Does positive set theory $\sf GPK^+_\infty$ prove the existence of a set $K= \{x \mid x \text { is von Neumann ordinal } \lor x=K\}$
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asked Mar 18, 2023 at 8:54
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$\begingroup$ I would guess no, "$x$ is a von Neumann ordinal" is not positive, otherwise $\{x\mid x\text{ is a von Neumann ordinal and }\forall y∈x∀f:y\to y\;((∀z∈y∀w∈y\;f(z)=f(w)⇒z=w)⇒∀z∈y∃w∈y\;f(w)=z)\}$ would have define $ω$ in $GPK^+$ (where $f:y\to y$ is "$∀p∈f∃z∈y∃w∈y\;(p=\{\{z\},\{z,w\}\})$ and $∀z∈y∀w∈y∀u∈y\;((\{\{z\},\{z,w\}\}∈f∧\{\{z\},\{z,u\}\})⇒w=u)$") $\endgroup$– HoloCommented Mar 18, 2023 at 17:12
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$\begingroup$ But you have the closure schema, so just close on von Neumanns and you get a set in which all von Neumanns are elements, but I don't know if you can get that closure to be the set $K$. $\endgroup$– Zuhair Al-JoharCommented Mar 18, 2023 at 18:25
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$\begingroup$ The closure only gives you a minimal super-set of each class, on face value there is no reason this set won't be e.g. "all transitive sets" (I meant, there is a reason, but my point is that the closure can result with a huge set, it doesn't tells you a lot) $\endgroup$– HoloCommented Mar 18, 2023 at 18:43
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$\begingroup$ @Holo, yes I know, but there might be another way to do that, not just closures. $\endgroup$– Zuhair Al-JoharCommented Mar 18, 2023 at 18:54
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$\begingroup$ @ZuhairAl-Johar: also take a look at the Stanford Encyclopedia of Philosophy entry , "Alternative Axiomatic Set Theories" It should be helpful as well. $\endgroup$– Thomas BenjaminCommented Mar 22, 2023 at 21:07
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1 Answer
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I came to know that this question had been answered to the positive by the founder of $\sf GPK^+_\infty$ himself at an article of him titled: Inconsistency of GPK + AFA. At Journal Math.Log.Quart 1996: vol. 42, issue 1, page 107
answered Mar 22, 2023 at 11:34