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Prove that V is infinite-dimensional if and only if there is a sequence v1,v2,... of vectors in V such that v1,...,vm is linearly independent for every positive integer m.

I already know that whether a vector space is finite-dimensional depends on whether there exist some list of vectors in it spans it, but I cannot give a rigorous proof on the problem above, please help me, thanks!

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  • $\begingroup$ Recall that for finite dimensional vector spaces, linearly independent sets always have fewer elements than spanning sets (this is how one defines the dimension of a vector space). $\endgroup$ May 5, 2015 at 1:07
  • $\begingroup$ yes you are right, but I want a relatively rigorous proof for this issue. $\endgroup$
    – When
    May 5, 2015 at 1:15

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Note: I assume the axiom of choice, so that every vector space has a basis. Otherwise, I'm not sure how we should define dimension.

Suppose that $V$ is finite dimensional. Let $n$ be the dimension of $V$. Then $v_1,\dots,v_m$ cannot be linearly independent if $m > n$.

For the converse: if $V$ is infinite dimensional, then $V$ has an infinite basis. Any countable subset of this basis satisfies the hypothesis on $\{v_1,v_2,\dots\}$.

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The problem is easy if you know that every vector space has a basis, and that the dimension is the cardinality of that basis. In case a vector space is finite-dimensional, say of dimension $56$ then any set of 57 vectors is linearly dependent. That shows sufficiency of your condition for infinite dimensionality.

For necessity take a basis of your infinite dimensional vector space. It cannot have finite number of elements.

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