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Hello !

The Urysohn's Lemma assert that in every topological spaces which is normal two closed subset may be separated by a real valued function. It's proof use axiom of countable choice (but not the law of excluded middle).

I would like to find a counterexample to this theorem in the internal logic of a topos in which the axiom of countable choice does not hold (for exemple, the topos of smooth action of some non discrete locally pro-finite group, or the topos of sheaf on [0,1].)

I need a counterexample which is compact, but If you have an example involving not a topological space but a local (an example of compact regular local which does not have enough functions with value in the Dedekind real) it's perfectly fine for me.

Thank you !

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    $\begingroup$ AFAICS, the usual proof of Urysohn’s lemma actually uses dependent choice, not just countable choice. $\endgroup$ Apr 26, 2012 at 15:23
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    $\begingroup$ How do you know the axiom of countable choice is sufficient for Urysohn's Lemma? The usual proof uses dependent choice. There is a variation of the proof, pointed out to me by David Pincus 35 years ago (boy, am I getting old!) that uses multiple choice instead. And I mixed the two proofs (also in ancient times) to show that dependent multiple choice suffices. But I'm not sure countable choice suffices. $\endgroup$ Apr 26, 2012 at 15:26
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    $\begingroup$ Emil's comment arrived while I was typing mine. Mine was addressed to Simon Henry. $\endgroup$ Apr 26, 2012 at 15:27
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    $\begingroup$ Yes, my Mistake: It's indeed dependent choice, not just countable choice which is used in the classical proof. Thank you for your answer, If someone know an example in a Grothendick topos I would be interested too. $\endgroup$ Apr 26, 2012 at 15:49
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    $\begingroup$ My instincts tell me that in the effective topos there are two closed subsets of the real line which are disjoint but cannot be separated by a real-valued function. Would that be of interest to you? I could think about it a bit. (Note: if by "closed" you mean closed under limits of convergent sequences, then I am not sure that every closed set is a zero-set in the effective topos.) $\endgroup$ Apr 26, 2012 at 17:33

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Urysohn's Lemma is not provable in ZF (without the axiom of choice but with classical logic), so a suitable model of ZF will provide a topos of the sort you want. Checking the standard reference for such questions, "Consequences of the Axiom of Choice" by Paul Howard and Jean Rubin, I find the following permutation model (of ZF with atoms), due to Läuchli, in which Urysohn's Lemma is false. Begin with a countable set of atoms ordered like the rationals, take the group of all order-automorphisms, and take as supports the sets $E$ of atoms such that $E$ has only finitely many accumulation points and every infinite subset of $E$ has an accumulation point.

A permutation model of ZF with atoms suffices to give a topos of the sort you want, but if you'd rather have a model of full ZF (i.e, without atoms), the Jech-Sochor embedding theorem lets you eliminate the atoms from Läuchli's example.

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