Timeline for Continuous relations?
Current License: CC BY-SA 3.0
51 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 14, 2023 at 22:28 | answer | added | Wlod AA | timeline score: 1 | |
| Feb 13, 2023 at 20:06 | answer | added | Emily | timeline score: 2 | |
| Dec 9, 2018 at 12:20 | comment | added | John Forkosh | I was just looking for this kind of definition myself (hence the long-delayed comment), and along with your question here, found Blyth's definition in Exercise 6.27, page 101, books.google.com/books?id=Ll0JXd11SW0C&pg=PA101 (which seems pretty intuitively obvious). Note that if such a continuous relation also happens to be a function, it will also be a continuous function. | |
| Dec 3, 2018 at 16:29 | answer | added | painday | timeline score: 1 | |
| Mar 16, 2018 at 21:32 | history | edited | Qfwfq | CC BY-SA 3.0 |
added 42 characters in body
|
| Mar 16, 2018 at 17:35 | answer | added | user57432 | timeline score: 1 | |
| Jul 7, 2016 at 9:54 | vote | accept | Lehs | ||
| Sep 23, 2014 at 14:56 | comment | added | Peter Arndt | Here is a bunch of proposals for a notion of smoothness for set-valued functions. Most can be downgraded to continuity: mathoverflow.net/q/38059/733 | |
| Sep 23, 2014 at 11:51 | answer | added | Daniele Zuddas | timeline score: 4 | |
| Sep 12, 2014 at 9:53 | history | edited | Lehs | CC BY-SA 3.0 |
deleted 836 characters in body
|
| Sep 10, 2014 at 11:18 | answer | added | Dominic van der Zypen | timeline score: 9 | |
| Sep 10, 2014 at 7:33 | history | edited | Lehs | CC BY-SA 3.0 |
added 70 characters in body
|
| Sep 9, 2014 at 17:04 | history | edited | Lehs | CC BY-SA 3.0 |
added 126 characters in body
|
| Sep 1, 2014 at 18:09 | history | edited | Lehs | CC BY-SA 3.0 |
added 76 characters in body
|
| Aug 28, 2014 at 10:42 | comment | added | Michał Masny | another math.se question: math.stackexchange.com/questions/110501/… | |
| Aug 24, 2014 at 22:27 | comment | added | Włodzimierz Holsztyński | @Eric: Theorem If $Y$ is Hausdorff then the graph $G(f) := \{(x\ y): f(x) = y\}\ $ of any continuous $\ f : X\rightarrow Y\ $ is closed in $\ X\times Y.\ $ **Proof** $\ G(f) = (f\times 1_Y)^{-1}(\Delta_Y),\ $ where $\ \Delta_Y := \{(y\ y):y\in Y\}$. | |
| Aug 24, 2014 at 10:11 | history | edited | Lehs | CC BY-SA 3.0 |
added 87 characters in body
|
| Aug 24, 2014 at 8:50 | comment | added | Lehs | @Mike Benfield: Is a continuous function, with the union of two disjoint closed intervalls as domain, a connected subspace? | |
| Aug 23, 2014 at 20:28 | answer | added | Lehs | timeline score: 5 | |
| Aug 23, 2014 at 19:35 | answer | added | Lehs | timeline score: 0 | |
| Aug 23, 2014 at 19:14 | answer | added | Dimitri Chikhladze | timeline score: 4 | |
| Aug 23, 2014 at 17:14 | history | edited | Lehs | CC BY-SA 3.0 |
added 191 characters in body
|
| Aug 23, 2014 at 17:11 | answer | added | Nik Weaver | timeline score: 10 | |
| Aug 23, 2014 at 16:50 | comment | added | Michael Benfield | It's possible the concept you are after is that R is a connected subspace of $X\times Y$. | |
| Aug 23, 2014 at 15:40 | answer | added | Eric Wofsey | timeline score: 17 | |
| Aug 23, 2014 at 14:22 | answer | added | Michal R. Przybylek | timeline score: 6 | |
| Aug 23, 2014 at 13:43 | answer | added | Joonas Ilmavirta | timeline score: 15 | |
| Aug 23, 2014 at 13:29 | history | reopened |
Joel David Hamkins Eric Wofsey Nik Weaver Stefan Kohl♦ Benjamin Steinberg |
||
| Aug 23, 2014 at 10:19 | comment | added | Michał Kukieła | Related math.SE question: math.stackexchange.com/questions/409902/…. | |
| Aug 23, 2014 at 10:05 | comment | added | Michał Kukieła | See en.wikipedia.org/wiki/Hemicontinuity. | |
| Aug 23, 2014 at 9:57 | comment | added | Lehs | Thanks! I found users.physik.fu-berlin.de/~kleinert/public_html/kleiner_reb11/… with 3 chapters with repetation of tensors and stuff and with the mv-functions from chapter 4. It seems to be relevant. | |
| Aug 23, 2014 at 7:29 | comment | added | Włodzimierz Holsztyński | There are various papers/definitions etc for continuous multi-valued functions--continuous in one sense or another (some of these publications are interesting). One may perhaps allow also empty set of values for some arguments but it would not be common. | |
| Aug 23, 2014 at 5:27 | history | edited | Lehs | CC BY-SA 3.0 |
added 11 characters in body
|
| Aug 23, 2014 at 5:01 | history | edited | Lehs | CC BY-SA 3.0 |
added 584 characters in body
|
| Aug 22, 2014 at 22:19 | comment | added | Lehs | I claim that my suggestion is equivalent with inverse relation preserving openness and mine "shows" what a continuous relation would be for intuition: if x is close to the inverse image of a set M, then y must be close to M. But is it consistent meaningful and useful? | |
| Aug 22, 2014 at 20:59 | comment | added | Lehs | It should be equivalent to mine above. | |
| Aug 22, 2014 at 20:50 | comment | added | Joel David Hamkins | Another natural possibility would be to say that $R^{-1}U$ is open for any open set $U\subset Y$, if $R\subset X\times Y$, where $R^{-1}U=\{x\in X\mid \exists y\in U, xRy\}$. How does this line up with the other concepts? | |
| Aug 22, 2014 at 20:17 | comment | added | მამუკა ჯიბლაძე | One possibility, frequently used, is to recast it as a continuous map $X\to{\mathscr P}Y$ where $\mathscr P$ stands for various versions of powerset (e. g. the Vietoris space in the topological case). | |
| Aug 22, 2014 at 20:16 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
clarify what kind of "relation" is meant
|
| Aug 22, 2014 at 19:57 | comment | added | Joel David Hamkins | Yes, I had thought along the same lines. So let's open the question and have some fuller answers. Note that the answers should also aim to generalize the concepts of continuity for partial functions. | |
| Aug 22, 2014 at 19:44 | comment | added | Eric Wofsey | The obvious definition would be to consider relations that are closed as a subset of $X\times Y$, though this fails to generalize continuous functions unless $Y$ is compact Hausdorff. More generally, "relations" can be defined in any category with finite products as subobjects of the product $X\times Y$; this coincides with closed relations in the category of compact Hausdorff spaces (but means "arbitrary relation together with a topology finer than the product topology" in the category of all topological spaces). | |
| Aug 22, 2014 at 19:37 | review | Reopen votes | |||
| Aug 22, 2014 at 23:18 | |||||
| Aug 22, 2014 at 19:29 | comment | added | Joel David Hamkins | I voted to re-open this question, since I would be interested to see what are the various notions we have for continuous relations and how robust they are. Of course, all the definitions should have continuous functions as a special case, but for multi-valued relations, there do seem to be various distinct concepts one might use. | |
| Aug 22, 2014 at 19:12 | history | edited | Lehs | CC BY-SA 3.0 |
final revision
|
| Aug 22, 2014 at 19:04 | history | closed |
abx j.c. Eric Wofsey José Figueroa-O'Farrill Chris Godsil |
Needs details or clarity | |
| Aug 22, 2014 at 18:47 | comment | added | Lehs | Correction: $\alpha M =\{u\in X|\exists v\in M:u\underline{\alpha}v\}$ | |
| Aug 22, 2014 at 18:38 | comment | added | Lehs | Well, perhaps a relation $\alpha\subseteq X\times Y$ such that $x\underline\alpha y \wedge x\in\overline{\alpha M}\Rightarrow y \in \overline{M}$, where $M\subseteq Y$ and $\alpha M=\{u\in X|\exists v\in Y: u\underline\alpha v\}$. Or, some categorical construction based on Rel and topology. | |
| Aug 22, 2014 at 18:37 | comment | added | Tobias Fritz | @EricWofsey: Presumably the idea is that continuous relations should generalize continuous functions in the same way as ordinary relations generalize functions. | |
| Aug 22, 2014 at 18:18 | comment | added | Eric Wofsey | What does "continuous relations" mean? | |
| Aug 22, 2014 at 17:21 | review | Close votes | |||
| Aug 22, 2014 at 19:04 | |||||
| Aug 22, 2014 at 17:00 | history | asked | Lehs | CC BY-SA 3.0 |