Timeline for Cantor's argument revisited
Current License: CC BY-SA 2.5
13 events
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| Nov 28, 2010 at 17:55 | comment | added | Andrés E. Caicedo | @Ewan : Yes, that is a natural question. I am not sure how one would formalize it, though. I'll think about it. | |
| Nov 28, 2010 at 17:52 | comment | added | Andrés E. Caicedo | @Ewan : Oh, I have no quarrel; all I meant was: "I do not see how to recover $F$ from the pair $(Y,Y')$." But my next comment addresses my question, so that's it. | |
| Nov 28, 2010 at 17:51 | comment | added | Ewan Delanoy | @ Andreas : another related question would be : we already know that there are definable subsets (A,B) witnessing noninjectivity, using Zermelo's argument. Is there a shorter, more "Cantor-like" argument, avoiding ordinals? Intuitively it looks like Cantor's diagonal argument should be "doubled", and A and B should resemble "mutually recursive" subets. | |
| Nov 28, 2010 at 17:45 | comment | added | Ewan Delanoy | @ Andreas : I do not understand your quarrel with "solving problem 2 solve problem 1". Indeed, it suffices to take $Y={\cal P}(X)$ and $Y'$ in the obvious way. I do not see what you mean when you say that $F$ is lost (both $Y'$ and the solution $B$ depend on $F$). | |
| Nov 28, 2010 at 17:43 | vote | accept | Andrés E. Caicedo | ||
| Nov 28, 2010 at 17:43 | comment | added | Andrés E. Caicedo | @Ewan : Well, very nice! Was not expecting the solution to go this way. | |
| Nov 28, 2010 at 17:36 | comment | added | Ewan Delanoy | @ Aaron : you asked "what is the point of the construction" ? Andreas' comments answer this : it shows that Andreas' definition does not exist in ZF (without AC), because if it existed we could prove AC. | |
| Nov 28, 2010 at 16:42 | comment | added | Andrés E. Caicedo | @Ewan : Hmm. If we can solve problem 2 "uniformly", then of course we can well-order any set, which gives a solution to problem 1. Is this how you meant the reduction to proceed? | |
| Nov 28, 2010 at 16:14 | comment | added | Andrés E. Caicedo | @Ewan : Thanks. Hope you do not mind the minor editing I made. I have a question: How does solving problem 2 solve problem 1? If you do something as Aaron suggests (or, say $Y={\mathcal P}(X)$ and $Y'$ as he suggests), I do not see how losing $F$ is not affecting definability in general. Anyway, I agree that your construction shows that solving problem 1 solves problem 2, which is not possible in general, thus showing problem 1 has a negative solution in general. | |
| Nov 28, 2010 at 16:06 | history | edited | Andrés E. Caicedo | CC BY-SA 2.5 |
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| Nov 28, 2010 at 15:47 | comment | added | Aaron Meyerowitz | I'm not sure I follow your reduction. Are you saying that we could take Y'=Y to be the set of all $B$ in ${\cal P}(X)$ such that $F(B)=F(A)$ but $B \ne A$? I am sure that that is not what you intend, but wouldn't that be a way to use your problem 2 to solve problem 1? Can you give an example where it is worth having $Y' \ne Y?$ Couldn't we always make that simplification? And then what is the point of the construction? (Although I do enjoy the construction!) | |
| Nov 28, 2010 at 9:57 | history | edited | Ewan Delanoy | CC BY-SA 2.5 |
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| Nov 28, 2010 at 9:52 | history | answered | Ewan Delanoy | CC BY-SA 2.5 |