Timeline for Cantor's argument revisited
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 28, 2010 at 23:07 | answer | added | Joel David Hamkins | timeline score: 9 | |
| Nov 28, 2010 at 17:43 | vote | accept | Andrés E. Caicedo | ||
| Nov 28, 2010 at 17:32 | comment | added | Ewan Delanoy | @ Aaron : more generally, call a subset $X'$ of $X$ stable if F maps ${\cal P}(X')$ to $X'$ (thus $X$ and $A$ are stable). Then the operation $\Phi(X')=\lbrace F(Z) | Z \subseteq X, F(Z) \not\in Z \rbrace $ preserves stable sets : $\Phi(X')$ stays stable if $X'$ is. Also, any intersection of stable sets is again stable, so we may consider for exemple the smallest stable set containing $F(A)$. However, if you want to use some limiting process you need Zorn's lemma and AC somewhere (my answer proves this). | |
| Nov 28, 2010 at 17:06 | comment | added | Aaron Meyerowitz | Just a question: Clearly the restriction of $F$ to $A$ maps $F: {\cal P}(A) \to A$ so we could take $$A'=\lbrace F(Z) | Z\subseteq A, F(Z) \not\in Z\rbrace.$$ Can one define the limit of this process? That is, some $Y \subseteq X$ so that $F: {\cal P}(Y) \to Y$ and $$Y=\lbrace F(Z) | Z\subseteq Y, F(Z) \not\in Z\rbrace?$$ | |
| Nov 28, 2010 at 9:52 | answer | added | Ewan Delanoy | timeline score: 13 | |
| Nov 27, 2010 at 19:25 | comment | added | Andrés E. Caicedo | @Ewan: If $B$ is unique, it is obviously definable, so the issue is really when this is not the case. Note I am not requiring $B$ to be first order definable on $({\mathcal P}(X)\sqcup X,\subseteq, F)$, simply to be definable from $F$ in the language of set theory; this is a much more generous notion of definability. | |
| Nov 27, 2010 at 14:32 | comment | added | Ewan Delanoy | One thing that makes me pessimistic about this problem is that it is easy to construct (non-injective) examples where the solution B is unique. Contrary to the usual Cantor proof of non-surjectivity where there is a plethora of solutions, which allows a "gross", "greedy" diagonalization construction to work, whatever "definable" construction we find must be sharp in many cases. | |
| Nov 26, 2010 at 20:41 | comment | added | Andrés E. Caicedo | A small comment: If the answer is positive, I expect it will be an explicit construction, and that's what I'm hoping for. It also explains the "combinatorics" tag. If the answer is negative, though, perhaps the tag was misplaced after all. I apologize for the possible inaccuracy. | |
| Nov 26, 2010 at 20:38 | history | asked | Andrés E. Caicedo | CC BY-SA 2.5 |