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A value of $g\ne 1$ is a convex lattice subgroup maximal among those omitting $g$. Zorn's lemma supplies a value: the union of a chain of convex lattice subgroups omitting $g$ still omits $g$. Such subgroups are prime. Their unique cover is the smallest convex lattice subgroup properly containing them.

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.^[ref]

Use the poset of linearly independent subsets containing $A$: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:

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.

Order the linearly independent subsets of a vector space by inclusion. A chain's union is independent, since every finite relation is contained in one chain member. Zorn's lemma gives a maximal independent subset. A vector outside its span could be adjoined while preserving independence, so it spans. Sending finite-support coefficient families to their linear combinations gives an isomorphism from the free module on that subset to $V$. This supplies an arbitrary, possibly infinite-dimensional basis using the axiom of choice; no pre-existing basis theorem is invoked.

Let $(P,\le )$ be a nonempty partially ordered set. If every chain in a partial order $C\subseteq P$ has an upper bound in $P$, then $P$ has a maximal element. The hypothesis meansThe upper bound may depend on $C$ and need not belong to $C$. A least upper bound is not required. The conclusion asserts that some $m\in P$ satisfies $m\le x⟹x=m$; it does not assert a greatest element, which would satisfy $x\le m$ for every $x\in P$.^[ref]

Here a chain in a partial order may be the empty set. Every member of $P$ is an upper bound of the empty chain in a partial order, so including it in the hypothesis already forces $P$ to be nonempty. If “chain” means only nonempty chains in a partial order, the separate nonempty-$P$ assumption is essential: the empty poset otherwise satisfies the hypothesis vacuously and has no maximal element. These are two equivalent conventions for the theorem.

The strength of Zorn's lemma is that compatible partial constructions can have a completed, unextendible member without an explicit procedure for finding it. The next examples separate the upper bound calculation from the reason a maximal element has the desired property. All chains in a partial order must be covered, including uncountable ones, rather than just increasing sequences. The condition is sufficient, not necessary: a poset can have a maximal element and still have an unbounded chain in a partial order elsewhere.

Consider the following finite set of subsets, ordered by subset containment:Every nonempty chain in a partial order has a greatest element within that chain in a partial order, hence an upper bound in $P$. For the empty chain in a partial order, choose $\varnothing \in P$. Thus the hypothesis of Zorn's lemma holds. The two maximal elements are $\{a,b\}$ and $\{b,c\}$: neither is a proper subset of another member of $P$. There is no greatest element, because it would have to contain both, and no member of $P$ contains $\{a,b,c\}$.

For the chain in a partial order $C=\{\varnothing ,\{b\}\}$, the member $\{b\}$ is already an upper bound, although it is not a maximal element of $P$. Both orange members are also upper bounds for $C$. A chain in a partial order need not have a unique upper bound, and the upper bound used to check the hypothesis need not be the eventual maximal element.

The restriction to chains in a partial order matters. The subset $\{\{a\},\{c\}\}\subseteq P$ has no upper bound in $P$, but its two members are incomparable, so it is not a chain in a partial order. Requiring every subset of $P$ to have an upper bound would be much stronger: taking that subset to be $P$ itself would force a greatest element.