closure of a vector subspace is a vector subspace

Theorem 1.

In a topological vector spaceMathworldPlanetmath the closure ( of a vector subspace is a vector subspace.


Let X be the topological vector space over 𝔽 where 𝔽 is either or , let V be a vector subspace in X, and let V¯ be the closure of V. To prove that V¯ is a vector subspace of X, it suffices to prove that V¯ is non-empty, and


whenever λ,μ𝔽 and x,yV¯.

First, as VV¯, V¯ contains the zero vectorMathworldPlanetmath, and V¯ is non-empty. Suppose λ,μ,x,y are as above. Then there are nets (xi)iI, (yj)jJ in V converging to x,y, respectively. In a topological vector space, additionPlanetmathPlanetmath and multiplication are continuousPlanetmathPlanetmath operationsMathworldPlanetmath. It follows that there is a net (λxk+μyk)kK that converges to λx+μy.

We have proven that λx+μyV¯, so V¯ is a vector subspace. ∎

Title closure of a vector subspace is a vector subspace
Canonical name ClosureOfAVectorSubspaceIsAVectorSubspace
Date of creation 2013-03-22 15:00:19
Last modified on 2013-03-22 15:00:19
Owner loner (106)
Last modified by loner (106)
Numerical id 8
Author loner (106)
Entry type Theorem
Classification msc 46B99
Classification msc 15A03
Classification msc 54A05
Related topic ClosureOfAVectorSubspaceIsAVectorSubspace
Related topic ClosureOfSetsClosedUnderAFinitaryOperation