Timeline for Is Replacement motivated by ranked iterative conception of sets?
Current License: CC BY-SA 4.0
29 events
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| Nov 25, 2018 at 18:58 | comment | added | Zuhair Al-Johar | I think Randall Holmes article that I've cited, addresses Kanamori's article you've mentioned. | |
| Nov 23, 2018 at 21:35 | vote | accept | Zuhair Al-Johar | ||
| Nov 23, 2018 at 21:26 | comment | added | Zuhair Al-Johar | One needs to also mention that Randall Holmes also sides by Boolos on this issue, and that he doesn't think at all that Replacement is built in the cumulative hierarchy, and that it exceed it. | |
| Nov 23, 2018 at 21:13 | comment | added | Zuhair Al-Johar | Thanks really, I must correct the posting as to address this mis-understanding and present the rest of my argument as being not related to Boolos, but rather about being related to the first point mentioned in the Wikipedia | |
| Nov 23, 2018 at 21:11 | comment | added | Noah Schweber | @ZuhairAl-Johar I've rewritten for clarity. Short summary: the theory Boolos has in mind does prove replacement without difficulty; you're correct that your theory doesn't; I waffle on whether I agree with you philosophically. | |
| Nov 23, 2018 at 21:08 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| Nov 23, 2018 at 20:53 | comment | added | Zuhair Al-Johar | ah, I see now, I thought by "this" you mean the theory that I've presented here. | |
| Nov 23, 2018 at 20:52 | comment | added | Noah Schweber | @ZuhairAl-Johar Ah, I was saying that Boolos' system proves replacement (that is, that Boolos' claim was correct); at a glance, I think your system doesn't prove replacement. | |
| Nov 23, 2018 at 20:50 | comment | added | Zuhair Al-Johar | I'm speaking about my system and not what Boolos meant. In my system your $R$ is not a ranking function? | |
| Nov 23, 2018 at 20:45 | comment | added | Noah Schweber | @ZuhairAl-Johar I'm not sure which step in my argument you're dubious of. Do you not understand my definition of $R$? Or why the $s$ we get from applying Boolos' principle to that $R$ is an upper bound on each $\omega_{\omega_{...}}$ (finitely many times)? Or why the existence of such an upper bound implies that the least fixed point of $\alpha\mapsto\omega_\alpha$ exists? I didn't write out all the details, but I don't think I left anything substantive out. | |
| Nov 23, 2018 at 20:43 | comment | added | Zuhair Al-Johar | ah, I see what you mean, you mean we add the assumption that every relation from sets to stages, then for every set there must exist..., Ok I need to think about that really. But still as for this theory, you didn't really formally prove that the first fixed point of the $\omega$ function is a set. You made a claim, but you didn't present a proof. | |
| Nov 23, 2018 at 20:39 | comment | added | Noah Schweber | @ZuhairAl-Johar That's an idiom. I mean "no additional hypotheses." | |
| Nov 23, 2018 at 20:39 | comment | added | Zuhair Al-Johar | where is the full stop, I see a comma? | |
| Nov 23, 2018 at 20:37 | comment | added | Noah Schweber | @ZuhairAl-Johar Right, fixed. (I apparently couldn't decide between "$s$" and "$F(z)$.") | |
| Nov 23, 2018 at 20:36 | comment | added | Noah Schweber | Again, we're not adding ranking functions, that's not what Boolos is talking about. In modern terms he's considering the principle "Every class function from sets to ordinals is "locally bounded,"" or "Ord is regular." | |
| Nov 23, 2018 at 20:34 | comment | added | Noah Schweber | @ZuhairAl-Johar No, it doesn't. You're conflating things. Boolos' principle is: "Whenever $F$ is a class map sending sets to stages - full stop - then for every set $z$ there is some stage $s$ such that for each $x\in z$ we have $F(x)$ is earlier than $s$." Note that there are no assumptions placed on $F$ at all here. I really have no idea where you're getting the claim that Boolos is restricting attention to correlations which satisfy the additional ranking property here. | |
| Nov 23, 2018 at 20:34 | comment | added | Zuhair Al-Johar | no you didn't present a proof, you made a claim that we can simply find .... , but I don't see how we can do that simply? from where we get this alleged large enough stage you are alluding to? I mean in my above mentioned formulation. Actually I don't see how we can get this even if we add all ranking functions. | |
| Nov 23, 2018 at 20:32 | comment | added | Zuhair Al-Johar | he doesn't use "rank" yes, but his second condition implies that! | |
| Nov 23, 2018 at 20:29 | comment | added | Zuhair Al-Johar | you need to complete his condition "...then for any set z there is a stage s such that for each member w of z, s is later than some stage with which w is correlated", this means that a ranking function is a subset of this relation he is speaking about. so your function if not adjoined with a ranking function, then it doesn't satisfy Boolos. | |
| Nov 23, 2018 at 20:29 | comment | added | Noah Schweber | In fact, Boolos doesn't use the word "rank" at all in the quoted passage, so I don't see how you're justified in assuming that the "correlations" Boolos mentions have to satisfy the additional ranking property. | |
| Nov 23, 2018 at 20:27 | comment | added | Noah Schweber | You are right, however, that my version of the axiom does misuse the term "ranking function." Rather, the one-axiom version should be: "If $R$ is any class relation assigning sets to stages, then for any set $z$ there is a stage $s$ such that each element of $z$ is $R$-related to a stage $t<s$." | |
| Nov 23, 2018 at 20:25 | comment | added | Noah Schweber | @ZuhairAl-Johar As to how this gives you the least fixed point of $\alpha\mapsto\omega_\alpha$, I explained exactly that in my second-to-last paragraph. And I explained how you get full replacement in the last paragraph. | |
| Nov 23, 2018 at 20:25 | comment | added | Noah Schweber | @ZuhairAl-Johar Note that Boolos' principle "'If each set is correlated with at least one stage (no matter how), then for any set z there is a stage s such that for each member w of z, s is later than some stage with which w is correlated'." does not say that the correlation in question needs to be a ranking function! That is, Boolos is considering arbitrary (class) maps from sets to stages. Similarly, I never describe the map given by my $\varphi$ as a ranking function. | |
| Nov 23, 2018 at 20:13 | comment | added | Zuhair Al-Johar | you mean to add on top of this theory the axiom: "For any class $R$, if $R$ is a ranking function..." but this would be excessive buildup of hierarchies, I thought he meant at least one ranking function per theory, and then say that one of those theories must imply replacement. But even if we add all ranking functions, would that prove Replacement? | |
| Nov 23, 2018 at 20:06 | comment | added | Zuhair Al-Johar | yet it is more into the heart of this issue. About your remark that "this" theory implies replacement, then do you mean by "this" the theory I've formulated here or Boolos's theory, if you mean the former, then please tell me how, how for example I can prove that $\omega_{\omega_{\omega_{...}}}$ (i.e. $\omega$ times) is a set in this theory? | |
| Nov 23, 2018 at 20:04 | comment | added | Zuhair Al-Johar | continuation..., but notice that the function you've spoken about is not fulfilling Boolos, since it is not membership sensitive, I can have a singleton of a doubleton set, and by then the member of that singleton is sent to $\omega_{\omega}$ while the singleton is sent to $\omega$, however the union of it with a ranking function would be a function that satisfy Boolos. However this branching by supersetting a ranking function is not desirable at all, and it is not related to the iterative principle, as Boolos himself admits. And so although my formlation might have not captured Boolos's ... | |
| Nov 23, 2018 at 20:02 | comment | added | Noah Schweber | @ZuhairAl-Johar That still doesn't seem strong enough to me: the point of Boolos is that all definable ranking functions need to be considered. And if you make $R$ a many-many relation, then it doesn't necessarily contain any information (e.g. $R=V\times V$). The right axiom, if you want to work in class theory, is "For any class $R$, if $R$ is a ranking function then ...." | |
| Nov 23, 2018 at 19:59 | comment | added | Zuhair Al-Johar | Yes, I understand the schematic point you'r raising, but here I'm speaking in term of "classes" and so we can reduce schemas into single axioms. Anyhow, you are right regarding the "at least one" point that Boolos had mentioned, this can be translated here by changing $H_1$ into R is a relation (i.e. a set of ordered pairs) that is a superclass of a ranking function on V, this way R can be one-many relation, or even many-many relation, and we can have part of it being the function you've mentioned. to be continued.... | |
| Nov 23, 2018 at 17:57 | history | answered | Noah Schweber | CC BY-SA 4.0 |