Timeline for A question on the consistency of a (seemingly) very weak set theory
Current License: CC BY-SA 3.0
25 events
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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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| S Mar 23, 2016 at 4:18 | history | bounty ended | CommunityBot | ||
| S Mar 23, 2016 at 4:18 | history | notice removed | CommunityBot | ||
| S Mar 15, 2016 at 2:59 | history | bounty started | Frode Alfson Bjørdal | ||
| S Mar 15, 2016 at 2:59 | history | notice added | Frode Alfson Bjørdal | Draw attention | |
| Mar 14, 2016 at 23:00 | comment | added | Frode Alfson Bjørdal | Thanks. In $19.2 you use principles way stronger than those offered by SU. | |
| Mar 14, 2016 at 18:08 | comment | added | Flash Sheridan | I’m not sure how much this will help, but I encountered an inconsistency in a theory with a universal set, which had a couple of commonalities with yours, in my own work. I’ve got a proof schema in my forthcoming Logique et Analyse article; unabridged preprint at logic-center.be/Publications/Bibliotheque/… §19.2. Adjunction is not used in the proof, but was part of the original motivation (footnote 7, p. 8, and p. 23.) | |
| Mar 13, 2016 at 16:13 | comment | added | Frode Alfson Bjørdal | @Noah Schweber I added some in an update. | |
| Mar 13, 2016 at 15:36 | comment | added | Frode Alfson Bjørdal | @Andrej Bauer I added some in an update. | |
| Mar 13, 2016 at 15:34 | history | edited | Frode Alfson Bjørdal | CC BY-SA 3.0 |
In accordance with what I mentioned in discussion, I added an axiom for universal set as well as two further inference rules. I also added the Leibnizian-Russellian account of identity and stressed that SU allows all set terms.
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| Mar 13, 2016 at 2:28 | comment | added | Frode Alfson Bjørdal | For the record, the further principle I seem to need is the existence of a universal set. In the interest of purity, the identity theory I want to presuppose is given by the Leibnizian-Russellian definition of $a=b$ as $\forall u(a\in u\rightarrow b\in u)$ and thence the schema $\vdash a=b\rightarrow (\alpha(a)\rightarrow\alpha(b))$. | |
| Mar 13, 2016 at 1:24 | comment | added | Frode Alfson Bjørdal | In the penultimate comment above I presuppose that SU has an underlying identity theory at its disposal. | |
| Mar 13, 2016 at 1:20 | comment | added | Frode Alfson Bjørdal | Aside from that, it seems that I may want some further principles to get this to do the work I want. | |
| Mar 13, 2016 at 1:16 | comment | added | Frode Alfson Bjørdal | SU has adjunction, so e.g. $\forall x(x\in\{y|y\in\emptyset\vee y=\emptyset\}\leftrightarrow (x\in\emptyset\vee x=\emptyset))$ is an axiom. So the system cannot be interpreted in such a way as you suggest. | |
| Mar 13, 2016 at 0:53 | comment | added | Noah Schweber | Is there any reason we can't just have $\{x: \alpha(x)\}$ be the emptyset for every formula $\alpha$? If not, do you really not want any other rules governing such terms? | |
| Mar 12, 2016 at 13:53 | comment | added | Frode Alfson Bjørdal | I had, for a peculiar reason, thought that I should need only closed set terms; so I do not allow parameters. Inconsistency arises immediately if the rules are changed into axioms. | |
| Mar 12, 2016 at 8:01 | comment | added | Andrej Bauer | I think the worry about inconsistency comes from the fact that changing your inference rule to the axiom $\forall a . (\alpha(a) \Leftrightarrow a \not\in \lbrace x \mid \lnot \alpha(x) \rbrace)$ opens the road to Russell's paradox. | |
| Mar 12, 2016 at 7:53 | comment | added | Andrej Bauer | May terms $\lbrace x \mid \alpha(x) \rbrace$ contain parameters? | |
| Mar 12, 2016 at 0:35 | comment | added | Frode Alfson Bjørdal | Of course I do not have that axiom schema, and I fail to understand why you would not be surprised if SU is inconsistent. | |
| Mar 12, 2016 at 0:14 | comment | added | Noah Schweber | OK, then I am worried your theory is inconsistent: you have a term $\{x: \neg (x\in x)\}$! Now, you don't have an axiom scheme saying $\forall x[\alpha(x)\iff x\in \{x: \alpha(x)\}]$, so we don't immediately hit a contradiction - but at this point I wouldn't be surprised if there is a contradiction. | |
| Mar 12, 2016 at 0:10 | comment | added | Frode Alfson Bjørdal | The addition you are alluding to comprise inference rules and not axioms. Yes, $\{x|\lnot\alpha(x)\}$ is a term. In fact, all terms $\{x|\alpha(x)\}$ are terms of SU. I would believe that SU is consistent relative to much weaker theories than $ZFC$, but any result is of interest. | |
| Mar 11, 2016 at 23:58 | comment | added | Noah Schweber | I don't quite understand the additional axiom in SU - is "$\{x: \neg \alpha(x)\}$" actually a term in your language? If not, what exactly is it shorthand for? (For example, are you saying that there is a set of all $x$ such that $\neg\alpha(x)$?) Also, what assumptions are you willing to make - e.g. if SU is consistent relative to ZFC, are you happy? | |
| Mar 11, 2016 at 23:45 | history | edited | Frode Alfson Bjørdal | CC BY-SA 3.0 |
I added the P.S.
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| Mar 11, 2016 at 22:50 | history | edited | Frode Alfson Bjørdal | CC BY-SA 3.0 |
I added the last question.
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| Mar 11, 2016 at 22:03 | history | asked | Frode Alfson Bjørdal | CC BY-SA 3.0 |