# “Paradox” with the Axiom of Dependent Choice.

Here "paradox" only refers to the fact that it conflicts with our real-life intuition. For example, one is able to prove the Banach–Tarski paradox as a theorem under ZF+AC(Axiom of Choice).

There are weaker versions of choice axioms, and among them, DC(Axiom of Dependent Choice) is heavily used in, for example, recursive defining sequences. There are a number of scenarios under which one must deal with a countable number of objects and elements in it, and in my opinion, DC is more basic and more "reasonable" than AC. So I start to wonder: Is there a "paradox" under ZF+DC that conflicts our geometric intuition or other types of common sense?

• Could you please leave a reason for the downvote so that I can make the question better-stated? – William Sun Mar 31 at 0:48
• Well,I didn't downvote but so far as I can tell you don't have any question at all and it's completely unclear what you want. I seriously have no idea. – fleablood Mar 31 at 1:05
• @fleablood The final sentence is "I start to wonder: Is there a "paradox" under ZF+DC that conflicts our geometric intuition or other types of common sense?" That seems to be a decent question. – Noah Schweber Mar 31 at 1:10
• There are paradoxes that don't use choice at all, for example $\aleph_0+\aleph_0=\aleph_0$. – bof Mar 31 at 2:04

I would argue that the answer to your question is (according to current knowledge) no.

The theory ZF + DC + "Every set of reals is Lebesgue measurable (and has the property of Baire and the perfect set property)" is consistent$$^1$$ by a theorem of Solovay, and this rules out the whole Banach-Tarski "flavor." It's inherently difficult to answer a question like yours negatively, but I think this is a strong argument against the possibility of DC-only geometric paradoxes.

$$^1$$Well, under a mild assumption: we need the theory ZFC + "There is an inaccessible cardinal" to be consistent (and in fact Shelah showed that this consistency assumption is optimal for both measurability and for the perfect set property, while no assumption beyond the consistency of ZFC is needed for the Baire property).

• This answer is not at all accurate. – Asaf Karagila Mar 31 at 7:12
• @AsafKaragila Wait, what'd I screw up? – Noah Schweber Mar 31 at 13:12
• "Every set of reals is Lebesgue measurable" proves $|\mathbb R|<|\mathbb R/\mathbb Q|$ which is not really nice – ℋolo Mar 31 at 13:19
• @Holo Sure, but is that really a geometric paradox? I would say no, personally. It's also not provable in ZF+DC, and the question was whether ZF+DC alone gave rise to paradoxes. – Noah Schweber Mar 31 at 13:20
• Ah, I missed the "geometric" part in the question, I still think that "dividing the reals into more parts than the reals themselves" is against geometric intuition, but I can see why you don't think on it like that. I commented about your "ZF + DC + Every set of reals is Lebesgue measurable rules out the whole Banach-Tarski "flavor"" only. Sorry I should have made it clear – ℋolo Mar 31 at 13:25

While $$\sf ZF+DC$$ does not prove the Division Paradox, it is certainly consistent with it, and in the models that we know of where the Banach–Tarski Paradox is false, the Division Paradox holds.

So, what is the Division Paradox? Your geometric intuition tells you that if you break a stick into five parts, then the amount of "stuff" on the stick is at least $$5$$. In more mathematical words, if you partition a line into non-empty parts, then the number of parts is less or equal to the number of points on the line.

In the models where all sets are measurable, which are the models where Banach–Tarski fails, we can partition the real line into more parts than points. We can also partition the real line into a completely incomparable number of parts, so there's no comparison between the number of parts and the points on the real line.

Arguably, this is not a geometric intuition. And arguably, this is not an answer since $$\sf ZF+DC$$ does not outright prove the Division Paradox. But in our current understanding of the universe of sets, one of the two paradoxes hold.

Of course, if you want to be more "grown up" about geometry, you might argue that the Hahn–Banach theorem is intuitive, or at least that if $$B$$ is a Banach space, then its dual space (of continuous functionals) is nontrivial, i.e. we can separate points using functionals, which is a statement equivalent to the Hahn–Banach theorem. And in that case, you're in trouble. Hahn–Banach is sufficient to prove the Banach–Tarski paradox. So $$\sf ZF+DC$$ proves that you have to choose between Banach–Tarski or Banach spaces with trivial duals.

(Caveat lector: paradoxes and intuition are subjective and over time we deconstruct and reconstruct our intuition. What is intuitive to one might as well be paradoxical to another.)

• Just being curious: I have seen the partition of the real line to some parts such that their size is incomparable to $2^{\aleph_0}$ in some of your posts here, since this is a somewhat amusing fact, can you give some or any references on this, please? – Shervin Sorouri Mar 31 at 11:25
• @ShervinSorouri look here just instead of singletons in $\Bbb R\setminus(0,1)$, take $\Bbb R\setminus(0,1)$ itself to be in the partition, then you have $\aleph_1$ size partition in a model where $\aleph_1$ and $2^{\aleph_0}$ are incomparable – ℋolo Mar 31 at 13:07
• @Holo, Thanks. Actually what Asaf proves there is a little bit more amusing since he actually constructs a stricly larger partition! – Shervin Sorouri Mar 31 at 13:47
• @Shervin: It is a standard ZF fact that the reals can be mapped on $\omega_1$. But it is consistent that there is no injection from $\omega_1$ into $\Bbb R$. This gives us a partition of $\Bbb R$ which is incomparable. Now do what Holo suggested, and we get a partition which is strictly larger. – Asaf Karagila Mar 31 at 13:57
• @ShervinSorouri A different way to get strictly larger partition: let $\sim$ be Vitali relation, then define $f:\mathbb (0,1)\to \mathbb R/\sim$ with $f(x)=\left[0.x_10x_1x_200x_1x_2x_3...\right]_\sim$(where $x_i$ is the digits of $x$), this is injective so $|\mathbb R/\sim|\ge|\mathbb R|$. And from here you get that if $|\mathbb R/\sim|=|\mathbb R|$ you have non-Lebegue measurable set so in Solovay's model $|\mathbb R/\sim|>|\mathbb R|$ – ℋolo Mar 31 at 13:59