## Every Vector Space Has a Basis

**Point of post:** In this post we will give a careful proof that every vector space does, in fact, have a basis. This is mainly so that when I reference this fact in further posts, I can link back to this one, for the interested reader.

**Theorem:** Let be a vector space (or more generally and -module over a division ring) and any linearly independent subset of . Then, there exists a basis for containing .

**Proof: **Let

Then, is a partially ordered set. So, let be a chain in . Clearly, if we have that is an upper bound for . So, to see that is, in fact, in it suffices to prove that it is linearly independent. So, let . Then, there exists such that . But, since is a chain we may conclude that for some . But, this means that and thus is a subset of a linearly independent set (namely ) and so it itself is linearly independent. It follows that any finite subset of is linearly independent, and thus by definition is, itself, linearly independent.

From this we see that every chain in has an upper bound, and so by Zorn’s Lemma, we are able to conclude that has some maximal element, call it . Now, since is linearly independent (since it’ in ) to prove it’s a basis it suffices to prove that . To do this, suppose to the contrary that there is some . We claim that is in . Since it suffices to prove that is linearly independent. To see this let be any finite subset of . If then linear independence of follows immediately. If not, then for some , assume without loss of generality that . Then, if

we must have that otherwise

contradicting that . Thus, and so the above equation may be rewritten as

but since is linearly independent it follows that . Thus, since the finite subset was arbitrary it follows that is linearly independent, and so . But, this contradicts the maximality of since . Thus, the assumption that must have been incorrect, in other words . But, this is only true if , but since we may conclude that . Thus, is a basis.

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