Timeline for Order homomorphism functions on $\omega_1$
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21 events
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| Apr 26, 2021 at 22:05 | comment | added | Mirko | Def.2.1 for $\omega_1$ could be "translated" as follows. Let $Q=\{(\gamma,\lambda):\gamma<\lambda<\omega_1, \lambda\ \mathrm{limit}\}$. For each $q=(\gamma_q,\lambda_q)\in Q$ there is $\delta_q\in[\gamma_q,\lambda_q)$ such that if $P\subset Q$ and $\mu<\omega_1$ with $\delta_p<\mu\le\lambda_p$ for each $p\in P$ then there is $R\subseteq P$ and $\nu<\mu$ such that (i) $\gamma_r\le\nu$ for all $r\in R$ and (ii) for each $p\in P$ there is $r\in R$ with $\gamma_r\le\delta_p<\lambda_p\le\lambda_r.$ The order-theoretic question was to help answer the topological one,but we ended up with the opposite | |
| Apr 26, 2021 at 17:29 | comment | added | SSequence | Regarding the answer you posted, is it possible to cast definition-2.1 and theorem-2.2 purely in terms of ordinals? My general maths knowledge is quite limited so I am not familiar (or fully re-call/understand) with many of the terms used there. At any rate, a proof of a negative answer (for $\omega_1$) def. sounds interesting. | |
| Apr 26, 2021 at 17:26 | comment | added | SSequence | @Mirko It has been some time since the posting of this answer, so I don't remember the exact statement of the question. However, I do re-call that at the time that my feeling was that one could perhaps give a generalized way of giving positive answer for all countable ordinals. Even if the answer is in negative for $\omega_1$, it would be of some interest (I think) to describe a streamlined procedure for producing the positive answer for every countable ordinal. | |
| Apr 26, 2021 at 3:27 | comment | added | Mirko | Thank you for your answer and interesting ideas, even if in the end the answer turned out to be negative, I just posted details. | |
| Apr 26, 2021 at 3:22 | vote | accept | Mirko | ||
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| Aug 21, 2020 at 15:04 | comment | added | Mirko | Hello SSequence, thank you again for the answer that you posted. Ultimately my question was answered in the negative by Gary Gruenhage (and for now I just posted a brief sketch at the top of the question). Thank you and best wishes, Mirko | |
| Oct 29, 2018 at 10:08 | comment | added | SSequence | @Mirko I had one other rough idea (and that's all I can add for now). Perhaps it might be a bit of mistake to focus too much on the maximum value $v_f$ of a function $f \in \mathcal F$ itself, but rather focus on the smallest point at which the value was achieved. For example, define $u_f$ as the smallest value for which $f(u_f)=v_f$. So, if we denote $ \mathcal A_\alpha$ as the set of all $f \in \mathcal F$ for which $u_f < \alpha$, then it might be more natural to consider functions from $ \mathcal A_\alpha$ to $\mathcal K$. | |
| Oct 29, 2018 at 7:46 | history | edited | SSequence | CC BY-SA 4.0 |
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| Oct 29, 2018 at 7:38 | history | edited | SSequence | CC BY-SA 4.0 |
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| Oct 29, 2018 at 7:18 | comment | added | SSequence | Small correction in last comment ..... "$f(x)=v_f$ for $x>\omega^2$" should be "$[\psi_\alpha(f)](x)=v_f$ for $x>\omega^2$" (obviously we are considering the case $\alpha \geq \omega^2+1$). | |
| Oct 29, 2018 at 4:49 | comment | added | SSequence | @Mirko Just one small point. Let's assume for a moment (for simplicity) that functions in $\mathcal F$ are required to be smooth. Then beyond $\psi_{\omega^2+1}$, there is an easy uniform way of dealing with all the functions $f$ for which $v_f < \omega^2$. For example, just set $f(x)=v_f$ for $x>\omega^2$. And furthermore, for some $f$ and $g$ ($v_g \geq \omega^2$ possibly), now just compare $f(\omega^2)$ and $g(\omega^2)$ and use these for mapping the values below $\omega^2$. This is the kind of issue that might be looked into more detail perhaps (and I haven't admittedly). | |
| Oct 29, 2018 at 3:42 | comment | added | Mirko | You may change the first line in the definition of $F$ to: $F(x)=\max\{f(x),\omega\cdot(a+1)\}$ for $x>\omega\cdot(a+1)$. This may perhaps avoid the necessity to change the definition, going to $\psi_{\omega^2+1}$ (and make possible to use Cantor normal form and recursion?). Re skipping, if $x$ is limit and $f(x+1)=x$ then $f$ skips the values between $f(x)$ and $x$. I have to think, re changing values of $\psi_{\omega^2+1}$, you are right they change (in your def). The answer by @NoahS seems related (he seems to redefine his function, as he inductively proceeds to larger ordinals, repeatedly) | |
| Oct 29, 2018 at 3:04 | comment | added | SSequence | Also, finally, the generic idea (that I was thinking along) seems to be whether we can define $\mathcal F_{\alpha}$ generically or not. And if the answer is "yes", then do the following two conditions hold(?): (a) We can do it in a somewhat uniform/systematic way (given the description of sequences below $\alpha$). (b) Do the definitions need to be adjusted constantly (the issue discussed in comment above). If the answer to (a) is yes and (b) is no, then the answer to question in OP is positive (in ZFC). But I will add that this is highly speculative (since it might be negative as well). | |
| Oct 29, 2018 at 2:54 | comment | added | SSequence | "Would $\psi_{\omega^3}$ agree with $\psi_{\omega^2}$ on $\mathcal F_{\omega^2}$?" It seems that the definitions for functions with maximum value less than $\omega^2$ would certainly need to be changed when moving from $\psi_{\omega^2}$ to $\psi_{\omega^2+1}$. That's because if for some $f_1$ we had $v_{f_1}=\omega^2$ and $v_{f_2}=\omega \cdot 10$. But on the other hand $f_1(\omega^2)<\omega \cdot 10$, then we might get an issue. But for values greater than $\omega^2+1$, I can't say without writing out the details a bit. | |
| Oct 29, 2018 at 2:45 | comment | added | SSequence | @Mirko You raised many points in the comments. Let me try to address them one by one. Yes, you are absolutely right regarding the issue with my def. of $F$. I think, subconsciously I was taking the additional assumption that the functions $f \in \mathcal F$ (and hence $\mathcal F_{\omega^2}$) increase smoothly (that is, $f$ can't skip on values less than $v_f$). It seemed easier to me to first see what we can do under that assumption. You would note that my definition is correct in that case. If we don't assume that $f$ increases smoothly, then some adjustment to it is needed definitely. | |
| Oct 28, 2018 at 22:00 | comment | added | Mirko | You may need to correct the definition of $F$. If $f(\omega\cdot(n+1))>\omega\cdot n$ then the condition $F(x)\ge f(x)$ fails at $x=\omega\cdot(n+1)$. This problem will be resolved if you define $F(x)$ to be the maximim of $f(x)$ and your current definition of $F(x)$, for each $x$. To illustrate, let $f(n)=0$ if $n<\omega$, $f(\omega+n)=n$, $f(\omega\cdot2)=\omega+5$, $f(\alpha)=\omega+10$ if $\alpha>\omega\cdot2$. Then your $F(\omega\cdot2)=\omega<\omega+5=f(\omega\cdot2)$, but we need $F(x)\ge f(x)$ for all $x$. Again, setting $F(x)=\max\{f(x),current F(x)\}$ resolves this problem. Thank you | |
| Oct 28, 2018 at 20:27 | comment | added | Mirko | I think I follow your description of $\psi_{\omega^2}$ (if $a=0$ take $F=f$). Would $\psi_{\omega^3}$ agree with $\psi_{\omega^2}$ on $\mathcal F_{\omega^2}$? Would $\psi_{\omega^\omega}$ agree with them? We could redefine $\psi_{\omega^n}$ to make it agree with $\psi_{\omega^\omega}$, but would we have to redefine $\psi_\alpha$ infinitely often? Unrelated: Let $\mathcal G_1=\{f\in\mathcal F:f(v_f)=0\}$. If $f\in\mathcal G_1$ let $\beta_f=\min\{\alpha:f(\alpha)>0\}$. Let $F(\alpha)=v_f$ if $\alpha\ge\beta_f$, else $F(\alpha)=0$. Works for $\mathcal G_1$. What about $\mathcal G_n=\{f:n_f=n\}$? | |
| Oct 28, 2018 at 19:15 | comment | added | Mirko | thank you for your answer. I believe that $\psi_\alpha$ would exist, I may have thought of this, don't remember anything specific, but tend to believe the answer for $\psi_\alpha$ is yes. There are other ways to partition $\mathcal F$, e.g. $\mathcal F_n=\{f\in\mathcal F:n_f=n\}$, $n_f$ as in Partial Answer (B), one could show there is $\psi$ that works for $\mathcal F_0\cup \mathcal F_1$ (thanks to Lynne for def $\mathcal F_n$). A vague idea, that I feel since yesterday, might be relevant, if $\psi$ exists for all of $\mathcal F$ try to relate to ordinal subtraction and get a contradiction | |
| Oct 28, 2018 at 13:39 | history | edited | SSequence | CC BY-SA 4.0 |
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| Oct 28, 2018 at 13:12 | comment | added | SSequence | Sorry I think there is a mistake in the specific construction that needs correction. I will edit it soon. | |
| Oct 28, 2018 at 5:15 | history | answered | SSequence | CC BY-SA 4.0 |