Re: Scott like trick? A question
David Libert
zuhair (zalj...@gmail.com) writes:
> I shall repeat this question here under a separate topic.
>
> Actually the important question is precisely the following:
>
> We know from ZF that for every set x there exist a transitive closure
> set TC(x).
>
> Now does ZF prove or refute the following?
>
> For all x Exist y
> (y equinumerous to x &
> ~ Exist z (z equinumerous to x &
> TC(z) strictly subnumerous to TC(y)))
>
> were
>
> x subnumerous to y <-> Exist f (f:x-->y, f is injective)
> x equinumerous to y <-> Exist f (f:x-->y, f is bijective)
>
> x strictly subnumerous to y <->
> (x subnumerous to y & ~ x equinumerous to y)
>
> If the above is a theorem of ZF in which I think regularity would play
> the most important rule in proving it, then we can use this trick to
> build cardinals for sets in ZF without the assumption of choice.
>
> This approach might turn to be a type of Scott trick.
>
> Zuhair
ZFC proves the statement above. Namely let y = the usual ZFC cardinality of x.
So y is equinumerous with x. Also y is an initial von Neumann ordinal, and hence
transitive, so TC(y) = y.
If z were equinumerous with x, then TC(z) would be a superset of z, and so
supernmerous to x.
So TC(z) would also be supernumerous to y, since y is equinerous to x, so
TC(z) would be supernumerouis to TC(y), since TC(y) = y.
So TC(z) being supernumrous to TC(y), would not be strictly subnumerous to TC(y).
So y is as required.
So, since ZFC proves this statement, ZF does not refute it if ZF is consistent.
(By Godel's relative consistency of ZFC over ZF).
On the other hand, I think by a separate proof I can also prove if ZF is consistent,
it does not prove that statement.
I will be using methods of
[1] David Libert "Cohen symmetric choiceless ZF models"
sci.logic July 6, 2000
http://groups.google.com/group/sci.logic/msg/b4271c2585d2f1e5
Using permutations on atoms, I will create an omega sequence of sets
<A_i | i in omega> where the created ZF model will also contain a sequence
of surjectiions f_i : A_i ->> A_i+1 i in omega.
These surjections as usual contravariantly induce injections of powersets
P(A_i+1) >-> P(A_i) by pulling back subset of A_i+1 to their preimages
in A_i by f_i.
I will arrange the construction so this omega sequence of injections upward
in the powerset tower has an infinite inverse limit. This inverse limit
will be the constructed x in the negating the desired property:
> For all x Exist y
> (y equinumerous to x &
> ~ Exist z (z equinumerous to x &
> TC(z) strictly subnumerous to TC(y)))
So x can be isomorphically realized as a quotient of P(A_i) for
each i in omega induced by the projection of P(A_i) onto its
inverse limit x. As i varies over omega these quotients give
omega many alternative z equinumerous with x.
Call the P(A_i) quotient above z_i.
So TC(z_i) will include A_i.
The construction will arrange that <#A_i | i in omega> is a
strictly decreasing tower of Scott-Potter casrinals.
Such an infinite descending squence of cardinals is impossible
in ZFC, but possible in ZF ~AC models, as discussed in
[2] David Libert "'Partial' AC/Well-Ordering?"
sci.math June 20, 2000
http://groups.google.com/group/sci.math/msg/16e537e0dd63ddc1
z_i is a quotient of P(A_i), and so subnumerous to P(A_i).
The construction will arrange that not merely is #A_i > #A_i+1,
but also #A_i > #TC(P(A_i+1)).
So it follows that <#TC(z_i) | i in omega> is a strictly
descending seqeunce of cardinals.
Finally, it would be argued that for any y equinumerous to
x, #TC(y) must be >= #TC(A_i) for some i in omega.
We arranged by construction that a quotient of each P(A_i)
injects into x. We blew up x by in various alternatives
throwing in things built from A_i.
This claim is that any other y making #x many elements
must have the complexity built into it of one of these
z_i pieces.
So any y isomorphic to x must have #TC(y) bounded
somewhere below in the <#z_i | i in omega> tower.
So there is no y isomorphic to x with #TC(y) minimal
among all $TC(z) with z ~ x.
So the statement under duscussion is false in such a
constructed model:
Statement:
> For all x Exist y
> (y equinumerous to x &
> ~ Exist z (z equinumerous to x &
> TC(z) strictly subnumerous to TC(y)))
The trickiest part in all this is showing any y isomorphic
to x must have #TC(y) >= some #TC(z_i).
I think that's ok.
Anyway, if everything here is ok, it would show that quoted
statement is neither proved nor refuted in ZF, if ZF is
consistent.
--
David Libert ah...@FreeNet.Carleton.CA
zuhair
Well that is a little bit complex to me.
I think it is better if one take matters simply first, illustrate
examples and then you go build up things from those simple first
steps.
Why I am asking that question?
The title of the post gives a hint to it.
Now it is known that ZF (without choice) can prove that
there exist a minimal rank for Equinumerous sets.
In Exact terms, ZF(without choice) proves the following:
for every set A there is a least rank VA in the cumulative hierarchy
when any set of the same cardinality as A appears.
The proof for that require Regularity, but it doesn't require
choice.
Now my question is related to that fact of ZF.
Now can there exist two equinumerous sets of the same rank with the
transitive closure of one of them being strictly subnumerous to the
transitive closure of the other?
i.e. can there exist sets A and B were
A equinumerous to B and
Rank(A)= Rank(B) and
TC(A)strictly subnumerous to TC(B)
It would be quite helpful if you give me an example of such sets.
and Suppose that such two sets exist, the question is Can that go
infinitely, i.e in specific terms
can we have a non empty set S such that:
For all x,y e S ( x equinumerous to y & Rank(x)= Rank(y))
and
For all x e S Exist y e S (TC(y) strictly subnumerous to TC(x))
Can you give me an example of such a set.
Now the way how I assume things is that such set S would violate
Regularity, however I am not sure, since I don't have a proof of that,
that's why I asked this question.
So my point is that even if we don't have axiom of choice, still the
above set would violate Regularity, wouldn't it?
Zuhair
The Unbearable Likeness of Being
Zuhair:
Regularity is such a banal word.
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Thank you,
M. Michael Musatov
Cryptic as ever. Right as rain. Sloppy - sometimes. Precise - this
time.