Is ZF + CH + not-CC consistent?
Herman Jurjus
More generally, the question is: 'how much' of AxC is implied by CH?
(As AC is equivalent to the trichotomy of cardinals, and CH imply
trichotomy for sets not larger than P(N), someone might expect that
ZF+CH perhaps implies some weak forms of choice.)
--
Cheers,
Herman Jurjus
David Hartley
<hjm...@hetnet.nl> writes
First, how do you state CH in ZF without AC? Is it:
there is no cardinal k such that aleph_0 < k < c,
there is no uncountable cardinal less than c,
c = aleph_1,
or every uncountable cardinal is greater than or equal to c
At first glance, none of the first three rules out infinite,
Dedekind-finite sets, so you may not have trichotomy for sets not larger
than P(N). (All but the first obviously give it for sets smaller than or
comparable to P(N).)
--
David Hartley
William Elliot
> Subject says it (CC stands for 'countable choice').
>
No I didn't hear the subject say "ouch", "uncle", "your magesty" nor "it"
nor "CC stands for countable choice". I hear the subject ask a question.
Tim Little
> CH imply trichotomy for sets not larger than P(N)
I don't see how CH rules out any set being incomparable in cardinality
with P(N). The hypothesis merely asserts that no set is both smaller
than P(N) and larger than N.
- Tim
Herman Jurjus
I was in a terrible hurry when I wrote that post - sorry for that.
Forget about all the baloney.
The question is:
Is it known whether CH implies CC, DC, or any other weak forms of AC?
--
Cheers,
Herman Jurjus
Herman Rubin
Essentially nothing. In all of the standard
Fraenkel-Mostovski models, the classical sets are
unaffected. So the GCH can hold for all ordinals,
but not necessarily for other objects, and there
can even be infinite sets which are not comaparable
even with N.
I am not sure how much of this goes over to Cohen
models; the Fraenkel-Mostovski models are for
ZFU, where elements are allowed which are not sets,
but at least some of this works.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
David Libert
On Nov 18 Herman Rubin posted to this thread about Fraenkel-Mostowski
models of ZFU + CH + not-CC :
http://groups.google.com/group/sci.logic/msg/0f956ac2e786a6d3
Herman's article appeared in google groups but not at my posting site,
so I am doing this as a followup article to the one above.
Herman wrote about these Fraenkel Mostowsku models copying over to
corresponding Cohen models.
I note that in fact such a transfer is indeed possible, by the
Jech Sochar Transfer Theorem. I wrote about these models generally
and that teransfer theorem in
[1] David Libert "Cohen symmetric choiceless ZF models"
sci.logic July 6, 2000
http://groups.google.com/group/sci.logic/msg/b4271c2585d2f1e5
In _Set Theory and the Continuum Hypothesis_ , Paul Cohen gives
a ZF model with a countable sequence of 2 element sets having no
choice function.
Copy this over to an F-M model, and use Jech Sochar to put that
sequence at high rank, with the universe at rank reals and below
same as the ground model (start from CH ground model).
Basically: you can make high rank sets be very different from
low rank sets.
Some other related points of discussion. The original question
above was for CH. But for a related question, consider GCH.
In the absence of AC as an assumption, there are even different
reasonable versions of what GCH might mean. Fior a couple of these
we can get results.
One GCH variation not assuming AC to start is for every set
A there are no sets of cardinality strictly between #A and
#P(A). (Make a reasonable ~AC generalization of cardinality
to all sets: for example Scott's trick).
In _Set Theory and the Continuum Hypothesis_ , Cohen
goves a proof that ZF proves that GCH version implies AC.
Another reading of GCH could be that for all von Neumann
cardinals, that is all initial ordinals, in other words
the cardinalities of well-orderable sets, the powerset
has cardinality the successor cardinal.
I think I have found a proof that ZF proves this statement
-> AC.
So that's GCH.
Regarding CH again, above were ~AC models with ~AC at high rank.
But we could also ask about a local version of AC : can the
reals be well-ordered.
This comes down to the question, as was raised by David Hartley
indirectlty quoted above, which version of CH we use in the
absence of AC.
One possibility, as David raised is c = aleph_1. So for this
version, of course the reals are well-orderable.
Another version was there are no cardinal k such that
aleph_0 < k < c .
I found online in Kanamori's book _The Higher Infinite_
that Solovay's famous model with everything Lebesgue measurable
also satisifes that every uncountable set of reals has the perfect
set property.
A perfect set is defined to be a closed set with no isolated
points. The perfect set property is that the set has a non-empty
perfectsubset.
Every perfect set has cardinality c.
So Solovay's model has no cardinal k s.t.
aleph_0 < k < c .
Also, if the reals are well-orderable, we can doagonalize to
contruct an uncountable set of reals without the perfect set
property. So the reals are not well-orderable in Solvay's
model.
(We already knew that: if the reals are well-orderable then
we can do the usual contruction of a non-measurable set in ZF,
so from everything being Lebesgue measurable we already knew
the reals are not well-orderable).
Solovay's model started with an inaccessible cardinal in the
ground model, and collapsed it by forcing to aleph_1.
Kanamori in the book mentions there was a theorom of Specker
in ZF : if every uncountable set of reals has the perfect
set property and aleph_1 is regular, then aleph_1 is
inaccessible in L.
In Solovay's model aleph_1 is regular. (Even though
in some ~AC models aleph_1 might not be regular, in
Solovay's particular model it is).
So Solovay really did need the inaccessible.
I think I found a proof of the Specker result, which
gives a bit more information than the statement.
As noted above if every uncountable set of reals has
the perfect set property, then every uncountable set
of reals is c size, and the reals are not well-orderable.
I think the proof actually shows these conclusions are
sufficient. So if every uncountable set of reals is c size
and the reals are not well-orderable and aleph_1 is regular,
then aleph_1 is inaccessible in L.
So this is saying there can be the no k
with aleph_0 < k < c reading of CH with the reals
not well-orderable, but this is in both directions has the
level of an inaccessible cardinal.
Also, AD gets all this. Every uncountable set has
the perfect set property and the reals are not well-orderable.
But AD is at a level way beyond one inaccessible.
The other possibility David raised above to phrase CH without
assuming AC is every uncountable cardinal is greater than or
equal to c.
I don't know how this version works out. I don't know if
Solovay's model satisfies this. And I don't even know if
the Jech Sochar models above satisfy this.
So that version is all unsettled for me for now.
--
David Libert ah...@FreeNet.Carleton.CA
Herman Jurjus
> Herman Jurjus (hjm...@hetnet.nl) writes:
>> The question is:
>> Is it known whether CH implies CC, DC, or any other weak forms of AC?
>>
>> --
>> Cheers,
>> Herman Jurjus
>
>
> On Nov 18 Herman Rubin posted to this thread about Fraenkel-Mostowski
> models of ZFU + CH + not-CC :
>
> http://groups.google.com/group/sci.logic/msg/0f956ac2e786a6d3
Yes, found!
> Herman's article appeared in google groups but not at my posting site,
> so I am doing this as a followup article to the one above.
Thanks very much to all respondents!
(And I'm glad that I included sci.math.)
> I found online in Kanamori's book _The Higher Infinite_
> that Solovay's famous model with everything Lebesgue measurable
> also satisifes that every uncountable set of reals has the perfect
> set property.
Online? Where?
BTW, is there an online account of Solovay's construction?
Perhaps in Kanamori's book?
--
Cheers,
Herman Jurjus
David Libert
> David Libert wrote:
>> Herman Jurjus (hjm...@hetnet.nl) writes:
>
>>> The question is:
>>> Is it known whether CH implies CC, DC, or any other weak forms of AC?
>>>
>>> --
>>> Cheers,
>>> Herman Jurjus
>>
>>
>> On Nov 18 Herman Rubin posted to this thread about Fraenkel-Mostowski
>> models of ZFU + CH + not-CC :
>>
>> http://groups.google.com/group/sci.logic/msg/0f956ac2e786a6d3
>
> Yes, found!
>
>> Herman's article appeared in google groups but not at my posting site,
>> so I am doing this as a followup article to the one above.
>
> Thanks very much to all respondents!
> (And I'm glad that I included sci.math.)
>
>> I found online in Kanamori's book _The Higher Infinite_
>> that Solovay's famous model with everything Lebesgue measurable
>> also satisifes that every uncountable set of reals has the perfect
>> set property.
>
> Online? Where?
Just now, I didn't find the exact page I looked at last time, but
found a similar one:
This is a sample from Kaanmori's book. Unfortuantely it is only a sample, to entice you
to buy the book. It does include discussion of Solovay's and Speckers results though. But not in
detail.
I found that by googling: perfect set property AD .
> BTW, is there an online account of Solovay's construction?
> Perhaps in Kanamori's book?
I don't know offhand. From what I saw of the samples, it looks like Kanamori will have more
discussion of Solovay's proof.
> --
> Cheers,
> Herman Jurjus
--
David Libert ah...@FreeNet.Carleton.CA
David Hartley
<ah...@FreeNet.Carleton.CA> writes
>The other possibility David raised above to phrase CH without assuming
>AC is every uncountable cardinal is greater than or equal to c.
>
> I don't know how this version works out. I don't know if Solovay's
>model satisfies this. And I don't even know if the Jech Sochar models
>above satisfy this.
>
> So that version is all unsettled for me for now.
I threw that one in as a way of getting trichotomy for all cardinals not
larger than c, which Herman had mentioned (though perhaps he really
meant 'smaller than or equal to c'). I too don't know anything about how
it works out (not that I know much about the other versions either).
--
David Hartley
Gc
I remember that G.A Edgar (Or D.Renfro) posted a link to the original
paper of Solovay in sci.math a month or couple of months ago. I tried
to find the link on the google group archives, but they seem to be
totally useless now.
David Libert
David's third version of CH was every uncountable cardinal is >= c .
I realized later, this means aleph_1 >= c, so in particular the reals
inject into aleph_1, so the reals can be well-ordered.
So c is an aleph. But since alpeh_1 >= c, the only possible aleph
for c is c = aleph_1 .
So this 3rd reading of CH implies the first version c = aleph_1.
But this thurd reading is stroner than that. Not merely is
c = aleph_1, but every uncountable cardinal is >= alpeh_1.
That last is an extra assumption. If ZF is consistent then there are
choiceless ZF models with c = aleph_1 but some uncontable cardinals
incomparable to c.
So what can we say about this thrid version of CH?
c = alpeh_1, so the reals are well-orderable, and in Solovay's
model for everything Lebesgue measurable, the reals are not
well-orderable.
So this version of CH fails in Solovay's model, answering a
question I posed quoted above.
What about the Jech Sohcar model from my earlier article in this
thread?
That model took Cohen's excample of an omega sequence of 2 element
sets with no choice function, copied it to FM atoms, then pushed those
to high rank well founded sets by Jech Sochar.
Consider the union of those omega many 2 element sets. If it were
countable, then we could get a choice function on the 2 element sets
by picking the least of each pair of elements in an omega ordering
counting it.
That contra the model. So that set is uncountable.
So by the assumed property, if it were true in that model,
alpeh_1 would inject into that set.
If you collapse that injection into the union back to the
omega set of pairs, ie send each ordinal alpha < aleph_1
to the least pair set according to the omega ordering on pairs
sets containing as member the injection on alpha, you get
a function from alpeh_1 into a coubtable set which is
at most 2 to 1.
This is a contradiction even over ZF. For example, the injection
from aleph_1 to the union of pairs has well orderable range,
well-ordering induced by map from alpeh_1, and is sujected by
omega x omega, so it is countable, jence not injeced from
alpeh_1.
So we conclude the Jech Sochar model with countable choice
failing on pair sets, as from my prevous article, also doesn't
satisfy this 3rd CH version.
A fairly easy generalization of Cohen's orginal ~AC models, is
you can make AC hold for choices below a regular cardinal, and
fail at the regular cardinal. Ie making a few choices is ok,
more isn't. In fact you can make the innder Ciken model have
all short seqwuences from the outer AC model.
In these ways we can arrange aleph_1 many choices AC style
ok, and AC fails for more choices.
By the stronger property actually, inheriting all alpeh_1
sequences from an outer AC model, we can make aleph_1 inject
into every uncountble cardinal.
Ig our starting AC model also has CH, we can make the ~AC
construction add no new reals.
So we can get ~AC and c = alpeh_1 and aleph_1 <=
every uncountable carindal.
So the 3rd version of CH doesn't -> AC, by techniques
close to Cohen's original.
Ths still doesn't answer the 3rd version CH part of the
original questuiobn here: CC.
My Jech Sochar ~CC model above didn't work for this
3rd version of CH.
What else can be said?
Suppose we have an omega swquence of countable sets with
no choice fuction. Could that coexist with this CH version?
Above I argued that 2 element sets wouldn't work.
The point about 2 element sets, is if you inject alpeh_1
into the union, since each part is only size 2, injecting
alpeh_1 into the union must spread the injectiion across
many parts, contadicting that there are on;y countably many
parts.
So what if instead you have an omega sequence of countable
sets?
Could aleph_1 inject onto those omega many pieces, each
countable?
One difficulty is what if aleph_1 has cofinality omega.
This can't happen in AC models but can in ~AC ones.
Then aleph_1 is a countable union of countable sets,
so inject these in turn maybe onto each part. Actually that
itself might require some choice to do, but at least it
would show it is hard to rule out.
But then I remembered: old theorem, even though aleph_1
might have confinality omega, ZF proves c never has confinatiy
omega. And our assumption is c = aleph_1.
So I think this allows a generalizatiion of my 2 element set
argument above, so say if there is an omega sequence of
countable sets then ~ (3rd version CH).
And in fact we can similarly argue, if there is any omega
sequence of sets with no choice function, then the subsequence
of countable sets has choice fuctions restricted to it.
So given any such example, you can throw away all the countable
sets, and the reduced result still has not choice function.
So, all the sets would be uncountable, and by the 3rd version
CH if true, would inject aleph_1.
So if there are ~CC models with 3rd vereion CH, they have
~CC examples with each set injecting aleph_1.
Is that possible? I think so. My quoted article in this
thread, do an FM model with aleph_1 support on atoms instead
of the more common finite support. Use basic atom sets of
ground model size alpeh_2, with aleph_1 support.
Redo Cohen's ~CC model that way. The Jech Sochar all that.
Do that over ground model AC + CH. Then the FM step with
aleph_1 support stil injects aleph_1 to everty uncountable
set. And it doesn't add any reals so c = alpeh_1 still.
Then Jech Sochar as i review the proof I think preserves
c = alpeh_1 and this aleph_1 <= all uncountable cardinals.
So I think this finally gives a no answer to 3rd version
CH -> CC.
--
David Libert ah...@FreeNet.Carleton.CA