Timeline for Is there an associative metric on the non-negative reals?
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
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| S Jul 15, 2014 at 23:51 | history | suggested | dvitek | CC BY-SA 3.0 |
Fixed a broken link (to Mazurkiewicz's argument).
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| Jul 15, 2014 at 23:42 | review | Suggested edits | |||
| S Jul 15, 2014 at 23:51 | |||||
| S Jul 14, 2014 at 17:40 | history | suggested | Sebastien Palcoux | CC BY-SA 3.0 |
Typos, minor edits
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| Jul 14, 2014 at 17:32 | review | Suggested edits | |||
| S Jul 14, 2014 at 17:40 | |||||
| Feb 28, 2010 at 21:47 | comment | added | Konrad Swanepoel | @t3suji: OK, right, so it works for all subfields of $\mathbb{R}$. | |
| Feb 26, 2010 at 13:40 | comment | added | t3suji | @Konrad Swanepoel. Perhaps I am missing something, but why can't we use the same argument for $\mathbb{Q}$? Order it in order type $\omega$, and start the recursion (which is not even transfinite anymore). On each step, choose rational $k$ between $0$ and $1$, since only finitely many $k$ are prohibited, this can be done. | |
| Feb 26, 2010 at 10:37 | comment | added | Konrad Swanepoel | This is a very nice solution now. So this seems to work for $[0,\infty)$ in any uncountable subfield of $\mathbb{R}$. So the question remains open for countable subfields of $\mathbb{R}$, in particular $\mathbb{Q}$. | |
| Feb 26, 2010 at 2:21 | comment | added | t3suji | @Jacques Carette: Thanks for pointing this out (the error only appears in final version, not in math preview). Should be better now. | |
| Feb 26, 2010 at 2:18 | history | edited | t3suji | CC BY-SA 2.5 |
added 2 characters in body; added 28 characters in body
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| Feb 26, 2010 at 2:13 | history | edited | t3suji | CC BY-SA 2.5 |
added 2 characters in body; added 8 characters in body; deleted 18 characters in body
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| Feb 26, 2010 at 0:52 | comment | added | Jacques Carette | Recent edits have some mistakes in the Latex (so that it shows up mangled), but I can't edit it. This is a fun bit of math, someone should make sure it is properly readable. | |
| Feb 26, 2010 at 0:40 | history | edited | t3suji | CC BY-SA 2.5 |
cleaned up set-theoretic argument
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| Feb 25, 2010 at 16:33 | comment | added | Pandelis Dodos | Very nice construction Dima. | |
| Feb 25, 2010 at 16:30 | comment | added | t3suji | @Joel David: Thanks for the comment. I thought that Zorn's Lemma can be avoided, but I did not see the neat argument that you suggest (I am much more used to Zorn's Lemma than to recursion). I'll incorporate the changes when I have time (later today?) @Pandelis Dodos: The name is Dima Arinkin (I somehow got so used to t3suji that I used it as a nickname without much thought). | |
| Feb 25, 2010 at 16:13 | vote | accept | aorq | ||
| Feb 25, 2010 at 15:05 | comment | added | Joel David Hamkins | Continuing my earlier comment, there is no need to consider the poset of all triples (I,S,star). Rather, you well order the reals in a order type c=continuum, and then you are defining the group S_alpha and star_alpha for each alpha less than c. Your construction describes how to build S_{alpha+1} and star_{alpha+1} from S_alpha and star_alpha, and at limit ordinals, you take unions. Indeed, I would say just that you are defining star itself by transfinite recursion, and omit the formality of star_alpha. That is, You define S_alpha and star on S_alpha by recursion on alpha. | |
| Feb 25, 2010 at 15:01 | comment | added | Pandelis Dodos | @t3suji nice but please edit properly the details. It could also be a good idea if you let us know your name... | |
| Feb 25, 2010 at 14:35 | comment | added | Joel David Hamkins | Your argument can be simplified into a straight transfinite recursion. Once you've fixed the well order, you don't need to apply Zorn's lemma as you do. Rather, you are giving a transfinite recursion: at limit ordinals, you take unions of what you've done so far (this is the content of your Zorn argument), and at successor ordinals, you have the main part of your construction. | |
| Feb 25, 2010 at 14:27 | comment | added | François G. Dorais | You might want to add that 0 is the first element under $\prec$, otherwise it's not necessarily easy to see that your collection of triples $(I,S,\star)$ is nonempty. (You might also want to add a proof of associativity and mention that 0 is the group identity.) | |
| Feb 25, 2010 at 13:53 | history | edited | t3suji | CC BY-SA 2.5 |
fixed terminology, added the inspirational link.
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| Feb 25, 2010 at 13:41 | comment | added | t3suji | @Konrad Swanepoel: Yes, that's what I meant. @ Pandelis Dodos: No, I mean that $S$ has a set of generators which is an initial segment of the order on all of ${\mathbb R}^{\ge 0}$. If you prefer, you can consider the largest initial segment that is contained in $S$ and require that it generates $S$ as a group; this way, we won't need to mention $I$ explicitly. | |
| Feb 25, 2010 at 11:41 | comment | added | Pandelis Dodos | Konrad's comment seems to me to be correct. What I don't understand is something in the construction. If you have $(I,S,\ast)$ as above, then I don't see how $I$ can be a subset of $S$ and an initial interval under $\succ$. Do you mean that $I$ is an initial interval of $S$? | |
| Feb 25, 2010 at 10:57 | comment | added | Konrad Swanepoel | Assuming $x\star y\leq x+ y$, $x\star x=0$, commutativity and associativity of $\star$, then $x\star z=x\star y\star y\star z\leq x\star y + y\star z$. Right? | |
| Feb 25, 2010 at 6:42 | comment | added | fedja | Erm... The original goal was to find $\star$ such that $x\star z\le x\star y+y\star z$, which is not quite the same as what you achieved in your construction. Am I missing something? | |
| Feb 25, 2010 at 0:01 | history | answered | t3suji | CC BY-SA 2.5 |