fom
Posts:
1,968
Registered:
12/4/12


Re: Grothendieck universe
Posted:
May 21, 2013 10:44 PM


On 5/21/2013 9:21 PM, William Elliot wrote: > On Tue, 21 May 2013, fom wrote: >> On 5/21/2013 3:51 AM, William Elliot wrote: > >>> A Grothendieck universe is a set G of ZFG with the axioms: >>> >>> for all A in G, B in A, B in G, >>> for all A,B in G, { A,B } in G, >>> for all A in G, P(A) in G, >>> I in G, for all j in I, Aj in G implies \/{ Aj  j in I } in G. >> >> ... and closure under unions, right? See FOM post listed >> at bottom. > > It's a theorem. A,B, { A,B } in G, \/{ A,B } in G. > In fact, the last in equivalent to for all A in G, \/A in G. > >>> The following are theorem of G: >>> >>> for all A in G, {A} in G, >>> for all B in G, if A subset B then A in G, >>> for all A in G, A /\ B, A / B in G, >>> for all A,B in G, (A,B) = { {A,B}, {B} } in G. >>> >>> Are the following theorems or need they be axioms? >>> If theorems, what would be a proof? > >>> For all A,B in G, AxB = { (a,b)  a in A, b in B } in G. >> >> Are not Cartesian products sometimes explained >> as subsets of P(P(A \/ B))? > > (a,b) = { {a,b}, {b} }. Ok, that's simple. > > Also if a,B in G, then {a}xB = \/{ (a,b)  b in B } in G > and if also A in G, then AxB = \/{ {a}xB  a in A } in G > >>> I in G, for all j in I, Aj in G implies prod_j Aj in G. > >> Along the same lines, wouldn't this involve >> an application of replacement to form the {A_ii in I} >> Then prod_j A_j would be a subset of P(P(\/{A_ii in I})) >> > No, P(P(I \/ \/{ Aj  j in I })) > >>> For all A in G, A < G. > >> Wouldn't the closure axiom on power sets make this true? >> >> Closure under power sets. Elements of an element >> is an element. So, if any element were equipollent >> with the universe, the universe would have the >> cardinal of its own power set. Right? > > Ok > >>> Is there any not empty Grothendieck universes. >> >> ...any favorite set theory >> >> http://en.wikipedia.org/wiki/Grothendieck_universe >> >> There are two simple universes discussed (empty set >> and V_omega). The rest are associated with the >> existence of strongly inaccessible cardinals. > > The latter don't exist in ZFO. So V_omega0 is the only > nontrivial Grothendeick universe. Doesn't V_omega0 = aleph_omega0
I believe this is correct.
> which is almost always big enough for mathematics? >
Well, that depends on what you mean by "mathematics".
I wrote a set theory that includes a universal class. I believe it is minimally modeled by an inaccessible cardinal (when the axiom of infinity is included). My argument for such a structure is that the philosophy of mathematics ought to be responsible for the ontology of its objects. So, I reject predicativist views that take "numbers" as given.
My views, however, are nonstandard and I am still working at how to understand them in relation to standard paradigms.

