Every set is F_sigma-delta-sigma in the Feferman-Levy model?
Dave L. Renfro
(AC) are needed to obtain a non-measurable subset of the reals.
In particular, it is consistent (assuming also the consistency
of a certain large cardinal hypothesis) that even in the presence
of the countable axiom of choice, every subset of the reals can
be Lebesgue measurable. [In fact, this is still possible under a
slightly stronger assumption, the Axiom of Dependent Choice). Just
about every graduate real analysis text quotes something along
these lines. However, I'm curious as to how much "worse" things can
be when no version AC is assumed (except for the finite Axiom of
Choice, which can be proved in ZF).
On page 208 of Gregory H. Moore's "Zermelo's Axiom of Choice: Its
Origins, Development, and Influence", Springer-Verlag, 1982
it is claimed that in the Feferman-Levy model of ZF, every subset
of the reals is F_sigma-delta-sigma. [This is the model in which
the set of reals is not a countable union of countable sets.]
Does anyone know of a precise literature reference to this and/or
more about it? My guess that without any infinite version of AC to
fall back on, even the notion of an F_sigma-delta-sigma set might
have more than one interpretation. So perhaps saying that every subset
of the reals is F_sigma-delta-sigma might be an oversimplification of
the actual result.
Dave L. Renfro
G. A. Edgar
<dlre...@gateway.net> wrote:
> On page 208 of Gregory H. Moore's "Zermelo's Axiom of Choice: Its
> Origins, Development, and Influence", Springer-Verlag, 1982
> it is claimed that in the Feferman-Levy model of ZF, every subset
> of the reals is F_sigma-delta-sigma. [This is the model in which
> the set of reals is not a countable union of countable sets.]
Of course on the other hand if the reals IS a countable union
of countable sets, then every subset of the reals is
F_sigma-sigma.
--
G. A. Edgar http://math.ohio-state.edu/~edgar/
D. Ross
| It's well known that rather strong forms of the Axiom of Choice
| (AC) are needed to obtain a non-measurable subset of the reals.
Strictly speaking, a nonprincipal ultrafilter on the integers is good
enough, though obviously this is not really the kind of weakening of AC
which you want to consider.
| more about it? My guess that without any infinite version of AC to
| fall back on, even the notion of an F_sigma-delta-sigma set might
| have more than one interpretation. So perhaps saying that every subset
| of the reals is F_sigma-delta-sigma might be an oversimplification of
| the actual result.
Lack of AC shouldn't affect this definition, since an element of this class
is obtained by a concrete operation on a function from N^3 to the closed
sets.
Can't help of the rest of the question.
- David R.
G. A. Edgar
> Lack of AC shouldn't affect this definition, since an element of this class
> is obtained by a concrete operation on a function from N^3 to the closed
> sets.
Is it? This is related to the "countable union of countable sets
is countable" problem. Certainly there is no problem when you
have a union operation on a function from N^2 to sets. But
that does not (without choice) cover all "countable unions of
countable sets".
D. Ross
|
| > Lack of AC shouldn't affect this definition, since an element of this class
| > is obtained by a concrete operation on a function from N^3 to the closed
| > sets.
|
| Is it? This is related to the "countable union of countable sets
| is countable" problem.
The problem with the countable union of countable sets is that nothing
guarantees the existence in the model of the bijection which 'witnesses'
that the resulting set is countable. That doesn't mean that the set itself
doesn't exist. I don't see that this is an issue with the F_sigadeltasigma,
since the usual definition does not require that the set be verified as such
after it is constructed.
Of course, the verification that a *given* set is an F_sds might be
problematic, if the way the set is given is not explicitly in terms of a
presentation as a union of an intersection of a union.
- David R.