= Zorn's lemma in a finite poset
{c}
Consider the following <finite set> of <subsets>, ordered by <subset> containment:
$$
P=\{\varnothing,\{a\},\{b\},\{c\},\{a,b\},\{b,c\}\}.
$$
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 $\{a,b,c\}$.
\Image[/zorn-finite-poset-two-maximal-elements.png]
{title=A finite inclusion poset with two maximal elements and no greatest element}
{description=Lines show immediate upward inclusions; following upward paths gives all other strict comparisons. The blue path is a chain, and the orange nodes are the two maximal elements.}
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>.
Back to article page
