Timeline for Order homomorphism functions on $\omega_1$
Current License: CC BY-SA 4.0
43 events
| when toggle format | what | by | license | comment | |
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| Apr 26, 2021 at 3:22 | vote | accept | Mirko | ||
| Apr 26, 2021 at 3:22 | vote | accept | Mirko | ||
| Apr 26, 2021 at 3:22 | |||||
| Apr 26, 2021 at 3:22 | vote | accept | Mirko | ||
| Apr 26, 2021 at 3:22 | |||||
| Apr 26, 2021 at 3:21 | answer | added | Mirko | timeline score: 2 | |
| Apr 26, 2021 at 3:20 | history | edited | Mirko | CC BY-SA 4.0 |
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| Aug 21, 2020 at 14:57 | history | edited | Mirko | CC BY-SA 4.0 |
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| May 21, 2019 at 4:49 | history | edited | Mirko | CC BY-SA 4.0 |
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| Oct 28, 2018 at 5:15 | answer | added | SSequence | timeline score: 1 | |
| Oct 19, 2018 at 16:19 | history | edited | Mirko | CC BY-SA 4.0 |
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| S Jun 1, 2017 at 6:55 | history | bounty ended | CommunityBot | ||
| S Jun 1, 2017 at 6:55 | history | notice removed | CommunityBot | ||
| May 29, 2017 at 5:26 | history | edited | Mirko | CC BY-SA 3.0 |
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| May 29, 2017 at 5:14 | comment | added | Mirko | @AlexRavsky I added an edit: Fix any ordinals $0\le\beta<\delta<\nu<\omega_1$. Let $f(\alpha)=g(\alpha)=0$ if $0\le\alpha<\nu$. Let $f(\alpha)=\beta$ and $g(\alpha)=\delta$ if $\alpha\ge\nu$. Clearly $f\sqsubseteq g$. Then $\psi(f)(\alpha)=\beta$ if $\alpha>\beta$, and $\psi(f)(\alpha)=0$ if $0\le\alpha\le\beta$ (where $\psi$ is as in partial answer $B$). While $\psi(g)(\alpha)=\delta$ if $\alpha>\delta$, and $\psi(g)(\alpha)=0$ if $0\le\alpha\le\delta$. In particular, if $\beta<\alpha\le\delta$ then $\psi(g)(\alpha)=0<\beta=\psi(f)(\alpha)$, so $\psi(f)\not\sqsubseteq \psi(g)$. | |
| May 29, 2017 at 5:10 | history | edited | Mirko | CC BY-SA 3.0 |
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| May 29, 2017 at 0:51 | comment | added | Alex Ravsky | @Mirko I didn’t read all the thread, so my question may be stupid, but I don’t see why the function $\psi$ constructed in Partial answer B is not $\sqsubseteq$-non-decreasing map. As for me, its $\sqsubseteq$-non-decreasity should easlily follow from the construction of $\psi(f)$. Can you provide a counterexample or it is already provided somewhere in the thread? | |
| S May 24, 2017 at 4:54 | history | bounty started | Mirko | ||
| S May 24, 2017 at 4:54 | history | notice added | Mirko | Draw attention | |
| May 23, 2017 at 0:36 | history | edited | Mirko | CC BY-SA 3.0 |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 12, 2015 at 20:05 | comment | added | Mirko | @YairHayut aah, well, ok, I think this might have been addressed somewhere in the comments, the problem is that $h(f)$ as you define it need not be regressive. It would be regressive on a tail, and usually one is satisfied with that, but I insist that it is regressive everywhere, i.e. $h(f)(\alpha)<\alpha$ for all $\ 0<\alpha<\omega_1$. If $h(f)(\alpha)=\gamma$, a constant, then whenever $\alpha\le\gamma$ the condition $h(f)(\alpha)<\alpha$ is violated. Also, you convinced me that $\sup f<\omega_1$ well, ok seems $\max f =\sup f$ as $f$ is not decreasing, you are right about that, thank you. | |
| Apr 12, 2015 at 19:49 | comment | added | Yair Hayut | As in Noah's answer, since $f$ is regressive and non-decreasing, it is eventually constant, and therefore $\max f = \sup f < \omega_1$. | |
| Apr 12, 2015 at 19:13 | comment | added | Yair Hayut | What about $h(f)(\alpha) = \max_{\beta < \omega_1} f(\beta)$ (a constant function with the eventual value of $f$)? | |
| Apr 12, 2015 at 15:25 | history | edited | Mirko | CC BY-SA 3.0 |
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| Apr 29, 2014 at 1:11 | history | edited | Mirko | CC BY-SA 3.0 |
clarified statement and current status
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| Apr 28, 2014 at 20:47 | history | edited | Mirko | CC BY-SA 3.0 |
unaccepted the answer (again) since it is incomplete, to give a chance to people to take a look again
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| Mar 29, 2014 at 16:50 | history | edited | Todd Trimble | CC BY-SA 3.0 |
moved an edit on Noah S by Mirko to Mirko's post
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| Mar 25, 2014 at 2:52 | vote | accept | Mirko | ||
| Apr 28, 2014 at 20:44 | |||||
| Mar 20, 2014 at 14:20 | history | edited | Mirko | CC BY-SA 3.0 |
clarified/rearranged statement, emphasizing that an additional requirement was shown essential in the comments
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| S Mar 20, 2014 at 8:34 | history | suggested | smyrlis | CC BY-SA 3.0 |
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| Mar 20, 2014 at 8:19 | review | Suggested edits | |||
| S Mar 20, 2014 at 8:34 | |||||
| Mar 20, 2014 at 5:14 | comment | added | Noah Schweber | Let me just say that my recent string of complete failures shows that this question is trickier than I at first assumed! :P | |
| Mar 19, 2014 at 21:46 | vote | accept | Mirko | ||
| Mar 20, 2014 at 2:10 | |||||
| Mar 19, 2014 at 21:26 | answer | added | Noah Schweber | timeline score: 3 | |
| Mar 19, 2014 at 20:28 | history | edited | Mirko | CC BY-SA 3.0 |
clarified statement
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| Mar 19, 2014 at 18:33 | review | Close votes | |||
| Apr 1, 2014 at 16:26 | |||||
| Mar 19, 2014 at 18:19 | comment | added | Noah Schweber | @Francois, you're right, I read the requirement as "$h(f)\le f$," since (as you point out) the other inequality is trivially unsatisfiable if "regressive" is meant; and if "decreasing" is meant, then all such functions have finite range, so the question is trivial in the other direction. | |
| Mar 19, 2014 at 17:58 | comment | added | François G. Dorais | Since there are regressive functions with unbounded range but every function in $K$ has bounded range, you can't always have $f \leq h(f)$. (Assuming regressive is what you meant.) | |
| Mar 19, 2014 at 17:58 | review | First posts | |||
| Mar 19, 2014 at 17:58 | |||||
| Mar 19, 2014 at 17:55 | comment | added | François G. Dorais | @NoahS: That doesn't satisfy $f \leq h(f)$. | |
| Mar 19, 2014 at 17:54 | comment | added | Noah Schweber | Sure - map everything to the zero map. Presumably you want an injective homomorphism, though. | |
| Mar 19, 2014 at 17:54 | history | edited | François G. Dorais | CC BY-SA 3.0 |
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| Mar 19, 2014 at 17:50 | comment | added | François G. Dorais | Do you mean decreasing or regressive? (You wrote the former but the definition you give is the latter.) | |
| Mar 19, 2014 at 17:40 | history | asked | Mirko | CC BY-SA 3.0 |