Timeline for Is this theory synonymous with PA?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 29 at 10:58 | comment | added | Jade Vanadium | Took me way too long to realize this theory proves $<$ is unbounded via Cantor's theorem, so I write it here for the sake of accessibility: given any $x$, by the Sets axiom we can find some $S=\{n\leq x : n\notin n\}$. Having $S\leq x$ would imply $S\in S \iff S\notin S$, impossible, hence $x<S$. | |
| Jan 5 at 11:46 | vote | accept | Zuhair Al-Johar | ||
| Jan 5 at 10:24 | answer | added | Ali Enayat | timeline score: 9 | |
| Jan 5 at 9:54 | history | became hot network question | |||
| Jan 5 at 9:54 | answer | added | Emil Jeřábek | timeline score: 14 | |
| Jan 5 at 8:53 | comment | added | Emil Jeřábek | Ah, I see. The argument I had in mind needs that every non-minimal element has a predecessor, and I didn’t realize this no longer follows from the simplified axiom. I guess that if it’s explicitly added, it’s not a simplifaction any more. | |
| Jan 5 at 8:46 | comment | added | paste bee | @EmilJeřábek Take $V_{\omega\cdot 2}$, and well-order it, first by rank and for sets of the same rank by an arbitrary well-ordering. This satisfies the well-ordering axioms, the simplified finiteness axiom ($x$ is larger than all of its elements), and the sets axiom (the resulting set has a rank at most one more than the rank of $n$, so it is in $V_{\omega\cdot 2}$), but not the original finiteness axiom (the set $\omega$ has no maximum element). | |
| Jan 5 at 8:45 | comment | added | Emil Jeřábek | What $\omega$? There is no $\omega$ anywhere in the axioms. | |
| Jan 5 at 8:17 | comment | added | Zuhair Al-Johar | @EmilJeřábek, but how to get rid of $\omega$? I thought this should be $\exists n \in x \forall m \in x \ m \leq n$. | |
| Jan 5 at 8:14 | answer | added | paste bee | timeline score: 5 | |
| Jan 5 at 6:39 | comment | added | Emil Jeřábek | Using the other axioms, you can simplify Finiteness to $\exists n\,\forall m\in x\,m<n$. | |
| Jan 5 at 5:01 | history | edited | Martin Sleziak |
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| Jan 5 at 1:10 | answer | added | Joel David Hamkins | timeline score: 4 | |
| Jan 4 at 20:32 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |