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In his answer to a Math Stack Exchange question of Katlus, Asaf Karagila wrote the following:

"It is a theorem that from $ZF+DC+$"$\aleph_1$$\le$$|$$\mathbb R$$|$" 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."

Three questions regarding this statement:

i) Since "it is a theorem that from $ZF+DC+$"$\aleph_1$$\le$$|$$\mathbb R$$|$" we can prove that there is an unmeasurable set of real numbers" could someone provide an explicit construction of this set in the aforementioned system?

ii) Are there models of $ZF$$+$"Every set of reals is Lebesgue measurable"$+$$\lnot$$CH$?

iii) In the models mentioned in ii), does $DC$ fail?

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Thomas Benjamin
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  • I tried to develop a link to Math Stack Exchange question #210587 but was unable to do so--I must have misunderstood the suggestion in the help box. How does one establish a link in Math Overflow, anyway? This was my first attempt. – Thomas Benjamin Jan 08 '15 at 16:26
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    The result i) is due to Raisonnier http://link.springer.com/article/10.1007%2FBF02760523 . – Emil Jeřábek Jan 08 '15 at 16:26
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    @ThomasBenjamin the main issue with your link was simply that the URL was wrong. There was a missing dot between math and stackexchange. –  Jan 08 '15 at 16:28
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    @quid: Oh... Thanks for fixing the URL. – Thomas Benjamin Jan 08 '15 at 16:29
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    Quick question: what do you mean by "$\neg CH$?" That there is an uncountable set of reals of size strictly less than that of $\mathbb{R}$, or just that $\vert\mathbb{R}\vert\not=\aleph_1$? – Noah Schweber Jan 08 '15 at 17:20
  • @NoahS: $\lnot$$CH$ means, in the context of my question, that there is an uncountable set of reals of cardinality strictly less than that of $\mathbb R$. – Thomas Benjamin Jan 08 '15 at 18:52
  • (cont.) I will also require (in light of the comments I received) that no uncountable subset of $\mathbb R$ be able to be well-ordered. In light of this restriction, how much real analysis will fall (so to speak) by the wayside? – Thomas Benjamin Jan 08 '15 at 19:01
  • You can always make $\sf DC$ fail so very high above the reals, that it has no effect on any set of reals. So the question you ought to be asking is rather different whether or not $\sf DC$ for trees whose domain is a subset of $\Bbb R$ is true. In any case, as pointed out by others, we don't know if that's at all possible so we don't know if there are such models or not. – Asaf Karagila Jan 08 '15 at 19:09
  • @Emil: Either you post this as an answer (perhaps just the reference, perhaps elaborating a little bit), or we close this as a duplicate of Ali Enayat's question linked below. – Asaf Karagila Jan 08 '15 at 19:13
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    Note that in 1991 Foreman and Wehrung proved from ZF + Hahn Banach theorem that there is a Lebesgue non-measurable set. The paper is interesting also because it develops Lebesgue measure without AC. Here is the reference: https://hal.archives-ouvertes.fr/hal-00004713/document – Avshalom Jan 08 '15 at 19:38
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    @Avshalom: If we are giving references for measure theory without choice, then Fremlin's book gets a mandatory mentioning. – Asaf Karagila Jan 08 '15 at 19:44
  • @Asaf: That's all I know about it, and it's too little for a proper answer IMO. Feel free to vote to close if you think it should be closed. – Emil Jeřábek Jan 09 '15 at 12:12

1 Answers1

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For (i) see here. For (ii), (iii) see this.

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