Timeline for "Uniformly (co)well-powered" categories?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jul 21, 2013 at 23:18 | comment | added | Andreas Blass | @SergeiAkbarov Being away from home, I can't easily look up references, but I believe Azriel Lévy's book "Basic Set Theory" tends to minimize the use of the axiom of choice. So I suspect that it would include a treatment of Scott's trick. | |
| Jul 21, 2013 at 6:10 | vote | accept | Sergei Akbarov | ||
| Jul 21, 2013 at 5:27 | comment | added | Sergei Akbarov | Andreas, which reading do you recommend? | |
| Jul 21, 2013 at 4:37 | comment | added | Andreas Blass | Addenda to my preceding comment: (1) By "partitioned into isomorphism classes" I mean to put two monomorphisms into the same subclass iff they have the same codomain and commute with an isomorphism between their domains. (2) Because Scott's trick uses the notion of rank, it requires the axiom of regularity. (3) Scott's trick gives a quite natural definition of cardinal numbers in the absence of the axiom of choice: The cardinality of $x$ is the set of all sets that admit bijections to $x$ and, subject to that requirement, have the lowest possible rank. | |
| Jul 21, 2013 at 4:33 | comment | added | Andreas Blass | @SergeiAkbarov OK, I've copied my comment into an answer. What Eric mentioned is often known as "Scott's trick" (named after Dana Scott). The idea is that, if you have a (definable) class (definably) partitioned into subclesses, then there is a (defniable) way to shrink each of those subclasses to a nonempty subset of itself, namely the set of those members of the subclass that have the smallest rank (in the sense of the cumulative hierarchy of sets) among the members of that subclass. In your situation, you'd have the class of monomorphisms partitioned into isomorphism classes. | |
| Jul 21, 2013 at 4:28 | answer | added | Andreas Blass | timeline score: 3 | |
| Jul 20, 2013 at 21:35 | history | edited | Sergei Akbarov | CC BY-SA 3.0 |
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| Jul 20, 2013 at 21:30 | comment | added | Sergei Akbarov | Yes, I found this result about well-ordering of the class of all sets in A.Levy's book. Andreas, I think you should put what you wrote in form of answer, and I will close this question. I'll make corrections in the question so that your answer will be natural. Or I do not know what to do with this. Eric, I still do not understand what you wrote. That is interesting, please, give me a reference. | |
| Jul 20, 2013 at 18:12 | comment | added | Sergei Akbarov | @Eric and Andreas, I need a reference. This is the first time I hear this. The class of all sets can be well-ordered? And this trick with sending X to the set of all skeleta... Is there a text to look how people do this? | |
| Jul 20, 2013 at 17:39 | comment | added | Eric Wofsey | More explicitly, using regularity you can get the map you ask for by sending $X$ to the set of all skeleta of ${\sf Mono}(X) $ which have minimal rank. | |
| Jul 20, 2013 at 17:27 | comment | added | Andreas Blass | @SergeiAkbarov In the presence of the other usual axioms of set theory, including especially the axiom of regularity, the axiom of global choice gives you a well-ordering of the class of all sets. Given such a well-ordering, you can define $S_X$ to contain just the first (with respect to the well-ordering) representative of each isomorphism class of monomorphisms into $X$. | |
| Jul 20, 2013 at 17:20 | comment | added | Sergei Akbarov | To apply the axiom of choice (i.e. to choose this map $X\mapsto S_X$) we must already have a map which to every $X$ assigns a set which contains some $S$ as element. But we do not have such a map. Or you mean something else? | |
| Jul 20, 2013 at 16:58 | comment | added | Sergei Akbarov | @Eric Wolfsey: this must be a misunderstanding... I don't see how this follows from global choice. | |
| Jul 20, 2013 at 16:40 | comment | added | Eric Wofsey | If you assume global choice, then this is equivalent to being well-powered, so you're not really going to be able to exhibit an example that you want in (2) (since global choice cannot be disproven from the usual axioms). The best you can hope for is to construct a model of set theory in which you have a counterexample. Or do you want to demand some sort of compatibility between the choices of $S_X$ for different objects X? | |
| Jul 20, 2013 at 15:59 | history | asked | Sergei Akbarov | CC BY-SA 3.0 |