3
$\begingroup$

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 !

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ AFAICS, the usual proof of Urysohn’s lemma actually uses dependent choice, not just countable choice. $\endgroup$
    – Emil Jeřábek
    Commented Apr 26, 2012 at 15:23
  • 3
    $\begingroup$ 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. $\endgroup$
    – Andreas Blass
    Commented Apr 26, 2012 at 15:26
  • 1
    $\begingroup$ Emil's comment arrived while I was typing mine. Mine was addressed to Simon Henry. $\endgroup$
    – Andreas Blass
    Commented Apr 26, 2012 at 15:27
  • 1
    $\begingroup$ 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. $\endgroup$
    – Simon Henry
    Commented Apr 26, 2012 at 15:49
  • 1
    $\begingroup$ 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.) $\endgroup$
    – Andrej Bauer
    Commented Apr 26, 2012 at 17:33

1 Answer 1

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .