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I'm confused about how to apply the separation axiom.

1) How does it follow that for a set $A$, $\{x\in A:x\not\in x\}$ is a set?

(a) Before I even start, what is $\{x\in A:x\not\in x\}$ in the axiomatic approach? In the usual life it is the set of all elements of the set $A$ such that $x\notin x$. Should I still treat it like this, or does it have some more formal definition? And how to formally justify that it is "the same" as (does it mean "equal to"?) the class $\{x: x\in A\land x\notin x\}$?

(b) By the separation axiom, there exists a set $B$ such that $x\in B\iff x\in A\land x\notin x$. I suppose the next steps should be $x\in \{x\in A: x\notin x\}$ iff $x\in A\land x\notin x$ iff $x\in B$, and so by the extensionality axiom $B=\{x\in A:x\notin x\}$, so the RHS is a set since the LHS is known to be a set. Is that the right reasoning? I'm not sure about the iff in bold above. In ordinary reasoning I have no doubt about that, but since I'm working in formal axiomatic system, I'm not sure what exactly it follows from. (If I knew that $\{x\in A:x\not\in x\}$ is equal to the class $\{x: x\in A\land x\notin x\}$, I guess that would clarify things.)

2) How does it follow that $\{x:x\in A\}=A$? Again, the separation axiom says that there is a set $B$ such that $x\in B\iff x\in A$. By extensionality, $A=B$. But I don't see how to deduce what's needed.

Andrés E. Caicedo
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user557
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  • If the axiomatic set theory you're working in is ZFC, then {$x∈A:x∉x$} is just $A$ because in ZFC no set is an element of itself. – alex811 Mar 17 '19 at 19:41

1 Answers1

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Set builder notation isn't actually part of the language of set theory; it's just a useful system of abbreviations. So reasoning about it is inherently informal, because it's actually not there fundamentally. I think the confusion you're having mostly results from trying to treat set builder notation as more fundamental than it actually is.

Generally we read "$\{x:\varphi(x)\}$" as referring to the set $A$ satisfying $$\forall x(x\in A\iff \varphi(x)).$$ Note that this involves an implicit claim - namely, that such an $A$ exists in the first place. (There's also a uniqueness claim, but by extensionality if one such $A$ exists then there is exactly one such $A$; so it's really only the existence claim which matters.)

The notation "$\{x\in A:\varphi(x)\}$" refers to the same thing as "$\{x: x\in A\wedge\varphi(x)\}$." However, it comes with a useful detail: the "$x\in A$" clause at the beginning guarantees existence via the separation axiom. So in some sense - once we accept the separation axiom (and extensionality) - the notation "$\{x\in A:\varphi(x)\}$" doesn't make any implicit assumptions. In particular:

For every formula $\varphi(x,y_1,...,y_n)$, ZF proves that for every set $A$ and all sets $a_1,...,a_n$ (the "parameters") there is a unique set $X$ such that $$\forall x(x\in X\iff x\in A\wedge \varphi(x; a_1,...,a_n)).$$

This set $X$ is denoted "$\{x\in A: \varphi(x, a_1,...,a_n)\}$."

Noah Schweber
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  • But the notation ${x: \varphi(x)}$ can also refer to proper classes, not necessarily sets, right? – user557 Mar 17 '19 at 19:48
  • @logic Yes, hence my "generally." In my experience that usage is somewhat rarer, though. – Noah Schweber Mar 17 '19 at 19:49
  • @logic I think the important takeaway though is that the notation $$"{x\in A: \varphi(x)}"$$ (and its "with-parameters" version) is completely unproblematic: it has built directly into it the proof that it refers to a unique set. – Noah Schweber Mar 17 '19 at 19:52
  • As far as classes go, I personally find it easier (at least, when working in ZF as opposed to something like NBG) to think of classes as formulas with parameters. Notation like "$x\in C$" for $C$ a class is shorthand for "$x$ has property $C$" - which would more accurately but maybe less clearly be denoted "$C(x)$." Some classes have corresponding sets - that is, for some but not all classes $C$ there is a set $X$ such that $\forall x(x\in X\iff C(x))$. In that case we conflate $C$ with that set $X$. – Noah Schweber Mar 17 '19 at 19:57
  • Incidentally, if you're familiar with model theory, a class is really just a definable subset of the model of ZF we're working in (the overloading of the term "set" is unfortunate here, but oh well); see e.g. my description here. The correspondence between ZF-language and model theory-language is: "set" is "element of the model" and "class" is "definable subset of the model." (cont'd) – Noah Schweber Mar 17 '19 at 20:01
  • Now suppose $M$ is a model of ZF. Regularity + replacement give us a way to assign to every element $x$ of $M$ an $M$-ordinal (that is, an element of $M$ which $M$ thinks is an ordinal) called the "rank" of $x$ (= $rk(x)$). This lets us break the classes (= definable subsets) of $M$ into two pieces. Suppose $\varphi$ is a formula (with parameters from $M$). If there is some $\alpha\in Ord^M$ such that $M\models \forall x(\varphi(x)\implies rk(x)<\alpha)$, then $\varphi$ "is a set" in the sense that there is a (unique!) $m\in M$ such that $M\models\forall x(x\in m\iff \varphi(x))$ (cont'd) ... – Noah Schweber Mar 17 '19 at 20:04
  • Conversely, if there is some $m\in M$ such that $M\models\forall x(x\in m\iff\varphi(x))$, then there is some $\alpha\in Ord^M$ such that $M\models\forall x(\varphi(x)\implies rk(x)<\alpha)$. Thinking along these lines (and using the Mostowski collapse) in fact gives us a "size" condition for being a set: if $\varphi$ is a formula with parameters from $M$, then either $\varphi$ is a set in $M$ in the sense above or there is an $M$-definable surjection from $\varphi^M$ to $Ord^M$ - roughly, a class is proper iff it surjects onto the ordinals. – Noah Schweber Mar 17 '19 at 20:07
  • Thanks for the detailed comments. Your notation for classes is the notation Moschovakis uses. But frankly the derivation that ${x\in A: \varphi(x)}$ is a set from the separation axiom still remains murky for me. Originally we don't know whether this is a set, so first of all we need to interpret that "object" somehow. Do we just say that by definition, that is "the collection of all sets and atoms that are elements of $A$ and satisfy $\varphi(x)$" and also that by definition ${x: x\in A\land \varphi(x)}$ is a synonymous notation for it? If so then my proof in the question should go through – user557 Mar 17 '19 at 20:08
  • I'll try to look into your interpretation about definable subsets, but I'm not very familiar with it... – user557 Mar 17 '19 at 20:09
  • @logic Yes, that's exactly right. Again, set-builder notation is an abbreviation - you shouldn't treat it too formally, since it doesn't actually exist in the formal syntax. (And you may reasonably ignore the model theory stuff if you're not familiar with some model theory already, it might only add confusion.) – Noah Schweber Mar 17 '19 at 20:10
  • Basically, "${x\in A:x=x}=A$" is shorthand for "$A$ is the unique set $X$ such that $\forall x(x\in X\iff (x\in A\wedge x=x)$," and this sentence should be very easy to prove (that $A$ has that property follows immediately from the laws of first-order logic itself, while uniqueness uses the extensionality axiom). – Noah Schweber Mar 17 '19 at 20:13