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I'm confused a little bit about this, I've been told many times that Zorn's lemma is equivalent to the axiom of choice.

Is it an axiom or is it lemma, I mean is there a proof of Zorn's lemma or we just accepte it as an axiom.

Also, Zorn's lemma has lots of applications in modern mathematics, for example we use Zorn's lemma in the proof of one of the most famous Theorems in Analysis, The Hahn-Banach theorem.
One of my professors once said, If we take Zorn's lemma out of mathematics we will lose a very big portion of modern mathematics.

Is it possible for some one in the future to show that Zorn's lemma is not correct?! Thank you.

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@BrianM.Scott So your are saying Zorn's lemma is an axiom, is it possible for some one to show that this axiom is not correct? –  i.a.m Feb 13 '13 at 23:08
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It is known to be consistent with the other axioms of set theory. If those axioms together with AC are inconsistent, then so are those axioms without AC. In other words, if someone found an inconsistency, it wouldn’t depend on AC. –  Brian M. Scott Feb 13 '13 at 23:15
    
Re "we will lose a very big portion of modern mathematics": Nonlinear systems, non-Euclidean and non-commutative geometry are studied on their own merit; funny how no one feels they lose the large portion of mathematical results that hold in resp. linear systems, Euclidean and commutative geometry. –  alancalvitti Feb 14 '13 at 0:21
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3 Answers

up vote 4 down vote accepted

If you take the axiom of choice as an axiom, then Zorn’s lemma becomes a provable consequence; if, on the other hand, you take Zorn’s lemma as an axiom, the axiom of choice becomes a provable consequence. Since they are provably equivalent, it really doesn’t matter which version you take as part of your system of axioms. The fact that one is actually called an axiom and the other a lemma is a historical accident.

Yes, a great deal of mathematics depends on the axiom of choice; some results that depend on it are listed in the answers to this question. It is conceivable (though in my opinion unlikely) that at some point someone will discover that the usual axioms of set theory, including the axiom of choice, are mutually inconsistent; that, however, could happen only if the axioms of set theory not including the axiom of choice are already inconsistent.

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I see what you mean, Thank you. –  i.a.m Feb 13 '13 at 23:16
    
Old joke: What is yellow and equivalent to the Aximom of choice? –  Harald Hanche-Olsen Feb 13 '13 at 23:23
    
@Harald: Ah, yes, right up there with What’s purple and commutes? –  Brian M. Scott Feb 13 '13 at 23:26
    
@HaraldHanche-Olsen: Bananach spaces! :-) –  Asaf Karagila Feb 13 '13 at 23:34
    
@i.a.m: You’re welcome. –  Brian M. Scott Feb 13 '13 at 23:35
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The terms "axiom" or "lemma" or "theorem" are just names. All these are simply sentences in the language of mathematics.

When something is first formulated it can be a lemma which is used to prove something, and later turns out to be quite useful. An axiom is something which is initially taken as a true statement (often something like this is formulated out of an observation of certain objects, and requiring a property to hold).

We can prove axioms from other axioms. The pairing axiom, which states that given $a$ and $b$ the ordered pair $\langle a,b\rangle$ exists, can be proved from other axioms. Zorn's lemma can be proved from the axioms of ZF+Choice (even less, actually). Which is fine.

But it turns out that if we assume that ZF+Zorn's lemma are all true statements, then the axiom of choice must be true as well. From this we infer that Zorn's lemma is equivalent to the axiom of choice.

Again, "axiom", "theorem" and "lemma" are just names which are initially given and stick with us.

As for "not correct", it is possible that someday in the future mathematicians will find severe inconsistencies in the current foundations of mathematics, and then Zorn's lemma will be incorrect as stated, or at least the proofs will be incorrect or incompatible with what we will know at the time. But in such case a lot of mathematics will fail anyway.

Very relevant:

What is the difference between an axiom and a postulate?


Edit:

Being "true" or "false" or "correct" or "incorrect" requires context, it requires us to have some way of interpreting objects like sets, or partially ordered sets, and conclude that some statement about them is incorrect.

Yes. It is possible that an axiom is incorrect, consider the axiom "There is some $x$ such that $x\cdot x=1+1$", you can take this statement as an axiom, and it holds if and only if $2$ has a square root. So it fails in the natural numbers, and in the rational numbers, but it holds just fine in the real numbers. This is what I mean when I say context.

If we fix ZF as a foundation of our mathematics then we want to work inside universes of sets. The axiom of choice is "correct" if the universe satisfies it, and it is incorrect if the universe doesn't satisfy it. Similarly Zorn's lemma is true when the axiom of choice is assumed, and it is false when the axiom of choice fails.

It was proved by Godel and Cohen that if we assume that ZF is consistent, then adding the axiom of choice or adding its negation will not result in an inconsistency. That is to say, we cannot add a contradiction by deciding the truth value of the axiom of choice to be true or by setting it to be false.

Because the axiom of choice is so useful (in part because Zorn's lemma is so useful) we choose to assume it. We choose to use it. But not everyone chooses to do so, and some people prefer to work without it. Others, like me, investigate what happens - or could happen - when it fails. In such contexts the axiom of choice is false and Zorn's lemma is therefore "incorrect".

Let me finish with a remark. Using the words "incorrect" or "correct" implies that there is some canonical, or even Platonist, universe of mathematics in which statements are true or false and that's what's really matters. We cannot prove or disprove such claims, and our perception of the physical universe is far too dull to even discuss infinite objects for which these axioms are even needed. ZF and ZFC do not have canonical models like the natural numbers or the real numbers or the rationals. So in some sense asking whether or not some set theoretical statement is "correct" lacks contexts and is often hard to answer. Of course this is my personal view of mathematics, and you can just as well ignore it.

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I dont have a problem of them being equivalent, is it possible for some one to show that this axiom is not correct in some sense? –  i.a.m Feb 13 '13 at 23:10
    
@i.a.m This question is void of context. Sure, the axiom of choice is very much not true when you are doing mathematics where the axiom of choice is not true. You can do mathematics in such context. In this context Zorn's lemma fails as well. Yes. You can choose your own context in mathematics. –  Asaf Karagila Feb 13 '13 at 23:12
    
@i.a.m: I tried to address your question in my edit. Let me know if it makes sense, or if you wish me to clarify things further. –  Asaf Karagila Feb 13 '13 at 23:34
    
Yes thank you for your answer –  i.a.m Feb 13 '13 at 23:36
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Let ZF = The Axioms of Zermelo-Fraenkel Set Theory, C = The Axiom Of Choice, L = Zorn's Lemma.

Then using ZF+C, one can prove L. And using ZF+L, one can prove C. Thus, using ZF, one can prove that "C if and only if L."

However, the equivalence of C and L may fail in other approaches to set theory (i.e. other than ZF).

So to answer your question, "Is it possible for some one in the future to show that Zorn's lemma is not correct?!" The answer is, "Yes, but only in a set theory different from ZF+C."

EDIT: As an aside, the axiom of choice is (in my opinion) intuitively sensible. It merely says, "If you have a collection of bags, each of which contains at least one fruit, you can fill a new bag with precisely one fruit from each bag in the collection." You'd sure hope so!

Furthermore, the important question isn't whether we "believe" the axioms of ZF, or the axioms of ZF+C. It's whether we believe they're consistent. Turns out that ZF and ZF+C are equiconsistent, (assuming first-order PA), so if you believe in the consistency of ZF, then you should believe in the consistency of ZF+C.

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