Timeline for What is the strength of this strict constructible iterative hierarchy?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26, 2018 at 5:43 | vote | accept | Zuhair Al-Johar | ||
| Nov 26, 2018 at 5:43 | comment | added | Zuhair Al-Johar | I see, there is no countable stage that can contain all subsets of $\omega$. Thanks | |
| Nov 25, 2018 at 22:07 | comment | added | Andreas Blass | @ZuhairAl-Johar To answer your latest comment, have you really looked at all the quantifiers in "$x$ is a well-ordering on $\omega$"? In either of the two formulations "every nonempty subset has a first element" or "there is no infinite decreasing sequence", there is a quantifier which, if interpreted as ranging only over $L_{\omega+1}$, is too weak to ensure that $x$ is really well-founded, i.e., is order-isomorphic to an ordinal. | |
| Nov 25, 2018 at 18:28 | comment | added | Zuhair Al-Johar | ...I see the formula $( x \in L_{\omega+1} \wedge x \text{ is a well ordering on } \omega) $ with all of its quantifiers bounded by $L_{\omega+1}$ to be a legitimate formula from $\langle L_{\omega+1}, \in\rangle$ world, so it can be used to construct a set in $L_{\omega+2}$ stage. There must be a problem with the formulation, but honestly I cannot figure it out. | |
| Nov 25, 2018 at 18:28 | comment | added | Zuhair Al-Johar | you must be right but I'm failing to know my error. you say that the length of every well ordering on $\omega$ that is in $L_{\omega_1^{CK}}$ must be a recursive ordinal, I'm not doubting what you are saying, but I'm rather failing to see how this is the case. I'm failing to see how the set of well orderings on $\omega$ that are elements of $L_{\omega+1}$ is not an element of $L_{\omega_1^{CK}}$, .... | |
| Nov 25, 2018 at 12:45 | comment | added | Andreas Blass | If, on the other hand, you use "set" to mean "set in $L_{\omega+2}$", then the resulting definition of "well-ordering" will include some relations that are not order-isomorphic to ordinals and therefore don't make sense in the context of using them to determine the extent of a hierarchy. | |
| Nov 25, 2018 at 12:44 | comment | added | Andreas Blass | @ZuhairAl-Johar The set of well-orderings on $\omega$ that are elements of $L_{\omega+1}$ and the set of equivalence classes of such well-orderings are not elements of $L_{\omega+5}$ or even of $L_{\omega_1^{CK}}$. The same is true if you use the word "set" (in the definition of well-ordering) to mean "set in the constructible hierarchy". (continued in next comment) | |
| Nov 25, 2018 at 11:04 | comment | added | Zuhair Al-Johar | .. to the $L_{\omega+5}$ would be the Scott ordinal for some ordinal $d \geq \omega_1^{CK}$. Which would be an element of $L_{\omega+6}$. Now I don't know how this can conform with what you are saying, it appears to negate it since your answer seem to imply that there is no well ordering relation that is a set in $L_{\omega_1^{CK}}$ that is not hyper-arithmetical, but clearly any element of $d$ is not hyper-arithmetical and yet it is in $L_{\omega_1^{CK}}$? | |
| Nov 25, 2018 at 10:50 | comment | added | Zuhair Al-Johar | I use the word "set" to mean an element of the stages of the constructible hierarchy that I've described. obviousely that is not the way you are using it. Now I thought that the formula "x is an equivalence class of well orderings on $\omega$ that are elements of $L_{\omega+1}$, under equivalence relation "isomorphism"" seems to me to be a formula of the $L_{\omega+2}$ world, so it can be used to construct elements of the stage $L_{\omega+3}$, so the set of all those equivalence classes is an element of $L_{\omega+3}$ stage. Now the set of all well orderings on the latter set belonging ... | |
| Nov 24, 2018 at 23:10 | comment | added | Andreas Blass | @ZuhairAl-Johar The set of equivalence classes that you described in your first comment is indeed a set, but it is not in $L_{\omega_1^{CK}}$. | |
| Nov 24, 2018 at 23:08 | comment | added | Andreas Blass | @ZuhairAl-Johar The class of relations on $\omega$ that are elements of $L_{\omega+1}$ is not definable over $L_{\omega+1}$. The reason is that there are recursive relations that are linear orderings, that have no recursive (or even hyperarithmetical) decreasing sequences, yet are not well-orderings. Similarly, the well-founded part of a recursive linear order on $\omega$ is not necessarily hyperarithmetical. (It is $\Pi^1_1$, but that won't put it into $L_{\omega_1^{CK}}$.) | |
| Nov 24, 2018 at 20:09 | comment | added | Zuhair Al-Johar | I think the answer as it stands is wrong, or perhaps I'm missing something. | |
| Nov 24, 2018 at 20:07 | comment | added | Zuhair Al-Johar | the idea of my construction is that the height of the whole hierarchy must not be ordinally smaller than a set in the hierarchy, so any well ordering on a set that is in the hierarchy must be isomorphic to a well ordering on a sub-hierarchy of the whole hierarchy (a subhierarchy is an initial segment of levels of the hierarchy), see my comments to Andrés E. Caicedo. | |
| Nov 24, 2018 at 18:17 | comment | added | Zuhair Al-Johar | But the class of all equivalence classes of recursive well orderings of $\omega$ that are elements of $L_{\omega +1}$ under equivalence relation "isomorphism", is a set (I'm just using Scott's trick)! and it would be well orderable by a well ordering relation that is a set, and that would be a well ordering that is isomorphic to a well ordering on $\omega_1^{CK}$, clearly this theory is going way beyond $\omega_1^{CK}$ | |
| Nov 24, 2018 at 16:44 | history | answered | Andreas Blass | CC BY-SA 4.0 |