Let $u$ be a free ultrafilter on $\omega$.
I am interested in constructing an ultrafilters on the closed subsets of $\mathbb R$ which contain the collection $\mathcal C$$=${$\bigcup_{n\in A} [n,n+1]:A\in u$}.
If $(x_n)\in \mathbb R ^\omega$ is a sequence of reals with the property $x_n\in (n,n+1)$ for each $n\in\omega$, then I can show the filter generated by $\mathcal C\cup${$x_n:n\in \omega$} is an ultrafilter.
Question: It is not true that every ultrafilter of the type I want can be constructed in this manner, but I am having some difficulty seeing why. Could someone describe an ultrafilter containing $\mathcal C$ which can not be generated from $\mathcal C$ and a sequence (of the type I used)?
EDIT: What about this. For each $A\in u$, let $(s^A_n)_{n\in A}$ be a real-valued sequence on $A$. Then consider a collection $\mathcal S$$=${$(s^A_n):A\in u$} that has the finite intersection property. It looks like $\mathcal C\cup \mathcal S$ generates an ultrafilter?