Editing Topological vector space (section)

===Defining topologies using neighborhoods of the origin===

Since every vector topology is translation invariant (which means that 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]]), to define a vector topology it suffices to define a [[neighborhood basis]] (or subbasis) for it at the origin.

{{Math theorem|name=Theorem{{sfn|Narici|Beckenstein|2011|pp=67-113}}|note=Neighborhood filter of the origin|math_statement=
Suppose that <math>X</math> is a real or complex vector space. If <math>\mathcal{B}</math> is a [[Empty set|non-empty]] additive collection of [[Balanced set|balanced]] and [[Absorbing set|absorbing]] subsets of <math>X</math> then <math>\mathcal{B}</math> is a [[neighborhood base]] at <math>0</math> for a vector topology on <math>X.</math> That is, the assumptions are that <math>\mathcal{B}</math> is a [[filter base]] that satisfies the following conditions:
# Every <math>B \in \mathcal{B}</math> is [[Balanced set|balanced]] and [[Absorbing set|absorbing]],
# <math>\mathcal{B}</math> is additive: For every <math>B \in \mathcal{B}</math> there exists a <math>U \in \mathcal{B}</math> such that <math>U + U \subseteq B,</math>

If <math>\mathcal{B}</math> satisfies the above two conditions but is {{em|not}} a filter base then it will form a neighborhood {{em|sub}}basis at <math>0</math> (rather than a neighborhood basis) for a vector topology on <math>X.</math>
}}

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.{{sfn|Wilansky|2013|pp=40-47}}

{{anchor|String|Strings}}