Timeline for Characterizing $\mathbf{R}$ as an ordered group
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2020 at 17:16 | history | edited | YCor | CC BY-SA 4.0 |
edited tags, made title slightly more precise
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| Jun 11, 2018 at 7:37 | comment | added | Liviu Nicolaescu | @JochenGlueck Yes, you are correct. Zorich regards $\mathbb{R}$ as an ordered field. | |
| Jun 9, 2018 at 23:41 | comment | added | YCor | @JochenGlueck OK thanks. Anyway I was more interested by the second part of my comment, which you confirmed in your answer (Corollary 7). | |
| Jun 9, 2018 at 21:36 | comment | added | Jochen Glueck | @YCor: It seems to me that coudy's remark on "without smallest and largest element" did not refer to the process of completion, but to the countable densely and totally ordered set in your comment. The assertion "the completion of any countable dense totally ordered set $S$ is (isomorphic to) $\mathbb{R}$" is indeed not correct as coudy's example $S = [0,1] \cap \mathbb{Q}$ shows. The correct assertion is "If $S$ is a countable dense totally ordered set and $S$ does neither have a minimum nor a maximum, then the completion of $S$ is isomorphic to $\mathbb{R}$. | |
| Jun 9, 2018 at 17:02 | comment | added | Qfwfq | @Emil Jeřábek: oh I see, I thought "completely ordered" was just a synonym with "totally ordered". Clearly, I don't know much about ordered structures and the terminology in the field. | |
| Jun 9, 2018 at 16:55 | comment | added | Emil Jeřábek | @Qfwfq I do mean discrete wrt order topology. The order of $\mathbb Q$ is not complete. As explained in the answer below, the only completely ordered abelian groups are $0$, $\mathbb Z$, and $\mathbb R$. | |
| Jun 9, 2018 at 16:38 | comment | added | YCor | @coudy it was pretty clear that I meant without the upper and lower bound (just filling the holes). I heard plenty of times that $\mathbf{R}$ can be defined as the Dedekind completion of $\mathbf{Q}$, so I think this wording is valid too. | |
| Jun 9, 2018 at 15:34 | comment | added | Qfwfq | @Emil Jeřábek "the reals are the only nondiscrete completely ordered group" - What do you mean by nondiscrete? You must not mean "discrete" w.r.t. the order topology because otherwise we have $\mathbb{Q}$ | |
| Jun 9, 2018 at 12:26 | vote | accept | coudy | ||
| Jun 9, 2018 at 11:33 | comment | added | Jochen Glueck | @LiviuNicolaescu: Thanks for the reference! I think we missunderstood each other: Zorich's book derives the Archimedean property from the completeness for linearly ordered fields. But I think the same should be true for linearly ordered groups, too. I've just found a proof for this for commutative groups, but I'm not quite sure if it's true for non-commutative groups, too. | |
| Jun 9, 2018 at 11:16 | comment | added | Liviu Nicolaescu | You can find it in the book by V. A. Zorich Mathematical Analysis I, Sec. 2.2.3. | |
| Jun 9, 2018 at 11:13 | comment | added | Jochen Glueck | @LiviuNicolaescu: I think I have also seen this result somewhere, but I can't remember the reference right now, and I can't find it in Fuchs' book (see my answer below), either. Do you have a reference? | |
| Jun 9, 2018 at 10:55 | answer | added | Jochen Glueck | timeline score: 31 | |
| Jun 9, 2018 at 10:53 | comment | added | Emil Jeřábek | Yes, the reals are the only nondiscrete completely ordered group. I'm pretty sure something like that has been asked on MO before. | |
| Jun 9, 2018 at 10:34 | comment | added | Liviu Nicolaescu | I think that the archimedean property follows from the completeness axiom. | |
| Jun 9, 2018 at 10:11 | comment | added | YCor | The ordered set $(\mathbf{R},\le)$ can be characterized as the Dedekind completion of any countable dense total order. I expect that the ordered group $(\mathbf{R},+,\le)$ is (up to isomorphism of ordered groups) the only ordered abelian group whose underlying ordered set is isomorphic to $(\mathbf{R},\le)$. | |
| Jun 9, 2018 at 10:11 | comment | added | coudy | Yes, that's the standard one for linearly ordered groups. | |
| Jun 9, 2018 at 10:08 | comment | added | YCor | The only definition of Archimedean I can imagine here is: for all $x,y>0$ there exists $n\ge 1$ such that $nx\ge y$ where $nx=x+\dots+x$ ($n$ times). | |
| Jun 9, 2018 at 10:06 | comment | added | R. van Dobben de Bruyn | What definition of 'archimedean' do you use? | |
| Jun 9, 2018 at 9:48 | history | asked | coudy | CC BY-SA 4.0 |