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July 20

When mathematicians speak of "Zermelo–Fraenkel set theory", do they mean ZFC or ZF?

I had always assumed that mathematicians mean ZFC when they use the phrase "Zermelo–Fraenkel set theory". Historically, the axiom of choice (the "C" in "ZFC") was part of Ernst Zermelo's earliest forms of an axiomatization of set theory (see Zermelo set theory), and it was the one axiom that caused the most debate, making it perhaps the most prominent part of Zermelo's axiomatization. This is why I myself had assumed that the phrase "Zermelo–Fraenkel set theory" always meant ZFC. However, some time ago, when editing Dedekind-infinite set, I came to doubt this. Now I wonder: when mathematicians speak of "Zermelo–Fraenkel set theory", do they conventionally mean ZFC or ZF ("ZFC minus C") or even some other variant? Or do mathematicians avoid to talk about "Zermelo–Fraenkel set theory" at all, directly using the unambiguous terms ZFC and ZF and so on instead?

BTW: Just for fun, I have looked at the way the phrase "Zermelo–Fraenkel set theory" is used on wikipedia, checking some of the pages that link to the article with the title "Zermelo–Fraenkel set theory".

My results

Some articles refer to ZFC as "Zermelo–Fraenkel set theory with the axiom of choice included" and others as "Zermelo–Fraenkel set theory with the axiom of choice added".

Some articles carefully avoid this issue, for example:

Zermelo–Fraenkel set theory starts with: "In mathematics, Zermelo–Fraenkel set theory with the axiom of choice, named after mathematicians Ernst Zermelo and Abraham Fraenkel and commonly abbreviated ZFC,…" (I think the part "named after mathematicians Ernst Zermelo and Abraham Fraenkel and commonly abbreviated ZFC" may actually have been added by myself some time ago.)
Axiom of choice#Nomenclature ZF, AC, and ZFC lists ZF as "Zermelo–Fraenkel set theory omitting the Axiom of Choice", and ZFC as "Zermelo–Fraenkel set theory, extended to include the Axiom of Choice".
Gödel's completeness theorem#Relationship to the compactness theorem says: "Weak König's lemma is provable in ZF, the system of Zermelo–Fraenkel set theory without axiom of choice."
Boolean prime ideal theorem: "Although the various prime ideal theorems may appear simple and intuitive, they cannot be derived in general from the axioms of Zermelo–Fraenkel set theory without the axiom of choice (abbreviated ZF)."

These articles apparently use "Zermelo–Fraenkel set theory" to mean ZFC:

Set theory#Axiomatic set theory says: "This includes the most common axiomatic set theory, Zermelo–Fraenkel set theory (ZFC), which includes the axiom of choice."
Von Neumann universe: "This collection, which is formalized by Zermelo–Fraenkel set theory (ZFC), is often used to provide an interpretation or motivation of the axioms of ZFC."
Constructive set theory: "It can look very much like classical mathematics done on the most common foundations, namely the Zermelo–Fraenkel axioms (ZFC)."

And these appear to use "Zermelo–Fraenkel set theory" to mean ZF:

Zermelo set theory#Connection with standard set theory starts with: "The most widely used and accepted set theory is known as ZFC, which consists of Zermelo–Fraenkel set theory with the addition of the axiom of choice."
Well-ordering theorem: "The well-ordering theorem is equivalent to the axiom of choice, in the sense that either one together with the Zermelo–Fraenkel axioms is sufficient to prove the other, in first order logic."
Zorn's lemma: "Zorn's lemma is equivalent to the well-ordering theorem and the axiom of choice, in the sense that any one of them, together with the Zermelo–Fraenkel axioms of set theory, is sufficient to prove the others."
Ordinal number: "The equivalence class is too large to be a set in the usual Zermelo–Fraenkel (ZF) formalization of set theory."
Continuum hypothesis#Independence from ZFC says: "Kurt Gödel showed in 1940 that the continuum hypothesis (CH for short) cannot be disproved from the standard Zermelo–Fraenkel set theory (ZF), even if the axiom of choice is adopted (ZFC)."
The lede of Dedekind-infinite set says: "In the early twentieth century Zermelo–Fraenkel set theory (ZF), today the most commonly used form of axiomatic set theory…")
First-order logic says: "Mathematical theories, such as number theory and set theory, have been formalized into first-order axiom schemata such as Peano arithmetic and Zermelo–Fraenkel set theory (ZF) respectively."
Russell's paradox says: "Zermelo's axioms went well beyond Frege's axioms of extensionality and unlimited set abstraction, and evolved into the now-canonical Zermelo–Fraenkel set theory (ZF)."
Kurt Gödel: "Gödel showed that both the axiom of choice (AC) and the generalized continuum hypothesis (GCH) are true in the constructible universe, and therefore must be consistent with the Zermelo–Fraenkel axioms for set theory (ZF)."
Paul Cohen (mathematician): "Cohen is noted for developing a mathematical technique called forcing, which he used to prove that neither the continuum hypothesis (CH), nor the axiom of choice, can be proved from the standard Zermelo–Fraenkel axioms (ZF) of set theory."
Independence (mathematical logic): "Many interesting statements in set theory are independent of Zermelo-Fraenkel set theory (ZF)."

– Tobias Bergemann (talk) 10:26, 20 July 2014 (UTC)[reply]

(Partially) answering my own question: Apparently I am somewhat mistaken in my assumption that ZF universally refers to "ZFC minus C". Indeed, most works on set theory appear to use this convention (e.g. the canonical works by Jech or Kunen or Levy or Smullyan/Fitting or Fraenkel/Bar-Hillel/Levy/van Dalen, and of course those books that specifically talk about the axiom of choice like Gregory H. Moore's "Zermelo's Axiom of Choice"). So I think it is safe to say that this use is conventional. But: I also found a few books on set theory that explicitly use "ZF" to refer to the axioms of Zermelo–Fraenkel set theory with the axiom of choice included (e.g. Cohens "Set Theory and the Continuum Hypothesis", and Mary Tiles's "The Philosophy of Set Theory"), at least if I'm not mistaken. However, they are certainly a tiny minority.

And the term "Zermelo–Fraenkel set theory" indeed appears to be ambiguous, with some authors including the axiom of choice and others not. On the one hand, for example, Keith Devlin's "The Joy of Sets" says (on p. 45): "The theory whose Axioms are 1–8 above is usually denoted by ZF. If we add Axiom 9, we denote the resulting theory by ZFC. This is at slight variance with the fact that 'Zermelo–Fraenkel set theory' has all nine axioms as its basic assumptions, but the nomenclature is now standard." On the other hand, Azriel Levy's "Basic Set Theory" says (chapter 5.24, p. 23): "The system consisting of the axioms of extensionality, union, power-set, replacement, infinity, and foundation is called the Zermelo–Fraenkel set theory and is denoted by ZF."

I am somewhat confused as several authors explicity or implicitly claim that the axiom of choice was added later to Zermelo–Fraenkel set theory. E.g. Smullyan/Fitting (Part I, Chapter 1 §9 "Zermelos set theory") talk about a "Zermelo set theory" without the axiom of choice (basically ZFC without choice, substition and replacement), then present substition and replacement (the additions of Fraenkel and, independently, Skolem) and then refer to the resulting system as "Zermelo–Fraenkel set theory" (ZF), without even mentioning choice, and (at least to me) implicitly suggesting that the axiom of choice was a much later addition to the axiomatization of set theory when it was in fact Axiom VI in Zermelo's 1908 article "Untersuchungen über die Grundlagen der Mengenlehre. I". (And, of course, Zermelo explicitly used choice in his 1904 proof of the well-ordering theorem, and he explicitly formulated and used the choice principle again in his modified proof of the well-ordering theorem presented in his 1908 article "Neuer Beweis für die Möglichkeit einer Wohlordnung".)

At least the meaning of "ZFC" is unambiguous.

– Tobias Bergemann (talk) 13:22, 20 July 2014 (UTC)[reply]

I have heard some vague claims that Zermelo's original theory, as he conceived it, was sort of implicitly in second-order logic, which could be part of the confusion. If you take Z to have full second-order separation and consider it as a theory in second-order logic, then it's categorical up to the first inaccessible cardinal. (More precisely, the only thing that can differ between two models is their height.) So if you think AC is true, then you think that all models of Z in this sense satisfy AC, so it maybe doesn't seem so important whether it is explicitly included or not.

All these ZF-like set theories have the same "picture" of the universe of sets (namely the von Neumann universe). Workers who study theories with a significantly different picture of the universe (like New Foundations) are sometimes heard to use the term "ZF" to refer, not to a precise formal theory, but to all set theory that views sets in that standard way. --Trovatore (talk) 05:23, 21 July 2014 (UTC)[reply]

Great circles through antipodal points

Is there any discontinuity-free procedure for choosing a great circle through any given pair of antipodal points on Earth -- i.e. so that the chosen great circle never suddenly jumps with a very small movement of the antipodal points? I feel that this may be impossible, but I'm not certain. Assume that the Earth is a perfect sphere. 86.167.19.47 (talk) 19:49, 20 July 2014 (UTC)[reply]

If your conditions are met - a pair of antipodal points on a perfect sphere - there are an infinity of identical great circles through them. Roger (Dodger67) (talk) 20:13, 20 July 2014 (UTC)[reply]

I am aware of that (though they are not "identical"). I want to know whether there is a systematic procedure for choosing ONE great circle through each pair of antipodal points so that no discontinuities occur. For example, a simple procedure would be to choose the great circle that also passes through the North Pole. However, this creates a discontinuity when the antipodal points are the poles. 86.167.19.47 (talk) 20:34, 20 July 2014 (UTC)[reply]

How about this:

1) For the initial pair of poles, choose any third random point on the equator between the poles. Use that point and the two poles to construct the great circle.

2) Find the midpoint of one of the two semi-great circles between the poles (choose the semi-great circle which is closest to the last midpoint), and use that as the third point for the next great circle. Repeat step 2 indefinitely. StuRat (talk) 20:32, 20 July 2014 (UTC)[reply]

Given any two antipodal points, the procedure, or algorithm, needs to independently return a unique great circle without any reference to any previous history. Also, I am interested in solving this in an exact mathematical sense, not anything related to precision of variables. 86.167.19.47 (talk) 20:37, 20 July 2014 (UTC)[reply]

Then no, I don't think that's possible. Note that if you aren't worried about precision of variables, then the chance of either point being exactly coincident with your random 3rd point is zero, however. StuRat (talk) 20:44, 20 July 2014 (UTC)[reply]

However, that is no consolation to me. Except in pretty exotic cases, mathematical discontinuities do occur only at single points. 86.167.19.47 (talk) 20:49, 20 July 2014 (UTC)[reply]

Also, you probably won't be interested in this "trick", but if you can limit the selection of the antipodal points to a certain set, say if they can only specify each coord up to a trillionth of a degree, then just choosing a random 3rd point not on that grid of points would work. StuRat (talk) 20:44, 20 July 2014 (UTC)[reply]

Right, I'm only interested in solutions that are completely theoretically valid. I'm not looking for practical fudges or kludges. 86.167.19.47 (talk) 20:54, 20 July 2014 (UTC)[reply]

Then I think you're out of luck. May I ask what this is for ? StuRat (talk) 21:00, 20 July 2014 (UTC)[reply]

It is just for interest. 86.167.19.47 (talk) 22:43, 20 July 2014 (UTC)[reply]

This isn't wholly rigorous, but:

Assume that the method you're looking for exists.
This means that for any point A on the sphere, a function f produces a circle f(A). (We have f(A) = f(B) iff B=A or B=-A.)
Let's define a vector v along f(A) from A, such that v also varies continuously. (This appears to follow from the definition of f, that it is continuous and unique.)
v for all A is a vector field.
The hairy ball theorem applies. Therefore the initial assumption is false - v cannot be continuous, so either v's continuity doesn't follow from f's, or f isn't continuous. If anyone can show that f can be continuous without v being so as well, I'll be surprised. AlexTiefling (talk) 23:15, 20 July 2014 (UTC)[reply]

I like it! I must be careful not to confuse with the hairy balls theorem... 86.167.19.47 (talk) 00:29, 21 July 2014 (UTC)[reply]

One clarification: If f is continuous, one could obviously choose a discontinuous v (but there's no reason to); but if you can't make v continuous, that's either because there's something peculiar about the choice of v (that I've missed) or else f isn't continuous either. Hence why I say this isn't rigorous. I can easily see an argument that if f(A) isn't unique for each A, v can't be guaranteed continuous, but not if f(A) is unique. (For example - distinguish the hemispheres into which f(A) divides the sphere; require v to be a unit along f(A) from A with a specified hemisphere on its left. If f(A) is continuous, the identity of the chosen hemisphere should be too.) AlexTiefling (talk) 11:38, 21 July 2014 (UTC)[reply]

The fundamental polygon of the projective plane

This problem is clearly what one might call a variant of the hairy ball theorem on a different space: the elliptic plane, topologically the real projective plane, since antipodal points are identified, and we are trying to find a field of tangent lines over this manifold that does not have topological discontinuities. A tangent line is equivalent to the pair of oppositely pointing units vectors, which is to say we can tolerate a 180° flip when we get back to the point via any path in the projective plane. Or: can one draw a set of non-crossing lines that covers the real projective plane? Looking at the fundamental polygon of the real projective projective plane, if we rule it with vertical lines, it would suggest that it might be possible. There is an inherent 180° shift in the vector field, but that is accommodated by the lines being non-directed. If one examines this more closely, this corresponds to the fields making two U-turns around the tips of the arrow B, creating two singularities in the field. This corresponds to two pairs of antipodes where small circular path around them will result in the vector direction spinning 180°, or if one shrinks them to one point, this becomes one pair of antipodes where there will be a spin of 360° (corresponding to diagonal lines in the diagram). While this is not rigorous, it does lend support to AlexTiefling's argument from a topological perspective. —Quondum 17:54, 21 July 2014 (UTC)[reply]

I think I agree. But is there necessarily a mapping from any ruling (even partial) of the fundamental polygon of the real projective projective plane onto a set of great circles on the corresponding (hemi)sphere? Projective geometry's not my strong point, I'm afraid. AlexTiefling (talk) 08:57, 22 July 2014 (UTC)[reply]

No, there isn't. The projective plane is a topological construct, so what is a straight line in the fundamental polygon isn't necessarily anything very nice in the projective plane (or in the sphere when taking preimages). For this, you need more structure than a topology. YohanN7 (talk) 14:14, 22 July 2014 (UTC)[reply]

That is to say, the concepts "straight line" and "great circle" both lack meaning in this topological setting. YohanN7 (talk) 14:37, 22 July 2014 (UTC)[reply]

Agreed, regarded as a topology, though this doesn't detract from the argument. (Actually, the real projective plane does have straight lines, and they do map to great circles in our case, but we are not interested in these, only in the topology. These lines are not straight in the diagram here, though.) If the real projective plane cannot be ruled with wavy lines that are locally parallel everywhere, this is a proof that there is no procedure as outlined in the original question. —Quondum 15:58, 22 July 2014 (UTC)[reply]

How do you define "straight" in the real projective plane? Its elements, in one construction, are straight lines but these are points in the space, not lines, so you mean something else obviously. I guess you could probably use the projection from the sphere (using its standard embedding in

R 3

) to define straight in one incarnation of the real projective plane. YohanN7 (talk) 16:21, 22 July 2014 (UTC)[reply]

The geometry of the projective plane is off-topic here. —Quondum 18:42, 22 July 2014 (UTC)[reply]

Don't bring it up then. YohanN7 (talk) 19:27, 22 July 2014 (UTC)[reply]

Alex, can you please further explain your uniqueness condition. As I understand it, nothing in the definition of the original problem explicitly prevents f(A) = f(B) where A ≠ B and A ≠ -B, but that the uniqueness is allowing you some method of creating a continuous v from a continuous f. I don't understand that method -- your "distinguish the hemispheres" example -- and I don't see how uniqueness comes into play.

Trying to map the tangent lines of a continuous f() to a vector field, it seems to me that given an A, and choosing an arbitrary hemisphere of f(A) to be on the left of the vector derived from f(A), and calling the point in the center of that hemisphere X, the continuous nature of f means that there is a neighborhood of A in which for all elements C of that neighborhood, f(C) will be sufficiently far from X that we can assign a vector where the hemisphere containing X is to its left, and that the resulting vector field in that neighborhood will be continuous. Does that work, and if f is uniformly continuous, should it be possible to paste together a continuous vector field for the entire sphere? But what if f is continuous, but not uniformly so? -- ToE 14:29, 22 July 2014 (UTC)[reply]

I was worried that I was out of my depth, and it looks like I might be. My attempt to clear up as much as I can:

You're right about more than two points being capable of producing the same circle; however, the uniqueness is not essential to the argument. I was over-specifying. The original problem only requires that f(A) is uniquely determined by choice of A, not that it is actually unique.
'Distinguishing hemispheres' was simply my way to try and visualise consistently assigning a vector to each A. But yes, I imagined that if f were continuous, small changes in A would result in small changes to v, and my original draft included identifying a point equivalent to your X - such that small changes to A would result in equivalent-sized small changes to X.
Your argument about neighbourhoods of A makes sense to me. But this is a proof by contradiction...
I'm weak on uniform continuity, but I'm not sure it's essential to the argument. Let's say I think my argument is true for uniform continuity, and might well be otherwise.
The argument about neighbourhoods of A implies that the function can be locally continuous, but the hairy ball theorem implies that there must be point discontinuities somewhere in the global function.

Does that make any sort of sense? AlexTiefling (talk) 14:58, 22 July 2014 (UTC)[reply]

I haven't followed the reasoning closely ITT, but it seems like the Hairy ball theorem is incorrectly interpreted. It says there is no nonvanishing continuous vector field on the two-sphere. Continuous vector fields do exist. YohanN7 (talk) 14:44, 22 July 2014 (UTC)[reply]

The definition of v(A) that I was using assumed it has unit magnitude everywhere. If it's capable of having zero magnitude, it can't correspond to the orientation of a uniquely specified circle through A. AlexTiefling (talk) 15:00, 22 July 2014 (UTC)[reply]

I think that we are on the same page, Alex. Clearly, any combing of a hairy ball would yield the f() asked for. But since no such combing exists, if we can show that a continuous f() can be used to construct a continuous unit vector field, then such an f() does not exist.

S² is compact, so I suppose that any continuous f() is uniformly continuous, so we should be able to piece together a continuous unit vector field by covering the sphere in finitely many neighborhoods. So yes, I think that the hairy ball theorem does preclude a continuous f(). -- ToE 15:39, 22 July 2014 (UTC) (I *know* that I'm out of my depth here, but this sounds reasonable to me.)[reply]

Fantastic question! In my mind, the question is precisely asking for a global section of the projectivised tangent bundle of $\mathbb {RP} ^{2}$ (for each pair of antipodal points, choose a direction (tangent line at either point), through which the great circle then goes). Nonexistence of such a section does not, to my eyes, immediately follow from the hairy ball theorem. Crucially, there is a difference between the following two circle bundles on the sphere $S^{2}$ :

the unit circle bundle associated to the tangent bundle, which has no global sections precisely by the hairy ball theorem, and
the projective bundle associated to the tangent bundle, which is the quotient of the previous circle bundle by the antipodal map (on the fibers).

The first bundle has Euler class $2\in H^{2}(S^{2},\mathbb {Z} )$ , as the Euler characteristic of $S^{2}$ is 2. This precludes the existence of a global section (essentially, this is the proof of the hairy ball theorem via Poincaré—Hopf). The second bundle will then have attached Euler class 1 and the same argument applies to rule out a global section. (Note that I don't have to worry about twisted Euler classes as $S^{2}$ is simply connected so every vector bundle is orientable). This was for the projectivised tangent bundle of $S^{2}$ , but as I noted you really care about the projectivised tangent bundle of $\mathbb {RP} ^{2}$ , in which case we do need to worry about the orientation sheaf to get a twisted Euler class, but the same argument should apply. --137.73.15.6 (talk) 16:12, 22 July 2014 (UTC)[reply]

Actually, here's another argument: if such a global section of the projectivised tangent bundle of $\mathbb {RP} ^{2}$ existed, we could lift it by the covering map to $S^{2}$ to get a global section of the projectivised tangent bundle of $S^{2}$ . Now $S^{2}$ is simply connected, hence it is always possible to consistently choose orientations, turning the continuous choice of tangent line inside the tangent space at each point (which is precisely what a global section of the projectivised tangent bundle is) into a choice of unit length tangent vector, and hence a continuous nowhere zero vector field on the sphere. This contradicts the hairy ball theorem, hence it is impossible to choose great circles as desired. --137.73.15.6 (talk) 16:03, 23 July 2014 (UTC)[reply]

A simpler argument: such a section if it existed would define a pair of sections of the unit tangent bundle: a double cover of the sphere. The sphere, being simply connected, has only trivial covers. So the double cover is the trivial one, and the unit tangent bundle has a section. Sławomir Biały (talk) 17:41, 23 July 2014 (UTC)[reply]