# Zorn's lemma

(Redirected from Zorn's Lemma)
For the film by Hollis Frampton, see Zorns Lemma (film).
Zorn's lemma can be used to show that every connected graph has a spanning tree. The set of all sub-graphs that are trees is ordered by inclusion, and the union of a chain is an upper bound. Zorn's lemma says that a maximal tree must exist, which is a spanning tree since the graph is connected. Zorn's lemma is not needed for finite graphs, such as the one pictured here.

Zorn's lemma, also known as the Kuratowski–Zorn lemma, after mathematicians Max Zorn and Kazimierz Kuratowski, is a proposition of set theory that states that a partially ordered set containing upper bounds for every chain (that is, every totally ordered subset) necessarily contains at least one maximal element.

Proved by Kuratowski in 1922 and independently by Zorn in 1935,[1] this lemma occurs in the proofs of several theorems of crucial importance, for instance the Hahn–Banach theorem in functional analysis, the theorem that every vector space has a basis, Tychonoff's theorem in topology stating that every product of compact spaces is compact, and the theorems in abstract algebra that every nonzero ring has a maximal ideal and that every field has an algebraic closure.[2]

Zorn's lemma is equivalent to the well-ordering theorem and the axiom of choice, in the sense that any one of them, together with the Zermelo–Fraenkel axioms of set theory, is sufficient to prove the others.[3] An earlier formulation of Zorn's lemma is Hausdorff's maximum principle which states that every totally ordered subset of a given partially ordered set is contained in a maximal totally ordered subset of that partially ordered set.[4]

## Background

Zorn's lemma can be stated as:

Lemma — Suppose a partially ordered set P has the property that every chain has an upper bound in P. Then the set P contains at least one maximal element.

## Empty chain as boundary case

In the formulation of Zorn's lemma above, the partially ordered set P is not explicitly required to be non-empty. However, the empty set is a chain (trivially), hence is required to have an upper bound, thus exhibiting at least one element of P. An equivalent formulation of the lemma is therefore:

Suppose a non-empty partially ordered set P has the property that every non-empty chain has an upper bound in P. Then the set P contains at least one maximal element.

The distinction may seem subtle, but proofs involving Zorn's lemma often involve taking a union of some sort to produce an upper bound. The case of an empty chain, hence empty union is a boundary case that is easily overlooked.

## Example application

Zorn's lemma can be used to show that every nontrivial ring R with unity contains a maximal ideal. In the terminology above, the set P consists of all (two-sided) ideals in R except R itself, which is not empty since it contains at least the trivial ideal {0}. This set is partially ordered by set inclusion. Finding a maximal ideal is the same as finding a maximal element in P. The ideal R was excluded because maximal ideals by definition are not equal to R.

To apply Zorn's lemma, take a non-empty totally ordered subset T of P. It is necessary to show that T has an upper bound, that is, there exists an ideal IR which is bigger than all members of T but still smaller than R (otherwise it would not be in P). Take I to be the union of all the ideals in T. Because T contains at least one element, and that element contains at least 0, the union I contains at least 0 and is not empty. To prove that I is an ideal, note that if a and b are elements of I, then there exist two ideals J, KT such that a is an element of J and b is an element of K. Since T is totally ordered, we know that JK or KJ. In the first case, both a and b are members of the ideal K, therefore their sum a + b is a member of K, which shows that a + b is a member of I. In the second case, both a and b are members of the ideal J, and thus a + bI. Furthermore, if rR, then ar and ra are elements of J and hence elements of I. Thus, I is an ideal in R.

Now, an ideal is equal to R if and only if it contains 1. (It is clear that if it is equal to R, then it must contain 1; on the other hand, if it contains 1 and r is an arbitrary element of R, then r1 = r is an element of the ideal, and so the ideal is equal to R.) So, if I were equal to R, then it would contain 1, and that means one of the members of T would contain 1 and would thus be equal to R – but R is explicitly excluded from P.

The condition of Zorn's lemma has been checked, and thus there is a maximal element in P, in other words a maximal ideal in R.

Note that the proof depends on the fact that our ring R has a multiplicative unit 1. Without this, the proof wouldn't work and indeed the statement would be false. For example, the ring with ${\displaystyle \mathbb {Q} }$ as additive group and trivial multiplication (i. e. ${\displaystyle ab=0}$ for all ${\displaystyle a,b}$) has no maximal ideal (and of course no 1): Its ideals are precisely the additive subgroups. The factor group ${\displaystyle \mathbb {Q} /A}$ by a proper subgroup ${\displaystyle A}$ is a divisible group, hence certainly not finitely generated, hence has a proper non-trivial subgroup, which gives rise to a subgroup and ideal containing ${\displaystyle A}$.

## 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 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 ai 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 aw = b({av: v < w}). Because the av 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 greater than or equal to x. That is, there is a maximal element which is comparable to x.

## History

The Hausdorff maximal principle is an early statement similar to Zorn's lemma.

Kazimierz Kuratowski proved in 1922[5] a version of the lemma close to its modern formulation (it applied to sets ordered by inclusion and closed under unions of well-ordered chains). Essentially the same formulation (weakened by using arbitrary chains, not just well-ordered) was independently given by Max Zorn in 1935,[6] who proposed it as a new axiom of set theory replacing the well-ordering theorem, exhibited some of its applications in algebra, and promised to show its equivalence with the axiom of choice in another paper, which never appeared.

The name "Zorn's lemma" appears to be due to John Tukey, who used it in his book Convergence and Uniformity in Topology in 1940. Bourbaki's Théorie des Ensembles of 1939 refers to a similar maximal principle as "le théorème de Zorn".[7] The name "Kuratowski–Zorn lemma" prevails in Poland and Russia.

## Equivalent forms of Zorn's lemma

Zorn's lemma is equivalent (in ZF) to three main results:

A well-known joke alluding to this equivalency (which may defy human intuition) is attributed to Jerry Bona: "The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?"[8]

Moreover, Zorn's lemma (or one of its equivalent forms) implies some major results in other mathematical areas. For example,

1. Banach's extension theorem which is used to prove one of the most fundamental results in functional analysis, the Hahn–Banach theorem
2. Every vector space has a Hamel basis, a result from linear algebra (to which it is equivalent[9])
3. Every commutative unital ring has a maximal ideal, a result from ring theory
4. Tychonoff's theorem in topology (to which it is also equivalent[10])

In this sense, we see how Zorn's lemma can be seen as a powerful tool, especially in the sense of unified mathematics[clarification needed].

## In popular culture

The 1970 film, Zorns Lemma, is named after the lemma.

This lemma was referenced on The Simpsons on the episode "Bart's New Friend".[11]

## Notes

1. ^ Moore 2013, p. 168
2. ^ Jech 2008, ch. 2, §2 Some applications of the Axiom of Choice in mathematics
3. ^ Jech 2008, p. 9
4. ^ Moore 2013, p. 168
5. ^ Kuratowski, Casimir (1922). "Une méthode d'élimination des nombres transfinis des raisonnements mathématiques" [A method of disposing of transfinite numbers of mathematical reasoning] (pdf). Fundamenta Mathematicae (in French) 3: 76–108. Retrieved 2013-04-24.
6. ^ Zorn, Max (1935). "A remark on method in transfinite algebra". Bulletin of the American Mathematical Society 41 (10): 667–670. doi:10.1090/S0002-9904-1935-06166-X.
7. ^ Campbell 1978, p. 82.
8. ^ Krantz, Steven G. (2002), Handbook of Logic and Proof Techniques for Computer Science, Springer, pp. 121–126, doi:10.1007/978-1-4612-0115-1_9.
9. ^ Blass, Andreas (1984). "Existence of bases implies the Axiom of Choice". Contemp. Math. 31: 31–33. doi:10.1090/conm/031/763890.
10. ^ Kelley, John L. (1950). "The Tychonoff product theorem implies the axiom of choice". Fundamenta mathematica 37: 75–76.
11. ^