle M} of a TVSX{\displaystyle X} hasthe extension property if any continuous linear functional onM{\displaystyle M} can be extended to a continuous linear functional onX.{\displaystyle X.}[22] Say that a TVSX{\displaystyle X} has theHahn-Banach extension property (HBEP) if every vector subspace ofX{\displaystyle X} has the extension property.[22]

TheHahn-Banach theorem guarantees that every Hausdorff locally convex space has the HBEP. For complete metrizable TVSs there is a converse:

Theorem (Kalton)Every complete metrizable TVS with the Hahn-Banach extension property is locally convex.[22]

If a vector spaceX{\displaystyle X} has uncountable dimension and if we endow it with thefinest vector topology then this is a TVS with the HBEP that is neither locally convex or metrizable.[22]

See also

[edit]

Notes

[edit]
  1. ^In fact, this is true for topological group, for the proof doesn't use the scalar multiplications.
  2. ^Not assumed to be translation-invariant.

Proofs

  1. ^Suppose the netiIri =def limAFiniteSubsets(I) iAri=lim{iAri:AI,A finite }{\textstyle \textstyle \sum \limits _{i\in I}r_{i}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\textstyle \lim \limits _{A\in \operatorname {FiniteSubsets} (I)}}\ \textstyle \sum \limits _{i\in A}r_{i}=\lim \left\{\textstyle \sum \limits _{i\in A}r_{i}\,:A\subseteq I,A{\text{ finite }}\right\}} converges to some point in a metrizable TVSX,{\displaystyle X,} where recall that this net's domain is thedirected set(FiniteSubsets(I),).{\displaystyle (\operatorname {FiniteSubsets} (I),\subseteq ).} Like every convergent net, this convergent net of partial sumsAiAri{\displaystyle A\mapsto \textstyle \sum \limits _{i\in A}r_{i}} is aCauchy net, which for this particular net means (by definition) that for every neighborhoodW{\displaystyle W} of the origin inX,{\displaystyle X,} there exists a finite subsetA0{\displaystyle A_{0}} ofI{\displaystyle I} such thatiBriiCriW{\textstyle \textstyle \sum \limits _{i\in B}r_{i}-\textstyle \sum \limits _{i\in C}r_{i}\in W} for all finite supersetsB,CA0;{\displaystyle B,C\supseteq A_{0};}this implies thatriW{\displaystyle r_{i}\in W} for everyiIA0{\displaystyle i\in I\setminus A_{0}} (by takingB:=A0{i}{\displaystyle B:=A_{0}\cup \{i\}} andC:=A0{\displaystyle C:=A_{0}}). SinceX{\displaystyle X} is metrizable, it has a countable neighborhood basisU1,U2,{\displaystyle U_{1},U_{2},\ldots } at the origin, whose intersection is necessarilyU1U2={0}{\displaystyle U_{1}\cap U_{2}\cap \cdots =\{0\}} (sinceX{\displaystyle X} is a Hausdorff TVS).For every positive integernN,{\displaystyle n\in \mathbb {N} ,} pick a finite subsetAnI{\displaystyle A_{n}\subseteq I} such thatriUn{\displaystyle r_{i}\in U_{n}} for everyiIAn.{\displaystyle i\in I\setminus A_{n}.} Ifi{\displaystyle i} belongs to(IA1)(IA2)=I(A1A2){\displaystyle (I\setminus A_{1})\cap (I\setminus A_{2})\cap \cdots =I\setminus \left(A_{1}\cup A_{2}\cup \cdots \right)} thenri{\displaystyle r_{i}} belongs toU1U2={0}.{\displaystyle U_{1}\cap U_{2}\cap \cdots =\{0\}.} Thusri=0{\displaystyle r_{i}=0} for every indexiI{\displaystyle i\in I} that does not belong to the countable setA1A2.{\displaystyle A_{1}\cup A_{2}\cup \cdots .}{\displaystyle \blacksquare }

References

[edit]
  1. ^Narici & Beckenstein 2011, pp. 1–18.
  2. ^abcNarici & Beckenstein 2011, pp. 37–40.
  3. ^abSwartz 1992, p. 15.
  4. ^Wilansky 2013, p. 17.
  5. ^abWilansky 2013, pp. 40–47.
  6. ^Wilansky 2013, p. 15.
  7. ^abSchechter 1996, pp. 689–691.
  8. ^abcdefghijklmnoWilansky 2013, pp. 15–18.
  9. ^abcdSchechter 1996, p. 692.
  10. ^abSchechter 1996, p. 691.
  11. ^abcdefghijklNarici & Beckenstein 2011, pp. 91–95.
  12. ^abcdefghijklmnopqrstJarchow 1981, pp. 38–42.
  13. ^abNarici & Beckenstein 2011, p. 123.
  14. ^abcdefghNarici & Beckenstein 2011, pp. 156–175.
  15. ^abcSchechter 1996, p. 487.
  16. ^abcSchechter 1996, pp. 692–693.
  17. ^Köthe 1983, section 15.11
  18. ^Schechter 1996, p. 706.
  19. ^Narici & Beckenstein 2011, pp. 115–154.
  20. ^Wilansky 2013, pp. 15–16.
  21. ^Schaefer & Wolff 1999, pp. 91–92.
  22. ^abcdeNarici & Beckenstein 2011, pp. 225–273.
  23. ^abcdGabriyelyan, S.S."On topological spaces and topological groups with certain local countable networks (2014)
  24. ^abSchaefer & Wolff 1999, p. 154.
  25. ^Schaefer & Wolff 1999, p. 196.
  26. ^abcSchaefer & Wolff 1999, p. 153.
  27. ^Schaefer & Wolff 1999, pp. 68–72.
  28. ^abTrèves 2006, p. 201.
  29. ^Wilansky 2013, p. 57.
  30. ^Jarchow 1981, p. 222.
  31. ^abcdNarici & Beckenstein 2011, pp. 371–423.
  32. ^Narici & Beckenstein 2011, pp. 459–483.
  33. ^Köthe 1969, p. 168.
  34. ^Wilansky 2013, p. 59.
  35. ^abSchaefer & Wolff 1999, pp. 12–35.
  36. ^Narici & Beckenstein 2011, pp. 47–50.
  37. ^Schaefer & Wolff 1999, p. 35.
  38. ^Klee, V. L. (1952)."Invariant metrics in groups (solution of a problem of Banach)"(PDF).Proc. Amer. Math. Soc.3 (3):484–487.doi:10.1090/s0002-9939-1952-0047250-4.
  39. ^Wilansky 2013, pp. 56–57.
  40. ^abNarici & Beckenstein 2011, pp. 47–66.
  41. ^Schaefer & Wolff 1999, pp. 190–202.
  42. ^Narici & Beckenstein 2011, pp. 172–173.
  43. ^abRudin 1991, p. 22.
  44. ^Narici & Beckenstein 2011, pp. 441–457.
  45. ^Rudin 1991, p. 67.
  46. ^abNarici & Beckenstein 2011, p. 125.
  47. ^abcdNarici & Beckenstein 2011, pp. 466–468.

Bibliography

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