Timeline for product spaces of rationals
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2012 at 13:52 | comment | added | Gerald Edgar | OK, I am happy with this version. | |
| Oct 5, 2012 at 1:06 | comment | added | Todd Trimble | Your suggestion also seems reasonable. :-) | |
| Oct 5, 2012 at 1:05 | comment | added | Todd Trimble | Thanks Gerald. I think it's fixed now, but what I wrote is inelegant. If I had to do it again, I'd instead use the countable dense linear order $Q$ of quadratic surds between 0 and 1, whose continued fractions are eventually periodic, and interleave those. That seems to me to be a more elegant, less hack-y solution. | |
| Oct 5, 2012 at 0:47 | comment | added | Gerald Edgar | Maybe you can interleave using decimals for something like $\{x+\sqrt{2} : x \in \mathbb Q\}$. | |
| Oct 4, 2012 at 21:02 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added 719 characters in body; added 17 characters in body; added 4 characters in body
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| Oct 4, 2012 at 19:57 | comment | added | Todd Trimble | Bah. Yes. I can't delete my answer though unless nemisiso unaccepts it. Thanks for pointing out the problem, Gerald. Maybe there's some simpleminded fix, but I'm otherwise occupied now. | |
| Oct 4, 2012 at 19:31 | comment | added | Gerald Edgar | "Forbidding tails of zeros" ... a slight problem. No pair of numbers interleaves to u = 0.40404040404040... despite its being eventually periodic and not having a tail of zeros. | |
| Oct 4, 2012 at 16:08 | comment | added | Sidney Raffer | @ Emil: Whoops! | |
| Oct 4, 2012 at 16:01 | comment | added | Emil Jeřábek | @SJR: Sets of the form $\{a\}\times(b,c)$ are open in the lexicographic order topology, but not in the product topology. | |
| Oct 4, 2012 at 15:46 | comment | added | Sidney Raffer | Isn't the product topology on QxQ the same as the order topology on QxQ determined by the lexicographic order? This, together with back-and-forth, would do the trick. | |
| Oct 4, 2012 at 15:33 | comment | added | Todd Trimble | I have substantially edited my answer. Apologies again. If there is still an error, then please (nemesiso) unaccept this answer and I will delete it. I really just wanted something semi-constructive. | |
| Oct 4, 2012 at 15:31 | history | edited | Todd Trimble | CC BY-SA 3.0 |
deleted 121 characters in body
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| Oct 4, 2012 at 15:13 | comment | added | Todd Trimble | Sorry! You are right. I will edit and try to fix this thing. | |
| Oct 4, 2012 at 14:51 | comment | added | Gerald Edgar | Am I confused? $f : \mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}(\sqrt{2})$ sending $(r,s)$ to $r+s\sqrt{2}$ is a homeomorphism? Surely you can have $r_n \to \infty$ and $s_n \to -\infty$ in such a way that $r_n+s_n\sqrt{2} \to 0$? | |
| Oct 4, 2012 at 14:22 | vote | accept | Marcus | ||
| Oct 4, 2012 at 14:20 | history | answered | Todd Trimble | CC BY-SA 3.0 |