Every vector space has a basis
September 24, 2011 6 Comments
If you haven’t seen a Zorn’s lemma argument before, this will probably be bewildering. It is here for the sake of completeness.
Every vector space has a basis.
Consider the collection of linearly independent sets ordered by inclusion. So if and are linearly independent sets, and , We would say . Of course, it may be that we have sets neither of which is a subset of the other. This is totally fine, since we are in a partially ordered set, or poset.
Zorn’s lemma says that if every chain has an upper bound, then there is a maximal element. That is, if we take a bunch of linearly independent sets that are all comparable (one contains the other), and they have an upper bound, then there is a set which is maximal in the sense that nothing contains it. Such a maximal set must have , for if it didn’t, we could add another element to it from , so it wouldn’t be maximal. By virtue of being in the partially ordered set to begin with, must be linearly independent. As is linearly independent and spanning, it is by definition a basis.
So we only need to show that every chain has an upper bound. If I have a bunch of linearly independent sets (possibly infinitely many, and a really big infinity too) and they form a chain of containment, we want to find a really big linearly independent set that contains all of the . It turns out that
works. Why? We just need to show that it is linearly independent. Well, the trick is that linear combinations have to be finite. So if I had some linear dependence
There must be some for which the . But then, since is linearly independent, each of the must be zero as desired.
Zorn’s lemma is always weird if you haven’t done it before. Luckily, you can try your own. Another theorem is that if we have a linearly independent set , we can extend it to a basis for our vector space. Try to recreate Zorn’s lemma argument. Instead of starting with all linearly independent sets, start with all linearly independent sets which contain .