# Counterexemple to Urysohn's lemma in a topos without denombrable choice ?

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 !

• How do you know ACC is necessary for Urysohn's Lemma? – François G. Dorais Apr 26 '12 at 15:09
• AFAICS, the usual proof of Urysohn’s lemma actually uses dependent choice, not just countable choice. – Emil Jeřábek Apr 26 '12 at 15:23
• 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. – Andreas Blass Apr 26 '12 at 15:26
• This is Form 78 in Consequences of the Axiom of Choice consequences.emich.edu/conseq.htm It is known to be false in the models $\mathcal{N}3$ and $\mathcal{N}8$; these are two permutation models. (I think the first is one of the original Mostowski models and the other is due to Laüchli, but I don't have the book handy right now.) – François G. Dorais Apr 26 '12 at 15:26
• 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.) – Andrej Bauer Apr 26 '12 at 17:33

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.