# Zorn's Lemma

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## Theorem

### Formulation 1

Let $\struct {S, \preceq}, S \ne \O$ be a non-empty ordered set such that every non-empty chain in $S$ has an upper bound in $S$.

Then $S$ has at least one maximal element.

### Formulation 2

Let $\struct {S, \preceq}, S \ne \O$ be a non-empty ordered set.

Let $T \subseteq \powerset S$ be the set of subsets of $S$ that are totally ordered by $\preceq$.

Then every element of $T$ is a subset of a maximal element of $T$ under the subset relation.

## Zorn's Lemma and Axiom of Choice

### Axiom of Choice implies Zorn's Lemma

Let the Axiom of Choice be accepted.

Then Zorn's Lemma holds.

### Zorn's Lemma implies Axiom of Choice

Let Zorn's Lemma be accepted.

Then the Axiom of Choice holds.

## Warning

The statement of Zorn's Lemma supposes the existence of an upper bound in **$S$** 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:

- $\forall a \in S: a \le x \implies a = x$

## Proof Outline

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.

- $(1): \quad$ By the Axiom of Choice, every partition has a transversal.

- $(2): \quad$ For every set of non-empty sets $S$ there is a choice function $f$.

Consider the choice function $f$ of partition $\set {\set s \times s: s \in S}$.

Partition since each pair of $\set s$ and hence each pair of $\set s \times S$ is disjoint.

This transversal 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.

- $(3): \quad$ 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, that is 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 Theorem (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.

- $(4): \quad$ 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.

- $(5): \quad$ 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.

- $(6): \quad$ 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.

- $(7): \quad$ 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.

## Also see

- Results about
**Zorn's Lemma**can be found**here**.

## Source of Name

This entry was named for Max August Zorn.

## Historical Note

Zorn's Lemma was published by Max August Zorn in $1935$.

However, it had previously been reported by Kazimierz Kuratowski in $1922$, in a slightly different (and simpler) form.

Hence this result is sometimes known as the Kuratowski-Zorn Lemma.

Zorn himself fully acknowledged Kuratowski in his own publication of this result.

He was apparently somewhat embarrassed to have had it named after him, rather than Kuratowski.

Kuratowski himself had not attached any particular importance to it, considering it a straightforward corollary of Hausdorff's Maximal Principle.

## Sources

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- 1922: Kazimierz Kuratowski:
*Une méthode d'élimination des nombres transfinis des raisonnements mathématiques*(*Fund. Math.***Vol. 3**: pp. 76 – 108)