Timeline for What's the difference between 2 and 3? [closed]
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 18, 2011 at 19:19 | comment | added | Robert Israel | Given any two meromorphic functions $f_1, f_2$ on an open disk $D$, there is a meromorphic function $g$ on $D$ that avoids them (i.e. $g(z) \ne f_1(z)$ and $g(z) \ne f_2(z)$ at all points of $D$. But there are three meromorphic functions $f_1, f_2, f_3$ on $D$ such that no meromorphic function $g$ can avoid them all (and in fact you can take $f_1 = 0$, $f_2 = \infty$ and $f_3$ a rational function). | |
| Apr 18, 2011 at 18:07 | comment | added | Qiaochu Yuan | I think these phenomena are just a special case of the general phenomenon described in mathoverflow.net/questions/5372/dimension-leaps restricted to the particular case of a dimension leap from $2$ to $3$. | |
| Apr 18, 2011 at 16:02 | history | closed |
Andrés E. Caicedo Felipe Voloch Charles Siegel darij grinberg Zev Chonoles |
not a real question | |
| Apr 18, 2011 at 15:55 | answer | added | Roland Bacher | timeline score: 1 | |
| Apr 18, 2011 at 15:52 | answer | added | Roland Bacher | timeline score: 4 | |
| Apr 18, 2011 at 15:44 | answer | added | Roland Bacher | timeline score: 2 | |
| Apr 18, 2011 at 15:41 | comment | added | Pietro Majer | So, it seems by now we have enough experimental evidence that 2 and 3 are actually different... Here's another one: The free complete lattice on three generators is a proper class (A.W.Hales, On the non-existence of free complete Boolean algebras). | |
| Apr 18, 2011 at 15:40 | answer | added | Roland Bacher | timeline score: 1 | |
| Apr 18, 2011 at 15:39 | comment | added | Roland Bacher | A correct answer to the question in the title would be 1 I believe. | |
| Apr 18, 2011 at 15:35 | comment | added | user9072 | While others have essentially said it already I need to say it too: this is so so broad and vague; hundreds of answers could be given, and I have the strong feeling that they will be given and the couple of interesting ones will be hard to find in this flood, while the question will be on the front page for way too long. | |
| Apr 18, 2011 at 15:33 | answer | added | Michael Renardy | timeline score: 2 | |
| Apr 18, 2011 at 15:33 | comment | added | Tony Huynh | The Banach-Tarski paradox holds in $\mathbb{R}^3$, but not in $\mathbb{R}^2$. | |
| Apr 18, 2011 at 15:30 | answer | added | Tom De Medts | timeline score: 2 | |
| Apr 18, 2011 at 15:29 | answer | added | subshift | timeline score: 1 | |
| Apr 18, 2011 at 15:25 | comment | added | Tara Brough | I don't think it's just you being in a bad mood, Pete. It would at least help if $3$ was replaced by $n\geq 3$ in the places where it can be, I guess. | |
| Apr 18, 2011 at 15:25 | answer | added | Hans-Peter Stricker | timeline score: 1 | |
| Apr 18, 2011 at 15:19 | comment | added | Pete L. Clark | I'm sorry. I may just be in a poor mood (those who follow other parts of the internet math Q&A community will catch an allusion here) but at the moment your question strikes me as somewhat superficial: $2$ is not equal to $3$, so there are going to be a lot of instances where changing $2$ to $3$ makes a big difference. But perhaps there is a good question lurking in here somewhere, something like: what common explanations can be given for these examples? It might be worth thinking about how to rephrase it. | |
| Apr 18, 2011 at 15:18 | comment | added | user8594 | All the prime numbers less than or equal to 2 are even, and all the prime numbers greater than or equal to 3 are odd :) | |
| Apr 18, 2011 at 15:09 | history | asked | Guntram | CC BY-SA 3.0 |