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From: zuhair on 30 Mar 2010 17:39
If we have a set theory T that disprove choice, and we
have two sub-theories T1, T2 of T, each consistent with choice (do
not disprove, neither prove choice) is there a possibility that
the theory "T1+T2" disprove choice?

Zuhair
From: Rotwang on 30 Mar 2010 18:08
zuhair wrote:
> If we have a set theory T that disprove choice, and we
> have two sub-theories T1, T2 of T, each consistent with choice (do
> not disprove, neither prove choice) is there a possibility that
> the theory "T1+T2" disprove choice?

I think so. Let N be the negation of AC, and let B be any closed
sentence such that B is independent of ZF, and AC is independent of ZF +
B (I /think/ I've read that the continuum hypothesis is such a sentence,
but I'm not sure; all I've found with a quick web search is that the GCH
implies AC, which doesn't help). Then consider T = ZF + B + N, T1 = ZF +
B, T2 = ZF + (B => N). T1 is does not prove N by hypothesis. If T2
proves N then ZF proves (B => N) => N, but ZF also proves �(B => N) =>
N, which would mean that ZF proves N, a contradiction. So T1 and T2 are
both consistent with choice, but T1 + T2 is not.
From: Rotwang on 30 Mar 2010 19:17
Rotwang wrote:
> zuhair wrote:
>> If we have a set theory T that disprove choice, and we
>> have two sub-theories T1, T2 of T, each consistent with choice (do
>> not disprove, neither prove choice) is there a possibility that
>> the theory "T1+T2" disprove choice?
>
> I think so. Let N be the negation of AC, and let B be any closed
> sentence such that B is independent of ZF, and AC is independent of ZF +
> B (I /think/ I've read that the continuum hypothesis is such a sentence,
> but I'm not sure; all I've found with a quick web search is that the GCH
> implies AC, which doesn't help). Then consider T = ZF + B + N, T1 = ZF +
> B, T2 = ZF + (B => N). T1 is does not prove N by hypothesis. If T2
> proves N then ZF proves (B => N) => N, but ZF also proves �(B => N) =>
> N,

Sorry, this is nonsense.
From: Rotwang on 30 Mar 2010 20:05
Rotwang wrote:
> Rotwang wrote:
>> zuhair wrote:
>>> If we have a set theory T that disprove choice, and we
>>> have two sub-theories T1, T2 of T, each consistent with choice (do
>>> not disprove, neither prove choice) is there a possibility that
>>> the theory "T1+T2" disprove choice?
>>
>> [nonsense]
>
> Sorry, this is nonsense.

Right, let's try again (and hopefully I won't write something completely
stupid this time): suppose there exists a closed sentence B such that

ZF + AC does not prove B
ZF + B does not prove AC
ZF + B does not prove ~AC

Then T = ZF + B + ~AC is consistent and disproves AC. T1 = ZF + B is a
subtheory which does not disprove AC, and nor does T2 = ZF + (B => ~AC);
for otherwise ZF proves (B => ~AC) => ~AC, which means that ZF + AC
proves B, contrary to hypothesis. But T1 + T2 proves ~AC.

Conversely, if restrict our attention to extensions of ZF, then the
existence of T1, T2 and T as in your question would imply the existence
of a sentence B which satisfies the three conditions above. For if T1 +
T2 proves ~AC then any such proof uses a finite set of sentences of T1
and T2 as axioms; since T1 or T2 proves a sentence s iff it proves the
universal closure of s, we may assume wolog that all the sentences in
question are closed. Let B be the conjunction of all those sentences of
T1 used in the proof of ~AC, and let C be the conjunction of all those
sentences of T2 used in the proof of ~AC. Then

ZF + B does not prove ~AC
ZF + C does not prove ~AC
but ZF + B&C does prove ~AC.

If ZF + B proves AC then ZF proves B => AC, and ZF also proves B => (C
=> ~AC). But then ZF proves B => ~C, so that ZF + B + C is inconsistent;
this contradicts our assumption that T1 and T2 are both contained in the
consistent theory T.

Now since ZF proves B&C => ~AC, it follows that ZF + AC proves ~(B&C) If
ZF + AC also proves B, then it must prove ~C, so ZF proves AC => ~C. But
then ZF + C proves ~AC, contrary to hypothesis, so ZF + AC does not prove B.

So the answer to your question is "yes" iff there exists a B satisfying
the three conditions above. I don't have a definitive answer on whether
such a B exists, sorry.
From: Kai-Uwe Bux on 30 Mar 2010 20:19
zuhair wrote:

> If we have a set theory T that disprove choice, and we
> have two sub-theories T1, T2 of T, each consistent with choice (do
> not disprove, neither prove choice) is there a possibility that
> the theory "T1+T2" disprove choice?

What about

T0 = ZF without AC (axiom of choice) and without CH (continuum hypothesis).
T1 = T0 and CH
T2 = T0 and (CH => not AC)
T = T1 and T2


Best

Kai-Uwe Bux
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