# Trouble w/ Maximal Principle (Zorn's Lemma)

1. Sep 13, 2004

### genxhis

The author introduces the "Maximal Principle" in order to prove that every vector space has a basis. For reference, I'll restate it as he wrote it: "Let F be a family of sets. If, for each chain C [subset of] F, there exists a member of F that contains each member of C, then F contains a maximal member".

If a chain C of F is finite, then by comparing all the members of C, we should be able to find a member in C (hence in F) that contains all the others. Therefore, if the hypothesis are to fail anywhere, it should be possible for an infinite chain to not have a member that contains all the others. But in his proof that every vector space has a basis, the author states: "But since C is a chain, one of these [members of C], say A, contains all the others" without knowing whether C is finite.

Could someone clear my head for me?

2. Sep 14, 2004

### matt grime

The maximal set does not have to be in the chain. If the chain is finite, we may assume the maximal set is, but if the set is infinite it may not be the case.

Consider the power set of the natural numbers P(N), then there is an obivious chain of sets:

{1}<{1,2}<{1,2,3}<{1,2,3,4}....

the set which contains all of these subsets is N, which isn't part of the chain.

Are you sure you've not missed something out in the proof. In particular is the C in the proof a certain kind of chain?

3. Sep 14, 2004

### HallsofIvy

Your statement as given doesn't make any sense because members of an arbitrary chain do not have to be sets so that "contains all the others" doesn't make sense. You need to specify that, in this particular proof, your chain consists of sets.

The proof that every vector space has a basis goes something like this:
(or, more generally, that if A is a set of independent vectors in the vector space, then there exist a basis containing A).

Let A be as given in the hypothesis, or without being given A, take A to contain a single non-zero vector. Consider the collection of all sets of independent vectors that have A as a subset. It can be partially ordered by inclusion (X<= Y if and only if X is a subset of Y). Let Let C= {Xn[\sub]} be a chain: Every Xn is a subset of Xn+1. Take X to be the union of all members of {Xn}.
It is easy to show that X is itself a set of independent vectors (given any finite collection of vectors {vi} in X, because of the inclusion, there exist some XN that contains all of them: if $\Sigma c_i v_i= 0$, then, since XN is itself independent, all ci= 0.) Of course X itself contains A and so is itself in the collection of "independent vectors that contains A as a subset" and so is the "member of F that contains each member of C". That is, I think, what was intended by "But since C is a chain, one of these [members of C], say A, contains all the others". If the author really said that, he was being very vague. X exists, not just because C is a chain but because of the specific way C is defined.

Of course, now that we know every chain has a maximum, Zorn's lemma tells us that the entire set of "sets of independent vectors that contain A as a subset" has a "maximal element": call it "Z". Now you need to prove that Z spans the space- if there were some vector v that could not be written as a linear combination of vectors in Z, we could add v iteslf to Z, getting another set of independent vectors, containing A as a subset, that is not as subset of Z, contradicting the fact that Z is maximal.

4. Sep 14, 2004

### mathwonk

I would like to remark that this sort of thing is pretty useless for anything except passing prelims. You will literally never use it again except when you write your own algebra book for future prelim takers.

5. Sep 14, 2004

### matt grime

Come on, what about Mittag Leffler conditions! Very important and useful especially in algebra: every module (for a group algebra) has a maximal projective summand.

6. Sep 14, 2004

### mathwonk

Thank you for proving again that statements with "never" and "always" in them, are immediately proven wrong. (always?)

And what about the existence of maximal, hence prime ideals, in any commutative ring?!

7. Sep 17, 2004

### genxhis

ah Hurkyl, Thanks. The author's proof is quite nearly the same (and correct now that I see what is going on). I guess his wording just threw me off.

8. Sep 18, 2004

### mathwonk

reflecting again on my statement that zorn's lemma has no use after prelims, I have been able to think of roughly one use in each subject I know, but no more. i.e. after proving the existence of maximal ideals, i cannot recall another use of it in commutative algebra.

in sheaf theory i recall using it once in the first homework set proving that if the kernel of a surjective sheaf map is flabby, then the induced map on global sections is surjective too. e.g. this argument is needed on page 48 of kempf's book on algebraic varieties but never again in the book, to my knowledge.

i suppose in point set topology one needs it to prove the product of an arbitrary family of spaces is non empoty, and in set theory to rpove that a surjective map has a right inverse.

I will make a new conjecture, that zorn's lemma is needed at most once in every subject. counterexamples are welcome.

9. Sep 19, 2004

### matt grime

I suppose that could well come down to semantics, but a very important use is in deciding when direct limits of exact objects are exact, or if you like it is trying to choose an infinite number of compatible maps. This appears as the Mittag Leffler condition, and is actually your flabby sheaf thing or perhaps more accurately a generalization of it, but it is used to show where derived functors of Lim vanish and so on. But this might of course be taken as proving your statement since I've just taken a whole load of uses and said they're essentially the same.
NB I'm doing this from memory and could be talking rubbish.

10. Sep 19, 2004

### mathwonk

by the way, in commutative algebra, noetherian rings eliminate even the use of zorn to find maximal ideals. perhaps hilbert introduced "noetherian" conditions for this purpose.