Showing that a net is universal

Question

Let $\mathcal{U}$ be an ultrafilter on $\mathbb{N}$ and let $x:\mathcal{U} \rightarrow \mathbb{N}$ be a net such that $x(U) \in U, \forall U \in \mathcal{U}$. Show that $x$ is universal.

I tried some ways to prove this statement: e.g. considering the "filter of tails" seems to lead to a dead end. Another way I tried is to consider a universal subnet of $x$, say $x'=\{ x(\psi(U))\}$, where $\psi$ is the map defining the subnet, and then taking the filter generated by $\{ \psi(U) \}$, say $\mathcal{U}'$ we can see (using the maximality of $\mathcal{U}$), that $\mathcal{U}=\mathcal{U}'$. But then I don't know how to go on.

Answer 1

For $x$ to be universal we need to show that for all $A \subseteq \mathbb{N}$ we have that $x$ is eventually in $A$ or $x$ is eventually in $\mathbb{N} \setminus A$.

For an ultrafilter $\mathcal{U}$ on $\mathbb{N}$ we know a very similar property: $$\forall A \subseteq \mathbb{N}: A \in \mathcal{U} \text{ or } \mathbb{N} \setminus A \in \mathcal{U}$$

The universality of $x$ quite directly follows from this last property of ultrafilters. Do you see how? You use in an essential way that $\mathcal{U}$ itself (in reverse inclusion order) is used as the index set as well!

This last "set or its complement"-property follows from the maximality of $\mathcal{U}$, which is also the cause of another property of ultrafilters:

Let $\mathcal{U}$ be an ultrafilter on a set $X$ and suppose that for some $A \subseteq X$ we know that $A$ intersects all members of $\mathcal{U}$, then we can conclude that $A \in \mathcal{U}$. (This property in fact characterises ultrafilters among all filters on $X$.)

Assuming this, if $A \notin \mathcal{U}$, this implies that it has empty intersection with some $B \in \mathcal{U}$. But then $B \subseteq X\setminus A$ and then $X\setminus A \in \mathcal{U}$, as filters are closed under enlargements.