Timeline for Graph on the set of all functions $f:\mathbb{N}\to\mathbb{N}$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| May 27, 2015 at 3:05 | review | Close votes | |||
| May 27, 2015 at 8:18 | |||||
| May 26, 2015 at 9:10 | comment | added | YCor | If you replace the target set $\mathbf{N}$ with a 2-element set, the components of the resulting graph are all isomorphic and called "hypercubes". When the target set is $\mathbf{N}$ or $\mathbf{Z}$, it is also natural to put an edge only when $f-g$ is $\pm$ the Dirac function at some point, so that the resulting distance is the $\ell^1$-distance. | |
| May 26, 2015 at 8:33 | history | edited | Dominic van der Zypen | CC BY-SA 3.0 |
deleted 10 characters in body
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| May 26, 2015 at 8:31 | comment | added | Dominic van der Zypen | Good point, have modified the post accordingly. | |
| May 26, 2015 at 7:29 | comment | added | David Roberts♦ | It may be just me, but your non-reflexive unordered graph would be much easier to grasp if it were defined as: vertices functions N->N, and there is an edge between f and g if they differ for exactly one input. Of course, you may be trying to emphasise the logical form of the set of edges in the language of some fragment of set theory... | |
| May 21, 2015 at 12:09 | comment | added | Joel David Hamkins | Please go ahead! | |
| May 21, 2015 at 12:06 | comment | added | Dominic van der Zypen | Joel, I think this question is worthwhile to put it as a MO question! | |
| May 21, 2015 at 12:04 | comment | added | Joel David Hamkins | This graph is Borel, and it would be interesting to ask whether there can be a Borel $\mathbb{N}$-coloring. If your edge relation connected points that differed finitely, then I believe there can be no Borel coloring, since it would give rise to a Borel selector on $E_0$, which is impossible. (For example, there can be no Borel coloring of the type that Chris describes.) But your relation is finer than $E_0$ and so a coloring needn't color the entire $E_0$ equivalence class differently. | |
| May 21, 2015 at 8:29 | review | Close votes | |||
| May 21, 2015 at 10:36 | |||||
| May 21, 2015 at 8:21 | vote | accept | Dominic van der Zypen | ||
| May 21, 2015 at 8:11 | answer | added | Chris Lambie-Hanson | timeline score: 9 | |
| May 21, 2015 at 7:54 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |