Timeline for Are there fragments of set theory which are axiomatized with only bounded (restricted) quantifiers used in axioms?
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| Aug 20, 2014 at 9:24 | comment | added | Ioachim Drugus | My congratulations! This 100th answer was really great. Thank you! I know that good axioms are needed and I have some: to these ones, I add this one. Here is shown how to inductively define union for hereditarily finite sets, but I believe, due to "Epsilon induction" the union can be defined for general case. (Yes, I mistyped the formula) | |
| Aug 19, 2014 at 20:20 | comment | added | Goldstern | This was my 100th answer. Where are the fireworks? | |
| Aug 19, 2014 at 20:19 | comment | added | Goldstern | @IoachimDrugus: if you are interested in working with infinite sets, adjunction won't get you very far. There is a good reason why we need a union axiom, an infinity axiom (in your theory that could be done by a constant) and (usually) a power set axiom (that could be a unary operation). (Btw, you meant $(x;y)=x$ iff $y\in x$, not if $x\in y$.) | |
| Aug 19, 2014 at 20:11 | comment | added | Goldstern | @AndreasBlass and @ Francois: An obvious drawback of ZFU is that it says nothing about the objects that are not in $U$. A minimalistic strengthening would be to demand $\forall x: x\in U \vee x=U$. | |
| Aug 19, 2014 at 14:29 | comment | added | Ioachim Drugus | Now I see why constants are needed in restricted quantification, and the constant $U$ of ZFU sounds as one of most needed for what I have in mind - this is to develop a set theory with only one binary operation (rather than membership relationship). Boolos called "adjunction" the operation denoted with semicolon here: $(x;\ y) = x \cup \ ${$y$}. In applications to natural languages, I called it "qualification" and read $(x;\ y)$ as "$x$ as qualifier of $y$". Notice, $(x;\ y)= x$ iff $x \in y$. My goal is also to present restricted quantification as qualification by certain constants. | |
| Aug 19, 2014 at 8:18 | comment | added | Andreas Blass | @FrançoisG.Dorais I must confess that I don't see much similarity between this answer and Ackermann set theory. In particular, important features of Ackermann's class of all sets are that it's not definable unsing only $\in$ and it's a member of various higher classes. Martin's $U$ might well be the only non-set in some model, so it might be definable and might not be a member of anything else. | |
| Aug 19, 2014 at 1:45 | comment | added | François G. Dorais | This is very similar in spirit to Ackermann Set Theory. | |
| Aug 18, 2014 at 20:45 | history | answered | Goldstern | CC BY-SA 3.0 |