Zorn's Lemma - Sketch of The Proof of Zorn's Lemma (from The Axiom of Choice)

Sketch of The Proof of Zorn's Lemma (from The Axiom of Choice)

A sketch of the proof of Zorn's lemma follows. 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 every element has a bigger one. 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 b, we need to employ the axiom of choice.

Using the function b, we are going to define elements a₀ < a₁ < a₂ < a₃ < ... in P. This sequence is really long: the indices are not just the natural numbers, but all 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, and the set P will be exhausted before long and then we will run into the desired contradiction.

The a_i are defined by transfinite recursion: we pick a₀ 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_w = b({a_v: v < w}). Because the a_v are totally ordered, this is a well-founded definition.

This proof shows that actually a slightly stronger version of Zorn's lemma is true:

If P is a poset in which every well-ordered subset has an upper bound, and if x is any element of P, then P has a maximal element that is greater than or equal to x. That is, there is a maximal element which is comparable to x.