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
11 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2018 at 9:34 | comment | added | Asaf Karagila♦ | @Zuhair: Well. You need that there is no bijection between the two sets. This follows from things like all sets are measurable or have the Baire property. | |
| Aug 11, 2018 at 9:33 | comment | added | Zuhair Al-Johar | @AsafKaragila can you explain what is this certain failure of choice, because that is one of the main motives behind this question, it is really about these certain types of failure of choice that are necessary for having such sets. | |
| Aug 11, 2018 at 6:57 | comment | added | Asaf Karagila♦ | (Note that in the case of $\Bbb R$ and $[\Bbb R]^\omega$, you need a certain failure of choice to ensure they are not of the same cardinality.) | |
| Aug 11, 2018 at 6:31 | comment | added | Asaf Karagila♦ | @Zuhair: That's correct. You just need two surjections. But you explicitly asked for $|X|<|Y|$, so there will be an injection from one to the other. And indeed, the sets need not be Dedekind-finite. It could just as well be $\Bbb R$ and $[\Bbb R]^\omega$. | |
| Aug 11, 2018 at 1:07 | vote | accept | Zuhair Al-Johar | ||
| Aug 11, 2018 at 0:44 | comment | added | Zuhair Al-Johar | this makes me think that if we have two sets $X,Y$ that are surjective onto each other and yet there do not exist any injection between them, then the same argument in the answer would be apply! isn't it? and this also violates choice! Now I think (I might be wrong really) that this doesn't need the sets to be Dedekind-finite sets. | |
| Aug 11, 2018 at 0:26 | comment | added | Zuhair Al-Johar | @AlecRhea thanks, I now know what Asaf is saying, I mean I understand that if this condition holds then there would be injections between the two powers in both directions and so the powers would be equal in size, but what I don't get is how in the first place we can have $X$ being strictly injective to $Y$ and yet also $X$ having a surjection onto $Y$? | |
| Aug 11, 2018 at 0:12 | comment | added | Alec Rhea | @Zuhair I believe $\leq^*=\{(Y,X): \text{there is a surjection from}\ X\ \text{onto}\ Y\}$. | |
| Aug 10, 2018 at 23:58 | comment | added | Zuhair Al-Johar | I want to understand this answer, first what is the definition of the relation $\leq^*$ | |
| Aug 10, 2018 at 23:40 | comment | added | Joel David Hamkins | Could you explain a little more how to construct the sets $X$ and $Y$ from a given Dedekind finite set? | |
| Aug 10, 2018 at 22:11 | history | answered | Asaf Karagila♦ | CC BY-SA 4.0 |