[FOM] Ordering a rank

[Marco Forti]

In fact, the main problem of the "Euclidean" theory of numerosity was  that
the "algebraic" total ordering of the ring of numerosities should have the
natural set theoretic characterization, namely:
   (#)             n(A)<n(B) if and only if exists C s.t. A (strictly)
included in C, and n(C)=n(B).
This fact has been established up to now, only for countable sets, by means
of selective ultrafilters, so a little beyond ZFC.
(on the other hand, also the totality of the Cantorian weak cardinal
ordering that
                     |A|<=|B| if and only if exists C s.t. A (weakly)
included in C, and |C|=|B|
had to wait till Zermelo's new axiom of choice!)
However  a proof in pure ZFC of the general statement (#) for all sets of
sufficiently comprehensive families, e.g. V_kappa or H(kappa) for any
cardinal kappa, has been found in the last couple of month, and will be
posted soon to the arxive.
Marco Forti
.


Marco Forti
Dipartimento di Matematica
Università di Pisa
Largo B. Pontecorvo 5
56123 PISA (Italy)
Tel. +39 050 2213876
email: marco.forti at unipi.it


On Wed, Jan 1, 2020 at 12:35 AM Joe Shipman <joeshipman at aol.com> wrote:

> Yes, I should have been clearer. I want the order not only to be total,
> but also to be compatible with ordered semiring operations as in the theory
> of “numerosities” that was referenced here recently — a way of measuring
> size of sets that is finer-grained than cardinality.
>
> Numerosities need not represent a total ordering of V or of some V_alpha
> because sets incomparable under inclusion can have the same numerosity;
> however, with choice one can, I think, get an appropriate total ordering
> from them. But the existence of numerosities even for V_(omega+1) seems to
> be independent of ZFC if I understand Mancosu correctly (also consistent
> without needing large cardinals, but I don’t know how far up you can go).
>
> — JS
>
> Sent from my iPhone
>
> On Dec 31, 2019, at 6:19 PM, Noah Schweber <schweber at berkeley.edu> wrote:
>
> 
> > For which ordinals alpha can there exist a total ordering < on V_alpha
> such that A<B whenever A is a proper subset of B?
>
> Unless I'm misunderstanding the question, the answer is "all of them"
> since every partial order can be extended to a linear order (this is
> Szpilrajn's Extension Theorem). Of course, this requires choice; without
> choice, I believe it is consistent with ZF that already $V_{\omega+1}$
> cannot be so ordered (I think Cohen's original model of the failure of
> choice witnesses this).
>
>
>
>  - Noah
>
> On Sun, Dec 29, 2019 at 2:48 PM Joe Shipman <joeshipman at aol.com> wrote:
>
>> For which ordinals alpha can there exist a total ordering < on V_alpha
>> such that A<B whenever A is a proper subset of B?
>>
>> — JS
>>
>> Sent from my iPhone
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