Talk:Zermelo–Fraenkel set theory

Some early questions

I'm confused: are the listed axioms for ZFC or von Neumann thingy lala? -Martin

They're ZFC; I've clarified. As a separate point, isn't the empty set axiom redundant here? It seems to follow from infinity and replacement. Matthew Woodcraft

It is redundant in some formulations, but not others. In any case, it's traditional to include it. I'll mention the redundancy on Axiom of the empty set. -- Toby 05:33 Feb 21, 2003 (UTC)

The singular, "Zermelo-Fraenkel axiom", does not make sense as the title of this article. It makes more sense to title an article "cat" than "cats", and is in accord with Wikipedia conventions, but we're not defining a general concept of a Zermelo-Fraenkel axiom; we're defining a short list of specific axioms and schemas; the whole phrase "Zermelo-Fraenkel axioms" is really a proper noun. It's as if ten separate articles were titled "Comandment" without any article titled "Ten Commandments". The plural in the title of this article makes sense for the same reason the plural in an article titled "Ten Commandments" would make sense. Michael Hardy 22:53 Jan 15, 2003 (UTC)

How about Zermelo-Fraenkel set theory? I'm finding that in metamathematics books more than anything else. -- Toby 05:33 Feb 21, 2003 (UTC)

Do most mathematicians believe anything about ZF? Most mathematicians use operations on sets, and the validity of those operations in in effect codified by ZF, but I don't think most mathematicians think about ZF, let alone believe anything about ZF. Michael Hardy 23:08 Jan 15, 2003 (UTC)

I agree; most mathematicians couldn't care less about ZF. Sure, the axiom of choice is interesting, but not the axiom schema of replacement or the axiom of well foundation. I've changed it to "metamathematicians", which may not be precisely the right term. -- Toby 05:33 Feb 21, 2003 (UTC)

"On the other hand, the consistency of ZFC can be proved by assuming the existence of an inaccessible cardinal." Does this refer to weakly or strongly inaccessible cardinals? -- Schnee 01:25, 10 Aug 2003 (UTC)

Weakly -- all inaccessible cardinals are weakly inaccessible, while only some of them are strongly so. -- Toby Bartels 22:33, 13 Feb 2004 (UTC)

Just noticed this comment; it's a couple years old, but should probably be addressed for the record. Actually the more common convention is that "inaccessible" means "strongly inaccessible", and if you mean "weakly" you say it explicitly, unless it's clear from context. However from the point of view of Schneelocke's question, it doesn't matter, because the existence of a weakly inaccessible cardinal has the same consistency strength as the existence of a strong inaccessible. --Trovatore 02:40, 29 April 2006 (UTC)[reply]

Shouldn't we state the axioms in their weakest forms, i.e. "if two sets are the same then they are equal" and "there is a set"?

list of set theory topics

Wikipedia has no list of set theory topics! Set-theory mavens, please help. Once it is created (or maybe even before it is created?), it should be added to the list of lists of mathematical topics. Michael Hardy 00:30, 13 Jun 2005 (UTC)

What does the exclamation point mean in logic?

The first order logic articles doesn't list it as one of the symbols, yet it is this article. --212.85.24.83

\exists !

means "there exists a unique". --Zundark 15:47, 6 December 2005 (UTC)[reply]

Consistency proof (partial, of course)

http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html reads: "Abian (1969) proved consistency and independence of four of the Zermelo-Fraenkel axioms". The original paper is available at http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.ndjfl/1093888220. I would really love to add a note about it to this article, but I fear I lack the necessary understanding on the topic.

Is the axiom of infinity consistent with the axiom of foundation?

Basically, we start with the set of all ordinals that do not contain themselves. This set, just like the set in Russell's paradox, is not well defined.The question that comes up is whether it is the set N of all finite ordinals or the set of Burali-Forti paradox.The following reasoning makes me think that it is the set N itself.

For, N={0}U{Sx:x eN},

This is true, but it is not the definition of N.--Aleph4 13:48, 27 January 2006 (UTC)[reply]

where S is the successor function ('e' is 'belongs to';! prefixed is the negation).By the axiom of foundation, x !e x for all x.

So, N is {0} U {Sx:x e N and Sx !e Sx}.

But {Sx:x e N and Sx !e Sx} ={y:y!=0,0 e y,y e N,and y !ey}

[The above equality holds only because N is infinite]

So, N={0}U{y:y!=0,0 ey,y e N,and y !ey}.

But the condition y e N is tautological.

No, it is not.--Aleph4 13:48, 27 January 2006 (UTC)[reply]

Please see the clarification below.--Apoorv1 06:59, 2 February 2006 (UTC)[reply]

So,N={0}U{y:0 e y and y !e y}.

But this set is not well defined.

--Apoorv1 06:55, 27 January 2006 (UTC)[reply]

Those are cryptic comments. A little more elaboration would help.--Apoorv1 05:16, 30 January 2006 (UTC)[reply]

In the absence of any further elaboration from your side, these are some additional comments.

'When we set

N={0}U{y:y!=0,0 ey,y e N,and y !ey}, it means

y e N iff y=0 or [y !=0 and 0 e y and y != y and y e N].

Now, for statements p and q, [p<-->p and q]<-->[p<-->q],so

I guess you mean [p <-->(p and q)] in the first bracket. Your formula is not a tautology. The first bracket is equivalent to [ p --> q ]. --Aleph4 10:28, 2 February 2006 (UTC) [reply]

The reverse implication is also true in this case, because the axiom of infinity by itself does not guarantee the existence of a set larger then N all of whose member contain 0.. So, in the system 'ZF less the powerset axiom', N is the set of all ordinals containig 0 that do not contain themselves. Alternatively, it is the set of all ordinals containing 0. In any case,in the system 'ZF less the powerset axiom'the set guaranteed to exist by the axiom of infinity will not satisfy the axiom of foundation.If the system 'ZF less the axiom of powerset' is inconsistent, will the system ZF be consistent?--Apoorv1 07:34, 6 February 2006 (UTC)[reply]

What do you mean by because the axiom of infinity by itself does not guarantee the existence of a set larger then N all of whose member contain 0.? In particular, what do you mean by "larger"?

ZF minus Power set does not guarantee the existence of an uncountable set.
ZF minus Power set does guarantee the existence of a set X with the following properties:

X contains N
X is not equal to N
all elements of X (except for 0) contain 0.

(Note that in the context of set theory, N is always understood to contain 0. I hope I have not misunderstood you there.)

Which precisely is this set X that you are referring to ?----Apoorv1 04:59, 18 March 2006 (UTC)[reply]

For example, the set omega+1.--Aleph4 08:21, 28 March 2006 (UTC)[reply]

X is supposed to be closed under the successor operation. So how is it w+1?In fact, as I have averred earlier, the repeated application of the successor operation can give you no set bigger than N, unles you assume that N !e N, or equivalently S(N) !=N.In the absence of the axiom of powerset and regularity, the membership of N in N cannot be decided.--Apoorv1 08:55, 28 March 2006 (UTC)[reply]

There is another way of approaching the issue. As you say,

ZF minus Power set does guarantee the existence of a set X with the following properties:

X contains N
X is not equal to N
all elements of X (except for 0) contain 0.

This means, that the axiom of infinity is actually two different axioms:

1A)The set N, containing 0 and closed under successor operation exists.

1B)Another set X, containing 0 and N and closed under successor operation exists.

Once again consider the system ZF minus powerset and regularity and only 1A part of the axiom of infinity.

Then N is the largest ordinal in this system and hence, N eN <--> N!eN .

The addition of axiom 1B or the axiom of regularity or powerset does not help us to resolve this basic contradiction.

--Apoorv1 11:14, 31 March 2006 (UTC)[reply]

N is never defined to be the set of all ordinals containing 0 that do not contain themselves. The clause "do not contain themselves" does not make much sense, because whenever x is an ordinal, then x is not an element of x. This follows from the definition of "ordinal".

If the system 'ZF less the axiom of powerset' is inconsistent, then of course ZF is also inconsistent. But you have not shown either of the two statements.

Aleph4 20:45, 16 March 2006 (UTC)[reply]

See remarks above.--Apoorv1 06:01, 28 March 2006 (UTC) I moved them to "below" Aleph4 08:21, 28 March 2006 (UTC)[reply]

y e N iff y =0 or [y !=0 and 0 e y and y !e y],so

N={0} u {y : 0 e y and y !e y} .

The analogy with

S={y:y e S and y=2} <-->S={y : y=2}

makes the above reasoning clearer'.--Apoorv1 06:59, 2 February 2006 (UTC)[reply]

The empty set will satisfy the left equality, but not the right one. --Aleph4 10:28, 2 February 2006 (UTC)[reply]

The point is well taken. However, see the comments above.--Apoorv1 07:34, 6 February 2006 (UTC)[reply]

Infinite sets

The discussion was getting a bit confusing, so I moved Apporv's question here.Aleph4 08:21, 28 March 2006 (UTC)Just to ensure readability, I have copied the relevant comments of Aleph4 below.--Apoorv1 10:50, 28 March 2006 (UTC) [reply]

'ZF minus Power set does not guarantee the existence of an uncountable set. ZF minus Power set does guarantee the existence of a set X with the following properties: X contains N; X is not equal to N;and all elements of X (except for 0) contain 0.' Aleph4

For the moment, let us say that X is some (as yet unidentified) countable set.Now X is infinite only if Sx!=x for all xeX.But Sx!=x iff x!ex.So X ={0 and all x containing 0 that do not contain themselves}. Now consider the system ZF less the powerset and regularity axioms.In the absence of the powerset axiom, P(X) does not exist. S(X) exists only if X !e X. But X is nothing but the set in Russell's paradox and the question whether X e X or X !e X cannot be answered.So X, which is countable by hypothesis,is not well defined. --Apoorv1 06:01, 28 March 2006 (UTC) [reply]

It seems that you are claiming various things that you cannot prove. For example, "Now X is infinite only if Sx!=x for all xeX". It seems to me that you are claiming

(A) if X is infinite, then Sx !=x for all x in X.

Or perhaps you meant to say

(B) if Sx != x for all x in X, then X is infinite.

I think we can agree that (B) is false. (e.g., take X empty).

The axiom of regularity implies that Sx != x for all x (because x not in x, for all x). So if you assume regularity, the clause "if X is infinite" is redundant.

Without the axiom of regularity, (A) cannot be shown.

Aleph4 08:21, 28 March 2006 (UTC)[reply]

I think we need to see my comments in the context of our discussion.The set X we are talking of is the set guaranteed to exist by the axiom of infinity and is closed under the successor operation. If X is finite, it could not be closed under the successor operation. If X is infinite and Sx=x for some xe X, then x =X and so X e X and SX=X for a countable set, directly in contradiction to the axiom of regularity.

So the only case of interest is the case when Sx!=x (i.e x!ex)for all x in X.

In this case, in the system ZF less powerset and regularity, the set S(X)!=X only if X !e X.Since X = {All x such that x !e x}, X , which is countable, is not well defined.Since X !e X <-->X e X in this system, the addition of the axiom of Regularity to this axiom system cannot remove this basic contradiction. --Apoorv1 10:35, 28 March 2006 (UTC)[reply]

Empty Set, Pairing, Subsets are redundant

I draw your attentions to the masterly exposition of Suppes (1972). Suppes dispenses with Empty Set by simply deriving the empty set as the extension of A not equal to A. He then sets out Pairing, and Subsets (Separation) very early on, and delays introducing Replacement as long as possible. But when he does so, he shows that Pairing and Subsets become easy theorems. Thus his definitive listing of the ZFC axioms, on p. nn, does not include Empty Set, Pairing, and Subsets.

It is indeed a revealing fact that the vast majority of working mathematicians don't know any axiomatic set theory and are not curious about it. They are not even interested in the foundations of mathematics. Taking set theory seriously seems limited nowadays to Berkeley, Tarski's students, a number of Israeli and Eastern European mathematicians, and the coterie studying Quinian set theory. The limited interest in set theory and metamathematics nowadays may be largely driven by the lack of interest in those subjects on the part of granting agencies.202.36.179.65 18:00, 27 February 2006 (UTC)[reply]

One might say that Paul Cohen killed the foundations of mathematics as classically conceived by answering (in the negative) in 1964 what remained post-1950 as its biggest open problem, whether Choice and GCH followed from the ZF axioms. Thereafter foundations went off in two directions. One direction, the primary outlet for which was then and still is JSL, the Association for Symbolic Logic's Journal of Symbolic Logic, continues to address the progressively more esoteric questions remaining within the ZF framework, of which there are plenty but which the average mathematician finds it harder to relate to as more of them get answered. The other direction is comprised of various new and not so new frameworks whose respective perspectives make the questions raised by classical foundations less well motivated and which instead substitute their own open problems motivated by their own perspective. Proof theory, modern or abstract algebra with an emphasis on universal algebra, and category theory are all active subjects today, each with at most one or two hundred actively contributing participants, each considering itself as addressing foundational concerns in mathematics. But even those are becoming old hat, and today we find a lot of interest in foundational studies of coalgebras (which arose from categorical thinking but which could as well have come from ZF), quantum programming languages (as a much needed sensitization of Birkhoff and von Neumann's old quantum logic to the more quantitative and dynamic aspects of Heisenberg uncertainty and entanglement), concurrency theory (broadly construed to cover any kind of concurrent behavior by fleets of vehicles, packet networks, parallel programs, corporations, orchestras, armies, etc.), and so on. Foundations is far from dead, it just isn't recognizable today if you define it narrowly to be ongoing research into the consequences of the ZF axioms. In view of this diversity, to say that the ZF axioms are the axioms on which mathematics is based today is a bit of a stretch. Sets and functions play a very important supporting role in modern mathematics, but the ZF axiomatization of binary membership, however popular with old-school foundationalists, is by no means the only approach to either foundations in general or sets and functions in particular. The language of ZF is not even mathematically natural: when did you last see anyone make use of a homomorphism that respected set membership? Monotone functions respect order, group homomorphisms respect the group operation, linear transformations respect linear combinations, and gangsters respect membership in the Cosa Nostra, but what morphism has ever respected membership in a set? It is sheer hubris for a relation that can't get no respect to claim to support mathematics. Vaughan Pratt 00:59, 21 August 2006 (UTC)[reply]

"Union" textual vs. equation

For any set x, there is a set y such that the elements of y are precisely the members of the members of x.

The text says (all) the members of (all) the members. However, the equation says ' $\exists D$ ', that is 'one member'. Is this correct ? I'm tagging the article as contradictory. --Hdante 08:13, 5 March 2006 (UTC)[reply]

The axiom of union is here correctly formulated. The axiom defines B := ∪A. B is called the union set of A. B collects precisely all sets C which are member of any set D in A. Otto ter Haar 12:02, 5 March 2006 (UTC)[reply]

That's true ! :-) --Hdante 17:26, 5 March 2006 (UTC)[reply]

Right arrow

What does the $\rightarrow$ mean ? Where is it defined ? Thanks. --Hdante 08:44, 5 March 2006 (UTC)[reply]

The right-arrow is the symbol for material implication in propositional logic. See Propositional logic#Soundness and completeness of the rules. Otto ter Haar 12:24, 5 March 2006 (UTC)[reply]

Symbolism

Where is the cheat sheet to explain what all the symbols mean? Kd4ttc 22:12, 10 March 2006 (UTC)[reply]

For those of use familar with mathematics but not expert the symbols in the text are opaque. Any reference that one can go to for the symbol meanings? Kd4ttc 23:02, 12 March 2006 (UTC)[reply]

Does first-order logic help? --Trovatore 23:14, 12 March 2006 (UTC)[reply]

Yes! I'm thinking of developing a compendium of symbols that the casual reader may browse. Kd4ttc 23:40, 12 March 2006 (UTC)[reply]

So there's already Table of mathematical symbols and List of operators (these are probably duplicative as it is). You might take a look at the best way to help people find these. --Trovatore 23:44, 12 March 2006 (UTC)[reply]

You are more than kind! Kd4ttc 01:12, 13 March 2006 (UTC)[reply]

Syntax and semantics

At the moment the article has syntax and semantics all mixed up. ZFC per se is purely syntax; a collection of strings of characters and rules for manipulating them. Therefore, for example, the first sentence from the introduction,

ZFC consists of a single primitive ontological notion, that of set, and a single ontological assumption, namely that all individuals in the universe of discourse (i.e., all mathematical objects) are sets.

is wrong; ZFC, in and of itself, has no ontology. The set-theoretic notions that (depending on your philosophy) interpret, justify, or motivate ZFC, do have an ontology, but they are distinct from ZFC itself.

(Really it's these set-theoretic notions that should be given the status of "the most common foundation of mathematics", not ZFC itself.) --Trovatore 03:41, 12 March 2006 (UTC)[reply]

Are ordinals essential?

I changed the sentence (referring to Zermelo's axioms)

This axiomatic theory did not allow the construction of the ordinal numbers, and hence was inadequate for all of ordinary mathematics.

because it is not true. Most of "ordinary" mathematics can be developed without using ordinals. Transfinite induction does appear sometimes in "ordinary mathematics" (which I understand in this context as "mathematics except set theory", or perhaps "mathematics except mathematical logic"), but usally any well-ordered set of the appropriate length will do, and in fact transfinite induction is usually done along any well-order of the set that is being investigated at the moment (field whose closure is to be computed, Banach space on which a functional is to be extended, etc). Often even transfinite induction is not used, and is replaced by using Zorn's lemma as a black box.

Set theory of course all the time uses von Neumann's cumulative hierarchy, where ordinals and the replacement axiom are quite natural or even necessary.

Aleph4 20:54, 16 March 2006 (UTC)[reply]

The silly consistency comment

I refer to the following comment

Because the axioms of Peano arithmetic are ZFC theorems, and the consistency of Peano arithmetic cannot be proved by virtue of Gödel's second incompleteness theorem, the consistency of ZFC cannot be proved using ordinary mathematics.

On its face, this sentence is misleading at best, false at worst. There are many known consistency proofs of PA, Godel's theorem notwithstanding. I plan to fix this soon. CMummert 22:37, 26 April 2006 (UTC)[reply]

Yet Another Silly Consistency Comment?

"Because of Gödel's second incompleteness theorem, the consistency of ZFC cannot be proved within ZFC itself. " Surely this should read something more like "Because of Gödel's second incompleteness theorem, the consistency if ZFC could be proved within ZFC itself, then ZFC would be inconsistent"?

Of course, ZFC is consistent, but we don't *know* that (do we?). SinghAgain 17:14, 04 Feb 2007 (UTC)

Distinctness condition missing in formal statement of Axiom of Choice?

Shouldn't there be an assertion that D != B in the statement of the axiom of choice? That is, two _distinct_ elements of A are disjoint. Otherwise, B and D can be the same element of A, in which case the consequent of the innermost implication fails and the axiom as a whole is vacuous. Or have I missed something...? Awmorp 11:33, 28 April 2006 (UTC)[reply]

Yes, you're right. CMummert 11:37, 28 April 2006 (UTC)[reply]

Limit Ordinals-a matter of faith?

The first limit ordinal is w (omega).It is defined as the union of all preceding ordinals. This presupposes that the set of all preceding ordinals exists-so the definition of w really presupposes the existence of and the definition of w (and is a non definition).

If we consider numbers as primary, then we know that given a number, there is a greater number,but there is no infinite number.Similarly,given a finite set of numbers, we can have a bigger set.But can we have an infinite set?

If we consider the seq. A1={1},A2={1,2},A3={1,2,3} and so on,what is the process by which the limit {1,2,3. . .}is obtained? In particular, there is no metric by which the successive members become closer to each other.So the axiomatic assertion,through the 'axiom of infinity' of the existence of a 'biggest' or 'infinite set w' appears to be a leap of faith,which is quite opposed to our assertion that an infinite number or magnitude does not exist. --Apoorv1 07:47, 12 May 2006 (UTC)[reply]

Well, an axiom basically is a formal leap of faith. You can prove that the axiom of infinity cannot be proven or disproven from the rest of ZFC. -Dan 17:11, 24 May 2006 (UTC)

Nitpick about finite axiomatizations

...proved that ZF (and hence a fortiori ZFC) cannot be ... finitely axiomatized.

Why a fortiori? Maybe this is backwards? -Dan 17:11, 24 May 2006 (UTC)

Good point. Anyone want to look up Montague's paper, and see what he actually proved? (We shouldn't reverse them unless he really did prove ZFC not finitely axiomatizable in 1957). --Trovatore 22:27, 24 May 2006 (UTC)[reply]

ZFC really is not finitely axiomatizable, because it proves that any finite subset of its axioms has a model. The sentence in the article is misleading. CMummert 11:40, 25 May 2006 (UTC)[reply]

I have now fixed that sentence and the false statement about Godel's theorem that I mentioned higher on the talk page. CMummert 12:00, 25 May 2006 (UTC)[reply]

Of course I knew that ZFC is not finitely axiomatizable. My question was about what Montague had proved. While it seems most natural that he would have proved the result for ZFC, I haven't actually seen his paper. Have you? --Trovatore 13:53, 25 May 2006 (UTC)[reply]

No, I haven't; I left in the reference to Montague only out of respect for the original author, and I would not mind if it were removed. I did look up some papers on Mathscinet before I edited the article this morning. The best bet seems to be MR0163840, which is dated 1961 instead of 1957. Here is a quote from its description of Montague's paper Fraenkel's addition to the axioms of Zermelo with ellipses to indicate where I pruned it.

Fränkel's addition is the replacement (or Ersetzungs-schema (RS). ... The author defines a (countable) subset of the cumulative type structure which can be proved to satisfy all instances of Zermelo's comprehension schema (SSF: schema of set formation); ... And if an additional finite set $A$ of sentences in the notation of set theory is added, a model satisfying both $A$ and SSF can be established. ... A consequence is that RS is not finitely axiomatisable over SSF, and the same holds for any consistent extension of RS ...

That quote indicates to me that Montague did prove the result that the article indicates, although the year may be wrong. Moving to the area of personal opinion, I generally feel that there is little reason to give attribution of results such as this one in wikipedia. Extremely important or extremely difficult results may deserve special attention, but this result does not have those properties. So I would vote in favor of not attributing the result at all in the article, and just pointing out the ZFC is not finitely axiomatizable. CMummert 14:36, 25 May 2006 (UTC)[reply]

Switch to Kunen's axioms

I changed the previous set of formal axioms, which I think were correct, to the exact set of axioms in Kunen's book. Here are my reasons:

Kunen was the first book I could find that gave symbolic forms of the axioms.
The other set of axioms was unsourced.
The other set of axiom was typeset poorly. The conventions were not those employed in contemporary literature. Several of the previous axioms were typeset at over 8 inches of width on my monitor. I could not easily tell whether the previous axioms were correct because the typesetting was too hard to read.

I also added references to Kunen and Jech's books. Once I figure out how to do proper inline citations I will fix that. I think that this article should somewhere mention the cumulative hierarchy, which is the fundamental motivation for the axioms of ZFC. CMummert 20:38, 23 June 2006 (UTC)[reply]

I have several objections concerning some details in Kunen's list:

The axiom of set existence is not really set-theoretical; it is a purely logical axiom (or a consequence of purely logical axioms). I suggest to either omit it altogether (as some axiomatizations do -- e.g. Jech or Fraenkel, Bar-Hillel, Levy), or to replace it by the "axiom of the empty set". The axiom of the empty set follows of course from the axiom of separation, but one needs the axiom of the empty set to deduce the axiom of separation from the replacement axiom.
I think that Kunen is unique in calling the "well-ordering theorem" an axiom. He does this only for his own convenience to speed up the development in his book. There are many theorems that are (over ZF) equivalent to the axiom of choice, but only few of them deserve the name axiom (rather than "theorem" or "lemma"). I suggest to use one (or several) of the customary formulations of AC (choice function for P(X), choice function for families of nonempty sets, choice function for disjoint families/disjoint sets of nonempty sets).

--Aleph4 18:39, 24 June 2006 (UTC)[reply]

I have made some changes that at least partially address your concerns.

A disadvantage of using Kunen's axioms would be that we are stuck with using them more or less exactly as he phrased them; if we change them, then they aren't Kunen's axioms any more. I did look at Jech's book; he gives the axioms in English, but not in symbolic form. I think it is nice to have a concrete reference for the specific formal axioms that we list, since the axioms are not completely canonical. It seems to fit WP:NOR better. On the other hand, I added English decriptions based on the previous article, and pointed out that those descriptions are not from Kunen.

I have no objection if someone else finds a book that gives formal symbolic statements of the axioms, puts those symbolic statements into the article correctly, and gives a correct citation to them. That is, I am not advocating Kunen over any other source. Kunen's book was just the first book I could find, and I thought it was sufficient.

I think that it would benefit the article more to explain the various ways in which the axioms are not canonical than it would to pick a different noncanonical choice of axioms.

My impression of the distinction between the well ordering principle and the axiom of chice is this. Cantor suggested the well ordering principle (Every set can be well ordered) as an axiom in the 1880s. Some opposition to the supposed obvious nature of this axiom arose, and Zermelo gave a proof which reduced the well ordering principle to the axiom of choice (Every sequence of nonempty sets has a choice function), which Zermelo believed to be conceptually simpler. So the well ordering principle has historical precedent as an axiom.

CMummert 19:14, 24 June 2006 (UTC)[reply]

As I understand history (although I do not have a reference at the moment), Cantor did not consider an "axiomatization" of set theory at all. (Does your "1880s" refer to "Über unendliche lineare Punktmannigfaltigkeiten"?)

On the other hand, the main point for Zermelo's 1904 and 1908 papers was to prove the well-ordering theorem as a theorem, and to isolate the axioms used in this proof, in particular the axiom of choice.

Jech's book (millenium edition) gives the axioms in English on the first page, and a few pages later gives formal versions. I like these version better than Kunen's -- not only because the axiom of choice is given in the customary form, but also because he uses lowercase and uppercase variables, which is more intuitive.

Aleph4 19:50, 24 June 2006 (UTC)[reply]

Ah. I saw the English ones at the start of Jech's book, looked at Kunen's book, and stopped looking. I have no objection if you would like to switch the article to Jech's axiomatization. CMummert 20:32, 24 June 2006 (UTC)[reply]

There are two different types of bi-implication arrows used in the Axioms as set out here. Is there a reason for this or should it be changed? 86.20.228.25 16:12, 4 February 2007 (UTC)[reply]

I cleaned that up some; it had drifted since the version originally added. I have no preference at all for Rightarrow vs. rightarrow, so I just chose one. You should feel free to make corrections like this yourself. CMummert · talk 18:03, 4 February 2007 (UTC)[reply]

Axiom of extensionality

The English description for this axiom reads "Two sets are the same if and only if they have the same members." But in fact, the axiom as it is presented states that "Two sets are the same if they have the same members" (a conditional rather than a biconditional). I'm not familiar with the original text from which this axiom was taken, but it's presented as an iff in the stand-alone article (axiom of extensionality). Whichever we decide to use on this page, the English description should match the mathematical description. Mathfreq 21:59, 15 August 2006 (UTC)[reply]

The fact that sets that are equal must have the same members is a property of equality which is an axiom of the underlying first order logic. Thus only the converse, that sets with the same members are equal, needs to be added as an axiom. More importantly, the formal axioms are directly quoted from Kunen; please don't change them unless you have a reference for the new ones. Anyone is free to change the English text, which is original here, to explain what is going on. CMummert 22:44, 15 August 2006 (UTC)[reply]

replacement implies comprehension

The article used to say that the ZFC axioms minus comprehension imply comprehension. Here is one proof that this is true when the empty set is assumed to exist. Let A be any set and assume we want to form the set $W=\{x\in A\mid \phi (x)\}$ . Let $Z$ be any set in A such that $\phi (Z)$ (if there is no such set Z then W is the empty set). Let $\psi (x,y)$ say that either $y=x$ and $\phi (x)$ or $y=Z$ and $\lnot \phi (x)$ . Thus $\psi$ defines a function from A to A such that any element satisfying $\phi$ maps to itself, and all the other elements map to $Z$ . Then the range of $\psi$ is exactly the set that we are trying to construct, and this exists by replacement on the formula $\psi$ . This proof doesn't need to be in the article, but I think the statement is interesting. CMummert 23:53, 4 September 2006 (UTC)[reply]

The problem is that you are assuming a version of the axiom of replacement which is not the one in the article. The one in the article says:

\forall A\,\forall w_{1},\ldots ,w_{n}[\forall x\in A\exists !y\phi \rightarrow \exists Y\forall x\in A\exists y\in Y\phi ].

Notice that there is no limit on how many extra elements one can put into Y and thus it could be that Y might just be the same as A in your example. One needs the axiom of separation to convert this version of replacement into the one which you are assuming. JRSpriggs 08:20, 5 September 2006 (UTC)[reply]

While I expect the version stated in the article to be equivalent to the one CM is thinking of, there remains a problem. The description of replacement still states that the generated set Y is the co-domain of the function, which is no longer true. Correct me if I'm wrong, but the cryptic mention of a restriction to avoid paradox can't possibly account for this discrepancy since making the set larger risks creating more paradoxes, if anything.

I rephrased the English description to make it agree with the formal axiom. The cryptic comment about a restriction is referring to the restriction on the free variables of the formula. I don't think the two versions of replacement are equivalent in the absence of comprehension, which is why JRSpriggs's comment settled the matter for me. CMummert · talk 23:52, 27 January 2007 (UTC)[reply]

Let me expand on what CMummert said about the restriction on free variables. Y should not appear free in φ. If it did, then the existential quantification over it on the right side of the implication would cause trouble because then the function implicitly defined by φ would change from what it was on the left. JRSpriggs 12:40, 28 January 2007 (UTC)[reply]

Axiom of Choice v. Well-ordering Theorem (v. Zorn's lemma)

I'm just wondering why we've labeled the well-ordering theorem with its ZF-logically equivalent "axiom of choice". The well-ordering theorem applies terms such as "minimal" which have no defined context.

Secondly, these axioms are supposed to allow an intuitive basis for ZFC logic. We very well could substitute the axiom of choice/well-ordering theorem with Zorn's lemma. This creates a structurally identical system; however, it would create more confusion.

I have only a little background in logic, so I don't feel comfortable changing this part of the page, but I think it should be reverted to a form more compatible to the name "axiom of choice," if only for the fact that it's more self-contained.

Mo Anabre 20:11, 22 February 2007 (UTC)[reply]