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Topic: ZFC really really really sucks -- really!
Replies: 20   Last Post: Jan 20, 2013 6:30 AM

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ross.finlayson@gmail.com

Posts: 1,218
Registered: 2/15/09
Re: ZF - Powerset + Choice = ZFC ?
Posted: Jan 12, 2013 11:36 AM
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On Jan 8, 6:45 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:
> 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 UTC-5, david petry wrote:
>

...

> > > > 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 two-elements 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 well-foundedness 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?
>


So, why axiomatize powerset if it's not an independent axiom?

Regards,

Ross Finlayson




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