most recent 30 from http://mathoverflow.net 2013-05-20T13:21:39Z http://mathoverflow.net/feeds/question/68335 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68335/what-do-coherent-topoi-have-to-do-with-completeness Akhil Mathew 2011-06-21T03:41:42Z 2013-05-17T04:03:16Z

<p>There is a theorem of Deligne in SGA4 that a "coherent" topos (e.g. one on a site where all objects are quasi-compact and quasi-separated) has enough points (i.e. isomorphisms can be detected via geometric morphisms to the topos of sets). I've heard it said that this is a form of Goedel's completeness theorem for first-order logic.</p> <p>Why is that? I get the intuition that having a model for a formula is supposed to be analogous to a point of a suitable topos, but this is very vague.</p> <p>I'm sorry for not providing more motivation, but I don't know enough about this connection to do so!</p> <p>(This was first posted on math.SE <a href="http://math.stackexchange.com/questions/45647/what-do-coherent-topoi-have-to-do-with-completeness" rel="nofollow">here</a>, where it did not (yet) receive a response.)</p>

http://mathoverflow.net/questions/68335/what-do-coherent-topoi-have-to-do-with-completeness/68342#68342 Torsten Ekedahl 2011-06-21T05:46:21Z 2011-06-21T10:34:38Z

<p>They are indeed formally equivalent. See for instance Johnstone: Topos theory, p. 243 but here is a quick explanation. Given a topos $T$ one may define a geometric theory associated to it consisting of formulas describing essentially the topos. More specifically a geometric morphism from another topos $S$ to $T$ is the same thing as a model for the theory in $S$. In particular if $S$ is the category of sets this a set-theoretical model fo the theory. In general the language of the theory has arbitrary disjunctions leading to theories that in general do not have models. However, if the topos is coherent only finite disjunctions are needed and we are in the realm of the Gödel completeness theorem which then can be interpreted as saying that a coherent topos has enough points. Conversely, given a geometric theory one can associate to it a syntactic site whose objects are the formulas. An implication from a disjunction of formulas to a formula is a covering. The topos of sheaves on this site will then be a classifying topos for the theory (i.e., geometric morphisms to it are the same as models). If the theory is finitary (i.e., uses only finite disjunctions) then the topology is coherent and there are models for the theory by Deligne's theorem.</p> <p>It is amusing that Deligne's fairly natural example of a topos without points, "measure sheaves" on a measure space for which all points have measure zero thus gives an example of a consistent geometric theory (with countable disjunctions) that doesn't have a model.</p>

http://mathoverflow.net/questions/68335/what-do-coherent-topoi-have-to-do-with-completeness/129875#129875 John Baez 2013-05-06T19:07:40Z 2013-05-06T19:07:40Z

<p>A good explanation of this topic can now be found here:</p> <ul> <li>Benjamin Frot, <a href="http://www.dtc.ox.ac.uk/people/12/frot/CoherentTopoi.pdf" rel="nofollow">G&ouml;del's Completeness Theorem and Deligne's Theorem</a>, talk at Paris VII, January 2012. </li> </ul>

http://mathoverflow.net/questions/68335/what-do-coherent-topoi-have-to-do-with-completeness/130906#130906 StephenK1 2013-05-17T04:03:16Z 2013-05-17T04:03:16Z

<p>Is this is equivalent to the SAT problem for an arbitrary Boolean formula to have an assignment that makes the formula "true"? <a href="http://mathworld.wolfram.com/SatisfiabilityProblem.html" rel="nofollow">http://mathworld.wolfram.com/SatisfiabilityProblem.html</a></p>