Thanks for responding. There are two possible answers to this.
(1) The existence of S is consistent with ZF, and the Lebesgue measurability of all subsets of R (of which S would be one) is consistent with ZF plus another axiom (such as AD or the existence of an inaccessible cardinal), so the question would be in a model of sets fulfilling the appropriate axioms to have such an S. In other words, I am excluding the possibility that S is something like a Vitali set. I do not know whether the existence of S is sufficient to answer the question; if not, go to (2).
(2) Any appropriate forcing technique, such as Cohen's forcing, will do.
as i understand it, you'll want to assume the axiom of choice is false as well (so that you can be sure S is measureable). i don't think you lose any generality by considering subsets of [0,1]. but i'm not convinced that this alone is enough to guarantee that the lebesgue measure of S will be 0. imagine you have uncountably many translates of a cantor set. can we stack enough of these together to get "some" measure, but not "mostly everything"? (if so, then perhaps models exist in which the lebesgue measure of S is any number between 0 and 1).
if a suitable model exists (and i suspect its construction would be rather complicated) what would it "mean"? would it be useful? from a philosophical point of view, there are compelling reasons for me to believe the axiom of choice should be true. every image should have a pre-image.
i think there are two distinct, but related problems here: one is related to cardinality, and one is related to measureability. i don't think assuming the negation of the continuum hypothesis is sufficient for assuming the measureability of a given uncountable subset of the reals. and it's not at all clear that ZF + ~C + ~CH only admits one model.
Although assuming ~AC and the existence of an inaccessible cardinal is a sure-fire way to make sure that there is a model in which S is measurable, it would be a further question as to the necessity of ~AC. That is, ZFC allows non-measurable sets, but I have not seen where it forbids the existence of a measurable uncountable set with cardinality less than that of the continuum.you'll want to assume the axiom of choice is false as well (so that you can be sure S is measureable).
Depends on the uncountability. Obviously you can stack a continuum-number of copies of any non-empty set to get something that gives a non-zero measure. But whether you can stack uncountable but less-than-continuum number of copies of a set of measure zero to get a non-zero measure is precisely my question.). i don't think you lose any generality by considering subsets of [0,1]. but i'm not convinced that this alone is enough to guarantee that the lebesgue measure of S will be 0. imagine you have uncountably many translates of a cantor set. can we stack enough of these together to get "some" measure, but not "mostly everything"?
If S did not have a unique measure, then it would not be considered measurable. Or do you mean that for every r in [0,1] there can be an S_r so that the measure of S_r = r?(if so, then perhaps models exist in which the lebesgue measure of S is any number between 0 and 1).
It might tell us something about the structure of the continuum.if a suitable model exists (and i suspect its construction would be rather complicated) what would it "mean"?
Depends what you mean by "useful"? Sometimes some odd spin-offs come from something that appears, at first glance, to be useless wondering. The continuum hypothesis is one such example.would it be useful?
You are perhaps aware of the joke, "The Axiom of Choice is obviously true, the Well Ordering Theorem is obviously false, and Zorn's Lemma is obviously undecidable. " But even if a proposition is accepted as true, seeing what its negation or its absence can bring can nonetheless be worthwhile.from a philosophical point of view, there are compelling reasons for me to believe the axiom of choice should be true. every image should have a pre-image.
I am asking how they are related. Probably the answer to my original question is positive, but there are often strange exceptions with things like Vitali sets, singular cardinals, and other weirdos, so I am cautious.i think there are two distinct, but related problems here: one is related to cardinality, and one is related to measureability.
Definitely not. In fact, I believe that ZF + ~CH & "there exists a Vitali set" is equiconsistent with ZF.i don't think assuming the negation of the continuum hypothesis is sufficient for assuming the measureability of a given uncountable subset of the reals.
Right, it admits an infinite number of models. Take any of them for which Lebesgue measurability is meaningful., and it's not at all clear that ZF + ~AC + ~CH only admits one model.
right. if you do not assume ~AC, then you have to prove your set S is measureable first.
and i think it's a HARD question. if we had a more explicit description of S, perhaps "S copies" would do the trick (i am thinking of some kind of quotient here).Depends on the uncountability. Obviously you can stack a continuum-number of copies of any non-empty set to get something that gives a non-zero measure. But whether you can stack uncountable but less-than-continuum number of copies of a set of measure zero to get a non-zero measure is precisely my question.
i meant the latter.If S did not have a unique measure, then it would not be considered measurable. Or do you mean that for every r in [0,1] there can be an S_r so that the measure of S_r = r?
this is possible. it is more likely to tell us something about measure.It might tell us something about the structure of the continuum.
perhaps i may be missing the point, but it seems to me that while all can agree that "countably infinite" and "uncountably infinite" are different, we haven't made much headway on where we should draw the boundary. i did not mean to imply such musings are not interesting in and of themselves, perhaps you might find some new insight on the nature of continuity in doing so.Depends what you mean by "useful"? Sometimes some odd spin-offs come from something that appears, at first glance, to be useless wondering. The continuum hypothesis is one such example.
Jerry Bona, i believeYou are perhaps aware of the joke, "The Axiom of Choice is obviously true, the Well Ordering Theorem is obviously false, and Zorn's Lemma is obviously undecidable. " But even if a proposition is accepted as true, seeing what its negation or its absence can bring can nonetheless be worthwhile.
i am inclined to believe you are correct almost everywhere.I am asking how they are related. Probably the answer to my original question is positive, but there are often strange exceptions with things like Vitali sets, singular cardinals, and other weirdos, so I am cautious.
my point exactly.Definitely not. In fact, I believe that ZF + ~CH & "there exists a Vitali set" is equiconsistent with ZF.
again, i don't think it is obvious that we will get the same measure in each model.Right, it admits an infinite number of models. Take any of them for which Lebesgue measurability is meaningful.
Thanks. Given your reference (Jerry Bona), I found the exact quote: "The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?"
As far as "for every r in [0,1] there can be an S_r so that the measure of S_r = r?", I don't think that the cardinality has anything to do with the non-zero size of a set. After all, even keeping the cardinality constant, that of the continuum, we have simply the usual measure on the collection of sub-intervals of [0,1] which fulfills this condition. So my question is simply whether all sets of less-than-continuum cardinality must have zero Lebesgue measure. True, different models will give different values for the measure, so I would like to know whether any of them will give a non-zero measure.
As far as S copies of S, how you would put the copies together would be crucial, since a straightforward bunch of copies would just give S X S, with the same cardinality of S, not getting us anywhere.
As far as a more explicit description of S: One could start by using the S from Cohen's proof, but if this gives a set of zero measure (and I believe that it does), then to find whether any S that satisfies ~CH AND that has a non-zero Lebesgue measure exists is part of the question. Yes, it is a hard question; that's why I don't know the answer.
You are right, the boundary between countable and uncountable is fuzzy, which is one of the reasons why CH has generated so much work.
I like your "correct almost everywhere". Now, I just have to find the right ultrafilter.... :-)