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Assume $M$ is a set, in which all axioms of $ZF - P + (V=L)$ hold. Does then $M$ believe that there exists an uncountable ordinal? I mean, why should the class of all countable ordinal numbers be a set?

Asaf Karagila
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Martin Brandenburg
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1 Answers1

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No. One cannot prove that $\omega_1$ exists in ZF - P, with or without V = L.

The set $HC$ of hereditarily countable sets always satisfies ZF - P. (This is straightforward to check axiom per axiom.) Of course, $\omega_1 \notin HC$ but moreover every set in $HC$ is countable or finite as witnessed by a function in $HC$. Therefore, $HC$ knows that every set in $HC$ is countable and hence $HC$ is a model of ZF - P + "every set is at most countable."

Throwing in V = L into the mix doesn't help. Indeed, $HC^L = L_{\omega_1^L}$ (see my answer to your recent MO question for a proof) is a model of ZF - P + V = L + "every set is at most countable."

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François G. Dorais
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  • Thank you :). I should have remembered these models! – Martin Brandenburg Sep 16 '10 at 14:04
  • çois: I'm wondering about a possible generalization of this idea. Can you take every regular ordinal and create a model of ZF[C]-P such that every set is smaller that the given ordinal? – Asaf Karagila Sep 16 '10 at 15:32
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    Yes, Asaf. If $\kappa > \omega$ is regular then $H(\kappa)$, the family of sets whose transitive closure has size less than $\kappa$, is a model of ZFC-P in which all sets have size at most $\kappa$. (Regularity is needed for the replacement axiom to hold.) – François G. Dorais Sep 16 '10 at 16:18
  • I only knew it was true for $H(\omega)$ for ZF-Inf. – Asaf Karagila Sep 16 '10 at 19:31