Editing Topological vector space (section)

===Closed and compact sets===

'''Compact and totally bounded sets'''

A subset of a TVS is compact if and only if it is complete and [[Totally bounded space|totally bounded]].{{sfn|Narici|Beckenstein|2011|pp=47-66}} Thus, in a [[complete topological vector space]], a closed and totally bounded subset is compact.{{sfn|Narici|Beckenstein|2011|pp=47-66}} 
A subset <math>S</math> of a TVS <math>X</math> is [[Totally bounded space|totally bounded]] if and only if <math>\operatorname{cl}_X S</math> is totally bounded,{{sfn|Schaefer|Wolff|1999|p=25}}{{sfn|Jarchow|1981|pp=56-73}} if and only if its image under the canonical quotient map <math display=block>X \to X / \operatorname{cl}_X (\{0\})</math> is totally bounded.{{sfn|Schaefer|Wolff|1999|pp=12-35}}

Every relatively compact set is totally bounded{{sfn|Narici|Beckenstein|2011|pp=47-66}} and the closure of a totally bounded set is totally bounded.{{sfn|Narici|Beckenstein|2011|pp=47-66}} 
The image of a totally bounded set under a uniformly continuous map (such as a continuous linear map for instance) is totally bounded.{{sfn|Narici|Beckenstein|2011|pp=47-66}} 
If <math>S</math> is a subset of a TVS <math>X</math> such that every sequence in <math>S</math> has a cluster point in <math>S</math> then <math>S</math> is totally bounded.{{sfn|Schaefer|Wolff|1999|pp=12-35}}

If <math>K</math> is a compact subset of a TVS <math>X</math> and <math>U</math> is an open subset of <math>X</math> containing <math>K,</math> then there exists a neighborhood <math>N</math> of 0 such that <math>K + N \subseteq U.</math>{{sfn|Narici|Beckenstein|2011|pp=19-45}}

'''Closure and closed set'''

The closure of any convex (respectively, any balanced, any absorbing) subset of any TVS has this same property. In particular, the closure of any convex, balanced, and absorbing subset is a [[Barrelled space#barrel|barrel]].

The closure of a vector subspace of a TVS is a vector subspace. Every finite dimensional vector subspace of a Hausdorff TVS is closed. 
The sum of a closed vector subspace and a finite-dimensional vector subspace is closed.{{sfn|Narici|Beckenstein|2011|pp=67-113}} 
If <math>M</math> is a vector subspace of <math>X</math> and <math>N</math> is a closed neighborhood of the origin in <math>X</math> such that <math>U \cap N</math> is closed in <math>X</math> then <math>M</math> is closed in <math>X.</math>{{sfn|Narici|Beckenstein|2011|pp=19-45}} 
The sum of a compact set and a closed set is closed. However, the sum of two closed subsets may fail to be closed{{sfn|Narici|Beckenstein|2011|pp=67-113}} (see this footnote<ref group=note>In <math>\R^2,</math> the sum of the <math>y</math>-axis and the graph of <math>y = \frac{1}{x},</math> which is the complement of the <math>y</math>-axis, is open in <math>\R^2.</math> In <math>\R,</math> the [[Minkowski sum]] <math>\Z + \sqrt{2}\Z</math> is a countable dense subset of <math>\R</math> so not closed in <math>\R.</math></ref> for examples). 

If <math>S \subseteq X</math> and <math>a</math> is a scalar then <math display=block>a \operatorname{cl}_X S \subseteq \operatorname{cl}_X (a S),</math> where if <math>X</math> is Hausdorff, <math>a \neq 0, \text{ or } S = \varnothing</math> then equality holds: <math>\operatorname{cl}_X (a S) = a \operatorname{cl}_X S.</math> In particular, every non-zero scalar multiple of a closed set is closed. 
If <math>S \subseteq X</math> and if <math>A</math> is a set of scalars such that neither <math>\operatorname{cl} S \text{ nor } \operatorname{cl} A</math> contain zero then{{sfn|Wilansky|2013|pp=43-44}} <math>\left(\operatorname{cl} A\right) \left(\operatorname{cl}_X S\right) = \operatorname{cl}_X (A S).</math>

If <math>S \subseteq X \text{ and } S + S \subseteq 2 \operatorname{cl}_X S</math> then <math>\operatorname{cl}_X S</math> is convex.{{sfn|Wilansky|2013|pp=43-44}}

If <math>R, S \subseteq X</math> then{{sfn|Narici|Beckenstein|2011|pp=67-113}} <math display=block>\operatorname{cl}_X (R) + \operatorname{cl}_X (S) ~\subseteq~ \operatorname{cl}_X (R + S)~ \text{ and } ~\operatorname{cl}_X \left[ \operatorname{cl}_X (R) + \operatorname{cl}_X (S) \right] ~=~ \operatorname{cl}_X (R + S)</math> and so consequently, if <math>R + S</math> is closed then so is <math>\operatorname{cl}_X (R) + \operatorname{cl}_X (S).</math>{{sfn|Wilansky|2013|pp=43-44}}

If <math>X</math> is a real TVS and <math>S \subseteq X,</math> then <math display=block>\bigcap_{r > 1} r S \subseteq \operatorname{cl}_X S</math> where the left hand side is independent of the topology on <math>X;</math> moreover, if <math>S</math> is a convex neighborhood of the origin then equality holds.

For any subset <math>S \subseteq X,</math> <math display=block>\operatorname{cl}_X S ~=~ \bigcap_{N \in \mathcal{N}} (S + N)</math> where <math>\mathcal{N}</math> is any neighborhood basis at the origin for <math>X.</math>{{sfn|Narici|Beckenstein|2011|pp=80}} 
However, <math display=block>\operatorname{cl}_X U ~\supseteq~ \bigcap \{U : S \subseteq U, U \text{ is open in } X\}</math> and it is possible for this containment to be proper{{sfn|Narici|Beckenstein|2011|pp=108-109}} (for example, if <math>X = \R</math> and <math>S</math> is the rational numbers). It follows that <math>\operatorname{cl}_X U \subseteq U + U</math> for every neighborhood <math>U</math> of the origin in <math>X.</math>{{sfn|Jarchow|1981|pp=30-32}}

'''Closed hulls'''

In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.{{sfn|Narici|Beckenstein|2011|pp=155-176}}

* The closed convex hull of a set is equal to the closure of the convex hull of that set; that is, equal to <math>\operatorname{cl}_X (\operatorname{co} S).</math>{{sfn|Narici|Beckenstein|2011|pp=67-113}}
* The closed balanced hull of a set is equal to the closure of the balanced hull of that set; that is, equal to <math>\operatorname{cl}_X (\operatorname{bal} S).</math>{{sfn|Narici|Beckenstein|2011|pp=67-113}}
* The closed [[Absolutely convex set|disked]] hull of a set is equal to the closure of the disked hull of that set; that is, equal to <math>\operatorname{cl}_X (\operatorname{cobal} S).</math>{{sfn|Narici|Beckenstein|2011|p=109}}

If <math>R, S \subseteq X</math> and the closed convex hull of one of the sets <math>S</math> or <math>R</math> is compact then{{sfn|Narici|Beckenstein|2011|p=109}} <math display=block>\operatorname{cl}_X (\operatorname{co} (R + S)) ~=~  \operatorname{cl}_X (\operatorname{co} R) + \operatorname{cl}_X (\operatorname{co} S).</math> 
If <math>R, S \subseteq X</math> each have a closed convex hull that is compact (that is, <math>\operatorname{cl}_X (\operatorname{co} R)</math> and <math>\operatorname{cl}_X (\operatorname{co} S)</math> are compact) then{{sfn|Narici|Beckenstein|2011|p=109}} <math display=block>\operatorname{cl}_X (\operatorname{co} (R \cup S)) ~=~ \operatorname{co} \left[ \operatorname{cl}_X (\operatorname{co} R) \cup \operatorname{cl}_X (\operatorname{co} S) \right].</math>

'''Hulls and compactness'''

In a general TVS, the closed convex hull of a compact set may {{em|fail}} to be compact. 
The balanced hull of a compact (respectively, [[totally bounded]]) set has that same property.{{sfn|Narici|Beckenstein|2011|pp=67-113}} 
The convex hull of a finite union of compact {{em|convex}} sets is again compact and convex.{{sfn|Narici|Beckenstein|2011|pp=67-113}}