Every vector space has a basis proof using zorn's lemma in linear algebra Every vector space has a basis proof  by using zorn's lemma
How will i prove this statement by using zorn's lemma 
 A: The poset involved is the set of all linearly independent subsets, ordered by inclusion, i.e. $A\le B$ iff $A\subseteq B$.  The problem is to show that there is a maximal such linearly independent subset.  Zorn's lemma says this is true if every chain of linearly independent subsets has an upper bound.  So the question is where to find such an upper bound.  I claim the union of the members of the chain will serve.  If a set of linearly independent subsets is linearly ordered by inclusion, then their union is also linearly independent.  To show this, one must recall that a set is linearly independent precisely if every finite linear relation among them is trivial.  I.e. linear independence is a property that has "finite character".
At this point, I will leave the details as an exercise.
Later note: In view of Brian M. Scott's comment below: If a linearly independent set fails to span the space, then there's some vector that is not a linear combination of finitely many members of that set.  Add it to the linearly independent set, getting a bigger linearly independent set.  It follows that every linearly independent set that fails to span the space is not maximal.  Hence every maximal one spans the space.
A: A dual approach to this proof is to consider the set of subsets of a vector space $V$ that generate $V$, ordered by reverse inclusion. Consider any chain and take the intersection to produce an upper bound. Since this is a nested intersection, it is still a generating set, so it is contained in our set. Thus by Zorn's lemma, there is a minimal generating set, which is an equivalent definition of a basis.
