most recent 30 from http://mathoverflow.net 2013-05-24T12:33:56Z http://mathoverflow.net/feeds/user/15147 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65774/definition-of-a-limit-of-a-type over 2011-05-23T17:42:16Z 2011-05-23T17:42:16Z

<p>I apologize if this is not the right forum to ask such a basic question... In model theory what does that mean that a type concentrates on one point?</p>

http://mathoverflow.net/questions/46156/is-every-closed-set-of-q-the-intersection-of-some-connected-closed-set-of-r-wit/65132#65132 over 2011-05-16T12:25:31Z 2011-05-16T12:25:31Z

<p>There might be something wrong with this suggestion but isn't it all right to enumerate all points of F and then connect two "consecutive" points the way Guillaume described in his question (by drawing two lines with irrational slope)?</p>

http://mathoverflow.net/questions/65774/definition-of-a-limit-of-a-type over 2011-05-30T02:11:15Z 2011-05-30T02:11:15Z

I know that this is equivalent to a third fact that is: any M-definable curve in X is completable. ($ X \subset M^n$ with M be an o-minimal structure) (a curve in X is a M-definable continuous embedding f: $(a,b) \stackrel{}{\rightarrow} X$ where (a,b)$\subset$M. It is said to be completable if $\lim_{x\rightarrow a^{+}} f (x) $ and $\lim_{x\rightarrow b^{-}} f (x)$ exist.) I've been told one can associate a definable type to such a curve. But I'm not sure of how. %aybe this would help to understand it?

http://mathoverflow.net/questions/65774/definition-of-a-limit-of-a-type over 2011-05-30T02:10:28Z 2011-05-30T02:10:28Z

Thank you for your ansswers. The type is supposed to be definable. I'm sorry I should have mentionned it maybe the notion of limit of a type makes sense otherwise. I've found this definition: M is an o-minimal structure. A point a of $M^n$ is a limit of p a definable type if for any definable neighborhood U of a (defined with parameters), p concentrates on U. The context is the following theorem: let M be an o-minimal structure, $ X \subset M^n$ then the fact that X is closed and bounded, is equivalent to the fact that any definable type concentrates on one point.

http://mathoverflow.net/questions/46156/is-every-closed-set-of-q-the-intersection-of-some-connected-closed-set-of-r-wit/65132#65132 over 2011-05-17T20:08:41Z 2011-05-17T20:08:41Z

oh yes! That's right! Thanks for your help!