Theorem
Every vector space has a basis.
Proof
The case is zero space is trivial, so we just consider the case is not zero.
We will use “Zorn’s lemma” in the following proof.
Let be the collection of all linearly independent subset of . Note that is partially ordered with respect to “inclusion”. Also note that is not empty, since it must contain for some .
Take a chain in . Denote , the union of all elements of the chain .
Now we need to show that is indeed an upper bound of in . It suffices to show that is linearly independent. For each finite subset of , we can find such that for . Since is a chain, there is some such that , which also implies . Notice that is linearly independent, the equality
holds only if all coefficients . Therefore, is linearly independent.
The hypothesis of Zorn’s lemma has been checked. Hence, there is a maximal element in . Let’s pick a maximal element from and call it . The next goal is showing that is a basis.
By definition, is a linearly independent subset of . To show is a basis of , it suffices to check that spans .
Suppose otherwise, does not span . Take some . Now is a linearly independent subset of and a superset of , which contradicts to the fact that is maximal. Therefore, indeed spans , and thus a basis of .