Let be a vector space (over a field), and let be the set of all sets of linearly independent vectors, partially ordered by set inclusion.

Any chain is bounded above by which is a linearly independent set of vectors because any finite subset of must be contained in some element of . Thus, by Zorn’s Lemma there exists at least one maximal element .

Suppose for contradiction that does not span. Then there is some which cannot be expressed as a finite linear combination of elements of . But this implies and , contradicting the maximality of M with respect to inclusion.

So must span, and since it is a linearly independent set, this makes it a basis.

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April 30, 2011 at 12:17 am

phanijvThanks for the post. A neat proof.

April 25, 2013 at 2:10 am

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