Timeline for Does Cantor-Bernstein hold for classes?
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Sep 6, 2010 at 20:51 | comment | added | Joel David Hamkins | Martin, the proof can be fixed by defining the sets $A_n$ internally, using finite sequence as I do in my proof, rather than defining them in the meta-theory as you have done. One way to think about it is that in a nonstandard model, you want your iteration to continue into the nonstandard natural numbers, which will only occur properly if you do the construction internally. But by defining a back-and-forth sequence (or something equivalent), as I do in my answer, one can carry this out. | |
| Sep 4, 2010 at 19:40 | comment | added | Martin Brandenburg | The point is that if we use the notation of Francois' proof, the range of $s$ lies in a fixed set, which is definable from $n$ (in fact, we may take it constant $V_\omega$. Then we may view $s$ as a set and the existence of $s$ is definable, and the proof goes through. But I don't know if the range is always such a set. | |
| Sep 4, 2010 at 19:37 | comment | added | Martin Brandenburg | I've played around with an example: If there is a injection g:$\mathbb{N}$→A, then there is a bijection h : A \ {g(0)} → A, namely x↦x if x is not in the image of g, and otherwise g(n)↦g(n−1). In the case of A=V, the universe, and g(n) = {...{0}...} (n braces), it is possible to write down a well-defined formula for h. | |
| Sep 4, 2010 at 8:43 | comment | added | Asaf Karagila♦ | Hmmm. Yes, I see the problem now. | |
| Sep 4, 2010 at 2:07 | comment | added | Martin Brandenburg | I don't understand. My problem is: We may build by "outer induction" a formula for $A_n$ for every $n$, but there does not seem to be a formula $\psi$ such that $\psi(n,x)$ iff $x \in A_n$. | |
| Sep 4, 2010 at 0:44 | comment | added | Asaf Karagila♦ | I think that if you look at the class $\omega\times A\times B$, then you can define $A_n$ and $B_n$ as sub-classes for which you have $\langle n, x, b\rangle$ where $x$ is the variable for the formula defining $A_n$. Then you can take intersection or union by writing a formula which says that for all/some $n\in\omega$ the pair $\langle a,b\rangle$ satisfies some property. Then you verify that it is in fact a function and you should be done. | |
| Sep 3, 2010 at 23:12 | comment | added | Martin Brandenburg | Right, this is another issue. | |
| Sep 3, 2010 at 23:07 | comment | added | Asaf Karagila♦ | Since you treat classes as formulae, I'm not sure you can always intersect or take a union of infinitely many, as this would translate to an infinitely long formula. So the definition for $h$ is a bit problematic. | |
| Sep 3, 2010 at 23:02 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Sep 3, 2010 at 22:48 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Sep 3, 2010 at 22:21 | history | answered | Martin Brandenburg | CC BY-SA 2.5 |