Theorem

Every vector space VV has a basis.

Proof

The case VV is zero space is trivial, so we just consider the case VV is not zero.

We will use “Zorn’s lemma” in the following proof.

Let PP be the collection of all linearly independent subset of VV. Note that PP is partially ordered with respect to “inclusion”. Also note that PP is not empty, since it must contain {v}\{v\} for some vVv \in V.

Take a chain CC in PP. Denote U=SCSU = \bigcup_{S \in C} S, the union of all elements of the chain CC.

Now we need to show that UU is indeed an upper bound of CC in PP. It suffices to show that UU is linearly independent. For each finite subset {v1,,vk}\{v_1, …, v_k\} of UU, we can find S1,,SkCS_1, …, S_k \in C such that viSiv_i \in S_i for i=1,,ki = 1, …, k. Since CC is a chain, there is some jj such that S1,,SkSjS_1, …, S_k \subseteq S_j, which also implies v1,,vkSjv_1, …, v_k \in S_j. Notice that SjS_j is linearly independent, the equality

a1v1++akvk=0a_1v_1 + … + a_kv_k = 0

holds only if all coefficients ai=0a_i = 0. Therefore, UU is linearly independent.

The hypothesis of Zorn’s lemma has been checked. Hence, there is a maximal element in PP. Let’s pick a maximal element from PP and call it MM. The next goal is showing that MM is a basis.

By definition, MM is a linearly independent subset of VV. To show MM is a basis of VV, it suffices to check that MM spans VV.

Suppose otherwise, MM does not span VV. Take some wVspan(M)w \in V \setminus \text{span}(M). Now {w}M\{w\} \cup M is a linearly independent subset of VV and a superset of MM, which contradicts to the fact that MM is maximal. Therefore, MM indeed spans VV, and thus a basis of VV.