Timeline for What is the smallest set of real continuous functions generating all rational numbers by iteration?
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 6, 2022 at 11:36 | vote | accept | Ivan Meir | ||
| Jan 5, 2022 at 0:03 | comment | added | wlad | I deleted my comment above, because I might have phrased it rudely. My point remains that I think $1/x$ is in fact a continuous function. Its behaviour at $x = 0$ is irrelevant because $0$ is not within its domain. | |
| Jan 4, 2022 at 14:18 | comment | added | Zerox | @wlad By continuity of a partial function on $\Bbb{R}$ I think it usually means whether the function can be extended to a continuous function defined all over $\Bbb{R}$. | |
| Jan 4, 2022 at 13:38 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Added 2 function example to second problem.
|
| Jan 2, 2022 at 4:19 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 2 characters in body
|
| Jan 1, 2022 at 21:45 | comment | added | RBarryYoung | Three works. Two, if you want only the positive rationals. | |
| Jan 1, 2022 at 19:11 | history | became hot network question | |||
| Jan 1, 2022 at 16:28 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Minor correction
|
| Jan 1, 2022 at 15:15 | answer | added | Martin M. W. | timeline score: 22 | |
| Jan 1, 2022 at 14:55 | comment | added | Saúl RM | I´m just saying $g(n)=q_n$, nothing about $g(q_n)$. You can define $g$ by linear interpolation for example in the intervals $[n,n+1]$ | |
| Jan 1, 2022 at 13:31 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Added clarification regarding USAMO solution.
|
| Jan 1, 2022 at 13:24 | answer | added | Saúl RM | timeline score: 33 | |
| Jan 1, 2022 at 13:03 | comment | added | Saúl RM | To do it with 2 functions continuous in all $\mathbb{R}$, you can pick $f(x)=x+1$ and, for an enumeration $q_n$ of the rationals, pick $g(x)$ continuous and with $g(n)=q_n$ for natural $n$ | |
| Jan 1, 2022 at 13:02 | comment | added | Ivan Meir | @JukkaKohonen Yes that is true but in the USAMO question you never apply the functions in this way since you are always trying to alternate say $\arctan$ followed by $\cos$ or $\arcsin$ followed by $\tan$ so I don't think you ever get into this scenario. I appreciate that isn't obvious so thanks for the comment I will clarify in the question. | |
| Jan 1, 2022 at 12:42 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Added the finite number of discontinuity case as a separate question
|
| Jan 1, 2022 at 12:38 | comment | added | Jukka Kohonen | @Ivan, are you sure you can just apply the inverse (in the USAMO scenario)? If your sequence went from a large $x$ to $y = \sin x$, then applying $z = \arcsin y$ will generally not give $z=x$. So it is not clear that you can reverse the chain. There may be a way around that problem but it is not obvious to me. | |
| Jan 1, 2022 at 12:36 | comment | added | Ivan Meir | @BrendanMcKay Thanks for your comment - yes I will update the question to include this. | |
| Jan 1, 2022 at 12:25 | comment | added | Brendan McKay | I think it was more interesting when continuity everywhere is required. Maybe you can ask for the answer with and without. What is an example with pure continuity? | |
| Jan 1, 2022 at 12:11 | comment | added | Ivan Meir | @Zerox I've just edited the question to include this. | |
| Jan 1, 2022 at 12:10 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Modified the range of possible functions to allow for a finite number of discontinuities.
|
| Jan 1, 2022 at 12:04 | comment | added | Ivan Meir | @Zerox Yes you are right - I guess we should change continuity to continuous except at a finite set of points which then includes at least the rational functions? | |
| Jan 1, 2022 at 11:58 | comment | added | Ivan Meir | @Kostya_I Once you have obtained a rational $q$ from your sequence of function applications you just apply the inverse of each function in reverse order to get back to 0. | |
| Jan 1, 2022 at 11:50 | comment | added | Zerox | In fact, the only continuous functions in the USAMO example are $\sin$, $\cos$ and $\arctan$. | |
| Jan 1, 2022 at 11:45 | comment | added | Kostya_I | In the case of the USAMO question they are equivalent - is it obvious? Let say I get a large number (e.g. using tan), then apply sin; how do you go back from there? | |
| Jan 1, 2022 at 11:42 | comment | added | Zerox | $f(x)=\dfrac{1}{x}$ is not continuous, so at least the process you described can not be achieved by $3$ continuous functions. | |
| Jan 1, 2022 at 11:11 | history | asked | Ivan Meir | CC BY-SA 4.0 |