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Aug 24, 2011 at 21:18 comment added James Freitag Ok, so my last comment is clearly wrong. Itai is right; it is the premise of the original question that is wrong.
Aug 24, 2011 at 15:23 comment added Itaï BEN YAACOV The problem with this approach is not stationarity. It is that the $\varphi$-definition of a type over $\emptyset$ (or over $\mathrm{acl}(\emptyset)$ - we do want it to be stationary) is not obtained type-definably from a realisation. In the example I give in the answer below, you have a sequence of types which represnts the formula $x = y$, but no accumulation point of the sequence represents it.
Aug 23, 2011 at 18:13 comment added James Freitag About fixing the variables. You are correct technically; I suppose that I am sort of proving that the fibers of the relation are type definable. Perhaps it can be completed by varying the definition in a type definable way with the first coordinate. I would have to think for a bit to write this down carefully...
Aug 23, 2011 at 18:11 comment added James Freitag Ok, good. I was not sure about what would happen without stationarity, though some people who work with general stable theories might comment on this. Representing a new formula is a criteria for forking. In, say, Pillay's small stability theory book, this sort of notion is definitely covered. This should be in Marker's model theory book as well. Perhaps the language is the issue. This terminology goes with the old style of describing heirs, etc.
Aug 23, 2011 at 16:15 comment added henry I don't understand what do you mean by 'tp(a/b) represents a new formula', but it seems to me that you fixing the variables a and b. p.s: for my purpose it suffice to assume that the empty set is Algebraically closed (so the type is stationary)
Aug 23, 2011 at 15:43 history edited James Freitag CC BY-SA 3.0
Comments at the end.
Aug 23, 2011 at 15:31 history answered James Freitag CC BY-SA 3.0