Timeline for isomorphism of Dedekind complete ordered fields
Current License: CC BY-SA 3.0
20 events
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| Jan 2, 2013 at 21:07 | history | edited | Brian M. Scott | CC BY-SA 3.0 |
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| Jan 2, 2013 at 20:47 | comment | added | Chris | I think I confused myself :P Yes, that is obvious (actually I had it written down, along with "this is clear"). | |
| Jan 2, 2013 at 20:43 | comment | added | Brian M. Scott | @user1296727: If $x,y\in\Bbb R$ with $x<y$, and if $u\in\operatorname{quot}_{\Bbb R}$, then $u<x\implies u<y$, so clearly $\Theta_x\subseteq\Theta_y$. | |
| Jan 2, 2013 at 20:41 | comment | added | Chris | As I wrote, in the question of why it is obvious. The "corresponding" statement if $\mathrm{quot}$ were actually $\mathbb{Q}$ would be obvious, like I noted. But they're not the same statement. | |
| Jan 2, 2013 at 20:41 | comment | added | Brian M. Scott | @user1296727: Your difficulty with it led me to misread the definition of $\Theta_x$. In fact my addition was completely unnecessary: there’s no isomorphism problem with the $x<y\implies\Theta_x\subset\Theta_y$ statement after all. Where do you think that there’s a problem? | |
| Jan 2, 2013 at 20:38 | history | edited | Brian M. Scott | CC BY-SA 3.0 |
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| Jan 2, 2013 at 20:34 | comment | added | Chris | No, no, I meant that you explicitly said $x, y \in \mathbb{R}$. And then you applied $h$ mapping $\mathbb{Q}$ to $\mathrm{quot}_{\mathbb{R}}$. (I'm assuming that's a typo, unless you say otherwise.) | |
| Jan 2, 2013 at 20:33 | comment | added | Brian M. Scott | @user1296727: And at this point Spivak is assuming that the constructions of $\Bbb Q$ and $\Bbb R$ have been done, and that we now apply the names $\Bbb N$ and $\Bbb Q$ to the appropriate subsets of $\Bbb R$. | |
| Jan 2, 2013 at 20:32 | comment | added | Chris | Okay, what I meant by that, most immediately, is that you have $x, y \in \mathbb{R}$ up above. Also: if you could elaborate on why this helps (just as an example), that would help me understand. | |
| Jan 2, 2013 at 20:30 | comment | added | Brian M. Scott | @user1296727: Not as I read the definition of $\Theta_x$. But sure, if you want, you can take $x$ in that definition to belong to $\Bbb Q$, in which case the fussy version is $$\Theta_x=\{f(y)\in F:y\in\operatorname{quot}_{\Bbb R}\text{ and }y<h(x)\}\;.$$ It really doesn’t matter: $\operatorname{quot}_{\Bbb R}$ and $\Bbb Q$ are the ‘same thing’ under different names. | |
| Jan 2, 2013 at 20:26 | comment | added | Chris | But $x$ and $y$ aren't in $\mathbb{Q}$? | |
| Jan 2, 2013 at 20:25 | comment | added | Brian M. Scott | @user1296727: No, I want the isomorphism in the direction that I gave: subscripts on $\Theta$ belong to $\operatorname{quot}_{\Bbb R}$, not to $\Bbb Q$. | |
| Jan 2, 2013 at 20:24 | history | edited | Brian M. Scott | CC BY-SA 3.0 |
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| Jan 2, 2013 at 20:20 | comment | added | Chris | I don't think that's right. I think you want the isomorphism mapping $\mathrm{quot}_{\mathbb{R}} \rightarrow \mathbb{Q}$. More to the point: could you tell me a bit about how this makes things work? | |
| Jan 2, 2013 at 20:19 | comment | added | Brian M. Scott | @user1296727: Does the example that I just added help? For induction on $\operatorname{ran}e\circ h$, just use $(e\circ h)^{-1}$ to transfer statements to $\Bbb N$, do the induction there, and use the isomorphism to transfer back to the copy of $\Bbb N$ in $\Bbb R$. These are isomorphisms: they preserve everything that you care about. | |
| Jan 2, 2013 at 20:17 | history | edited | Brian M. Scott | CC BY-SA 3.0 |
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| Jan 2, 2013 at 20:13 | comment | added | Chris | Alright, and I took that for granted. But how can I use it? The difficulty is, there are several instances where specific properties of those structures are used, and I don't know how to show they hold just as well in their isomorphisms. Like induction on the range of $e \circ h$. | |
| Jan 2, 2013 at 20:12 | comment | added | Brian M. Scott | @user1296727: You have some isomorphism $h:\Bbb N\to\Bbb Q$, and you have an isomorphism $e:\Bbb Q\to\Bbb R$, so you have an isomorphism $e\circ h:\Bbb N\to\Bbb R$. Given those, I honestly don’t see what the difficulty is. | |
| Jan 2, 2013 at 20:10 | comment | added | Chris | @BrianMScott I worked part of this out in Rudin last year; I saw how $\mathbb{Q}$ is isomorphic to a subfield of $\mathbb{R}$. I assume that it's pretty similar for $\mathbb{N}$. I genuinely do not see how to go about here; otherwise I wouldn't have spent the past hour typesetting. (i.e., it isn't true that "if I really wish to do so", I can; I don't see how to, and that's why I asked.) | |
| Jan 2, 2013 at 20:06 | history | answered | Brian M. Scott | CC BY-SA 3.0 |