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It is well known that Replacement gives us that $\omega + \omega$ is a set (provable by a simple recursion). It is vaguely intuitive that this could not be done without -- whilst we can 'see' $\{\omega, \omega + 1, \omega + 2, \ldots \}$, there is no guarantee that this is a set (Burali-Forti might make us wary).

I was wondering where I might find a proof of this fact -- i.e. that in $\textsf{ZF}$ without replacement, we are unable to prove that $\omega + \omega$ is a set.

I've seen some basic techniques for stuff like this -- by considering 'hereditarily finite' sets (sets $X$ which are finite, and which have $x$ finite for all $x \in X$) we can show that from Empty Set, Extensionality, Pairs, Unions, and Comprehension we can't get Infinity; by changing 'finite' to 'countable' we can show that we can't get Powerset. (The proofs proceed by considering the classes of all hereditarily finite/countable sets, and showing the first collection of axioms is met, whilst the last is not -- so there is no $\omega$ or $\mathcal{P}X$, resp., in them.) Is the proof for this proof similar or do the techniques get more advanced?

Thanks for any help

aidangallagher4
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1 Answers1

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The proof of this result is exactly the same: we find an ordinal $\alpha$ such that $V_\alpha\models\mathsf{ZC}$ (= $\mathsf{ZFC}$ without replacement; although you didn't ask for it, I'm including choice because it gives a stronger result but doesn't cost us anything) but $\omega+\omega\not\in V_\alpha$.

Note that this latter condition is equivalent to $\omega+\omega\ge\alpha$; does this suggest a choice of $\alpha$ to look at? (Think about how Powerset and Infinity restrict our options here.)

EDIT: OK fine, "exactly" might be unfair here. You mentioned constructions using hereditary cardinality, namely $H_\omega$ (= hereditarily finite sets) and $H_{\omega_1}$ (= hereditarily countable sets); here we want to use the $V$-hierarchy instead of the $H$-hierarchy. However, modulo this difference the arguments are the same: determine which levels of the relevant hierarchy satisfy which axioms of the theory in question, and then amongst those levels pick one which avoids the ordinal in question.

Noah Schweber
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  • "Exactly the same" might be overselling the similiarity, since the proof the OP is aware of uses a hereditary restriction of cardinality rather than a restriction of rank. Even though $V_\omega$ consists of exactly the hereditarily finite sets, the correspondence doesn't continue after that. – Troposphere May 29 '21 at 19:33
  • I would guess $\omega$ -- so the proof would essentially be 'the sets of $V_{\omega}$ satisfy $\textsf{ZC}$, and here we don't have $\omega + \omega$, thus we need replacement to get it'? – aidangallagher4 May 29 '21 at 19:33
  • @aidangallagher4: $V_\omega$ doesn't quite cut it, since that is the hereditarily finite sets, so it fails the axiom of infinity. – Troposphere May 29 '21 at 19:34
  • True -- I always forget we need to go 'one up' (ie that $\alpha \notin V_{\alpha}$). Presumably then $V_{\omega + 1}$? – aidangallagher4 May 29 '21 at 19:36
  • @aidangallagher4: Not that either -- $\alpha$ cannot be a successor ordinal, or $V_\alpha$ would fail the axiom of Pairing (as well as Power Set). But now there's only one ordinal $\le \omega\cdot 2$ left to try :-) – Troposphere May 29 '21 at 19:37
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    Oh yeah I realised the powerset issue just as I clicked 'add comment' -- $V_{\omega + \omega}$! – aidangallagher4 May 29 '21 at 19:40
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    @aidangallagher4 Yup, it's $V_{\omega+\omega}$. The point is that Infinity holds in $V_\alpha$ exactly when $\alpha>\omega$, and Powerset holds in $V_\alpha$ exactly when $\alpha$ is a limit ordinal. – Noah Schweber May 29 '21 at 19:43
  • @Troposphere Fair enough, I've edited to address this. – Noah Schweber May 29 '21 at 19:46