Timeline for A question about the Axiom of Choice
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Jul 8, 2013 at 17:29 | comment | added | Garabed Gulbenkian | I see what is wrong with my statement of PP. In the absence of AC, even though S can be mapped on to T, this does not imply (in ZF) that T can necessarily be mapped injectively into S (which is what is meant by saying that the cardinal number of T is less than or equal to the cardinal number of S). Thanks for pointing out my mistake. | |
| Jul 8, 2013 at 4:30 | comment | added | 喻 良 | @Asaf, it is still open. | |
| Jul 7, 2013 at 19:14 | comment | added | Asaf Karagila♦ | Garabed, no that is $\sf WPP$ (the Weak Partition Principle). The Partition Principle itself states that if $S$ can be mapped onto $T$ then the cardinal of $T$ is less or equal than the cardinal of $S$. | |
| Jul 7, 2013 at 18:51 | comment | added | Garabed Gulbenkian | Maybe I am missing something in my statement of PP but I cannot quite see what. My statement simply insures that the cardinal number of T (however cardinal numbers are defined) is never greater than the cardinal number of S. I thought that this is what PP states. | |
| Jul 7, 2013 at 18:40 | comment | added | Garabed Gulbenkian | Maybe I am missing something in my statement of pp but I cannot quite see what. My statement simply insures that the | |
| Jul 7, 2013 at 18:09 | comment | added | Asaf Karagila♦ | Don't press "enter" until you're done writing your comment. | |
| Jul 7, 2013 at 18:07 | comment | added | Garabed Gulbenkian | I don't seem able to enter any comments. I need help! | |
| Jul 7, 2013 at 18:03 | comment | added | Garabed Gulbenkian | I stated a version of PP which simply stipulates that the | |
| Jul 7, 2013 at 17:42 | vote | accept | Garabed Gulbenkian | ||
| Jul 7, 2013 at 8:32 | comment | added | Asaf Karagila♦ | @Liang: Is this problem still open? (I'm asking because I've toyed with it before.) | |
| Jul 7, 2013 at 8:01 | comment | added | 喻 良 | This reminds me the question whether PP for reals implies the existence of a well ordering of reals. We may prove that there is a $ZF$-universe in which there is a transversal for any countable Borel equivalence relation but no well ordering of reals. We are not sure whether over $ZF+DC$, the full PP for reals implies the existence. | |
| Jul 6, 2013 at 20:29 | comment | added | Asaf Karagila♦ | @Andres: And in that spirit, Second and third edits of this question are also related to this thread (and to yours). | |
| Jul 6, 2013 at 20:17 | comment | added | Andrés E. Caicedo | Related. | |
| Jul 6, 2013 at 19:03 | comment | added | Andreas Blass | Noah is right. The conclusion of PP should not be that S can be mapped onto T (which is trivial) but that T can be mapped one-to-one into S. | |
| Jul 6, 2013 at 18:54 | comment | added | Asaf Karagila♦ | Garabed, PP states that if $A\leq^\ast B$ then $A\leq B$, where $\leq$ means there is an injection and $\leq^\ast$ means there is a surjection. | |
| Jul 6, 2013 at 18:51 | comment | added | Noah Schweber | I think I'm missing something. Fix some $t\in T$; why isn't the following map a surjection? $f: x\mapsto s\in T$ if $x\in s$, $x\mapsto t$ if $\forall s\in T(x\not\in s)$. Since the sets in $T$ are pairwise disjoint, and $T$ is nonempty, this is well defined, and can be proved to exist by ZF, and since each element of $T$ is a subset of $S$ this map is clearly onto. (Unless $T$ happens to contain $\emptyset$, but then just take $t=\emptyset$.) What am I missing? | |
| Jul 6, 2013 at 18:50 | answer | added | Asaf Karagila♦ | timeline score: 12 | |
| Jul 6, 2013 at 18:47 | comment | added | François G. Dorais | PP+DC implies the existence of a nonmeasurable set - mathoverflow.net/questions/22927/… | |
| Jul 6, 2013 at 18:42 | history | edited | François G. Dorais |
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| Jul 6, 2013 at 18:38 | history | asked | Garabed Gulbenkian | CC BY-SA 3.0 |