Editing Topological vector space (section)

==Topological structure==

A vector space is an [[abelian group]] with respect to the operation of addition, and in a topological vector space the inverse operation is always continuous (since it is the same as multiplication by <math>-1</math>). Hence, every topological vector space is an abelian [[topological group]]. Every TVS is [[completely regular]] but a TVS need not be [[Normal space|normal]].{{sfn|Wilansky|2013|p=53}}

Let <math>X</math> be a topological vector space. Given a [[Subspace topology|subspace]] <math>M \subseteq X,</math> the quotient space <math>X / M</math> with the usual [[quotient space (topology)|quotient topology]] is a Hausdorff topological vector space if and only if <math>M</math> is closed.<ref group=note>In particular, <math>X</math> is Hausdorff if and only if the set <math>\{0\}</math> is closed (that is, <math>X</math> is a [[T1 space|T<sub>1</sub> space]]).</ref> This permits the following construction: given a topological vector space <math>X</math> (that is probably not Hausdorff), form the quotient space <math>X / M</math> where <math>M</math> is the closure of <math>\{0\}.</math> <math>X / M</math> is then a Hausdorff topological vector space that can be studied instead of <math>X.</math>

===Invariance of vector topologies===

One of the most used properties of vector topologies is that every vector topology is {{em|{{visible anchor|translation invariant}}}}:
:for all <math>x_0 \in X,</math> the map <math>X \to X</math> defined by <math>x \mapsto x_0 + x</math> is a [[homeomorphism]], but if <math>x_0 \neq 0</math> then it is not linear and so not a TVS-isomorphism.
Scalar multiplication by a non-zero scalar is a TVS-isomorphism. This means that if <math>s \neq 0</math> then the linear map <math>X \to X</math> defined by <math>x \mapsto s x</math> is a homeomorphism. Using <math>s = -1</math> produces the negation map <math>X \to X</math> defined by <math>x \mapsto - x,</math> which is consequently a linear homeomorphism and thus a TVS-isomorphism.

If <math>x \in X</math> and any subset <math>S \subseteq X,</math> then <math>\operatorname{cl}_X (x + S) = x + \operatorname{cl}_X S</math>{{sfn|Narici|Beckenstein|2011|pp=67-113}} and moreover, if <math>0 \in S</math> then <math>x + S</math> is a [[Neighborhood (topology)|neighborhood]] (resp. open neighborhood, closed neighborhood) of <math>x</math> in <math>X</math> if and only if the same is true of <math>S</math> at the origin.

===Local notions===

A subset <math>E</math> of a vector space <math>X</math> is said to be
* '''[[Absorbing set|absorbing]]''' (in <math>X</math>): if for every <math>x \in X,</math> there exists a real <math>r > 0</math> such that <math>c x \in E</math> for any scalar <math>c</math> satisfying <math>|c| \leq r.</math>{{sfn|Rudin|1991|p=6 §1.4}}
* '''[[Balanced set|balanced]]''' or '''circled''': if <math>t E \subseteq E</math> for every scalar <math>|t| \leq 1.</math>{{sfn|Rudin|1991|p=6 §1.4}}
* '''[[Convex set|convex]]''': if <math>t E + (1 - t) E \subseteq E</math> for every real <math>0 \leq t \leq 1.</math>{{sfn|Rudin|1991|p=6 §1.4}}
* a '''[[Absolutely convex set|disk]]''' or '''[[Absolutely convex set|absolutely convex]]''': if <math>E</math> is convex and balanced.
* '''[[Symmetric set|symmetric]]''': if <math>- E \subseteq E,</math> or equivalently, if <math>- E = E.</math>

Every neighborhood of the origin is an [[absorbing set]] and contains an open [[Balanced set|balanced]] neighborhood of <math>0</math>{{sfn|Narici|Beckenstein|2011|pp=67-113}} so every topological vector space has a local base of absorbing and [[balanced set]]s. The origin even has a neighborhood basis consisting of closed balanced neighborhoods of <math>0;</math> if the space is [[locally convex]] then it also has a neighborhood basis consisting of closed convex balanced neighborhoods of the origin.

'''Bounded subsets'''

A subset <math>E</math> of a topological vector space <math>X</math> is '''[[Bounded set (topological vector space)|bounded]]'''{{sfn|Rudin|1991|p=8}} if for every neighborhood <math>V</math> of the origin there exists <math>t</math> such that <math>E \subseteq t V</math>.

The definition of boundedness can be weakened a bit; <math>E</math> 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.{{sfn|Narici|Beckenstein|2011|pp=155-176}} Also, <math>E</math> is bounded if and only if for every balanced neighborhood <math>V</math> of the origin, there exists <math>t</math> such that <math>E \subseteq t V.</math> Moreover, when <math>X</math> is locally convex, the boundedness can be characterized by [[seminorm]]s: the subset <math>E</math> is bounded if and only if every continuous seminorm <math>p</math> is bounded on <math>E.</math>{{sfn|Rudin|1991|p=27-28 Theorem 1.37}}

Every [[totally bounded]] set is bounded.{{sfn|Narici|Beckenstein|2011|pp=155-176}} If <math>M</math> is a vector subspace of a TVS <math>X,</math> then a subset of <math>M</math> is bounded in <math>M</math> if and only if it is bounded in <math>X.</math>{{sfn|Narici|Beckenstein|2011|pp=155-176}}

===Metrizability===

{{Math theorem|name=[[Birkhoff–Kakutani theorem]]|math_statement=
If <math>(X, \tau)</math> is a topological vector space then the following four conditions are equivalent:{{sfn|Köthe|1983|loc=section 15.11}}<ref group=note>In fact, this is true for topological group, since the proof does not use the scalar multiplications.</ref>
# The origin <math>\{0\}</math> is closed in <math>X</math> and there is a [[countable]] [[neighborhood basis|basis of neighborhoods]] at the origin in <math>X.</math>
# <math>(X, \tau)</math> is [[Metrizable space|metrizable]] (as a topological space).
# There is a [[translation-invariant metric]] on <math>X</math> that induces on <math>X</math> the topology <math>\tau,</math> which is the given topology on <math>X.</math>
# <math>(X, \tau)</math> is a [[metrizable topological vector space]].<ref group=note>Also called a '''metric linear space''', which means that it is a real or complex vector space together with a translation-invariant metric for which addition and scalar multiplication are continuous.</ref>

By the Birkhoff–Kakutani theorem, it follows that there is an [[Equivalence of metrics|equivalent metric]] that is translation-invariant.
}}

A TVS is [[Metrizable TVS|pseudometrizable]] if and only if it has a countable neighborhood basis at the origin, or equivalent, if and only if its topology is generated by an [[Metrizable TVS|''F''-seminorm]]. A TVS is metrizable if and only if it is Hausdorff and pseudometrizable.

More strongly: a topological vector space is said to be '''[[normable]]''' if its topology can be induced by a norm. A topological vector space is normable if and only if it is Hausdorff and has a convex bounded neighborhood of the origin.<ref name="springer">{{SpringerEOM|title=Topological vector space|access-date=26 February 2021}}</ref>

Let <math>\mathbb{K}</math> be a non-[[Discrete space|discrete]] [[locally compact]] topological field, for example the real or complex numbers. A [[Hausdorff space|Hausdorff]] topological vector space over <math>\mathbb{K}</math> is locally compact if and only if it is [[finite-dimensional]], that is, isomorphic to <math>\mathbb{K}^n</math> for some natural number <math>n.</math>{{sfn|Rudin|1991|p=17 Theorem 1.22}}

===Completeness and uniform structure===
{{Main|Complete topological vector space}}

The '''[[Complete topological vector space|canonical uniformity]]'''{{sfn|Schaefer|Wolff|1999|pp=12-19}} on a TVS <math>(X, \tau)</math> is the unique translation-invariant [[Uniform space|uniformity]] that induces the topology <math>\tau</math> on <math>X.</math>

Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs into [[uniform space]]s. This allows one to talk{{clarify|date=September 2020}} about related notions such as [[Complete topological vector space|completeness]], [[uniform convergence]], Cauchy nets, and [[uniform continuity]], etc., which are always assumed to be with respect to this uniformity (unless indicated other). This implies that every Hausdorff topological vector space is [[Tychonoff space|Tychonoff]].{{sfn|Schaefer|Wolff|1999|p=16}} A subset of a TVS is [[Compact space|compact]] if and only if it is complete and [[totally bounded]] (for Hausdorff TVSs, a set being totally bounded is equivalent to it being [[Totally bounded space#In topological groups|precompact]]). But if the TVS is not Hausdorff then there exist compact subsets that are not closed. However, the closure of a compact subset of a non-Hausdorff TVS is again compact (so compact subsets are [[relatively compact]]).

With respect to this uniformity, a [[Net (mathematics)|net]] (or [[Sequence (mathematics)|sequence]]) <math>x_{\bull} = \left(x_i\right)_{i \in I}</math> is '''Cauchy''' if and only if for every neighborhood <math>V</math> of <math>0,</math> there exists some index <math>n</math> such that <math>x_i - x_j \in V</math> whenever <math>i \geq n</math> and <math>j \geq n.</math>

Every [[Cauchy sequence]] is bounded, although Cauchy nets and Cauchy filters may not be bounded. A topological vector space where every Cauchy sequence converges is called '''[[sequentially complete]]'''; in general, it may not be complete (in the sense that all Cauchy filters converge).

The vector space operation of addition is uniformly continuous and an [[Open and closed map|open map]]. Scalar multiplication is [[Cauchy continuous]] but in general, it is almost never uniformly continuous. Because of this, every topological vector space can be completed and is thus a [[Dense set|dense]] [[linear subspace]] of a [[complete topological vector space]].

* Every TVS has a [[Complete topological vector space|completion]] and every Hausdorff TVS has a Hausdorff completion.{{sfn|Narici|Beckenstein|2011|pp=67-113}} Every TVS (even those that are Hausdorff and/or complete) has infinitely many non-isomorphic non-Hausdorff completions.
* A compact subset of a TVS (not necessarily Hausdorff) is complete.{{sfn|Narici|Beckenstein|2011|pp=115-154}} A complete subset of a Hausdorff TVS is closed.{{sfn|Narici|Beckenstein|2011|pp=115-154}}
* If <math>C</math> is a complete subset of a TVS then any subset of <math>C</math> that is closed in <math>C</math> is complete.{{sfn|Narici|Beckenstein|2011|pp=115-154}}
* A Cauchy sequence in a Hausdorff TVS <math>X</math> is not necessarily [[relatively compact]] (that is, its closure in <math>X</math> is not necessarily compact).
* If a Cauchy filter in a TVS has an [[Filters in topology|accumulation point]] <math>x</math> then it converges to <math>x.</math>
* If a series <math display=inline>\sum_{i=1}^{\infty} x_i</math> converges<ref group="note">A series <math display=inline>\sum_{i=1}^{\infty} x_i</math> is said to '''converge''' in a TVS <math>X</math> if the sequence of partial sums converges.</ref> in a TVS <math>X</math> then <math>x_{\bull} \to 0</math> in <math>X.</math>{{sfn|Swartz|1992|pp=27-29}}