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 Incoming links: Zorn’s Lemma and Some Applications by Keith Conrad

Assume the axiom of choice and suppose a poset satisfying the hypothesis has no maximal element. Fix a choice function on its nonempty subsets. By transfinite recursion, choose at each ordinal stage an upper bound of the chain in a partial order already constructed, then choose a strictly larger member. Both choices are available, including the initial empty stage. This produces distinct members throughout the Hartogs ordinal of

P

, contradicting the Hartogs theorem. Thus a maximal element exists. Conversely, apply Zorn's lemma to partial choice functions on a family of nonempty sets, ordered by extension. A chain in a partial order has the compatible union as an upper bound, and a maximal element must be defined on the whole family, since one missing value could be adjoined.

Zorn's lemma was introduced in 1935 to shorten algebra proofs using choice or the well-ordering theorem.^[ref]

V

F

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.^[ref]

A

P_{A} = {S \subseteq V : A \subseteq S, S is linearly independent over F}, S \leq T ⟺ S \subseteq T .

It is a set, being a subset of the power set of

V

, and is nonempty because

A \in P_{A}

A short proof outline supplies both essential steps:

For a nonempty chain in a partial order $C \subseteq P_{A}$ , put $U = ⋃_{S \in C} S$ . Any finite linear relation in $U$ lies in one member of $C$ : among finitely many containing members, one contains the rest. Its linear independence makes every coefficient zero. Thus $U \in P_{A}$ 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.^[ref]

Taking

A = \emptyset

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

{2 v}

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.