Zorn's lemma, also known as the Kuratowski-Zorn lemma, is a theorem of set theory that states that:
Every partially ordered set in which every chain (i.e. totally ordered subset) has an upper bound contains at least one maximal element.
It is named after the mathematician Max Zorn.
The terms are defined as follows. Suppose (P,≤) is the partially ordered set. A subset T is totally ordered if for any s, t in T we have either s ≤ t or t ≤ s. Such a set T has an upper bound u in P if t ≤ u for all t in T. Note that u is an element of P but need not be an element of T. A maximal element of P is an element m in P such that the only element x in P with m ≤ x is x = m itself.
Like the well-ordering principle, Zorn's Lemma is equivalent to the axiom of choice, in the sense that either one together with the Zermelo-Fraenkel axioms of set theory is sufficient to prove the other. It is probably the most useful of all equivalents of the axiom of choice and 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 ring has a maximal ideal and that every field has an algebraic closure.
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2 Sketch of the proof of Zorn's lemma 3 History 4 References |
We will go over a typical application of Zorn's lemma, the proof that every ring R contains a maximal ideal. The set P here consists of all (two-sided) ideals in R, except R itself. This set is partially ordered by set inclusion. We are done if we can find a maximal element in P. The ideal R was excluded because maximal ideals by definition are not equal to R.
We want to apply Zorn's lemma, and so we take a totally ordered subset T of P and have to show that T has an upper bound, i.e. that there exists an ideal I in R which is bigger than all members of T but still smaller than R (otherwise it wouldn't be in P). We take I to be the union of all the ideals in T. I is an ideal: if a and b are elements of I, then there exist two ideals J and K in T such that a ∈ J and b ∈ K. Since T is totally ordered, we know that J is a subset of K or vice versa. 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 we conclude similarly that a + b is contained in I. Furthermore, if r is an element of R, then ar and ra are elements of J and hence elements of I. We have shown that I is an ideal in R.
Now comes the heart of the proof: why is I smaller than R? The crucial observation is that 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 we explicitly excluded R from P.
The condition of Zorn's lemma has been checked, and we thus get 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.
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, 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's are defined by transfinite induction: 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 works just fine.
This proof shows that actually a slightly stronger version of Zorn's lemma is true:
Zorn's lemma was first discovered by K. Kuratowski in 1922, and independently by Max Zorn in 1935.An example application
Sketch of the proof of Zorn's lemma
History