Timeline for Existence of Non-Borel sets in models of "All sets measurable"
Current License: CC BY-SA 3.0
16 events
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| Nov 1, 2012 at 23:53 | comment | added | Andrés E. Caicedo | Hi Asaf. It definitely becomes relevant when studying this problem whether one is looking at codes or at sets, as the answers may be different, as Andreas mentioned. For example, though $\mathbb R$ may be a countable union of null sets, it is never itself null. This is also related to whether we want to look at the ideal or the $\sigma$-ideal of null sets. Nice questions! | |
| Nov 1, 2012 at 18:02 | comment | added | Asaf Karagila♦ | Emil, I know you that. I wasn't 100% focused and just wrote a blurb of thought. I meant to say that I don't know what to answer exactly because it seems that the nontrivial failures of the null ideal to be $\sigma$-ideal might be a reasonable answer. Anyway, I tried to address this in the edit. | |
| Nov 1, 2012 at 18:01 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
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| Nov 1, 2012 at 14:35 | comment | added | Emil Jeřábek | Asaf, I probably didn’t make myself clear. I wasn’t asking for the truth value of the disjunction in my last question, but rather which of the three disjuncts holds. | |
| Nov 1, 2012 at 14:30 | comment | added | Andreas Blass | I also salute Joel's perseverance about "beg", but I'm not sure "raise" is the optimal replacement for it in this context. I think people raise questions, while facts, phenomena, and the like suggest questions, or lead to questions, or .... | |
| Nov 1, 2012 at 14:08 | comment | added | Asaf Karagila♦ | Emil, I want to say "yes", but I feel that I am walking into a trap. Namely, this ideal can be everything. On the other hand, I feel that this is might be what I am looking for as an answer. | |
| Nov 1, 2012 at 14:01 | comment | added | Emil Jeřábek | In view of Joel’s example, the definition is still unclear to me. The set $I$ of all sets that can be covered for every $\epsilon>0$ by a countable sequence of open intervals whose lengths sum to $\epsilon$ is an ideal, but not ZF-provably a $\sigma$-ideal. Is the “null ideal” $I$, the $\sigma$-ideal generated by $I$, or something else? | |
| Nov 1, 2012 at 13:56 | comment | added | HJRW | Joel - I salute your perseverance, but that horse bolted a long time ago. | |
| Nov 1, 2012 at 13:43 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
All this begging and whatnot...; deleted 4 characters in body
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| Nov 1, 2012 at 13:34 | comment | added | Joel David Hamkins | And by the way, it is a great question! | |
| Nov 1, 2012 at 13:30 | comment | added | Joel David Hamkins | Incidentally, Asaf, I think you mean to raise the question, rather than beg it. begthequestion.info | |
| Nov 1, 2012 at 13:29 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
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| Nov 1, 2012 at 12:48 | comment | added | Joel David Hamkins | There is always a set without a Borel code: the set of all Borel codes $b$ which are not elements of the set coded by $b$. | |
| Nov 1, 2012 at 12:00 | comment | added | Andreas Blass | While you're at it, could you say what you mean exactly by "Borel set"? I assume that you mean a set in the smallest $\sigma$-algebra containing the open sets, but, if you talk with enough set theorists, you might mean a set with a Borel code (i.e., a real that encodes how to build the set from open sets by iterated complementation and countable union). The two meanings can differ in the absence of choice. | |
| Nov 1, 2012 at 11:19 | comment | added | Joel David Hamkins | Could you say what you mean exactly by "all sets are Lebesgue measurable" in a context without DC? After all, the theory of Lebesgue measure in ZF can seem to break down (imagine that the reals are a countable union of countable sets). | |
| Nov 1, 2012 at 9:30 | history | asked | Asaf Karagila♦ | CC BY-SA 3.0 |