Timeline for What is known about size-restricted power set axioms?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2012 at 22:18 | vote | accept | Colin McLarty | ||
| Jul 29, 2012 at 21:59 | history | edited | Andreas Blass | CC BY-SA 3.0 |
added another model, to answer the original question
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| Jul 29, 2012 at 21:48 | comment | added | Andreas Blass | @Colin: I think my answer can be modified to get a model where the set of natural numbers has a power set but not every set has a set-of-countable-subsets. I'll edit to add this to my answer (so as to make it a real answer). | |
| Jul 29, 2012 at 21:47 | comment | added | Andreas Blass | @Zsban: I claim that not only the replacement scheme but full second-order replacement is true in this model. If $X$ is a set in the model and $Y$ is a subset of the model and is the "image" of $X$ under a "function", then $Y$ is in the model. The transitive closure of $Y$ consists of the elements of $Y$ (of which there are at most $\mathfrak c$ since $Y$ is an image of $X$) together with the union of the transitive closures of all these elements of $Y$. That's a union of at most $\mathfrak c$ sets, each of size at most $\mathfrak c$, so the union still has size at most $\mathfrak c$. | |
| Jul 29, 2012 at 15:42 | comment | added | Colin McLarty | Yes, I know this theory is weaker than ZF. But is it stronger than ZF[1]? (That is ZF without powerset but with an axiom saying the natural numbers have a power set.) There are really two parts (at least) to that question. Does ZF[1] actually not prove every set has a set of all countable subsets? I expect it does not, but I do not know. And if my expectation is true, does this theory have strictly higher consistency strength than ZF[1]? | |
| Jul 29, 2012 at 15:28 | comment | added | Zsbán Ambrus | Can you prove that the replacement axiom scheme is true in this model? | |
| Jul 29, 2012 at 14:47 | history | answered | Andreas Blass | CC BY-SA 3.0 |