nLab
basis theorem

Idea

The basis theorem for vector spaces states that every vector space $V$ admits a basis, or in other words is a free module over its ground field of scalars. It is a famous classical consequence of the axiom of choice (and is equivalent to it by a result of Andreas Blass, proved in 1984).

Statement and proofs

Basis theorem

If $V$ is a vector space over any field $K$, then $V$ has a basis.

Proof

We apply Zorn's lemma as follows: consider the poset consisting of the linearly independent subsets of $V$, ordered by inclusion (so $S\le S\prime$ if and only if $S\subseteq S\prime$). If $(S_{\alpha })$ is a chain in the poset, then

$$S=\bigcup _{\alpha }{S}_{\alpha }$$S = \bigcup_\alpha S_\alpha

is an upper bound, for each $v$ in the span of $S$ belongs to some $S_{\alpha }$ and is uniquely expressible as a finite linear combination of elements in $S_{\alpha }$ and in any $S_{\beta }$ containing $S_{\alpha }$, hence uniquely expressible as a finite linear combination of elements in $S$. Thus the hypothesis of Zorn’s lemma obtains for this poset; therefore this poset has a maximal element, say $B$.

Let $W$ be the span of $B$. If $W$ were a proper subspace of $V$, then for any $v$ in the set-theoretic complement of $W$, $B\cup \{v\}$ is a linearly independent set (for if

$$a_{0}v+a_1{w}_1+\dots +a_n{w}_n=0(w_1,\dots ,w_n\in B)$$a_0 v + a_1 w_1 + \ldots + a_n w_n = 0 \qquad (w_1, \ldots, w_n \in B)

we must have $a_0=0$ – else we can multiply by $1/a_0$ and express $v$ as a linear combination of the $w_i$, contradicting $\neg (v\in W)$ – and then the remaining $a_i$ are 0 since the $w_i$ are linearly independent). This contradicts the maximality of $B$. We therefore conclude that $W=V$, and $B$ is a basis for $V$.

I’ll write out a proof of the converse, that the axiom of choice follows from the basis theorem, as soon as I’ve digested it – Todd.

Generalisations

Given any linearly independent set $A$ and spanning set $C$, if $A\subseteq C$, then there is a basis $B$ with $A\subseteq B\subseteq C$; the theorem above is the special case where $A=\varnothing$ and $C=V$. The proof of this more general theorem is a straightforward generalisation of the proof above.