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Topic: question on set theory
Replies: 4   Last Post: Jan 4, 2013 2:17 AM

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Robert E. Beaudoin

Posts: 8
Registered: 5/25/09
Re: question on set theory
Posted: Jan 4, 2013 2:17 AM
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On 01/03/13 02:22, matmzc%hofstra.edu@gtempaccount.com wrote:
> On Wednesday, January 2, 2013 3:30:01 PM UTC-5, Dan Luecking wrote:
>> On Wed, 2 Jan 2013 17:37:20 +0000 (GMT),
>>

>
> Apologies everyone. I posted the question late at night and wrote
> the proposition backwards. I meant to say consider
>
> Prop: If X< Y then P(X)< P(Y)
>
> Can that be proved in ZFC? If not is it perhaps equivalent to
> generalized continuum hypothesis? It is easy to show in ZF with or
> without choice that that X< Y implies P(X)<=P(Y). But can it
> possibly be the case that X< Y but P(X)=P(Y)?
>
> Mike
>
>


No, and no (unless of course you want to consider the possibility that
ZFC is inconsistent, in which case anything would be provable in ZFC).
Assuming ZF(C) is consistent, so are ZFC plus the failure of the
continuum hypothesis plus either your proposition or its negation (where
in the failure X can be taken to be the set of natural numbers); this
follows from well-known (old) results of Easton about the behavior of
the continuum function (which maps the cardinality of a set to the
cardinality of its power set) at regular cardinals together with
well-known (old) results of Jensen constraining its behavior at singular
cardinals.

There are a few propositions about the continuum function easily
provable in ZFC: (1) Cantor's theorem that for any set X, |X| < |P(X)|
(where of course P denotes power set and |.| denotes cardinality), (2)
for any sets X and Y, |X| <= |Y| implies |P(X)| <= |P(Y)|, and (3) for
any set X, P(X) is not the union of |X| or fewer sets each of
cardinality less than |P(X)|. Easton essentially showed (assuming ZFC
is consistent) that nothing else can be proved in ZFC about the mapping
from |X| to |P(X)| for regular cardinalities |X| (i.e. for the case in
which X is not the union of fewer than |X| sets each of cardinality less
than |X|). In particular, we can either have |P(N)| = |P(Y)| for some
uncountable Y (one can take Y to be a subset of the reals, though by (1)
Y must then have cardinality less than that of the reals), in which case
your proposition and the continuum hypothesis both fail, or |P(N)| > |R|
(so the continuum hypothesis fails) and yet your proposition holds.

Cardinal arithmetic is interesting in that it has really elementary
results living cheek-by-jowl with incredibly deep theorems.

Robert E. Beaudoin





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