home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!dtix!darwin.sura.net!jvnc.net!princeton!phoenix.Princeton.EDU!carabalo
- From: carabalo@phoenix.Princeton.EDU (David G. Caraballo)
- Newsgroups: sci.math
- Subject: Choice and Measurability
- Message-ID: <1992Jul21.183209.8629@Princeton.EDU>
- Date: 21 Jul 92 18:32:09 GMT
- Sender: news@Princeton.EDU (USENET News System)
- Organization: Princeton University Mathematics Department
- Lines: 41
- Originator: news@ernie.Princeton.EDU
- Nntp-Posting-Host: phoenix.princeton.edu
-
- In article <1992Jul20.171849.17294@Princeton.EDU> I wrote
-
- > Let M be the axiom "All subsets of R^n are measurable". Then the
- > following is true:
- > M is consistent with ZF and Countable Choice, but not with ZF and
- > AC. Does someone have a reference?
-
- I have been given a reference that supports this claim. I will include
- it later in this post.
-
- In article <ARA.92Jul20200638@camelot.ai.mit.edu> Allan Adler wrote
-
- > David Carabalo believes that Solovay proved that the nonexistence of
- > nonmeasurable sets is consistent with ZF+DC.
-
- For the record, David Caraballo believes no such thing. The assumption
- that there exists an inaccessible cardinal is essential here, as Shelah
- proved. I wrote "Countable Choice", not DC ("Dependent Choice"). I believe
- it was Bernays (Is this the correct attribution?) who proved that
- AC => DC => Countable Choice. Consistency with ZF+DC is, of course, much
- more useful (DC allows us to prove all sorts of useful results, such as
- "The union of a countable number of countable sets is countable.").
-
- The relevant result that actually appears in Solovay's paper mentions
- DC, not Countable Choice. The complete reference, by the way, is
- Annals of Math, 92; 1970, pp. 1-56. However, consider the following
- reference which someone sent to me in response to my post:
-
- "Solovay in 1964 proved that the assertion 'all sets are Lebesgue
- measurable' is consistent with ZF and a restricted version of choice.
- This restricted version of choice, called the countable axiom of choice,
- asserts that every countable set of non-empty sets has a choice function."
- Malitz, _Introduction to Mathematical Logic_, p.49
-
- I had something similar in my own notes (which is why I posted something to
- this effect -- included above). I am not prepared to disregard my notes
- (and now, the above reference) without seeing an actual proof that my claim
- "M is consistent with ZF and Countable Choice" is false. If someone has a
- proof, I would love to see it. Thank you.
-
- David G. Caraballo
-