Timeline for Can ZFC commit cardinality errors?
Current License: CC BY-SA 4.0
22 events
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| Sep 27, 2019 at 21:39 | vote | accept | Zuhair Al-Johar | ||
| Sep 24, 2019 at 16:49 | comment | added | Zuhair Al-Johar | Yes it is the first option (with "c" not allowed to be used in the formulation of cardinal in equality and cardinal equality), you forgot that ZFC has identity relation also = with the first order identity axioms. | |
| Sep 24, 2019 at 16:39 | history | edited | Greg Kirmayer | CC BY-SA 4.0 |
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| Sep 24, 2019 at 16:23 | history | edited | Greg Kirmayer | CC BY-SA 4.0 |
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| Sep 24, 2019 at 6:21 | comment | added | Greg Kirmayer | Perhaps you mean that a model M of the language with a binary relation โ and a functon symbol c commits a cardinality error of the first kind if ZFC holds in M for the โ-formulas, the Cardinal Equality axioms hold in M, the Cardinal Inequality rules hold in M, and โ๐,๐:|๐|โ |๐|โง๐(๐)=๐(๐) holds in M. Also maybe you mean that a model M for the language with a binary relation symbol โ does not commit a cardinality error of the first kind if for all functions f:M-->M, (M,f) does not commit a cardinality error of the first kind. | |
| Sep 24, 2019 at 6:15 | comment | added | Greg Kirmayer | "if a set theory T extended with the above, proves" a certain statement, then it is "guilty of committing cardinaity error of the" first(or(with different statement) second) kind. What it means for a model for the language with a binary relation symbol, or a model for the language with a binary relation symbol and a function symbol, to suffer a cardinality error has not been defined. | |
| Sep 24, 2019 at 3:42 | comment | added | Zuhair Al-Johar | No! It doesn't need to be changed, the definition is clearly saying "committing" also "guilty" of, it doesn't say "..then it can commit cardinality error...", 'can' mean that some models of the theory can suffer this cardinality error, which I think the majority of models of ZFC does commit cardinality error of the second kind, but never of the first kind. | |
| Sep 23, 2019 at 19:06 | comment | added | Greg Kirmayer | The question poser wrote "notice that the question is "can" and not "does", it might be the case that ZFC doesn't commit cardinality errors but it "can" commit them". "if a set theory T extended with the above, proves that:โ๐,๐:|๐|โ |๐|โง๐(๐)=๐(๐) Then its guilty of committing cardinaity error of the first kind." Perhaps this needs to be changed. | |
| Sep 23, 2019 at 18:40 | comment | added | Greg Kirmayer | The question poser wrote "I didn't get your comment regarding the second c(X)=c(Y) and |X|โ |Y|". This is a proof that |X|=|Y| iff c(X)=c(Y), in the given extension of ZFC(with parameters). First we showed that if |X|=|Y|, then c(X)=c(Y). Then we assumed that c(X)=c(Y) and |X|โ |Y| and obtained a contradiction. b is the least cardinal such that there exist X and Y with |X|=b and c(X)=c(Y) and |X|โ |Y|. The conditions of the rule are met by this choice of ๐ and ๐. The conclusion of the rule yields a contradiction and so our assumption could not have held. | |
| Sep 23, 2019 at 18:28 | comment | added | Zuhair Al-Johar | your formulation include "c", which must not be used in formulations of axioms and inference rules, because they are not primitives of the tested theory. I didn't mention that right. But it should be like that. | |
| Sep 23, 2019 at 13:59 | comment | added | Zuhair Al-Johar | This proves that ZFC would not prove the existence of sets which possess cardinailty errors yes, but it doesn't say that ZFC is inconsistent with them. I think that ZFC is inconsistent with errors of the second kind, but some models of ZFC can commit errors of the first kind, so the theory "ZFC + ZFC commits error of the first kind", is I think consistent. Actually even some models of ZFC+V=L can commit cardinality error of the first kind | |
| Sep 23, 2019 at 11:04 | comment | added | Zuhair Al-Johar | notice that the question is "can" and not "does", it might be the case that ZFC doesn't commit cardinality errors but it "can" commit them, it means it doesn't prove that it doesn't commit them. I tend to think that it is provable in this extended theory that ZFC cannot commit cardinality error of the first kind, but definitely it can commit the second kind error, and that's why Monroe Eskew was speaking about in his comments. | |
| Sep 23, 2019 at 6:11 | comment | added | Zuhair Al-Johar | formulas must not use the symbol "c" because it is not a primitive of the tested theory. | |
| Sep 23, 2019 at 1:14 | history | edited | Greg Kirmayer | CC BY-SA 4.0 |
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| Sep 23, 2019 at 1:09 | history | edited | Greg Kirmayer | CC BY-SA 4.0 |
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| Sep 22, 2019 at 18:48 | comment | added | Zuhair Al-Johar | @what is the proof that there is a model of that theory in which c(X)=|X|? by the way I don't think the global choice function is a definable function, so how you'll introduce it in your argument? | |
| Sep 21, 2019 at 7:03 | comment | added | Zuhair Al-Johar | I didn't get your comment regarding the second c(X)=c(Y) and |X| $\neq$|Y|. are you suggesting that b=c(X) and also that b=|X|, why should those agree, there might be no set X assigned the same cardinal object value under both $c$ and || functions? and why the inference rule should result in c(X)$\neq$c(Y) | |
| Sep 20, 2019 at 20:37 | comment | added | Zuhair Al-Johar | what if we forbid $\phi(x,y)$ from having parameters? | |
| Sep 20, 2019 at 19:37 | comment | added | Greg Kirmayer | correction of previous comment:Suppose |X|=|Y|. Then there is a bijection f from X to Y. Let ๐(๐ฅ,๐ฆ,z) be the formula (xy)โz. Then โ๐ฅโ๐โ!๐ฆโ๐(๐(๐ฅ,๐ฆ.f))โงโ๐ฆโ๐โ!๐ฅโ๐(๐(๐ฅ,๐ฆ.f). Thus c(X)=c(Y). Suppose there are X and Y with c(X)=c(Y) and |X|โ |Y|. Let b be the least cardinal of such an X. Let ๐(๐) be the formula |X|=b. Let ๐(๐) be the formula |Y|โ b. By the inference rule c(X)โ c(Y) whenever ๐(๐) and ๐(๐). This contradicts our choice of b and thus c(X)=c(Y) iff |X|=|Y | |
| Sep 20, 2019 at 17:49 | vote | accept | Zuhair Al-Johar | ||
| Sep 24, 2019 at 17:05 | |||||
| Sep 20, 2019 at 16:28 | history | edited | Greg Kirmayer | CC BY-SA 4.0 |
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| Sep 20, 2019 at 16:18 | history | answered | Greg Kirmayer | CC BY-SA 4.0 |