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closure of a vector subspace is a vector subspace
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(Theorem)
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Proof. Let $X$ be the topological vector space over
 where
 is either
 or
 , let $V$ be a vector subspace in $X$ , and let $\overline{V}$ be the closure of $V$ . To prove that $\overline{V}$ is a vector subspace of $X$ , it suffices to prove that $\overline{V}$ is non-empty, and $$ \lambda x + \mu y \in \overline{V} $$ whenever
 and $x,y\in \overline{V}$ .
First, as $V\subseteq \overline{V}$ , $\overline{V}$ contains the zero vector, and $\overline{V}$ is non-empty. Suppose $\lambda,\mu,x,y$ are as above. Then there are nets $(x_i)_{i \in I}$ , $(y_j)_{j \in J}$ in $V$ converging to $x,y$ , respectively. In a topological vector space, addition and multiplication are continuous operations. It follows that there is a net $(\lambda x_k + \mu y_k)_{k \in K}$ that converges to $\lambda x + \mu y$ .
We have proven that $\lambda x + \mu y \in \overline{V}$ , so $\overline{V}$ is a vector subspace. 
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"closure of a vector subspace is a vector subspace" is owned by loner. [ full author list (2) | owner history (1) ]
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Cross-references: converges, operations, continuous, multiplication, addition, nets, zero vector, contains, closure, vector subspace, topological vector space
This is version 5 of closure of a vector subspace is a vector subspace, born on 2005-02-04, modified 2005-03-01.
Object id is 6710, canonical name is ClosureOfAVectorSubspaceIsAVectorSubspace2.
Accessed 1899 times total.
Classification:
| AMS MSC: | 15A03 (Linear and multilinear algebra; matrix theory :: Vector spaces, linear dependence, rank) | | | 46B99 (Functional analysis :: Normed linear spaces and Banach spaces; Banach lattices :: Miscellaneous) | | | 54A05 (General topology :: Generalities :: Topological spaces and generalizations ) |
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Pending Errata and Addenda
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