## I think, therefore Vector Basis Exists

This is the most fascinating mathematical fact among those that are simple enough for me to understand.

Every vector space has a basis. What’s fascinating is that

**This fact doesn’t have anything to do with vectors, linearity, etc.**It is derived directly from the very Foundations. To prove it, it is sufficient to know these 3 things:1. Any subset of a linearly independent set is linearly independent. (This is trivial).

2. A set is linearly independent if any finite subset of it is linearly independent. (This is really a definition, not even an axiom).

3. The Axiom of Choice, or its more convenient equivalent – Zorn’s Lemma.

Proof. V is a vector space. L is the set of all linearly independent subsets in V. L is partially ordered by inclusion. Every totally ordered chain C in L has an upper bound B simply given by the union of all sets in C. It may seem obvious that B is linearly independent. It’s not obvious but true. B is linearly independent because any finite subset of B must belong to some set from C which are all from L. Since all chains have upper bounds, by Zorn’s lemma, L has a maximal element B*. This maximal element is linearly independent because it belongs to L. And it is not contained in any other linearly independent subset of V because it’s maximal. So B* is a basis. QED.

It’s easy to see why any v from V can be represented as a (finite) linear combination of vectors from B*. If not, the set B* + v is linearly independent, so it belongs to L and contains B*, which is a contradiction since B* is maximal.

I don’t know how to prove that the basis “chosen” by this proof defines dimension. I.e. that all bases have the same cardinality.

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[…] solve the problem. I already wrote about the most fascinating example of this approach that I know: https://bbzippo.wordpress.com/2009/11/05/i-think-therefore-vector-basis-exists/ […]

Alternative theme for counting equivalence classes « Drawing Blanks02/15/2010 at 8:10 am

[…] Dehn’s proof is really cool, and I love Hamel’s basis (I wrote about it here https://bbzippo.wordpress.com/2009/11/05/i-think-therefore-vector-basis-exists/ ), but I was always kind of sad when I was thinking that this problem that is so “finitistic” […]

Hilbert’s 3rd problem and the Axiom of Choice « Drawing Blanks10/15/2010 at 5:01 am