Editing Topological vector space (section)

===Other properties===

'''Meager, nowhere dense, and Baire'''

A [[Absolutely convex set|disk]] in a TVS is not [[nowhere dense]] if and only if its closure is a neighborhood of the origin.{{sfn|Narici|Beckenstein|2011|pp=371-423}} 
A vector subspace of a TVS that is closed but not open is [[nowhere dense]].{{sfn|Narici|Beckenstein|2011|pp=371-423}}

Suppose <math>X</math> is a TVS that does not carry the [[indiscrete topology]]. Then <math>X</math> is a [[Baire space]] if and only if <math>X</math> has no balanced absorbing nowhere dense subset.{{sfn|Narici|Beckenstein|2011|pp=371-423}}

A TVS <math>X</math> is a Baire space if and only if <math>X</math> is [[nonmeager]], which happens if and only if there does not exist a [[nowhere dense]] set <math>D</math> such that <math display=inline>X = \bigcup_{n \in \N} n D.</math>{{sfn|Narici|Beckenstein|2011|pp=371-423}} 
Every [[nonmeager]] locally convex TVS is a [[barrelled space]].{{sfn|Narici|Beckenstein|2011|pp=371-423}}

'''Important algebraic facts and common misconceptions'''

If <math>S \subseteq X</math> then <math>2 S \subseteq S + S</math>; if <math>S</math> is convex then equality holds. For an example where equality does {{em|not}} hold, let <math>x</math> be non-zero and set <math>S = \{- x, x\};</math> <math>S = \{x, 2 x\}</math> also works.

A subset <math>C</math> is convex if and only if <math>(s + t) C = s C + t C</math> for all positive real <math>s > 0 \text{ and } t > 0,</math>{{sfn|Rudin|1991|p=38}} or equivalently, if and only if <math>t C + (1 - t) C \subseteq C</math> for all <math>0 \leq t \leq 1.</math>{{sfn|Rudin|1991|p=6}}

The [[convex balanced hull]] of a set <math>S \subseteq X</math> is equal to the convex hull of the [[balanced hull]] of <math>S;</math> that is, it is equal to <math>\operatorname{co} (\operatorname{bal} S).</math> But in general, <math display=block>\operatorname{bal} (\operatorname{co} S) ~\subseteq~ \operatorname{cobal} S ~=~ \operatorname{co} (\operatorname{bal} S),</math> where the inclusion might be strict since the [[balanced hull]] of a convex set need not be convex (counter-examples exist even in <math>\R^2</math>).

If <math>R, S \subseteq X</math> and <math>a</math> is a scalar then{{sfn|Narici|Beckenstein|2011|pp=67-113}} <math display=block>a(R + S) = aR + a S,~ \text{ and } ~\operatorname{co} (R + S) = \operatorname{co} R + \operatorname{co} S,~ \text{ and } ~\operatorname{co} (a S) = a \operatorname{co} S.</math> 
If <math>R, S \subseteq X</math> are convex non-empty disjoint sets and <math>x \not\in R \cup S,</math> then <math>S \cap \operatorname{co} (R \cup \{x\}) = \varnothing </math> or <math>R \cap \operatorname{co} (S \cup \{x\}) = \varnothing.</math>

In any non-trivial vector space <math>X,</math> there exist two disjoint non-empty convex subsets whose union is <math>X.</math>

'''Other properties'''

Every TVS topology can be generated by a {{em|family}} of [[F-seminorm|''F''-seminorms]].{{sfn|Swartz|1992|p=35}} 
<!--START: REMOVED INFO- If <math>f : X \to \R</math> is a subadditive function (that is, <math>f(x + y) \leq f(x) + f(y)</math> for all <math>x, y \in X</math>) such as a [[sublinear function]], [[seminorm]], or [[Linear form|linear functional]], then <math>f</math> is continuous at the origin if and only if it is uniformly continuous on <math>X.</math>{{sfn|Narici|Beckenstein|2011|pp=192-193}} If <math>f : X \to \R</math> is a subadditive and satisfies <math>f(0) = 0</math> then <math>f</math> is continuous if its absolute value <math>|f| : X \to [0, \infty)</math> is continuous. -END:REMOVED INFO-->

If <math>P(x)</math> is some unary [[Predicate (mathematical logic)|predicate]] (a true or false statement dependent on <math>x \in X</math>) then for any <math>z \in X,</math> <math>z + \{x \in X : P(x)\} = \{x \in X : P(x - z)\}.</math><ref group=proof><math display=block>z + \{x \in X : P(x)\} = \{z + x : x \in X, P(x)\} = \{z + x : x \in X, P((z + x) - z)\}</math> and so using <math>y = z + x</math> and the fact that <math>z + X = X,</math> this is equal to <math display=block>\{y : y - z \in X, P(y - z)\} = \{y : y \in X, P(y - z)\} = \{y \in X : P(y - z)\}.</math> [[Q.E.D.]] <math>\blacksquare</math></ref> 
So for example, if <math>P(x)</math> denotes "<math>\|x\| < 1</math>" then for any <math>z \in X,</math> <math>z + \{x \in X : \|x\| < 1\} = \{x \in X : \|x - z\| < 1\}.</math> Similarly, if <math>s \neq 0</math> is a scalar then <math>s \{x \in X : P(x)\} = \left\{x \in X : P\left(\tfrac{1}{s} x\right)\right\}.</math> The elements <math>x \in X</math> of these sets must range over a vector space (that is, over <math>X</math>) rather than not just a subset or else these equalities are no longer guaranteed; similarly, <math>z</math> must belong to this vector space (that is, <math>z \in X</math>).