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I know it is possible to construct a model with a non trivial ultrafilter u and I want to show u is such that for all fonction f from $\mathbb{N}$ to $\mathbb{N}$ there is an element of u such that f is constant or strictly monotonic on u (we will say that u is a selective ultrafilter). I know that we must go from a model of secondary order logic M with comprehension schema and dependant choice axiom and with set of individuals $\mathscr{P}(N)$ and consider N=M[G] where G is a generic which is a non trivial ultrafilter in N on $\mathbb{N}$ (G exists) but I don't know how to show u is selective.

Thanks for your help and suggestions.

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What kind of models are you talking about? The second-order structure of the reals is not a model of set theory. – Emil Jeřábek Apr 4 '12 at 10:40
In any case, yes, in ZFC one can add a selective ultrafilter by taking a generic ultrafilter in $\mathcal P(\omega)/\mathrm{fin}$, which is apparently the forcing notion you describe. Since every $\omega$-chain in this forcing has a lower bound, there are no new functions $f\colon\omega\to\omega$ (or subsets of $\omega$) in the extension, and for any $f\colon\omega\to\omega$ from the ground model, the set of infinite sets on which $f$ is constant or increasing is dense. This is a simple exercise. – Emil Jeřábek Apr 4 '12 at 11:02
Assuming CH, it is also possible to construct a selective ultrafilter by a transfinite recursion of length $\omega_1$. (Let $\langle f_\alpha \rangle_{\alpha\lt\omega_1}$ enumerate all functions $\omega\to\omega$. On the $\alpha$-th step, find a pseudointersection to the countably many sets added to $\mathcal{U}$ so far, and shrink it to an infinite set on which $f_\alpha$ is either constant or strictly increasing.) – François G. Dorais Apr 4 '12 at 13:13
Thanks for your reply and your attention, I was not very clear in my first post I didn't speak about a model of ZFC but M is a model of second order with comprehension schema. I add to M in M[G] a non trivial ultrafilter. I showed following your indication that the set of infinite sets in $\mathbb{N}$ on which f is constant or increasing is dense then G cuts this set and I conclude that there is an element of the generic G on which f is constant or strictly increasing. But I don't know justify why f is in M, it is not clear for me. Thanks. – user2664 Apr 5 '12 at 10:51
I precise that it is necessary to show that f is in M to conclude (but how yet ?) that D(f) is in M (where D(f) is the set of all infinite sets of $\mathbb{N}$ on which f is bounded or strictly increasing), then G meets D(f) because G is a generic in M thus it meets all the dense set in M. – user2664 Apr 5 '12 at 15:49

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