Timeline for Proofs of the uncountability of the reals
Current License: CC BY-SA 3.0
14 events
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| Aug 18, 2020 at 8:13 | comment | added | Martin Sleziak | This link to the Kanamori-Pincus was linked in another answer: math.bu.edu/people/aki/7.pdf (Wayback Machine). | |
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Sep 7, 2017 at 2:50 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jul 31, 2014 at 1:28 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
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| Feb 27, 2014 at 6:30 | comment | added | Andrés E. Caicedo | @MartinBrandenburg A very late reply, but here is a reason to bother: The argument shows that 1) The collection of well-orderable subsets of $X$ has strictly larger size than $X$. This is an improvement over Cantor's result in the context of $\mathsf{ZF}$. 2) Given any $f:\mathcal P(X)\to X$, we can find $A\subsetneq B$ with $f(A)=f(B)$. This is also a combinatorial strengthening (and it can be pushed further, see here). | |
| Nov 26, 2010 at 17:39 | comment | added | Andrés E. Caicedo | @François : I located a paper from Aki and Pincus where the argument is presented. It actually seems to go back to Zermelo. (And I used the paper in the course I gave, so most likely I originally read it there.) | |
| Nov 26, 2010 at 17:37 | comment | added | Andrés E. Caicedo | @Martin: I think rather than "coming to me", I read the argument in a paper by Kanamori-Pincus. I added a reference. | |
| Nov 26, 2010 at 17:36 | history | edited | Andrés E. Caicedo | CC BY-SA 2.5 |
Added a reference.
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| Nov 26, 2010 at 16:22 | comment | added | Andrés E. Caicedo | @François : Oh, I should email Aki and ask him, then. Thanks! | |
| Nov 26, 2010 at 15:25 | comment | added | François G. Dorais | Now that I see it, I think I had seen this proof in a talk by Aki Kanamori. If I remember correctly, Aki attributed the proof to Tarski. Since this is from a long time ago, my memory may be off... | |
| Nov 23, 2010 at 3:46 | comment | added | Andrés E. Caicedo | Hehe. As I mentioned, I found this argument while teaching a topics course; meaning: I was lecturing on ideas related to the arguments above, and while preparing notes for the class, it came to me that one would get a diagonalization-free proof of Cantor's theorem by following the indicated path; I looked in the literature, and couldn't find evidence of this being known. I wasn't explicitly looking for it at any point. Anyway, there is technical interest in the matter, precisely because the argument is so ubiquitous, as Joel's answer indicates. | |
| Nov 23, 2010 at 3:12 | comment | added | Martin Brandenburg | Although I think this is very interesting, I still wonder why we bother trying to avoid the diagonal argument by using tons of more advanced arguments, which perhaps in the end, when we enfold them into elementary arguments, use some sort of diagonal argument. | |
| Nov 23, 2010 at 0:01 | history | answered | Andrés E. Caicedo | CC BY-SA 2.5 |