Timeline for Continuous self-maps in the Golomb space that are neither increasing nor decreasing
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 21, 2017 at 14:30 | vote | accept | Dominic van der Zypen | ||
| Nov 21, 2017 at 1:53 | answer | added | Taras Banakh | timeline score: 3 | |
| Nov 16, 2017 at 16:22 | comment | added | Taras Banakh | By the way, the Golomb space is a topological semigroup with respect to the multiplication of natural numbers. So, at least the maps $f(x)=ax^n$ are continuous. But all such maps are increasing. | |
| Nov 13, 2017 at 19:42 | comment | added | Taras Banakh | Each (continuous) decreasing map $f:\mathbb G\to\mathbb G$ has finite image (and hence is constant) since the set $\mathbb G=\mathbb N$ is well-ordered. So, the question is equivalent to the existing a non-constant continuous map which is not increasing. | |
| Nov 13, 2017 at 11:13 | comment | added | YCor | Constant maps are also increasing in the sense indicated by Dominic | |
| Nov 13, 2017 at 10:33 | comment | added | Taras Banakh | The Golomb space admits many continuous increasing maps, in particular, for any $a\in\mathbb N$ the multiplication map $\mathbb G\to\mathbb G$, $x\mapsto ax$, is continuous. But I do not know other maps (shifts are no continuous). | |
| Nov 13, 2017 at 10:10 | comment | added | YCor | Related to a question in mathoverflow.net/a/285890/14094 (if there's no such map, then the homeomorphism group of $\mathbf{G}$ is trivial). | |
| Nov 13, 2017 at 9:58 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |