Editing Topological vector space (section)

===Non-Hausdorff spaces and the closure of the origin===

A topological vector space <math>X</math> is Hausdorff if and only if <math>\{0\}</math> is a closed subset of <math>X,</math> or equivalently, if and only if <math>\{0\} = \operatorname{cl}_X \{0\}.</math> Because <math>\{0\}</math> is a vector subspace of <math>X,</math> the same is true of its closure <math>\operatorname{cl}_X \{0\},</math> which is referred to as {{em|the closure of the origin}} in <math>X.</math> This vector space satisfies <math display=block>\operatorname{cl}_X \{0\} = \bigcap_{N \in \mathcal{N}(0)} N</math> so that in particular, every neighborhood of the origin in <math>X</math> contains the vector space <math>\operatorname{cl}_X \{0\}</math> as a subset. 
The [[subspace topology]] on <math>\operatorname{cl}_X \{0\}</math> is always the [[trivial topology]], which in particular implies that the topological vector space <math>\operatorname{cl}_X \{0\}</math> a [[compact space]] (even if its dimension is non-zero or even infinite) and consequently also a [[Bounded set (topological vector space)|bounded subset]] of <math>X.</math> In fact, a vector subspace of a TVS is bounded if and only if it is contained in the closure of <math>\{0\}.</math>{{sfn|Narici|Beckenstein|2011|pp=155-176}} 
Every subset of <math>\operatorname{cl}_X \{0\}</math> also carries the trivial topology and so is itself a compact, and thus also complete, [[Topological subspace|subspace]] (see footnote for a proof).<ref group="proof">Since <math>\operatorname{cl}_X \{0\}</math> has the trivial topology, so does each of its subsets, which makes them all compact. It is known that a subset of any uniform space is compact if and only if it is complete and totally bounded.</ref> In particular, if <math>X</math> is not Hausdorff then there exist subsets that are both {{em|compact and complete}} but {{em|not closed}} in <math>X</math>;{{sfn|Narici|Beckenstein|2011|pp=47-66}} for instance, this will be true of any non-empty proper subset of <math>\operatorname{cl}_X \{0\}.</math>

If <math>S \subseteq X</math> is compact, then <math>\operatorname{cl}_X S = S + \operatorname{cl}_X \{0\}</math> and this set is compact. Thus the closure of a compact subset of a TVS is compact (said differently, all compact sets are [[relatively compact]]),{{sfn|Narici|Beckenstein|2011|p=156}} which is not guaranteed for arbitrary non-Hausdorff [[topological space]]s.<ref group="note">In general topology, the closure of a compact subset of a non-Hausdorff space may fail to be compact (for example, the [[particular point topology]] on an infinite set). This result shows that this does not happen in non-Hausdorff TVSs. <math>S + \operatorname{cl}_X \{0\}</math> is compact because it is the image of the compact set <math>S \times \operatorname{cl}_X \{0\}</math> under the continuous addition map <math>\cdot\, + \,\cdot\; : X \times X \to X.</math> Recall also that the sum of a compact set (that is, <math>S</math>) and a closed set is closed so <math>S + \operatorname{cl}_X \{0\}</math> is closed in <math>X.</math></ref>

For every subset <math>S \subseteq X,</math> <math display=block>S + \operatorname{cl}_X \{0\} \subseteq \operatorname{cl}_X S</math> and consequently, if <math>S \subseteq X</math> is open or closed in <math>X</math> then <math>S + \operatorname{cl}_X \{0\} = S</math><ref group="proof" name="ProofSumOfSetAndClosureOf0">If <math>s \in S</math> then <math>s + \operatorname{cl}_X \{0\} = \operatorname{cl}_X (s + \{0\}) = \operatorname{cl}_X \{s\} \subseteq \operatorname{cl}_X S.</math> Because <math>S \subseteq S + \operatorname{cl}_X \{0\} \subseteq \operatorname{cl}_X S,</math> if <math>S</math> is closed then equality holds. Using the fact that <math>\operatorname{cl}_X \{0\}</math> is a vector space, it is readily verified that the complement in <math>X</math> of any set <math>S</math> satisfying the equality <math>S + \operatorname{cl}_X \{0\} = S</math> must also satisfy this equality (when <math>X \setminus S</math> is substituted for <math>S</math>).</ref> (so that this {{em|arbitrary}} open {{em|or}} closed subsets <math>S</math> can be described as a [[Tube lemma|"tube"]] whose vertical side is the vector space <math>\operatorname{cl}_X \{0\}</math>). 
For any subset <math>S \subseteq X</math> of this TVS <math>X,</math> the following are equivalent:

* <math>S</math> is [[Totally bounded space|totally bounded]].
* <math>S + \operatorname{cl}_X \{0\}</math> is totally bounded.{{sfn|Schaefer|Wolff|1999|pp=12-35}}
* <math>\operatorname{cl}_X S</math> is totally bounded.{{sfn|Schaefer|Wolff|1999|p=25}}{{sfn|Jarchow|1981|pp=56-73}}
* The image if <math>S</math> under the canonical quotient map <math>X \to X / \operatorname{cl}_X (\{0\})</math> is totally bounded.{{sfn|Schaefer|Wolff|1999|pp=12-35}}

If <math>M</math> is a vector subspace of a TVS <math>X</math> then <math>X / M</math> is Hausdorff if and only if <math>M</math> is closed in <math>X.</math> 
Moreover, the [[quotient map]] <math>q : X \to X / \operatorname{cl}_X \{0\}</math> is always a [[Open and closed maps|closed map]] onto the (necessarily) Hausdorff TVS.{{sfn|Narici|Beckenstein|2011|pp=107-112}}

Every vector subspace of <math>X</math> that is an algebraic complement of <math>\operatorname{cl}_X \{0\}</math> (that is, a vector subspace <math>H</math> that satisfies <math>\{0\} = H \cap \operatorname{cl}_X \{0\}</math> and <math>X = H + \operatorname{cl}_X \{0\}</math>) is a [[Complemented subspace|topological complement]] of <math>\operatorname{cl}_X \{0\}.</math> 
Consequently, if <math>H</math> is an algebraic complement of <math>\operatorname{cl}_X \{0\}</math> in <math>X</math> then the addition map <math>H \times \operatorname{cl}_X \{0\} \to X,</math> defined by <math>(h, n) \mapsto h + n</math> is a TVS-isomorphism, where <math>H</math> is necessarily Hausdorff and <math>\operatorname{cl}_X \{0\}</math> has the [[indiscrete topology]].{{sfn|Wilansky|2013|p=63}} Moreover, if <math>C</math> is a Hausdorff [[Complete topological vector space|completion]] of <math>H</math> then <math>C \times \operatorname{cl}_X \{0\}</math> is a completion of <math>X \cong H \times \operatorname{cl}_X \{0\}.</math>{{sfn|Schaefer|Wolff|1999|pp=12-35}}