Timeline for Are there known examples of sets whose power set is equal in size to power set of larger sets only in absence of choice?
Current License: CC BY-SA 4.0
22 events
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| Aug 12, 2018 at 16:21 | history | edited | Asaf Karagila♦ |
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| Aug 11, 2018 at 13:56 | answer | added | Andrés E. Caicedo | timeline score: 12 | |
| Aug 11, 2018 at 9:30 | comment | added | Zuhair Al-Johar | @bof Ok I've edited the title, to resolve this confusion. Thanks | |
| Aug 11, 2018 at 9:29 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
edited title
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| Aug 11, 2018 at 1:57 | comment | added | bof | The title of your question is confusing. Apparently you are using "power" to mean "power set", but traditionally "power" is a synonym for "cardinality". | |
| Aug 11, 2018 at 1:07 | vote | accept | Zuhair Al-Johar | ||
| Aug 11, 2018 at 0:38 | comment | added | Joel David Hamkins | Well, AC failing is fundamentally connected with the failure of AC, and so it seems my example formally fulfills what you have requested. But I agree that it is silly, and this is why I believe that the question would benefit from greater clarity. | |
| Aug 11, 2018 at 0:03 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins I'm asking of a theorem in a theory in which choice fails which the above condition holds but that makes an essential use of failure of choice, i.e. it cannot be done with choice, the examples you've provided can be easily done with choice "simply replace "and AC fails" with "and AC holds" and you get the same result" so your examples doesn't provide an essential use of failure of choice, it look like cheating really. | |
| Aug 10, 2018 at 22:11 | answer | added | Asaf Karagila♦ | timeline score: 16 | |
| Aug 10, 2018 at 22:07 | comment | added | Asaf Karagila♦ | Related: Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice? | |
| Aug 10, 2018 at 21:05 | comment | added | Joel David Hamkins | That's fine. But still, I don't really understand what the question is exactly. We can obviously provide definitions of sets $x$ and $y$ with that property, such that the definitions succeed only if AC fails. e.g. $x$ is "the unique object which is $\omega$, provided Luzin's hypothesis holds and AC fails," and $y=\omega_1$. But that also is silly. So what is the actual question? | |
| Aug 10, 2018 at 20:59 | comment | added | Andrés E. Caicedo | @Joel The "making essential use of the way choice fails" is the comment-sized way of saying that the question is not about such silly interpretations. :-) | |
| Aug 10, 2018 at 20:51 | comment | added | FWE | @Joel I find this kind of shocking. | |
| Aug 10, 2018 at 20:43 | comment | added | Joel David Hamkins | @FWE But you are in good company, for it seems that many people are surprised by the independence of this statement. See mathoverflow.net/a/6594/1946. | |
| Aug 10, 2018 at 20:35 | comment | added | FWE | oh my sorry ..your critics is both correct. I withdraw the comments. | |
| Aug 10, 2018 at 20:34 | comment | added | Joel David Hamkins | Zuhair, could you clarify your question? What does it mean to provide an "example...that necessitates the violation of choice"? Are you asking for a model, a theory, a definition, or what? | |
| Aug 10, 2018 at 20:32 | comment | added | Joel David Hamkins | @AndrésE.Caicedo For that interpretation of the question, one can start with a model with Luzin's hypothesis or some other violation of injectivity for the continuum function, and then force a violation of AC via a symmetric construction much higher up. But I'm not sure if this is what Zuhair has asked. | |
| Aug 10, 2018 at 20:19 | comment | added | Andrés E. Caicedo | @FWE Your comment seems irrelevant to the question. In the context of AC, it is consistent that there are sets $x,y$ with $|x|<|y|$ and yet $|\mathcal P(x)|=|\mathcal P(y)|$. The question is asking whether such examples can be (consistently) provided in the absence of choice, probably making essential use of the way choice fails. | |
| Aug 10, 2018 at 20:18 | comment | added | Joel David Hamkins | @FWE Your comment is mistaken. It is consistent with ZFC that $2^\omega=2^{\omega_1}$; this is known as Luzin's hypothesis, and it holds in Cohen's model for the negation of CH. | |
| Aug 10, 2018 at 20:06 | comment | added | FWE | The reason for your question might be, that you believe, that you need AC for Cantor's diagonal argument. However this is not the case. | |
| Aug 10, 2018 at 19:59 | comment | added | FWE | Cantor's diagonal argument should ensue that the power set of a set has always a strictly higher cardinality? | |
| Aug 10, 2018 at 19:05 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |