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My textbook claims (without proof, which I'm trying to see for myself) that for every class $C$ there exists a unique class HC such that

$$\forall x (x\in HC \iff (x \in C \wedge \forall y \in x, y\in HC)$$

And the text calls HC the class of "hereditarily C sets".

This is in $ZF^{--}_F$.

First, is this a typo, is it mean to say "hereditary C sets"?

According to wikipedia (https://en.wikipedia.org/wiki/Hereditary_set) in pure set theory all sets are hereditary because the only property we describe as hereditary is that of being a set, and our only objects of discourse are sets. Is the answer here obvious then by taking HC to be the universe class $V=\{x| x=x\}$?

Mike
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    As soon as $C$ is not the universal class, neither is $HC$ (because it is a subclass of $C$). And using the adverbial form "heriditarily" is not a typo because we are not only investigating hereditary sets, but stes that are in $C$ and are hereditarily in $C$ (hereditarily finite sets, hereditarily countable sets, hereditarily whatever). – Hagen von Eitzen Feb 22 '19 at 20:16
  • @HagenvonEitzen Thanks, that makes sense. I'll take another crack at it. – Mike Feb 22 '19 at 20:43
  • I’m not sure what $ZF^{ - -}_F$ is but I presume it is enough to define the transitive closure of a set. Then HC will be defined by the formula that says all sets in the transitive closure and the set itself are in C. – spaceisdarkgreen Feb 22 '19 at 22:22
  • @spaceisdarkgreen I was just coming here to propose that as the answer. Thanks! If you write that as the answer I'll accept it. – Mike Feb 23 '19 at 01:03
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    @Mike Sure, but first could you tell me what $ZF^{--}_F$ is? – spaceisdarkgreen Feb 23 '19 at 01:04
  • @spaceisdarkgreen In this textbook (I'd reference it if it were published) $ZF^{--}$ stands for set theory with the existence, pairing, union, extensionality, separation and replacement axioms. And $ZF^{--}_F$ also throws in the foundation axiom. – Mike Feb 23 '19 at 01:11
  • @Mike Thanks. I don't think foundation is necessary here (I could be wrong about this), but I think the lack of infinity complicates things a bit. – spaceisdarkgreen Feb 23 '19 at 01:20
  • @Mike In fact this is a bit of a head-scratcher to get right (I might just be confused for dumb reasons). I don't have a lot of time right now, so I'll write an answer later, but would invite anyone, including yourself to intervene in the meantime. – spaceisdarkgreen Feb 23 '19 at 01:31

1 Answers1

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EDIT Thanks to Eric Wofsey for correcting some errors I made in the original answer.

As discussed in the comments, we can use the transitive closure to make the definition explicit, defining HC as $$ x\in HC \iff x\in C\land (\forall y \in \operatorname{trcl}(x))(y\in C).$$ The usual definition of the transitive closure is by a recursion of length $\omega$ $$ G(x,0) = x\\G(x,n+1) = \cup G(x,n)\\ \operatorname{trcl}(x) = \bigcup_{n\in\omega} G(x,n)$$

The fact that we don't have infinity causes an issue here. The recursion theorem still works and produces the class function $G,$ but the union in the last line does not necessarily produce a set.

But as far as I can tell, we only need $\operatorname{trcl}(x)$ as a class to make this work, so the definition can be carried out in the theory your book calls ZF$^{--}$ (i.e. foundation isn't required).

(Note in my original answer I thought that foundation could also be used to resolve this via $\in$-recursion. It turns out that was wrong and that foundation has some subtleties in the absence of infinity. See the comments.)

spaceisdarkgreen
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  • Maybe I am missing something but I think there is a real issue here. To use $\in$-recursion, you need to already know that transitive closures exist as sets (see for instance the issue that comes up in this answer). – Eric Wofsey Feb 23 '19 at 22:21
  • However, to define $HC$, it is enough to have transitive closures just as classes, and your construction by recursion on $\omega$ works for that. Note that this works even in the absence of Infinity since we just need to be able to define the transitive closure as a class. – Eric Wofsey Feb 23 '19 at 22:23
  • @EricWofsey Regarding your first, I suspect you aren't missing something and will look into it. Regarding your second, yes that's the upshot of my second to last paragraph (which if your first comment is right, should probably be the entire answer). – spaceisdarkgreen Feb 23 '19 at 23:29
  • @EricWofsey Maybe I misremembered the 'definition of trcl via $\in$-recursion'. Poked around and found JDH here https://math.stackexchange.com/questions/2284992/existence-of-the-transitive-closure-for-sets This seems to prove the existence of a transitive closure from foundation w/o infinity. Does this resolve your concerns? – spaceisdarkgreen Feb 24 '19 at 00:06
  • JDH's argument there is wrong. In particular, the step "If there is a set without a transitive closure, then there is an $\in$-minimal such set" is wrong because Foundation only applies to sets, so you can't apply it to the class of all sets without a transitive closure. His second argument using the cumulative hierarchy is also circular, since the existence of transitive closures is usually used in order to prove every set is in some level of the cumulative hierarchy. – Eric Wofsey Feb 24 '19 at 02:34
  • @EricWofsey Okay, thank you, I should have thought about your point more carefully rather than believing the first corroborating evidence I found. Will correct the answer. So it's fair to say that $\in$-induction, the trcl function, and the rank function, $V=\bigcup_\alpha V_\alpha,$ all require the axiom of infinity? Seems counterintuitive since foundation implies you never actually have an infinite depth, but it also seems right that situations like this where we try to replace a construction of a set by a recursion of length $\omega$ with a recursion on rank would be too good to be true. – spaceisdarkgreen Feb 24 '19 at 03:14
  • @EricWofsey (I suppose you need either the rank function or power set to define $V_\alpha$ in the first place.) – spaceisdarkgreen Feb 24 '19 at 03:21
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    Yes, I believe that's correct. Note that when Infinity is dropped from ZFC, it is common to replace Foundation with an axiom schema for $\in$-induction, for exactly this reason. See this great answer for some discussion. – Eric Wofsey Feb 24 '19 at 03:23