Editing Banach space (section)

====Complete norms vs complete topological vector spaces====

There is another notion of completeness besides metric completeness and that is the notion of a [[complete topological vector space]] (TVS) or TVS-completeness, which uses the theory of [[uniform space]]s. 
Specifically, the notion of TVS-completeness uses a unique translation-invariant [[Uniformity (topology)|uniformity]], called the [[Complete topological vector space#Canonical uniformity|canonical uniformity]], that depends {{em|only}} on vector subtraction and the topology <math>\tau</math> that the vector space is endowed with, and so in particular, this notion of TVS completeness is independent of whatever norm induced the topology <math>\tau</math> (and even applies to TVSs that are {{em|not}} even metrizable). 
Every Banach space is a complete TVS. Moreover, a normed space is a Banach space (that is, its norm-induced metric is complete) if and only if it is complete as a topological vector space. 
If <math>(X, \tau)</math> is a [[metrizable topological vector space]] (such as any norm induced topology, for example), then <math>(X, \tau)</math> is a complete TVS if and only if it is a {{em|sequentially}} complete TVS, meaning that it is enough to check that every Cauchy {{em|sequence}} in <math>(X, \tau)</math> converges in <math>(X, \tau)</math> to some point of <math>X</math> (that is, there is no need to consider the more general notion of arbitrary Cauchy [[Net (mathematics)|nets]]).

If <math>(X, \tau)</math> is a topological vector space whose topology is induced by {{em|some}} (possibly unknown) norm (such spaces are called {{em|[[Normable space|normable]]}}), then <math>(X, \tau)</math> is a complete topological vector space if and only if <math>X</math> may be assigned a [[Norm (mathematics)|norm]] <math>\|{\cdot}\|</math> that induces on <math>X</math> the topology <math>\tau</math> and also makes <math>(X, \|{\cdot}\|)</math> into a Banach space. 
A [[Hausdorff space|Hausdorff]] [[locally convex topological vector space]] <math>X</math> is [[Normable space|normable]] if and only if its [[strong dual space]] <math>X'_b</math> is normable,{{sfn|Trèves|2006|p=201}} in which case <math>X'_b</math> is a Banach space (<math>X'_b</math> denotes the [[strong dual space]] of <math>X,</math> whose topology is a generalization of the [[dual norm]]-induced topology on the [[continuous dual space]] <math>X'</math>; see this footnote<ref group=note><math>X'</math> denotes the [[continuous dual space]] of <math>X.</math> When <math>X'</math> is endowed with the [[Strong topology (polar topology)|strong dual space topology]], also called the [[topology of uniform convergence]] on [[Bounded set (functional analysis)|bounded subsets]] of <math>X,</math> then this is indicated by writing <math>X'_b</math> (sometimes, the subscript <math>\beta</math> is used instead of <math>b</math>). When <math>X</math> is a normed space with norm <math>\|{\cdot}\|</math> then this topology is equal to the topology on <math>X'</math> induced by the [[dual norm]]. In this way, the [[Strong topology (polar topology)|strong topology]] is a generalization of the usual dual norm-induced topology on <math>X'.</math></ref> for more details). 
If <math>X</math> is a [[Metrizable topological vector space|metrizable]] locally convex TVS, then <math>X</math> is normable if and only if <math>X'_b</math> is a [[Fréchet–Urysohn space]].<ref name="Gabriyelyan 2014">Gabriyelyan, S.S. [https://arxiv.org/pdf/1412.1497.pdf "On topological spaces and topological groups with certain local countable networks] (2014)</ref> 
This shows that in the category of [[Locally convex topological vector space|locally convex TVSs]], Banach spaces are exactly those complete spaces that are both [[Metrizable topological vector space|metrizable]] and have metrizable [[strong dual space]]s.