Zorn's lemma

The lemma of anger, also known as Kuratowski -Zorn lemma or anger cal lemma is a theorem of set theory, more specifically, the Zermelo -Fraenkel set theory, which involves the axiom of choice. It is named after the German - American mathematician Max Zorn, who discovered it in 1935 (regardless of the discovery by Kuratowski 1922), and related to Hausdorff's maximum chain set from 1914.

Statement

The lemma of anger says:

Explanation:

• Given a partially ordered set. This means that the order relation must be transitive, reflexive and antisymmetric.

• We now consider special subsets of which have the following property: For all or always applies. Such subsets are called chains or totally ordered.

• For these special subsets ( and only these) are now required in addition that they must have an upper bound in. This means that for each chain of a exists, such that for all. Note that does not have to lie in.

• The statement of the lemma of Zorn is now: The set has a maximal element. This means that there exists an element for which there is no greater element. Therefore follows from always.

What makes the lemma of anger, therefore, is that one comes from relatively weak statements about very specific subsets of a pretty strong statement about the set itself.

Note: The conditions include not initially from that empty. However, the empty set automatically a chain. Their existing condition, according to upper bound then automatically supplies an element of. An equivalent formulation of the lemma of Zorn is accordingly:

Use

Like the well-ordering theorem is equivalent to the axiom of choice Wrath lemma, ie you can prove with one of these three sets together with the Zermelo -Fraenkel set theory, the other two. Wrath lemma is used in many critical evidence, for example,

An example of application

We prove as a typical application of the lemma of Zorn that each ring with 1, which is not the zero ring has a maximal ideal. The amount here consists of all ( two-sided ) ideals which do not include the 1. This set is not empty ( it contains the zero ideal, since it is assumed ) and partially ordered with respect to set inclusion. If we can find a maximal element of this set, then we are done, because this is a really ideal and included in every major ideal is not, therefore contains the 1 and thus also ideal as each element of, ie there is no greater in real included Ideal.

To apply Wrath lemma, we take a non-empty totally ordered subset of, and must demonstrate that it has an upper bound, ie, an ideal exists in that contains all ideals, but not equal to ( otherwise it would not ). We choose as the union of all elements of. Then it is not empty, for containing at least one ideal as an element that in turn is contained in the subset. is an ideal, as are elements of and, then there are ideals, so that in and lies. There is totally ordered, is one of the two ideals in others, we can assume without loss that is included. Then, and in both, so lie and for each in and in and thus also in. Thus so is actually an ideal of. Since none of the underlying ideals contains 1, also contains the 1 not so is. Thus, an upper bound of in lying.

Since the conditions for Wrath lemma are satisfied, we obtain the existence of a maximal element in, and that is a maximal ideal of.

This proof requires the assumption that the ring has a 1. Without that he would not be feasible and indeed the claim would be false. An example of a ring without maximal ideal (and without 1) is connected to the multiplication for all. Ideals are identical with ( additive ) subgroups in this ring and for every proper subgroup of the factor group is the same as the output group divisible, hence not finitely generated, thus has a non-trivial real (eg cyclic ) subgroup, and this provides a archetype containing a, proper ideal.

Conclusion of Lemma Wrath of the axiom of choice

Last we give a proof sketch of how one can deduce the lemma of anger from the axiom of choice.

Suppose that the lemma would be false. Then there would be a partially ordered set in which every totally ordered subset would have an upper bound, but still each element a genuinely larger would ( there is no maximal element in ). For every totally ordered subset, we now define an element that is larger than any element in by taking an upper bound of and rely on an element that is even greater than this bound. In order to thereby define a function, we need the axiom of choice ( because we do not say that upper bound and greater element which we assume ).

With this function, we then determine elements. This episode is really long: The indices are not only all natural numbers, but all ordinals. This sequence is too long for the amount, because there are more ordinal numbers may be included as elements in any amount, and we obtain a contradiction.

Which we define by transfinite induction: For every ordinal we set

This is because the are totally ordered by this construction.