You are currently browsing the tag archive for the ‘well-ordered sets’ tag.

Notational convention: As in Notes 2, I will colour a statement red in this post if it assumes the axiom of choice.  We will, of course, rely on every other axiom of Zermelo-Frankel set theory here (and in the rest of the course).  \diamond

In this course we will often need to iterate some sort of operation “infinitely many times” (e.g. to create a infinite basis by choosing one basis element at a time).  In order to do this rigorously, we will rely on Zorn’s lemma:

Zorn’s Lemma. Let (X, \leq) be a non-empty partially ordered set, with the property that every chain (i.e. a totally ordered set) in X has an upper bound.  Then X contains a maximal element (i.e. an element with no larger element).

Indeed, we have used this lemma several times already in previous notes.  Given the other standard axioms of set theory, this lemma is logically equivalent to

Axiom of choice. Let X be a set, and let {\mathcal F} be a collection of non-empty subsets of X.  Then there exists a choice function f: {\mathcal F} \to X, i.e. a function such that f(A) \in A for all A \in {\mathcal F}.

One implication is easy:

Proof of axiom of choice using Zorn’s lemma. Define a partial choice function to be a pair ({\mathcal F}', f'), where {\mathcal F}' is a subset of {\mathcal F} and f': {\mathcal F}' \to X is a choice function for {\mathcal F'}.  We can partially order the collection of partial choice functions by writing ({\mathcal F}', f') \leq ({\mathcal F}'', f'') if {\mathcal F}' \subset {\mathcal F}'' and f” extends f’.  The collection of partial choice functions is non-empty (since it contains the pair (\emptyset, ()) consisting of the empty set and the empty function), and it is easy to see that any chain of partial choice functions has an upper bound (formed by gluing all the partial choices together).  Hence, by Zorn’s lemma, there is a maximal partial choice function ({\mathcal F}_*, f_*).  But the domain {\mathcal F}_* of this function must be all of {\mathcal F}, since otherwise one could enlarge {\mathcal F}_* by a single set A and extend f_* to A by choosing a single element of A.  (One does not need the axiom of choice to make a single choice, or finitely many choices; it is only when making infinitely many choices that the axiom becomes necessary.)  The claim follows. \Box

In the rest of these notes I would like to supply the reverse implication, using the machinery of well-ordered sets.  Instead of giving the shortest or slickest proof of Zorn’s lemma here, I would like to take the opportunity to place the lemma in the context of several related topics, such as ordinals and transfinite induction, noting that much of this material is in fact independent of the axiom of choice.  The material here is standard, but for the purposes of this course one may simply take Zorn’s lemma as a “black box” and not worry about the proof, so this material is optional.

Read the rest of this entry »

Archives

RSS Google+ feed

  • An error has occurred; the feed is probably down. Try again later.
Follow

Get every new post delivered to your Inbox.

Join 3,885 other followers