= Vector space bases from Zorn's lemma
{title2=$\exists B\subseteq V:\ B\text{ is a basis of }V$}
Every <vector space> $V$ over a <field> $F$ has a <basis>, assuming the <axiom of choice>. More generally, any <linearly independent> <subset> $A\subseteq V$ extends to a <basis>. This is an algebraic <Hamel basis>: each representation uses a finite <linear combination>, even when the <basis> is infinite.<Zorn’s Lemma and Some Applications by Keith Conrad>[Zorn’s Lemma and Some Applications, §4, pp. 18–19, Theorem 4.1 and Corollary 4.2]{ref}
Use the <poset> of <linearly independent> <subsets> containing $A$:
$$
\mathcal P_A=\{S\subseteq V:A\subseteq S,\ S\text{ is linearly independent over }F\},
\qquad S\leq T\Longleftrightarrow S\subseteq T.
$$
It is a <set>, being a <subset> of the <power set> of $V$, and is nonempty because $A\in\mathcal P_A$.
A short proof outline supplies both essential steps:
* For a nonempty <chain in a partial order> $\mathcal C\subseteq\mathcal P_A$, put $U=\bigcup_{S\in\mathcal C}S$. Any finite <linear relation> in $U$ lies in one member of $\mathcal C$: among finitely many containing members, one contains the rest. Its <linear independence> makes every <coefficient> zero. Thus $U\in\mathcal P_A$ is an <upper bound>. For the empty <chain in a partial order>, use $A$; its empty <union> need not contain $A$.
* <Zorn's lemma> gives a <maximal element> $B$. If a <vector> $v$ lies outside its <linear span>, then $B\cup\{v\}$ is <linearly independent>: a <linear relation> with nonzero <coefficient> on $v$ would express $v$ in that <linear span>. This contradicts maximality, so $B$ is a <spanning set> and hence a <basis>.<Zorn’s Lemma and Some Applications by Keith Conrad>[Zorn’s Lemma and Some Applications, §4, pp. 18–19, chain-union proof of Theorem 4.1 and extension argument of Corollary 4.2]{ref}
Taking $A=\varnothing$ proves existence, including $V=\{0\}$, whose <basis> is the <empty set>. The <chain in a partial order> need not be indexed by the <natural numbers>: only the finitely many members needed for a particular <linear relation> are compared. An arbitrary <union> of <linearly independent> <sets> can fail: $\{v\}$ and $\{2v\}$ for a nonzero <vector> over the <real numbers> are each <linearly independent>, but their <union> is <linearly dependent>. The nesting of a <chain in a partial order> is what prevents this failure.
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