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 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_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 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/2008-March/012783.html

http://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory