Timeline for Binary operations compatible with the usual order on the reals
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2010 at 18:34 | answer | added | David Feldman | timeline score: 1 | |
| Dec 16, 2010 at 8:57 | answer | added | Aaron Meyerowitz | timeline score: 0 | |
| Dec 16, 2010 at 8:10 | answer | added | gowers | timeline score: 3 | |
| Dec 16, 2010 at 6:51 | history | edited | Bugs Bunny |
tags
|
|
| Dec 16, 2010 at 6:50 | comment | added | Bugs Bunny | Let $G$ be the group of order-preserving homeomorphisms of the reals. The group $G^3$ acts on the set of your operations by your key formula $\oplus = (f,g,h)\cdot \otimes$ if $f(x\oplus y) = g(x)\otimes h(y)$. A good question to ask is what the set of $G^3$-orbits is. | |
| Dec 16, 2010 at 6:41 | comment | added | Colin McQuillan | Also, any compatible function is measurable (even Borel measurable), because for all $x$ the inverse image of $[\-infty,x]$ is the region under the graph of a decreasing function. | |
| Dec 16, 2010 at 6:17 | comment | added | Colin McQuillan | On the other hand, the property seems to be preserved under convolution, so any compatible measurable function is a almost-everywhere pointwise limit of smooth compatible functions (e.g. by theorem 2.16 of Analysis, 2nd Edition, Lieb and Loss). | |
| Dec 16, 2010 at 5:53 | comment | added | Colin McQuillan | A nice way to write the definition is as an order-preserving function $\oplus:\mathbb{R}^2->\mathbb{R}$ where $\mathbb{R}^2$ is a poset with the product order. An example of a discontinuous such function is: $ x ⊕ y = \begin{cases} max(x,y)&\mbox{ if }x,y\geq 0\\\\ min(x,y)&\mbox{ if }x,y\leq 0\\\\ 0&\mbox{ otherwise} \end{cases}$ | |
| Dec 16, 2010 at 5:26 | answer | added | fedja | timeline score: 1 | |
| Dec 16, 2010 at 2:59 | comment | added | user6976 | The first thing to check is that $\oplus$ is a continuous function $\mathbb{R}^2\to \mathbb{R}$. This seems to be the case. Second, you take the set $\Omega$ of all functions $\oplus(f,g)$ obtained as in your example and take its closure (in some natural topology). It might coincide with the set of all possible $\oplus$'s. The proof should be similar to the standard proofs that the set of continuous functions is a closure of a subset (say, polynomial functions). That is you approximate your $\oplus$ on bigger and bigger finite subsets. | |
| Dec 15, 2010 at 23:04 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Dec 15, 2010 at 22:36 | history | asked | JBL | CC BY-SA 2.5 |