Timeline for The sum of two well-ordered subsets is well-ordered
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2021 at 6:29 | comment | added | nombre | There is also an older proof of this result in Bernhard Neumann's "On ordered division rings". | |
| Jan 19, 2021 at 5:33 | comment | added | მამუკა ჯიბლაძე | @EmilJeřábek Clear! I stupidly thought that you can destroy well-orderedness just by removing smallest element of some subset :D | |
| Jan 18, 2021 at 21:21 | comment | added | Gerald Edgar | Correct. If $A,B\subset G^+$, then $A+B \subseteq (A\cup B)^*$ and a subset of a well ordered set is well ordered. So (as LSpace said) in general, translate $A,B$ to get them inside $G^+$. | |
| Jan 18, 2021 at 20:28 | comment | added | LSpice | (I think that "$m$ is something below $A \cup B$" means to mean that $m$ is the element by which we translate $A \cup B$ to ensure that it's positive. Perhaps it's also worth remarking that a translate of a well ordered set is well ordered, so from the well ordering of a set containing $A + B + 2m$ we can conclude, as @EmilJeřábek says, the well ordering of $A + B + 2m$, and thence that of $A + B$.) | |
| Jan 18, 2021 at 19:38 | comment | added | Emil Jeřábek | @მამუკაჯიბლაძე I’m not sure what you mean by $m$, but note that any subset of a well-ordered set is well-ordered. | |
| Jan 18, 2021 at 18:27 | comment | added | მამუკა ჯიბლაძე | Is it easy to see that the theorem you cite implies what one needs? A priori it only says that $(A+m)\cup(B+m)\cup(A+A+2m)\cup(A+B+2m)\cup(B+B+2m)\cup(A+A+A+3m)\cup(A+A+B+3m)\cup(A+B+B+3m)\cup\cdots$ is well-ordered (where $m$ is something below $A\cup B$) | |
| Jan 18, 2021 at 16:12 | comment | added | Gerald Edgar | $G^+ := \{x \in G\;:\; x > 0\}$ | |
| Jan 18, 2021 at 16:10 | history | answered | Gerald Edgar | CC BY-SA 4.0 |