# Wellordering vs. Zorn's lemma

Many mathematicians outside mathematical logic dislike wellorderings, ordinals and corresponding transfinite arguments. They use zorn's lemma instead and claim one does not need ordinals at all. Examples are

• Every vector space has a basis
• Every filter can be extended to an ultrafilter
• Hahn-Banach theorem

However many cases are not that easy. I often have difficulties to convert a simple proof that uses ordinals and transfinite induction into one using zorn's lemma instead. For example, how would one proove the following assertions with zorn's lemma?

1. $\mathbb{R}^3$ is the union of pairwise disjoint unit-circles.
2. There is a set of reals of the cardinality of the continuum that has no perfect subset.
3. There is a non-determined set of reals.
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They do? Who are these mathematicians? –  Chris Eagle Mar 1 '13 at 20:28
Not sure why you'd have to choose, due to equivalence. I have a very vague memory that someone told me that you could do most of commutative algebra without choice by altering definitions to not reference maximal elements when you need them, but rather reference chains, thus essentially "bypassing" Zorn in some ways. But that was 20+ years ago, and even then I was only told such an idea existed. –  Thomas Andrews Mar 1 '13 at 20:34
@Thomas: The other way around. The equivalence between "every increasing chain is finite" to "every non-empty collection of ideals has a maximal element" requires some choice; but without choice it would be smarter to work with the definition guaranteeing the maximal element, rather the one with the chains. –  Asaf Karagila Mar 1 '13 at 20:38
If they are completely uninterested in logic and set theory, why would they care about these theorems? If they are interested, why would they have trouble with well-orderings? –  Thomas Andrews Mar 1 '13 at 20:55
@user64573: I would prove the well-ordering principle and deduce the wanted corollaries. But the point is that it is possible. It might be the case that there is even a very short and amazing proof for these facts using this choice equivalent. But It makes no sense to do so; similarly some things are better suited for Zorn's lemma and others are not as well suited for it. –  Asaf Karagila Mar 1 '13 at 20:59

Every proof with transfinite induction on a well-ordering can be essentially translated into a proof by Zorn's lemma. However this is an issue of simplicity. One can write a very long and difficult proof that a injective polynomial map from $\Bbb C^n\to\Bbb C^n$ is surjective, or one can use the correct tools from model theory and prove this quickly.

Sometimes things are easier to prove by well-ordering a set and going by induction, and sometimes things are easier to do with Zorn's lemma. Sometimes things are difficult in either case, and sometimes they are easy in either case. The idea is to identify the needed property for the proof and use the most suitable choice principle for that.

Equally you don't see people prove that there is a Bernstein set using the fact that every vector space has a Hamel basis; or using Tychonoff's theorem.

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The result is that an injective polynomial map is surjective. Your version is obviously wrong. –  Chris Eagle Mar 1 '13 at 20:38
@Chris: Thanks. –  Asaf Karagila Mar 1 '13 at 20:39
I don't know of any example where using zorn's lemma would be essentially easier than using wellordering. However i can't think of a remotely comprehensible proof using zorn for one of the three cases i mentioned. –  user64573 Mar 1 '13 at 20:53
@user64573: Hausdorff's Maximality Principle; Kurepa's Maximal Antichain Principle. –  Asaf Karagila Mar 1 '13 at 20:55
@Asaf: Very easy. Enumerate the set, proceed by induction, add elements to your (anti-)chain als long as possible. It's just as easy and essentially the same argument as the zorn version. This seems to be true of any example of proof by zorn i know of. But not the other way round. –  user64573 Mar 1 '13 at 21:02

I remember using Zorn's lemma to prove that every vector space has a basis, and I remember my professor at the time giving us a warning that not all mathematicians entirely agree with Zorn's lemma. As to one of your questions, I believe this paper does use Zorn's lemma, though I have not read it fully: http://www.math.chalmers.se/~wastlund/partitions.ps .

I'm not really sure where you got 'many mathematicians' from. I was only aware that the stubborn few did not like it, and not the majority of the community.

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This doesn't seem to have anything to do with the question at hand. We're talking about people who prefer Zorn to other forms of choice, not people who reject choice in all its variants. –  Chris Eagle Mar 1 '13 at 20:40
@Chris he asked a question for an example of a proof that $\mathbb{R}^3$ is the union of disjoint circles without using AOC. –  noobProgrammer Mar 1 '13 at 20:42
No, the OP asked for such a proof "with Zorn's lemma". –  Chris Eagle Mar 1 '13 at 20:43
The proof linked seems to be using both well-orderings and Zorn's lemma. –  Asaf Karagila Mar 1 '13 at 20:43
@AsafKaragila I'll edit my post accordingly, thanks. –  noobProgrammer Mar 1 '13 at 20:44