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Second-order ZFC is nearly categorical, except that it does not determine the 'height' of the cumulative hierarchy (intuitively speaking). However, additional axioms can be added to second-order ZFC to provide a categorical axiomatization, e.g. specifying that there are no inaccessible cardinals (or that there is exactly one, etc); see this previous answer for details on this background assumption [1]. However, once a particular maximum level has been specified for the cumulative hierarchy (by an additional axiom), does this risk creating an inconsistency with the axiom of replacement, since the range of a function (used in replacement) could potentially include sets arbitrarily high up in the stages of the cumulative hierarchy (where the height of the hierarchy has been fixed by the maximum inaccessible cardinal specified in the added axiom)? Admittedly, this is arguably already a problem for the iterative conception of sets even without an additional axiom, but I'm wondering if it's particularly problematic when an axiom is added (to achieve categoricity) which 'arbitrarily' restricts the height of the cumulative hierarchy (so that a set created by replacement which exists at some level of the hierarchy, does not exist in the 'cropped' hierarchy determined by the new axiomatization)?

[1] What axioms need to be added to second-order ZFC before it has a unique model (up to isomorphism)?

Community
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Andy
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2 Answers2

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There wouldn't be any problems, because the axiom schema of replacement applies only to definable mappings, and so you couldn't jump outside your hierarchy. Any definable mapping cannot send a set above the limit of the portion of the cumulative hierarchy in question, because the height being used already has to produce a collection closed under the operations of ZFC.

jack
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  • Thanks for the response. However, 2nd order ZFC does not involve an axiom schema for replacement? http://math.stackexchange.com/questions/311101/how-is-second-order-zfc-defined – Andy May 03 '15 at 00:17
  • Also, the setup did not involve sending a set 'above the limit of the portion of the hierarchy in question'; rather, the concern is that the set containing all of the sets in the function's range could be outside of the cropped hierarchy (even though the individual sets in the range are within the current hierarchy) – Andy May 03 '15 at 00:21
  • So, as you said, the range elements are within the hierarchy. Then the range is a class in the hierarchy that has at most the same cardinality as a set in the hierarchy. A strong limit ordinal is such that the corresponding universe has no proper class with the same cardinality of any set. Thus the range is a set. – jack May 03 '15 at 01:05
  • As for the axiom of replacement verses the schema, yes you could use a single full second order axiom, and that would be different than the schema approach I used. However, the response given including the above comment are the same either way. – jack May 03 '15 at 01:13
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As the answer by Jack indicates, the usual form of replacement quantifies only over definable functions. But it would be equally possible to make a genuinely second-order axiom of replacement that quantifies over all functions whose domain is a set in the model and whose range is contained in the model.

Suppose that there is an inaccessible cardinal κ in the metatheory, and let $M=V_\kappa$(so $M$ is a model of at least full second-order Zermelo set theory with the axiom of choice).

To show that $M$ satisfies the second-order replacement axiom, suppose that $F$ is any function in the metatheory from some set $A∈M$ to $M$. Then, again in the metatheory, the range of $F$ is also a set of rank less than $\kappa$, because $\kappa$ is inaccessible. So the metatheory sees that $F \in V_\kappa$. But that means that $F \in M$, by definition, so the range of $F$ is also in $M$. Hence $M$ does satisfy the second-order replacement axiom.

This shows that if there is an inaccessible in the metatheory then second-order Zermelo set theory with choice and with the second-order form of replacement mentioned above is consistent.

In particular, we could take $\kappa$ to be the first inaccessible in the metatheory. So the theory in the previous paragraph is consistent with "there is no inaccessible".

Carl Mummert
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  • Thanks for the explanation! Does the full proof that the set corresponding to the range of F must be in M require the axiom of replacement? And if so, is there a circularity concern if the argument justifying the 'safety' of including the axiom of replacement relies on the axiom of replacement? – Andy May 03 '15 at 01:31
  • No, the argument does not seem to use replacement in the metatheory, because $F$ is already a set in the metatheory. It does use the fact that $\kappa$ is a limit cardinal to ensure that if $F$ has rank less than $\kappa$ then the range of $F$ is also less than $\kappa$. The need for $\kappa$ to be inaccessible is to ensure $M$ satisfies Zermelo set theory. – Carl Mummert May 03 '15 at 11:16
  • I should have written: $\kappa$ should be a regular limit cardinal. – Carl Mummert May 03 '15 at 11:55