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Background:

Here's the original text from the book "A book of Set Theory" by Charles Pinter.

... Thus, a capital letter, such as A, may denote either an element or a class which is not an element, but a lower-case letter, such as x, may denote only an element. Intuitively, two classes should be called equal if they are elements of the same classes.

1.9 Definition: Let $A$ and $B$ be classes. We define $A = B$ to mean that every class that has $A$ as an element also has $B$ as an element, and vice-versa

$A=B\iff \forall X\Bigl( (A\in X\to B\in X)\wedge (B\in X\to A\in X)\Bigr)$

We have defined two classes to be equal if and only if they are members of the same class. Informally, we may therefore think of them as interchangeable. Equal classes have another property. If $A$ and $B$ are equal, we expect them to have the same elements. This property is stated as our first axiom.

A1. $A=B\iff \forall x \Bigl( (x\in A\to x\in B)\wedge (x\in B\to x\in A)\Bigr)$.

This axiom is sometimes called the Axiom of Extent.

Question

My question is why do we not use the Axiom of Extent as the definition of class equality?

It seems to me that we could just as well use the Axiom of Extent as the definition of equality of classes? In fact, it is more intuitive for me to understand equality from the perspective of the elements the classes contain.

The given class equality definition from the perspective "being elements of the same classes" seems to be only applicable to classes that can be an element. But it also previously in the introduction of the book said Von Neumann's system allow for classes that cannot be elements of any other classes. Following that given definition, wouldn't we conclude that all classes that are not elements of any other classes are equal?

Update

  • Mr Pinter did mention in the introduction that "In this text, we shall use a slightly modified form of von Neumann's system of axioms."

  • I have just started on the book and haven't reached far enough, but there was this phrase preceding the quoted section in my question.

...in order to avoid logical paradoxes, we have to distinguish between two kinds of classes - those that are called sets and those that are called proper classes. We shall say no more about this distinction now, but shall return to it later.

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  • $\begingroup$ Shouldn't A1 also have a $\forall x$ in it? $\endgroup$ Nov 10, 2023 at 19:02
  • $\begingroup$ @Arturo Magidin, yes you are right, sorry about that, I'll fix it. $\endgroup$ Nov 10, 2023 at 19:04

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Yes, you can reverse these, so that A1 is the definition and 1.9 is the axiom (although then you probably would number them differently). This seems more usual to me in my experience with axiomatic set theory.

Another common approach is to take equality as an undefined notion from logic satisfying some standard axioms (which would not be considered axioms of set theory, but just part of first-order logic-with-equality); then A1 is an axiom again, but it's still treated as more fundamental, because the additional logical properties of equality will now allow you (together with other axioms of set theory, principally the Axiom of Pairing) to prove 1.9. (And in fact, you then only have to state A1 in the $\Leftarrow$ direction, since the $\Rightarrow$ direction can also be proved.)

Well, except that you can't actually prove 1.9 for all classes, only for elements! (which are the only things that the Axiom of Pairing applies to). Indeed, for classes, 1.9 just seems straight-up incorrect. Maybe Pinter is doing something very unusual here where there is only one proper class (a class that's not an element), but that's not von Neumann's class theory, where we want any first-order statement (well, at least a broad swath of them) to define a class (the principle of Class Comprehension). So this really looks like an error to me, for the reason you say.

This is another reason to treat A1 (whether it's a definition of equality, or an axiom for an undefined equality) as the more fundamental concept. If A1 is a definition, then we really only need the $\Rightarrow$ direction of 1.9 as an axiom, since the $\Leftarrow$ direction can be proved (using Pairing) for elements and is incorrect for proper classes. And if A1 is an axiom (for an undefined equality), then the $\Rightarrow$ direction of 1.9 follows from the logical axioms of equality, while the $\Leftarrow$ direction again follows (using Pairing) for elements and is incorrect for proper classes. Only if you're doing a pure set theory with no proper classes, can you even attempt using both directions of 1.9, as either a definition or an axiom.

ETA: I see that Pinter's book is free online as a university-hosted PDF. After looking through it, I'm more convinced than before that this was a mix-up on Pinter's part; A1 should be the definition, and 1.9 (either only the $\Rightarrow$ direction, or stated only for elements) should be the axiom. Also, I should have realized that Pinter wouldn't need the Axiom of Pairing yet, since you can use Class Comprehension instead (what Pinter calls the axiom of class construction) when working with classes, but the key point about that is that you have some axiom that allows you to create $\{x : x = A\}$, which then allows you to prove the $\Leftarrow$ direction of 1.9 for elements.

PS: Here's a careful proof that Pinter's axioms are contradictory. First construct the universal class $ \mathcal U $ and the empty class $ \varnothing $ (which Pinter does shortly after A2). Then use the statement $ x \ne \varnothing $ to construct the class $ \{ x : x \ne \varnothing \} $ (where $ X \ne Y $ is defined to mean $ \neg ( X = Y ) $), which I'll call $ \mathcal V $ for short. (The important thing is to have two proper classes that aren't equal; it doesn't have to be these two.) Pinter's Axiom A4 (without which, as he correctly notes, we can't prove that any elements exist at all) says that $ \varnothing $ is an element, so $ \varnothing \in \mathcal U $ and $ \varnothing \notin \mathcal V $ (because $ \varnothing = \varnothing $ trivially by Definition 1.9, or you could A1 here), so $ \mathcal U \ne \mathcal V $ by A1. Now, Pinter shows (by Russell's Paradox) that $ \mathcal U $ is not an element. But essentially the same argument proves that $ \mathcal V $ is not an element either. Or you could use the later axioms to construct (in turn) the sets $ \{ \varnothing \} $, $ \{ \mathcal V , \{ \varnothing \} \} $, and $ \bigcup \{ \mathcal V , \{ \varnothing \} \} $, and note that this last one is equal to $ \mathcal U $ (since every element is either equal or unequal to $ \varnothing $). In any case, neither $ \mathcal U $ nor $ \mathcal V $ is an element. So trivially by Definition 1.9, $ \mathcal U = \mathcal V $, which contradicts the conclusion above (using A1) that $ \mathcal U \ne \mathcal V $.

But if instead, you adopt A1 as the definition of equality and only the $ \Rightarrow $ direction of 1.9 as the Axiom of Extent, then the $ \Leftarrow $ direction of 1.9 can still be proved for elements; that is, $$ a = B \iff \forall X \, \Bigl ( ( a \in X \Rightarrow B \in X ) \wedge ( B \in X \Rightarrow a \in X ) \Bigr ) \text , $$ where I have a lowercase $ a $ (showing that this one must be an element). We already have the $ \Rightarrow $ direction (that's been adopted as an axiom), so to prove the $ \Leftarrow $ direction, let $ X $ be $ \{ a \} $ (that is $ \{ x : a = x \} $, the class constructed from the statement $ a = x $) and notice that $ a \in \{ a \} $ (because $ a $ is an element and $ a = a $ follows trivially from A1). So if $ a \in X \Rightarrow B \in X $, then this means that $ B \in \{ a \} $, and therefore $ a = B $, proving the left-hand side. (So we don't even need the second hypothesis on the right-hand side, giving an even stronger theorem.)

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  • $\begingroup$ Thanks @Toby Bartels for the wonderfully detailed answer. I would've upvoted had I enough reputation points. For fear of framing the question out of context since I'm just learning the subject, I'll update my post with 2 additional things. $\endgroup$ Nov 10, 2023 at 19:49
  • $\begingroup$ You're welcome! On those two additions: since Pinter's axiom of class construction allows arbitrary class-variables (not just element-variables) in the statements used to construct classes, it's really Morse–Kelley set/class theory instead of Gödel–von Neumann–Bernays set/class theory (which is fine). And I find it weird that the axioms are stated about ‘elements’ instead of ‘sets’, when Pinter defines the two terms in the same way; most people would say ‘sets’ (but I guess that's OK too). $\endgroup$ Nov 11, 2023 at 3:44
  • $\begingroup$ I added a postscript with the details of the contradiction in Pinter's axioms, just to make sure that we didn't miss anything. $\endgroup$ Nov 11, 2023 at 4:41
  • $\begingroup$ Thanks again Toby for the proof of your verbal explanation. I'm fully convinced. Now the following is a bit off topic but I'd really appreciate your guidance here. It's a pity though because as I am starting on the subject on my own, I found really few texts that are as approachable as Pinter's book. And after attempting a read on one of the lecture notes using ZFC, I find myself struggling with understanding the axioms of ZFC on a intuitive level, sure if I follow the axioms I can probably think hard and use them to prove something, but after chancing on Pinter's book (TBC) $\endgroup$ Nov 12, 2023 at 4:23
  • $\begingroup$ I found NBG's axioms more approachable intuitively. I come from EECS background, so having some kind of types in the system speaks to me. I wonder if you have any recommendation on any set theory book that is suitable for self-study. And on that note, would it make sense for me to proceed with Pinter's book overlooking this mishap? (I would mentally substitute your suggestion for the section quoted from Pinter of course) $\endgroup$ Nov 12, 2023 at 4:26

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