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

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 William Elliot Posts: 2,490 Registered: 1/8/12
Re: Grothendieck universe
Posted: May 21, 2013 10:21 PM

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_i|i in I}
> Then prod_j A_j would be a subset of P(P(\/{A_i|i 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
non-trivial Grothendeick universe. Doesn't |V_omega0| = aleph_omega0
which is almost always big enough for mathematics?

> 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
>

Date Subject Author
5/21/13 William Elliot
5/21/13 fom
5/21/13 William Elliot
5/21/13 fom
5/22/13 William Elliot
5/22/13 fom