how to find an omega-complete non-principal ultrafilter on omega? Omega is the set of all natural numbers.Can anyboby give me an exact example of an omega-complete non-principal ultrafilter on omega?(I want an exact example,not just to prove the ultrafilter exist)
 A: An $\omega$-complete usually means closed under finite intersections. And every filter is $\omega$-complete by definition (and if your definition required that it is closed under intersection of two sets, then by induction it is closed under all finite intersections).
I'm not quite sure how you can give an explicit example. In details.


*

*An ultrafilter on $\omega$ has cardinality of $2^{\aleph_0}$. Writing one explicitly, set by set, not to mention that each set is infinite (since it's non-principal). That's gonna take a long time, and will overcrowd the server hard drives.

*We prove the existence of non-principal ultrafilters by appealing to Zorn's lemma. This use is fairly essential. It is consistent that the Zorn's lemma fails, and all the ultrafilters on $\omega$ are principal. In that case, you can't quite find such an ultrafilter.
This means that there's no formula $\varphi(X)$ in the language of set theory such that there is a unique ultrafilter on $\omega$ that $\sf ZFC$ proves $\varphi(X)$ defines. 
If you have meant ultrafilter which is closed under countable intersection (which is called $\sigma$-complete, or $\omega_1$-complete), then this is even less possible for two reasons:


*

*If $\cal U$ is a non-principal filter on $\omega$, then $\cal U$ includes all the end segments of $\omega$; so if $\cal U$ is closed under countable intersections, it must include their intersection which is the empty set.

*If you replace $\omega$ by another cardinal, a larger cardinal, then taking $\kappa$ to be the least cardinal on which there is a $\sigma$-complete ultrafilter, we can actually show that:


*

*There is an ultrafilter which is in fact $\kappa$-complete on $\kappa$.

*The existence of this $\kappa$ allows us to prove the consistency of $\sf ZFC$, and much, much, so much more than just that. Which means, however, that $\sf ZFC$ cannot prove its existence.
But we have yet to find a contradiction in the theory which includes $\sf ZFC$ and the axiom that such a cardinal exists. So while we cannot prove their existence from the axioms of $\sf ZFC$, we have yet to disproved them either.
A: A filter is said to be $\kappa$-complete if it's closed under intersections of fewer than $\kappa$ sets. This notion is normally used when $\kappa$ is an uncountable regular cardinal. "$\omega$-complete filter" is redundant, because closure under finite intersections is part of the definition of a filter. So every ultrafilter on $\omega$ is $\omega$-complete; however, it is not possible to give an explicit example of a nonprincipal ultrafilter on $\omega$.
Perhaps you meant "closed under countable intersections". An ultrafilter which is closed under countable intersections is said to be $\omega_1$-complete, or $\sigma$-complete, or countably complete. The only $\sigma$-complete ultrafilters on $\omega$ are the principal ultrafilters.
