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I've read that the axiom of dependent choice can't implies the Axiom of choice because the axiom of dependent choice is a "little" case of the axiom of choice. But i really don't know how to make a counterexample of that. I really appreciate your help. Thanks.

Asaf Karagila
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sango
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  • First I suggest stating or defining the two axioms you are concerned about, for clarity and to make the question self sustain itself, second to clarify what the question is. Are you asking for a counter example or proof that axiom of dependent choice implies the axiom of choice? The title and last sentence appear to contradict one another? – marshal craft May 20 '17 at 03:19
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    What makes you think that this is "elementary"? – Asaf Karagila May 20 '17 at 05:36
  • The famous Solovay model of ZF+DC+LM does what you want. LM says every set of reals is Lebesgue measurable, so the Banach-Tarski paradox sets cannot be constructed. This shows it does not obey AC. – Ross Millikan May 20 '17 at 14:45
  • @Ross but that construction relies on an inaccessible, and that's not needed for this. – Noah Schweber May 20 '17 at 15:05
  • @asaf to be fair, given how frequently it's said one might believe it's simple. – Noah Schweber May 20 '17 at 15:06

2 Answers2

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It is indeed the case that (over ZF) DC does not imply AC. However, this isn't easy to prove. Showing nonimplications between set-theoretic principles is difficult, because it requires us to build models in which one holds and the other fails, and building models of ZF and related systems is quite hard. The two main techniques here are inner models and forcing.

This puts me in an awkward position with regards to the rest of your question: you (quite reasonably) want a counterexample, but the only ways to build one rely on advanced techniques which you may not have seen before, and which are too involved for me to explain here. Below I give an extremely terse description of how to build one such counterexample, since I would rather include something than nothing at all, but this is definitely a very advanced topic; the text you should look to is Kunen's set theory book, especially chapters 6 and 7.

So how can we get a model of ZF + DC + not AC?

The fastest approach I know mixes inner models and forcing. On the inner model side, exercise 8F.13 in Moschovakis' descriptive set theory text shows that $L(\mathbb{R})$, the smallest inner model containing every real, satisfies DC. (More precisely, the exercise shows that if $M$ is a model of a large enough fragment of ZFC, then $L(\mathbb{R})^M$ satisfies ZF+DC.)

So it's enough to show that it is possible for $L(\mathbb{R})$ to not satisfy choice, and this we do by forcing. Specifically, it's not hard to show that $L(\mathbb{R})^M$ does not satisfy AC (in particular it has no well-ordering of $\mathbb{R}$) whenever $M$ satisfies "for every real $r$ there is a real $s$ which is Cohen relative to $r$," and this is straightforward to force.

Noah Schweber
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In order to show that a theory $T$ does not imply $\varphi$, the two easy methods are: (1) show that $T$ proves $\lnot\varphi$; or (2) find at least one model of $T$ in which $\varphi$ fails.

In the case of $\sf ZF+DC$ as our theory and $\sf AC$ as $\varphi$, (1) is blatantly false (since $\sf AC$ implies $\sf DC$, any model of $\sf ZFC$ is a model of both $\sf ZF+DC$ and $\sf AC$, so (1) would imply that $\sf ZFC$ is inconsistent). So we have to resort to (2).

Generating models of $\sf ZF$ is not very easy. As Noah mentioned, forcing and inner models are the common methods for this sort of proofs (often used in tandem), and both of them are technical and require some time to master.

For a slightly simpler proof, you can replace $\sf ZF$ by $\sf ZFA$, i.e., add atoms (or non-sets). This allows for slightly simplified proofs of independence for choice-related propositions. You can learn more about this in Jech's "Axiom of Choice" book, which also discusses forcing related constructions. You can even find answers to more refined questions on Chapter 8 of his book.

If you are already somewhat familiar with forcing and symmetric extensions, here is a nice outline of a model where $\sf DC$ holds (for a suitable choice of parameters) and the axiom of choice fails: Failure of Choice only for sets above a certain rank.

Asaf Karagila
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