```Date: May 21, 2013 8:54 PM
Author: fom
Subject: Re: Grothendieck universe

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 listedat 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 explainedas 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 involvean application of replacement to form the {A_i|i in I}Then prod_j A_j would be a subset of P(P(\/{A_i|i in I}))> For all A in G, |A| < |G|.>Wouldn't the closure axiom on power sets make thistrue?Closure under power sets.  Elements of an elementis an element.  So, if any element were equipollentwith the universe, the universe would have thecardinal of its own power set.  Right?> Is there any not empty Grothendieck universes....any favorite set theoryhttp://en.wikipedia.org/wiki/Grothendieck_universeThere are two simple universes discussed (empty setand V_omega).  The rest are associated with theexistence of strongly inaccessible cardinals.You might find this to be of interest,http://www.cs.nyu.edu/pipermail/fom/2008-March/012783.htmlhttp://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory
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