The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 1,968
Registered: 12/4/12
Re: Grothendieck universe
Posted: May 21, 2013 10:44 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 5/21/2013 9:21 PM, William Elliot wrote:
> 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
>> 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

I believe this is correct.

> which is almost always big enough for mathematics?

Well, that depends on what you mean by "mathematics".

I wrote a set theory that includes a universal class.
I believe it is minimally modeled by an inaccessible
cardinal (when the axiom of infinity is included). My
argument for such a structure is that the philosophy
of mathematics ought to be responsible for the ontology
of its objects. So, I reject predicativist views that
take "numbers" as given.

My views, however, are non-standard and I am still working
at how to understand them in relation to standard paradigms.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.