Editing Zorn's lemma (section)

==Proof sketch==
A sketch of the proof of Zorn's lemma follows, assuming the [[axiom of choice]]. Suppose the lemma is false. Then there exists a partially ordered set, or poset, ''P'' such that every totally ordered subset has an upper bound, and that for every element in ''P'' there is another element bigger than it. For every totally ordered subset ''T'' we may then define a bigger element ''b''(''T''), because ''T'' has an upper bound, and that upper bound has a bigger element. To actually define the [[function (mathematics)|function]] ''b'', we need to employ the axiom of choice (explicitly: let <math> B(T) = \{b \in P : \forall t \in T, b \ge t\} </math>, that is, the set of upper bounds for ''T''. The axiom of choice furnishes <math> b: b(T) \in B(T) </math>).

Using the function ''b'', we are going to define elements ''a''<sub>0</sub> < ''a''<sub>1</sub> < ''a''<sub>2</sub> < ''a''<sub>3</sub> < ... < a<sub>ω</sub> < a<sub>ω+1</sub> <…, in ''P''. This uncountable sequence is '''really long''': the indices are not just the [[natural number]]s, but all [[ordinal number|ordinals]]. In fact, the sequence is too long for the set ''P''; there are too many ordinals (a [[proper class]]), more than there are elements in any set (in other words, given any set of ordinals, there exists a larger ordinal), and the set ''P'' will be exhausted before long and then we will run into the desired contradiction.

The ''a<sub>i</sub>'' are defined by [[transfinite recursion]]: we pick ''a''<sub>0</sub> in ''P'' arbitrary (this is possible, since ''P'' contains an upper bound for the empty set and is thus not empty) and for any other ordinal ''w'' we set ''a''<sub>''w''</sub> = ''b''({''a''<sub>''v''</sub> : ''v'' < ''w''}). Because the ''a''<sub>''v''</sub> are totally ordered, this is a well-founded definition.

The above proof can be formulated without explicitly referring to ordinals by considering the initial segments {''a''<sub>''v''</sub> : ''v'' < ''w''} as subsets of ''P''. Such sets can be easily characterized as well-ordered chains ''S'' ⊆ ''P'' where each ''x'' ∈ ''S'' satisfies ''x'' = ''b''({''y'' ∈ ''S'' : ''y'' < ''x''}). Contradiction is reached by noting that we can always find a "next" initial segment either by taking the union of all such ''S'' (corresponding to the limit ordinal case) or by appending ''b''(''S'') to the "last" ''S'' (corresponding to the successor ordinal case).<ref>{{cite journal |last1=Lewin |first1=Jonathan W. |date=1991 |title=A simple proof of Zorn's lemma |url=https://digitalcommons.kennesaw.edu/cgi/viewcontent.cgi?article=2161&context=facpubs |journal=The American Mathematical Monthly |volume=98 |issue=4 |pages=353–354 |doi=10.1080/00029890.1991.12000768 }}</ref>

This proof shows that actually a slightly stronger version of Zorn's lemma is true:
{{Math theorem|If ''P'' is a [[poset]] in which every [[well-order]]ed subset has an upper bound, and if ''x'' is any element of ''P'', then ''P'' has a maximal element greater than or equal to ''x''. That is, there is a maximal element which is comparable to ''x''.|name=Lemma}}

Alternatively, one can use the same proof for the [[Hausdorff maximal principle]]. This is the proof given for example in Halmos' ''[[Naive Set Theory]]'' or in {{section link||Proof}} below.

Finally, the [[Bourbaki–Witt theorem]] can also be used to give a proof.