I'd like examples of toposes in which Countable Choice is true but Dependent Choice isn't. I'd prefer examples without Excluded Middle. It's hard to find a natural example.
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5$\begingroup$ Of course this isn't what you want - hence a comment as opposed to an answer - but any (classical) model of ZF+CC+$\neg$DC yields a topos of the appropriate (modulo the logic) type. I suspect that the usual forcing argument for building such models can be adapted to work over an appropriately-picked topos in which CC holds but LEM fails, but this is well outside my zone of competence. $\endgroup$– Noah SchweberApr 7, 2020 at 16:45
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2$\begingroup$ @ogogmad that’s just a product of copies of the category of sets, and (thus) satisfies whatever choice principle your category of sets does. $\endgroup$– Kevin CarlsonApr 7, 2020 at 19:28
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2$\begingroup$ @Andrej: You add an homogeneous tree of height $\omega$ and preserve countable bounded parts of the tree. This ensures there are no branches, so DC fails, but given a countable set of non-empty sets, we must have that these sets are supported jointly by some bounded set, and utilise this to get a choice function. It's not a simple argument, but that's the idea, if my memory serves me right. $\endgroup$– Asaf Karagila ♦Apr 7, 2020 at 20:12
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1$\begingroup$ @Andrej: I will look for it tomorrow. Jech's AC book, Felgner's "Models of ZF set theory", both are good starting points. $\endgroup$– Asaf Karagila ♦Apr 7, 2020 at 21:25
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3$\begingroup$ Theorem 8.12 In Jech's "Axiom of Choice", if I'm reading it correctly. It describes a topological group $\mathcal{G}$ such that sets with a continuous $\mathcal{G}$-action give the topos you want. $\endgroup$– David Roberts ♦Apr 8, 2020 at 5:35
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1 Answer
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Given any topological group $G$, the topos of sets with a continuous $G$-action is Boolean, and very often violates some choice principle or other. Under the translation between material and structural set theories, such toposes correspond to permutation (or Fraenkel–Mostowski) models of ZFA.
Theorem 8.12 in Jech's Axiom of Choice describes in terms in material sets a model of set theory in which countable choice (i.e. $\mathrm{AC}_{\aleph_0}$) holds, but DC doesn't. [In fact, Jech describes something more general, of which this is the "$<\aleph_1$ case"]
Consider the set $A := \aleph_1^{<\omega} = \bigcup_{n\in \omega} \aleph_1^n$ of finite sequences of countable ordinals. This carries a partial order where $s \leq t$ iff $t$ extends $s$, and is in fact a tree. Consider the automorphism group of this tree, call it $G$. This gets a topology by specifying a filter $F$ generated by an ideal $I\subset P(A)$. A subset $E\subset A$ is in $I$ precisely if it is a countable, bounded-height sub-tree. Then $F$ is the filter generated by the subgroups $\mathrm{Fix}(E) < G$ consisting of automorphisms that fix the subset $E \subset A$ pointwise.
Then the topos of continuous $G$-sets with this topological group is what you wanted.
answered Apr 11, 2020 at 5:28
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$\begingroup$ I think you've missed a
_0on that definition of $A$. $\endgroup$– Asaf Karagila ♦Apr 11, 2020 at 19:26 -
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1$\begingroup$ Still weird to see $\aleph_1$ and not $\omega_1$ there, by the way. :) $\endgroup$– Asaf Karagila ♦Apr 11, 2020 at 22:23
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1$\begingroup$ Well, I was not particularly aiming for consistency with traditional set-theoretic notation. I've already edited this a bunch of times, you're welcome to change it if you like. I think it would be more nicely expressed in any case if we could write $G$ as the inverse limit of a cofiltered diagram of discrete groups (namely, the quotients of $G$ by subgroups in the filter that are normal), but I'm not quite invested enough to dedicate the time to figure it out. $\endgroup$– David Roberts ♦Apr 11, 2020 at 22:37