Timeline for Does ZFC prove the universe is linearly orderable?
Current License: CC BY-SA 3.0
24 events
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| Nov 3, 2012 at 7:19 | vote | accept | Asaf Karagila♦ | ||
| Nov 3, 2012 at 3:47 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Fixed the forcing relation symbol
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| Nov 3, 2012 at 3:39 | comment | added | Joel David Hamkins | I've now fixed the argument, by moving up a level, which is what we should have expected by analogy with the Cohen real case. | |
| Nov 3, 2012 at 3:36 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 3, 2012 at 3:29 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Argument repaired
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| Nov 2, 2012 at 21:14 | comment | added | Gerhard Paseman | I'm mildly surprised that my intuition was correct, although I am not surprised that it was for the wrong reason. I was not consciously thinking of extending set partial orders to set linear orders. Gerhard "Not Doing Much Introspection These Days" Paseman, 2012.11.02 | |
| Nov 2, 2012 at 14:17 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
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| Nov 2, 2012 at 13:32 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 28, 2012 at 17:19 | vote | accept | Asaf Karagila♦ | ||
| Nov 2, 2012 at 13:53 | |||||
| Oct 28, 2012 at 16:02 | comment | added | Joel David Hamkins | Gerhard, perhaps you are thinking of the result that any partial order extends to a linear order. The issue is that although one can prove this for set orders in ZFC, using the axiom of choice, in order to extend the argument to proper class sized orders, we need a choice principle suitable for class sized objects, which is the global choice principle. In the model of my answer, although we have the axiom of choice for sets, we do not have the global choice principle, and there is no way to assemble all the set linear orders together into one class linear order. | |
| Oct 27, 2012 at 18:11 | comment | added | Gerhard Paseman | I may be able to write down the function which has the desired value for f(42), but then I have to convince myself that f is an automorphism. However, I will buy that P is such that there are a sufficiency of initial fragments such that there are a wealth of automorphisms that fix each fragment and mix the rest of P in nice fashions. Part of what I was hoping for were some order-theoretic characteristics of P. I also hope that Joel has a different take on my difficulty so as to provide a different comment. Gerhard "Will Think On Yours, Thanks" Paseman, 2012.10.27 | |
| Oct 27, 2012 at 17:44 | comment | added | Asaf Karagila♦ | To your quick question about $\pi$, we know it exists because we can describe it. We know how the forcing looks like, and we know how to write the definition of $\pi$. Ask yourself for functions from the natural numbers into $\lbrace0,1\rbrace$, can you actually write down the automorphism which switches the values at $42$, whereas the values of all functions before $42$ remain constant? Of course you can. The idea is similar here, only applied to a much larger scale. | |
| Oct 27, 2012 at 17:42 | comment | added | Asaf Karagila♦ | Gerhard, the well-ordering of $L$ is not only generated by the rank, but by the fact that we can identify each member not only its rank but also the defining formula and its parameters. By the combination of these two we can generate a well-ordering of the universe. Outside of $L$, in strange and complicated universes obtained by class forcing (for example) one cannot do that anymore. Now not everything is definable from ordinals in the sense that $V=HOD$ (which is true in $L$, obviously), and so we might not have an effective way of deciding who's on first. | |
| Oct 27, 2012 at 17:41 | comment | added | Gerhard Paseman | Granted I am thinking of Conway's Surreal Numbers as I write this, and that might lead my intuition astray. So my first naive question is, is there a simpler argument why rank functions or predicates don't provide such an order? Also, one problem I have with proof by contradiction is that it is easy for me to insert another wrong assumption and hard to find out when I did. Just to make this quick, e.g. why should I believe that your claimed specific automorphism pi exists? I'm hoping for an answer easier than 'Take Math 235 again.' Gerhard "Maybe If It's On Coursera..." Paseman, 2012.10.27 | |
| Oct 27, 2012 at 17:33 | comment | added | Gerhard Paseman | I am having a little difficulty with the above, and it will take me two or three comment blocks to explain. The source of the difficulty is the notion of rank functions definable in strong enough fragments of ZFC. I thought that the machinery to rank members of L not only implied a well order on L, but that such a rank extended to V, and I (possibly mis-) remember the ranking notion as absolute. (Despite Solovay's best efforts, I am still pretty weak on set theory.) Gerhard "Continued In The Next Block" Paseman, 2012.10.27 | |
| Oct 27, 2012 at 12:31 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
Two minor typos, I believe.
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| Oct 27, 2012 at 12:30 | comment | added | Asaf Karagila♦ | Joel, I believe you have a couple of typos there. I will correct them now, but feel free to undo my doing if I'm just misunderstanding. | |
| Oct 27, 2012 at 12:18 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 27, 2012 at 12:18 | comment | added | Asaf Karagila♦ | Yes, well-ordering I understand that is definable. I imagine this is about the same proof as the usual proof that AC holds in the ground model then it holds in the extension, but for GC. Your suggestion for a next question is good; I actually had in mind something else: can we prove this definable linear order is an extension of $(V,\in)$ or $(V,\subseteq)$? Essentially this is the same track as the usual independence of linear order theorems, but with the axiom of choice holding and global choice failing... | |
| Oct 27, 2012 at 12:06 | comment | added | Joel David Hamkins | Yes. If the ground model has a global well-order, then allowing $G$ as a predicate in $V[G]$ will allow us to define a well-order in $V[G]$: $x$ is before $y$ if $x$ has an earlier name than $y$. (For this, you have to define $V$ in $V[G]$, but this is possible by my theorem on the definability of the ground model after closure point class forcing.) | |
| Oct 27, 2012 at 12:02 | comment | added | Asaf Karagila♦ | So if we allow $G$ as a predicate, we should be able to define a linear ordering of he universe in ZFC? Or would we still have to talk about it in NBG? | |
| Oct 27, 2012 at 11:59 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 27, 2012 at 11:54 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Oct 27, 2012 at 11:47 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |