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Reference; Foundation for analysis without axiom of choice?

Please let me know if I'm misunderstanding something and I hope you explain this with relatively easy words. I am eager to learn, but I have no idea what forcing is like, so i couldn't understand this properly from past posts.

  1. Solovay's model says that "In ZF, there exists a model such that all subsets of $\mathbb{R}$ is measurable." (I believe I'm misunderstanding this.. Since if it is true, then it must be true in ZFC)

Why are Solovay's model and Dependent Choice related? Is this true in ZF+DC, but not true in ZFC? How is that possible?

  1. In the link, Asaf says, $\mathbb{R} = \bigcup_{i\in I} A_i$, where $A_i \subset \mathbb{R}$ and are mutually disjoint, nonempty and $|I|>|\mathbb{R}|$.

Really??

You can see, $|\mathbb{R}|=|\bigcup_{i\in I} A_i|=\sum_{i\in I} |A_i|≧\sum_{i\in I} 1=|I|$.

(I'm not sure if $≧$ part uses AC)

Thus $|I|≦|\mathbb{R}|$. How come $|I|>|\mathbb{R}|$?

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Katlus
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    "In ZF, there exists a model such that all subsets of $\mathbb R$ is measurable." There is a slight misunderstanding here. The correct statement is that "There is a model of ZF such that ..." The difference is that to say "In ZF, blah" means that ZF can prove blah. Which is not true in this case. – Andrés E. Caicedo Oct 11 '12 at 03:06

2 Answers2

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In Solovay's model the following holds: ZF, DC, All sets of real numbers are Lebesgue measurable (and more, much more).

It is a theorem that from ZF+DC+"$\aleph_1\leq|\mathbb R|$" we can prove that there is an unmeasurable set of real numbers, so in Solovay's model we have that there are no sets of real numbers which have size $\aleph_1$. In fact, in Solovay's model every uncountable set of reals is of size continuum, and in some sense the continuum hypothesis holds.

However the continuum can always be mapped onto $\aleph_1$. So if you map the interval $[0,1]$ onto $\aleph_1$, it forms a partition of $[0,1]$ to $\aleph_1$ many parts; add the singletons of $\mathbb R\setminus[0,1]$ and you have a partition of $\mathbb R$ into $2^{\aleph_0}+\aleph_1$ many sets. However since $\aleph_1\nleq2^{\aleph_0}$ we have that $\aleph_1+2^{\aleph_0}>2^{\aleph_0}$.

Your arguments about cardinality and sums fails because without the axiom of choice infinite sums of cardinals play by different rules. The fact there is no injection which chooses a point from every $A_i$ means that $I$ does not have to be of smaller cardinality.

Indeed $\sum_{i\in I}|A_i|\geq\sum_{i\in I} 1$ means that you can choose a point from every $A_i$, but in this case you cannot.

The importance of DC in Solovay's model is that under the assumptions of ZF+DC we can prove a lot of classical analysis and basic measure theory. This means that this is a reasonable model for mathematicians to consider. In fact in his paper from the 1970's about automatic continuity of linear operators Garnir foresaw functional analysis developed in Solovay's model naturally. While I am aware of some functional analysis being done in such models, I don't think it caught on strongly as Garnir hoped.

One more point to make about the paradoxical decomposition to more parts than elements, I should add that currently we do not know of any model of ZF+$\lnot$AC where such decomposition does not exist. Namely, as far as we know, in all models where choice fails there is some set which can be partitioned into more parts than elements.

Asaf Karagila
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I believe there might be a misinterpretation here. Solovay's model is a model of ZF in which the following hold:

  • The axiom of dependent choice;
  • Every set of reals in Lebesgue measurable; and
  • Every set of reals has the Baire property.

Due to the usual construction of a Vitali set from the Axiom of Choice it follows in particular that the full Axiom of Choice fails in this model. Since the Axiom of Choice fails, most of the intuition about cardinal arithmetic falls by the wayside.

(You should also note that the construction of Solovay's model does not begin with ZF or even a model of ZF+DC, but instead with a model of "ZFC + $\exists$ inaccessible cardinal".)

user642796
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  • Last sentence is exactly what i wanted to know. So to be clear, it's not a 'counter-intuitional' example when accepting DC? – Katlus Oct 10 '12 at 18:49
  • @Katlus: It is just a side-effect; the same could happen when the real numbers are a countable union of countable sets (e.g. Feferman-Levy model) where DC fails. This has nothing to do with DC, but rather something to do with the fact that there are no sets of real numbers of size $\aleph_1$. – Asaf Karagila Oct 10 '12 at 18:55
  • @Asaf You mean $\aleph_1$ in ZF? – Katlus Oct 10 '12 at 19:17
  • @Katlus: What do you mean by that? – Asaf Karagila Oct 10 '12 at 19:26
  • @Katlus: The fact that DC holds in this model makes it, in my opinion, somewhat a stronger counterexample than other models where AC "totally fails" (countable choice fails). But it is, again only my opinion, a counter-intuitive model that says that one's intuition about measurability must be checked when working with only DC without the benefit of AC also holding. Of course, as AC implies DC it does not mean that DC is incompatible with one's usual intuition about measurability. And one could probably arrange a failure of AC to happen high enough that it doesn't affect measurability. – user642796 Oct 10 '12 at 19:44
  • @Asaf Least well-orderable set such that cardinality is greater than $\mathbb{N}$? I asked because Arthur said the model begin with ZFC and you said $\aleph_1 \nleq 2^{\aleph_0}$ – Katlus Oct 10 '12 at 19:52
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    @Katlus: No... I didn't say that. I said that in Solovay's model we know that ZF+DC+... holds; to construct Solovay's model we begin with a model of ZFC+Inaccessible cardinal. – Asaf Karagila Oct 10 '12 at 19:57