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There is this theorem in my notes that says that in a finite-dimensional vector space any linearly independent list of vectors is shorter than or equal in length to every spanning list. I understand the proof but it doesn't appear to use the assumption that the space is finite-dimensional.

Could someone show me an example of an infinite dimensional space in which there exists a linearly independent list of vectors that is strictly longer than some spanning list?

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Could someone show me an example of an infinite dimensional space in which there exists a linearly independent list of vectors that is strictly longer than some spanning list?

No. Since it's also true in infinite dimensional spaces, if we replace "length" with "cardinality" and "list" with "set".

But the proof is more complicated in the infinite dimensional case (and may require the axiom of choice/Zorn's lemma, I don't remember).

If the proof in the lecture didn't seem to use the assumption of finite dimensionality, that just isn't obvious.

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  • $\begingroup$ existence of bases for vectorspaces is equivalent to choice, whilst equal cardinality of bases is equivalent to the ultrafilter lemma $\endgroup$ – citedcorpse Jun 26 '13 at 16:33
  • $\begingroup$ If I remembered what the ultrafilter lemma is, I wouldn't have to look it up :/ $\endgroup$ – Daniel Fischer Jun 26 '13 at 16:45

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