Zorn's Lemma An Example Application Sketch Of The Proof Of

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 theoremIn mathematics, Tychonoff's theorem states that the product of any collection of compact topological spaces is compact. For finite collections of compact spaces, this is not very surprising. The statement is in fact true for infinite collections of arbitr in topologyTopology is the study or science of places. It derives its name from the Greek words τοπος meaning place and λογος meaning study, talk. See also earth science, geography, human geography, g stating that every product of compact spacesIn mathematics, a compact space is a space that resembles a closed and bounded subset of Euclidean space R n in that it is "small" in a certain sense and "contains all its limit points". The modern general definition calls a topological space compact if e is compact, and the theorems in abstract algebraAbstract algebra Abstract algebra is the field of mathematics concerned with the study of algebraic structures such as groups, rings and fields. The term "abstract algebra" is used to distinguish the field from " elementary algebra" or "high school algebr that every ring has a maximal ideal and that every fieldIn abstract algebra, a field is an algebraic structure in which the operations of addition, subtraction, multiplication, and division (except division by zero) may be performed and the associative, commutative, and distributive rules hold, which are famil has an algebraic closureField theory In mathematics, an algebraic closure of a field K is an algebraic extension of K that is algebraically closed. It is one of many closures in mathematics. Using Zorn's lemma, it can be shown that every field has an algebraic closure, and that.

1 An example application

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 would not 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 is an element of J and b is an element of 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.

2 Sketch of the proof of Zorn's lemma

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 a_w = b({a_v: v < w}). Because the a_v are totally ordered, this works just fine.

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 that is greater than or equal to x. That is, there is a maximal element which is comparable to x.

3 History

Zorn's lemma was first discovered by K. Kuratowski in 1922, and independently by Max Zorn in 1935.

4 Joke

Q: What's yellow and equivalent to the Axiom of Choice?

A: Zorn's Lemon.

5 References

Set Theory for the Working Mathematician. Ciesielski, Krzysztof. Cambridge University Press, 1997. BooksEnthsiast.com

Set theory Order theory

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