[FOM] Cardinality Beyond Regularity and Choice!
Zuhair Abdul Ghafoor Al-Johar
zaljohar at yahoo.com
Tue Dec 15 05:58:13 EST 2009
I was specifically speaking of sets that are
singletons and hereditarily singletons
at the same time, better named
as "recursive singletons".
I said the above because of the lemma in
ZF minus Regularity that for every x for every y
such that y is a member of TC(x) then there must exist
a finite sequence x1,x2,...,xn were x1 is a member
of x and every xi+1 is a member of xi and
So the transitive closure of every
recursive singleton is countable.
so you cannot have more than
a countable set of recursive singletons.
Of course no doubt you can have
and uncountable set of sets of
recursive singletons, simply
by powering the set of all
recursive singletons and so on.
My point is that if one can have
a proper class of recursive singletons,
then these of course would post a threat
to defining Cardinality
beyond Regularity and Choice.
To me Ur-elements weather they are
distinct empty objects, or Quine atoms
or recursive singletons, all pose
a big threat towards any definition of
cardinality beyond Regularity and Choice, since
it seems that there can be sets of them
that are incomparable to any well founded set.
However if we work in a theory
with pure sets, i.e. sets that do not
have any of those Ur-elements in their
transitive closures, then my guess
is that we may have a general definition
of Cardinality even beyond Regularity and
Ur-elements are the culprits.(I assume)
--- On Tue, 12/15/09, T.Forster at dpmms.cam.ac.uk <T.Forster at dpmms.cam.ac.uk> wrote:
> From: T.Forster at dpmms.cam.ac.uk <T.Forster at dpmms.cam.ac.uk>
> Subject: Re: [FOM] Cardinality Beyond Regularity and Choice!
> To: "Zuhair Abdul Ghafoor Al-Johar" <zaljohar at yahoo.com>
> Date: Tuesday, December 15, 2009, 12:01 AM
> I would bet good money that sets that
> are herediotarily singletons would do just as well. Why do
> you say there can be only countably many of them?
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