Timeline for A "surnatural numbers" as a largest model of the natural numbers
Current License: CC BY-SA 4.0
20 events
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| Jul 5, 2023 at 22:23 | vote | accept | Mike Battaglia | ||
| Apr 4, 2022 at 21:23 | history | edited | Mike Battaglia | CC BY-SA 4.0 |
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| Mar 21, 2022 at 22:03 | answer | added | James E Hanson | timeline score: 5 | |
| Mar 21, 2022 at 20:15 | answer | added | nombre | timeline score: 4 | |
| Mar 18, 2022 at 8:36 | comment | added | Alec Rhea | @ZhenLin The ordinals only get fun algebraically when considered together with their natural operations — roughly speaking, expand an ordinal in Cantor normal form then treat it like a polynomial in the indeterminate $\omega$. The ordinals together with natural addition/multiplication are isomorphic to the polynomial ring over $\mathbb{N}$ in a proper class of variables. | |
| Mar 18, 2022 at 8:21 | history | edited | Mike Battaglia | CC BY-SA 4.0 |
cleaned up and updated
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| Mar 18, 2022 at 8:14 | comment | added | Mike Battaglia | @EmilJeřábek that is true. So basically, you are saying that there must be some element which is divisible by all standard positive integers, and if that element is $\omega$, then that means we would at least have to have $\omega \cdot q$ for all positive rationals $q$. I guess you could also easily build such an element which is a perfect square, a perfect cube, and a perfect n-th power for all standard positive n, such as $({1!}^{1!}, 2!^{2!}, 3!^{3!}, 4!^{4!}, ...)$. Then if $\omega$ is hat number, that would mean we would also have to have $\omega^q_1 \cdot q_2$ for rationals $q_1, q_2$. | |
| Mar 18, 2022 at 7:53 | comment | added | Emil Jeřábek | The example does not work whether you say “finitely” or “hyperfinitely” many. Let me put it differently: the fact that $\omega$ is divisible by all standard positive integers does not in any way contradict its being an element of a monster model of arithmetic. In fact, a monster model (or any nonstandard model, for that matter) of arithmetic must contain such an element. | |
| Mar 18, 2022 at 7:51 | history | edited | Mike Battaglia | CC BY-SA 4.0 |
clarification
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| Mar 18, 2022 at 7:47 | comment | added | Mike Battaglia | @EmilJeřábek yes, it would be "hyperfinitely" many primes, I guess. Well, your example is better, which is that there are two omnific integers whose quotient is sqrt(2), so I will put that instead. | |
| Mar 18, 2022 at 7:39 | comment | added | Emil Jeřábek | You can formulate something to that effect using sequence encoding, but this will not have the usual algebraic meaning of “unique factorization”. No nonstandard model of (sufficiently strong) arithmetic is a UFD: for example, a nonstandard power of $2$, or a product of all primes below a nonstandard number (as usually formalized using sequence encoding), cannot be written as a product of finitely many primes. | |
| Mar 18, 2022 at 7:20 | comment | added | Mike Battaglia | @ZhenLin that is why I referred to commutative addition and multiplication above (also called Hessenberg addition and multiplication). | |
| Mar 18, 2022 at 7:19 | comment | added | Mike Battaglia | @EmilJeřábek I thought this answer from Carl Mummert showed how to formalize it in first order logic? math.stackexchange.com/questions/532350/… | |
| Mar 18, 2022 at 6:42 | comment | added | Zhen Lin | Ordinal addition is not even commutative, so it is not really appropriate to speak of the Grothendieck group, but if we understand it to be the universal abelian group that the ordinals map into, well, since $1 + \omega = \omega$ then the image of $1$ must be $0$, so by induction the finite ordinals are all annihilated. | |
| Mar 18, 2022 at 6:35 | comment | added | Emil Jeřábek | Unique factorization is not a first-order property. But yes, there are other basic properties of standard natural numbers that fail in omnific integers, see mathoverflow.net/a/72942 . | |
| Mar 18, 2022 at 6:11 | comment | added | Mike Battaglia | I guess not! Good point. So the monster model must be larger than the Grothendieck group then, but smaller than the omnific integers. | |
| Mar 18, 2022 at 6:10 | history | edited | Mike Battaglia | CC BY-SA 4.0 |
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| Mar 18, 2022 at 5:40 | history | edited | Mike Battaglia | CC BY-SA 4.0 |
ring
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| Mar 18, 2022 at 5:17 | comment | added | Noah Schweber | In the Grothendieck group (or ring?) of the ordinals, is there an element $x$ satisfying $x+x=\omega\vee x+x+1=\omega$? | |
| Mar 18, 2022 at 5:05 | history | asked | Mike Battaglia | CC BY-SA 4.0 |