fom
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Re: Grothendieck universe
Posted:
May 21, 2013 8:54 PM


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.
> > 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?
As always, I am a little out of practice...
> > 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))?
> > 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}))
> 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?
> 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.
You might find this to be of interest,
http://www.cs.nyu.edu/pipermail/fom/2008March/012783.html
http://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory

