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Topic: Grothendieck universe
Replies: 5   Last Post: May 22, 2013 5:41 PM

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fom

Posts: 1,968
Registered: 12/4/12
Re: Grothendieck universe
Posted: May 21, 2013 8:54 PM
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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






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