Timeline for Peano axioms— mathematical induction and other axioms
Current License: CC BY-SA 4.0
23 events
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| Nov 3, 2020 at 18:00 | review | Close votes | |||
| Nov 9, 2020 at 3:06 | |||||
| Nov 3, 2020 at 17:42 | comment | added | Noah Schweber | I think this question would be more appropriate at math.stackexchange. | |
| Nov 3, 2020 at 17:25 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title, changed tag
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| Nov 3, 2020 at 17:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| S Oct 4, 2020 at 16:10 | history | suggested | gmvh |
Added top-level tag
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| Oct 4, 2020 at 15:21 | review | Suggested edits | |||
| S Oct 4, 2020 at 16:10 | |||||
| Oct 4, 2020 at 14:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 4, 2020 at 17:20 | comment | added | Pace Nielsen | @EmilJeřábek Ah, I meant to write "not the axioms of Peano arithmetic." Sorry for the inaccuracy. | |
| Sep 4, 2020 at 16:42 | comment | added | Emil Jeřábek | @PaceNielsen These are, in fact, more or less the axioms originally postulated by Peano (see archive.org/details/arithmeticespri00peangoog/page/n22/mode/2up). This is different from what later became to be known as the “Peano arithmetic”. | |
| Sep 4, 2020 at 13:51 | comment | added | Emil Jeřábek | You can rewrite the argument symbolically without using the graph-theoretical terminology (but I think this makes it less intuitive): if 4 fails, then (since $\mathbb N=\{s^{(k)}(1):k<\omega\}$ by 5) there exists $k>0$ such that $s^{(k)}(1)=1$. Then $s^{(k+l)}(1)=s^{(l)}(1)$ for all $l$, that is, $s^{(k)}(n)=n$ for all $n\in\mathbb N$. (Or if you prefer: then $s^{(k)}(n)=n$ for all $n\in\mathbb N$ by induction on $n$.) Thus, if $s(n)=s(m)$, then $n=s^{(k)}(n)=s^{(k-1)}(s(n))=s^{(k-1)}(s(m))=s^{(k)}(m)=m$. | |
| Sep 4, 2020 at 13:20 | history | edited | Andrés E. Caicedo |
edited tags
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| Sep 4, 2020 at 12:04 | answer | added | Gerald Edgar | timeline score: 0 | |
| Sep 4, 2020 at 11:20 | history | edited | Curiosity |
added set-theory tag
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| Sep 4, 2020 at 10:37 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
deleted 2 characters in body
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| Sep 4, 2020 at 10:29 | comment | added | Curiosity | This question came up in a discussion (not homework) on elementary set theory for first year university students using a book like Enderton's Elements of Set Theory. So, elementary means using only basic facts like logic, sets, what a function is (covered in the first 3 chapters of Enderton's book). | |
| Sep 4, 2020 at 10:22 | answer | added | JimN | timeline score: 0 | |
| Sep 4, 2020 at 10:15 | comment | added | JimN | I don't see why "an infinite path" must imply 3 and 4 to hold. | |
| Sep 4, 2020 at 9:47 | review | Close votes | |||
| Sep 6, 2020 at 20:42 | |||||
| Sep 4, 2020 at 9:46 | comment | added | Emil Jeřábek | If you consider cycles and paths advanced, then I really have no idea what you mean by elementary. | |
| Sep 4, 2020 at 9:40 | comment | added | Curiosity | @EmilJeřábek Thanks, I see now. However, since this statement is about basic set theory, is there a proof that uses just the bare-minimum of elementary logic/reasoning rather than more advanced notions of paths/cycles? It seems like there should be such a proof. | |
| Sep 4, 2020 at 9:30 | comment | added | Emil Jeřábek | Well, the induction axiom is equivalent to $\mathbb N=\{1,s(1),s(s(1)),\dots\}$ (that is, $\{s^{(k)}(1):k\in\omega\}$). Thus, the graph of $s$ consists of a single directed walk starting from $1$. Either this is an infinite path (in which case both 3 and 4 hold), or it eventually enters a cycle; if it is just a cycle, then 3 holds and 4 fails, whereas if there is a nonempty path leading to the cycle, then 3 fails and 4 holds. So, yes, 3 or 4 must hold. | |
| Sep 4, 2020 at 9:21 | review | First posts | |||
| Sep 4, 2020 at 10:38 | |||||
| Sep 4, 2020 at 9:12 | history | asked | Curiosity | CC BY-SA 4.0 |