# Peculiar examples with Axiom of Countable Choice ?

I've been going over the extremely interesting discussions about Axiom of Choice.

It looks to me like all the "weird" consequences of AC (Banach-Tarski etc) come from using it on uncountable collections of sets.

If, instead, we only believe the Axiom of Countable Choice, do we still get unintuitive consequences in the same sense ?

Apologies in advance if the question is vague.

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Presumably, you want to do all this in classical logic. If you stick to intuitionistic logic then, since the countable axiom of choice has a computable realization, everything proved from it will be computable in a suitable sense. And computable things are typically "intuitive" and not too surprising. –  Andrej Bauer May 2 '10 at 10:34
How much of Tychonoff and Hahn-Banach can we prove with countable choice? Perhaps we can show that a countable product of compact spaces is compact and that Hahn-Banach holds for separable spaces? But these would all be nice consequences, not unintuitive ones. –  Andrej Bauer May 2 '10 at 15:59
It is consistent with countable choice that a product of nonempty sets may be empty, which I find at least as unintuitive as, say, Banach-Tarski. (Sorry for being deliberately obtuse.) –  Dan Petersen May 2 '10 at 16:59
@Dan: The statement "All green cows are from Mars" is also consistent with the axiom of countable choice, but isn't a consequence of it (which is what Cosmonut asked for). Your example is a consequence of the statement "uncountable choice is false", not "countable choice is true". –  Paul Siegel May 2 '10 at 17:21

If you assume the existence of suitable large cardinals, then $L(\mathbb{R})$ is a model of the Axiom of Determinacy $AD$ and the Axiom of Dependent Choice $DC$. In particular, since $DC$ is stronger than the Axiom of Countable Choice $AC_{\omega}$, it follows that $AC_{\omega}$ is also true in $L(\mathbb{R})$. Since $L(\mathbb{R})$ satisfies $AD$, all subsets $X \subseteq \mathbb{R}$ and maps $f: \mathbb{R} \to \mathbb{R}$ are measurable, etc. So it seems extremely unlikely that you will find any unintuitive consequences of $AC_{\omega}$ in the more classical areas of mathematics.

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Unless, of course, you find it counterintuitive that all subsets of the reals are measurable. –  Gerry Myerson May 3 '10 at 0:21
Gerry, Simon wasn't saying that that statement (all sets are measurable) is a consequence of $AC_\omega$, but rather, only that it is consistent with $AC_\omega$. Thus, his argument shows that one cannot expect any of the paradoxes arising from non-measurability to arise as consequences of $AC_\omega$. –  Joel David Hamkins May 3 '10 at 0:43
Joel, many thanks. –  Gerry Myerson May 3 '10 at 12:40