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It couldn't be infinite, because that necessarily invokes notions of convergence and norm which do not necessarily apply to an arbitrary vector space?

BiP

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- Thread starter Bipolarity
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- #1

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It couldn't be infinite, because that necessarily invokes notions of convergence and norm which do not necessarily apply to an arbitrary vector space?

BiP

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Office_Shredder

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- #3

mathman

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You haven't defined V well enough. If V has a topology, then completeness is meaningful.

It couldn't be infinite, because that necessarily invokes notions of convergence and norm which do not necessarily apply to an arbitrary vector space?

BiP

Without a topology you need to be very precise in defining "spans". It may mean that every vector in V is expressible by a finite linear combination. Essentially the question is answered by "yes" by definition.

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For an infinite-dimensional vector space, a Hamel basis refers to a basis in the linear algebra sense: every element of the vector space can be expressed uniquely as a linear combination of a FINITE number of elements of the Hamel basis. Every vector space has a Hamel basis, as a consequence of Zorn's lemma, but in general it's not possible to specify one concretely.

As others have noted, if you want to allow infinite linear combinations, there needs to be a topology involved. For an infinite-dimensional topological vector space, one has the notion of a Schauder basis: http://en.wikipedia.org/wiki/Schauder_basis But not every topological vector space necessarily has such a basis. If you impose more structure, then you can have a guarantee: for example, every Hilbert space has a basis in this sense (orthonormal, even).

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