Timeline for What fragments of ZF are consistent with a set being equal in size to its power set?
Current License: CC BY-SA 4.0
6 events
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| Aug 26, 2018 at 8:53 | comment | added | Andrej Bauer | I am not familiar with NF, so thanks for explaining where things fail to work. From a category-theoretic point of view I find it odd that we'd still call something "function space" or "powerset" if it does not behave like one. | |
| Aug 26, 2018 at 3:00 | comment | added | Zuhair Al-Johar | as about the ability of the fragments that I've mentioned, the last one which is the one having non-extensionals is as strong as $\omega$-order arithmetic. I think (I'm not sure) that we can have much stronger fragments and yet compatible with existence of a set that is equipotent with its power. | |
| Aug 26, 2018 at 2:55 | comment | added | Zuhair Al-Johar | The definition of map $g$ is not stratified, you have $e(x)(x)$ so $x$ will receive the same type as $e(x)$ since $e(x)$ is the image of $x$ under $e$ and also $x$ appears one type lower than $e(x)$ since it is the argument of $e(x)$, so that won't work in the stratified fragments that I've mentioned. | |
| Aug 25, 2018 at 22:30 | comment | added | Malice Vidrine | Note that this is true even in NF as long as you look at it categorically. The reason you can have $V=\mathcal{P}(V)$ is because $\mathcal{P}(V)$ is not a categorical power object of $V$; it's a power object of $\{\{x\}:x\in V\}$. NF's stratification requirement prevents us from forming the necessary universal maps to make it the power object of $V$. Something similar will probably be going on in any such fragment: your "power sets" will lack a certain universal property. | |
| Aug 25, 2018 at 10:13 | history | edited | Andrej Bauer | CC BY-SA 4.0 |
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| Aug 25, 2018 at 10:05 | history | answered | Andrej Bauer | CC BY-SA 4.0 |