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closure of a vector subspace is a vector subspace
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(Theorem)
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Proof. Let  be the topological vector space over
 where
 is either
 or
 , let  be a vector subspace in  , and let
 be the closure of  . To prove that
 is a vector subspace of  , it suffices to prove that
 is non-empty, and
whenever
 and
 .
First, as
,
contains the zero vector, and
is non-empty. Suppose
are as above. Then there are nets
,
in converging to , respectively. In a topological vector space, addition and multiplication are continuous operations. It follows that there is a net
that converges to
.
We have proven that
, so
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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(view preamble)
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 1485 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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