Existence of a Basis for Every Vector Space

Existence of a Basis for Every Vector Space: Let $V$ be a vector space over a field $F$. Then there exists a subset $B\subseteq V$ such that:

• (Spanning) Every $v\in V$ can be written as a finite linear combination of elements of $B$, i.e. $v=\sum_{i=1}^n a_i b_i$ with $a_i\in F$ and $b_i\in B$.
• (Linear independence) If $\sum_{i=1}^n a_i b_i=0$ with $a_i\in F$ and distinct $b_i\in B$, then all $a_i=0$.

Such a set $B$ is called a basis of $V$.

In full generality, this theorem is equivalent to the axiom of choice (and hence also to Zorn's lemma ). The standard proof uses a maximality argument in a partially ordered set .

Proof sketch: Consider the collection $\mathcal{L}$ of all linearly independent subsets of $V$, ordered by inclusion. Every chain in $\mathcal{L}$ has an upper bound given by its union, so Zorn’s lemma gives a maximal linearly independent set $B$. If $B$ did not span $V$, one could add a vector outside its span to get a larger linearly independent set, contradicting maximality. Hence $B$ is both independent and spanning, i.e. a basis.