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.

Let $\mathbb X$ be the set of all chains in $\left({X, \preceq}\right)$.

Every element of $X$ is included in $\bar s \left({x}\right)$ for some $x \in X$.

$\mathbb X$ is a non-empty set of sets which are ordered (perhaps partially) by subset.

If $\mathcal C$ is a chain in $\mathbb X$, then:
 * $\displaystyle \bigcup_{A \in \mathcal C} A \in \mathbb X$

Since each set in $\mathbb X$ is dominated by some set in $\mathbb S$, going from $\mathbb S$ to $\mathbb X$ can not introduce any new maximal elements.

The main advantage of using $\mathbb X$ is that the chain hypothesis is in a slightly more specific form.

Instead of saying that each chain in $\mathcal C$ has an upper bound in $\mathbb S$, we can explicitly state that the union of the sets of $\mathcal C$ is an element of $\mathbb X$.

This union of the sets of $\mathcal C$ is clearly an upper bound of $\mathcal C$.

Another advantage of $\mathbb X$ is that, from Subset of Toset is Toset, it contains all the subsets of each of its sets.

Thus we can embiggen non-maximal sets in $\mathbb X$ one element at a time.

So, from now on, we need consider only this non-empty collection $\mathbb X$ of subsets of a non-empty set $X$.

$\mathbb X$ is subject to two conditions:
 * $(1): \quad$ Every subset of each set in $\mathbb X$ is in $\mathbb X$.
 * $(2): \quad$ The union of each chain of sets in $\mathbb X$ is in $\mathbb X$.

It follows from $(1)$ that $\varnothing \in \mathbb X$.

We need to show that there exists a maximal set in $\mathbb X$.

Let $f$ be a choice function for $\mathbb X$:
 * $\forall A \in \mathbb X: f \left({A}\right) \in A$

For each $A \in \mathbb X$, let $\hat A$ be defined as:
 * $\hat A := \left\{{x \in X: A \cup \left\{{x}\right\} \in \mathbb X}\right\}$

That is, $\hat A$ consists of all the elements of $X$ which, when added to $A$, make a set which is also in $\mathbb X$.

From its definition:
 * $\displaystyle \hat A = \bigcup_{x \in \hat A} \left({A \cup \left\{{x}\right\}}\right)$

where each of $A \cup \left\{{x}\right\}$ are chains in $X$ and so elements of $\mathbb X$.

Hence it follows from $(2)$ above that $\hat A$ is a chain in $X$ and so an element of $\mathbb X$.

Hence, from $(1)$ above and Set Difference Subset it follows that $\hat A \setminus A$ is in the domain of $f$.

It follows that the mapping $g: \mathbb X \to \mathbb X$ may validly be defined as:
 * $\forall A \in \mathbb X: g \left({A}\right) = \begin{cases}

A \cup \left\{{f \left({\hat A \setminus A}\right)}\right\} & : \hat A \setminus A \ne \varnothing \\ A & : \text{otherwise} \end{cases}$

From the definition of $\hat A$, it follows that $\hat A \setminus A = \varnothing$ iff $A$ is maximal.

Thus what we now have to prove is that:
 * $\exists A \in \mathbb X: g \left({A}\right) = A$

Note that from the definition of $g$:
 * $\forall A \in \mathbb X: A \subseteq g \left({A}\right)$

The property of $g$ that is crucial is the fact that $g \left({A}\right)$ contains at most one more element than $A$.

We (temporarily) define a tower as being a subset $\mathcal T$ of $\mathbb X$ such that:
 * $(1): \quad \varnothing \in \mathcal T$
 * $(2): \quad A \in \mathcal T \implies g \left({A}\right) \in \mathcal T$
 * $(3): \quad $ If $\mathcal C$ is a chain in $\mathcal T$, then $\displaystyle \bigcup_{A \in \mathcal C} A \in \mathcal T$

There is of course at least one tower in $\mathbb X$, as $\mathbb X$ itself is one.

It follows from its definition that the intersection of a collection of towers is itself a tower.

It follows in particular that if $\mathcal T_0$ is the intersection of all towers in $\mathbb X$, then $\mathcal T_0$ is the smallest tower in $\mathbb X$.

Next we demonstrate that $\mathcal T_0$ is a chain.

We (temporarily) define a set $C \in \mathcal T_0$ as comparable if it is comparable with every element of $\mathcal T_0$.

That is, if $A \in \mathcal T_0$ then $C \subseteq A$ or $A \subseteq C$.

To say that $\mathcal T_0$ is a chain means that all sets of $\mathcal T_0$ are comparable.

There is at least one comparable set in $\mathcal T_0$, as $\varnothing$ is one of them.

So, suppose $C \in \mathcal T_0$ is comparable.

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.