Timeline for Why should we believe in the axiom of regularity?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2016 at 21:12 | comment | added | Andreas Blass | @Nullachtfünfzehn I was using "circular" in the broader sense of a definition that refers to the term that is being defined. Strictly speaking, even a recursive definition is circular in this sense. (For example, the recursive definition of $n!$ includes the clause $(n+1)!=n!\cdot(n=1)$ so it defines ! in terms of !.) But we know how to remove this sort of circularity in the case of well-founded recursions. So "circular" is often limited to mean "non-well-foundedly circular", and this is what I intended in my answer. | |
| Nov 16, 2016 at 19:45 | comment | added | user99916 | @AndreasBlass: Why would the instructions "form all sets whose elements are at earlier levels" be circular without well-ordering? Without well-ordering, we could have for example this structure: $0,1,2,3\dots,\dots\infty-2,\infty-1,\infty,\infty+1,\infty+2\dots$ where we have a stage called "$∞+n$" for any $n∈\mathbb Z$. Then $\infty + 2$ is defined in terms of $\infty + 1$, which is in turn defined in terms of $\infty$, which in turn is defined in terms of $\infty - 1$. This goes on forever, but I think there's never a "circle". | |
| Nov 16, 2016 at 17:03 | comment | added | user99916 | @AndreasBlass: And where did I confuse something? You say "then every stage should involve formation not only of sets with elements from earlier stages but also of functions between them, ordered pairs of them". Isn't this exactly what I was trying to say in my comment? | |
| Nov 16, 2016 at 16:54 | comment | added | Andreas Blass | [continuation of previous comment] If one wants to regard the T world as a cumulative hierarchy, then every stage should involve formation not only of sets with elements from earlier stages but also of functions between them, ordered pairs of them (since ordered pairs would not be coded as sets), etc. | |
| Nov 16, 2016 at 16:51 | comment | added | Andreas Blass | @Nullachtfünfzehn You seem to be mixing two different ways of viewing the mathematical universe. The first, as described by "T" in my answer at mathoverflow.net/q/90945, would indeed say that functions are not sets; they are entities of another sort (or sorts). The second, the result of interpreting T in ZFC, would replace those functions with their set-theoretic encodings, and would say that functions are sets of ordered pairs. The present question is about the second of these viewpoints, as described by ZFC. [continued in next comment] | |
| Nov 16, 2016 at 16:26 | comment | added | user99916 | Also, functions do not lie at the botton of the hierarchy. Instead, on every stage $\alpha$ one can form all sets that lie on lower stages, and form all functions A --> B where A, B are sets created at a lower stage. | |
| Nov 16, 2016 at 16:25 | comment | added | user99916 | @AndreasBlass: You write "Begin with some non-set entities called atoms ("some" could be "none" if you want a world consisting exclusively of sets), then form all sets of these". But not every collection of urelements is a set, and not every urelement lies at the botton of the hierarchy. For example, functions are objects that aren't sets, and thus functions are urelements. But one can't form the set of all functions. | |
| Sep 30, 2015 at 22:48 | comment | added | Avshalom | Shoenfield also expresses a very similar view of sethood in his paper "Axioms of set theory", in Jon Barwise & H. Jerome Keisler (eds.), Handbook of Mathematical Logic. North-Holland Pub. Co. 90 (1977), although he does not restrict his exploration to the axiom of regularity. | |
| Sep 30, 2015 at 21:36 | comment | added | Joel David Hamkins | I agree very much with this answer. | |
| Sep 30, 2015 at 14:02 | comment | added | Wojowu | I must admit that cumultative hierarchy spanning whole universe is how I've been justifying to myself why regularity "should" hold - it puts a kind of "order" on sets, showing which ones are "simpler" than others. | |
| Sep 30, 2015 at 10:14 | history | answered | Andreas Blass | CC BY-SA 3.0 |