
ZF  Powerset + Choice = ZFC ?
Posted:
Jan 8, 2013 9:45 PM


On Jan 7, 7:13 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > On Jan 7, 2:30 pm, Dan Christensen <Dan_Christen...@sympatico.ca> > wrote: > > > > > > > > > > > On Monday, January 7, 2013 4:51:47 PM UTC5, david petry wrote: > > > On Monday, January 7, 2013 1:45:40 PM UTC8, Dan Christensen wrote: > > > > > On Monday, January 7, 2013 3:59:48 PM UTC5, david petry wrote: > > > > > > On Monday, January 7, 2013 11:14:57 AM UTC8, Dan Christensen wrote: > > > > > > > On Monday, January 7, 2013 8:50:09 AM UTC5, david petry wrote > > > > > > > > An article by Nic Weaver is worth a read: > > > > > > > >http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.1680v1.pdf > > > > > > I suppose it is surprising to classically trained mathematicians, but it is not necessary to define a set of all continuous functions, nor even a set of all real numbers, to develop the mathematics used in science. > > > > > That WOULD be surprising if it were true. > > > > Once again, I recommend that you read Nik Weaver's article. > > > IIUC, he hopes to do mathematics (e.g. real analysis) without sets (or any equivalent notion). If he wants to be taken seriously, he should just go ahead and do so. I have tried to do so for a number of years myself to no avail. I don't much care for the ZF axioms of regularity and infinity myself. I haven't found any use for them in my own work, and haven't incorporated them (or any equivalent) in my own simplified set theory. But I really don't see how you can do foundational work without a powerset axiom. > > > Dan > > Download my DC Proof 2.0 software athttp://www.dcproof.com > > Doesn't it follow from pairing and union? > > Basically we know that any element of a set is a set via union. > > Then, each of those as elements as a singleton subset, is a set, as > necessary via inductive recursion and building back up the sets. > > Via pairing, the twoelements subsets are sets, via pairing, the three > element sets are sets, ..., via induction, each of the subsets are > sets. > > Then all the subsets are each sets, via pairing, that's a set. > > It follows from pairing and union. >
Then, here it seems that there is a ready enough comprehension of the elements of the set, composed in each way, composing a set, that would be the set of subsets of a set, or its powerset, without axiomatic support, instead as a theorem of pairing and union, and into strata with countable or general choice.
In a set theory, then what sets have powersets that don't exist via union and pairing? In a theory without wellfoundedness perhaps those that are irregular, yet then the transitive closure would simply be irregular too and the powerset would be simply enough constructed (via induction and for the infinite, transfinite induction).
So, what sets have powersets not constructible as the result of induction over union and pairing, in ZF  Powerset or ZFC  Powerset?
Regards,
Ross Finlayson

