# 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

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A basis is a maximal linearly independent set under the $\subseteq$-ordering. –  Michael Greinecker Sep 1 '12 at 16:50
I downvoted. I think the asker should put some more effort. I will gladly reverse my downvote if the question gets edited accordingly. –  Giuseppe Negro Sep 1 '12 at 17:14
What have you tried? If this is homework, tag it as such. –  lhf Sep 1 '12 at 17:45

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.

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Showing that there is a maximal linear independent set is only half of if: one must still show that a maximal lin. indep. set spans the space. –  Brian M. Scott Sep 1 '12 at 17:39
@BrianM.Scott : I've added something on that above. –  Michael Hardy Sep 1 '12 at 18:05
Looks good now. +1 –  Brian M. Scott Sep 1 '12 at 18:07
I said "every finite linear relation among them is trivial". One could also say "Every finite linear dependence among them is zero". A linear dependence among vectors $\vec{v}_1,\ldots,\vec{v}_n$ is a tuple of scalars $c_1,\ldots,c_n$ such that $\sum_k c_k \vec{v}_k=\vec{0}$. One can show that two linear dependences are really the same dependence if one of them is a nonzero scalar multiple of the other; hence the set of linear dependences is a projective space. –  Michael Hardy Sep 1 '12 at 18:28
@BrianM.Scott : It seems to me that that "half" of the problem is much less than half: it's quite simple by comparison to the other "half". –  Michael Hardy Sep 1 '12 at 22:13