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Topological vector space
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===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}}
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