Zorn's Lemma

Theorem
Let $\left({X, \preceq}\right), X \ne \varnothing$ be a non-empty ordered set such that every non-empty chain in $X$ has an upper bound in $X$.

Then $X$ has at least one maximal element.

Proof
For each $x \in X$, consider the weak initial segment $\bar s \left({x}\right)$:
 * $\bar s \left({x}\right) = \left\{{y \in X: y \preceq x}\right\}$

Let $\mathbb S \subseteq \mathcal P \left({X}\right)$ be the image of $\bar s$ considered as a mapping from $X$ to $P \left({X}\right)$, where $P \left({X}\right)$ is the power set of $X$.

From Ordering Equivalent to a Subset Relation:
 * $\forall x, y \in X: \bar s \left({x}\right) \subseteq \bar s \left({y}\right) \iff x \preceq y$

Thus the task of finding a maximal element of $X$ is equivalent to finding a maximal set in $\mathbb S$.

Thus the statement of the result is equivalent to a statement about chains in $\mathbb S$:


 * Let $\mathbb S$ be a non-empty subset of $P \left({X}\right), X \ne \varnothing$ such that every non-empty chain in $\mathbb S$, ordered by $\subseteq$, has an upper bound in $\mathbb S$.


 * Then $\mathbb S$ has at least one maximal set.

Note
The statement of Zorn's Lemma supposes the existence of an upper bound in $X$ for any (non-empty) chain $A$.

It does not guarantee the existence of an upper bound for $A$ in $A$ itself.

The conclusion is that:
 * $\forall a \in X: a \le x \implies a = x$

Discussion
It can be shown that this follows from the Axiom of Choice and is in fact equivalent to it.

This quick very rough sketch indicates an appropriate chain of equivalences for others to elaborate.

1. Every partition has a transversal (axiom of choice) 2. For every set of non empty sets S there is a choice function f.

Consider traversal f of partition {{s} X s: s in S}. Partition since each pair of {s} and hence each pair of {s} X S is disjoint. This traversal exists by 1 and is a choice function since range is S from first elements of ordered pairs and each second element of pair has to be in the first element of the pair.

3. Any chain closed poset has a maximal element above any particular element. (Chain closed means every sub-chain has a least upper bound)

If not, every element has a non-empty set of strictly greater elements or strict upper bounds. Choice function exists by 2 since the sets of greater elements are all non empty. This choice function is an increasing map on a chain closed poset with no fixed point since strictly increasing as maps to strictly greater elements. Contradicts Bourbaki-Witt fixed point theorem that every increasing map on a chain closed poset has a fixed point. (Quick separate proof using injective map from class of all ordinals unless there is a fixed point. Start map from ordinals into the poset with ordinal 0 to any particular element. Defined for successor ordinals directly from the increasing map ie the choice function to set of upper bounds, defined for limit ordinals from the least upper bound of chain for elements mapped by previous ordinals. This LUB exists since chain closed poset. Converse relation would be map since injective - from subset onto the ordinals so ordinals as the range of a map from a set would be a set. But ordinals are a proper class.)

Note: Item 3 keeps the transfinite argument to separate proof of Bourbaki-Witt fixed point (which also has a longer proof not using transfinite). Other approaches mix a similar transfinite argument directly into proofs of zorns lemma being equivalent to other forms of choice.

4. Any chain of a poset extends to a maximal chain (Hausdorff). Consider poset of chains ordered by inclusion. This is chain closed since union of chain is a chain that is their least upper bound. Apply 3 so it has a maximal element which is a maximal chain.

5. Any chain bounded poset has a maximal element above any particular element (Zorn). It has a maximal chain which has an upper bound since it is a chain bounded poset. That upper bound of the maximal chain must be a maximal element.

6. Any set has a well-ordering. Consider the partial well orderings of the set ordered by being an initial segment. This is a chain bounded (and also chain closed) poset. So it has a maximal element by 5. If it was properly partial then any of the remaining elements could be added at end to refute maximality. So it must be a complete well ordering.

7. Consider any well ordering of the union of a partition. Define a transversal by choosing the least element of each block of the partition. So 6 implies 1 and all are equivalent to axiom of choice.