Why are the ultrafilters on $\mathbb{N}$ precisely the minimal Cauchyfilters? One can place an equivalence relation on the set of all Cauchyfilters in a uniform space: $\mathcal{U}\sim\mathcal{V}$ iff $\mathcal{U}\cap\mathcal{V}$ is a Cauchyfilter. In my topology course, we proved that there is a minimal Cauchyfilter in every equivalence class.
Now, consider $\mathbb{N}$ with the initial uniformity $\mathcal{C}^*$ for the source of all continuous functions from $\mathbb{N}$ to $[0, 1]$. I want to prove that the ultrafilters on $\mathbb{N}$ are precisely the minimal Cauchyfilters on $\mathbb{N}$. In one of the lectures, my professor proved that two different ultrafilters can't be equivalent, so they all form a separate equivalence class. Apparently, this settles the problem.
My question is: isn't it possible that an ultrafilter $\mathcal{U}$ is equivalent to a coarser non-ultra Cauchyfilter $\mathcal{F}$? In that case, I only see that $\mathcal{F}\subset \mathcal{V}$ for a certain ultrafilter $\mathcal{V}$ and that this implies $\mathcal{U}=\mathcal{V}$. But what is the problem in this situation?
 A: If $\mathscr{F}$ is Cauchy, then for every finite set $H$ of functions from $\Bbb N$ to $[0,1]$ and every $\epsilon>0$ there is an $F\in\mathscr{F}$ such that $|h(x)-h(y)|<\epsilon$ for all $x,y\in F$ and $h\in H$. Let $A\subseteq\Bbb N$ be arbitrary, and let $h$ be the indicator function of $A$; there is an $F\in\mathscr{F}$ such that $|h(x)-h(y)|<1$ for all $x,y\in F$. Clearly either $F\subseteq A$ or $F\subseteq\Bbb N\setminus A$, so $A\in\mathscr{F}$ or $\Bbb N\setminus A\in\mathscr{F}$, and $\mathscr{F}$ is therefore an ultrafilter.
A: In order to show that the ultrafilters on $\mathbb{N}$ are precisely the minimal Cauchyfilters, it is sufficient to prove that the equivalence classes of Cauchyfilters are singletons of ultrafilters.
First, note that every ultrafilter is a Cauchyfilter, since $\mathcal{C}^*$ is totally bounded. We will show that the intersection of two ultrafilters cannot be a Cauchyfilter, i.e. no two ultrafilters are equivalent. The idea is actually similar to Brian M. Scott's answer. Suppose that $\mathcal{U}$ and $\mathcal{V}$ are different ultrafilters where $\mathcal{U}\cap\mathcal{V}$ is Cauchy. Since $\mathcal{C}^*$ is initial, $f(\mathcal{U}\cap\mathcal{V})$ should be a Cauchyfilter for every continuous function $f\colon \mathbb{N}\to[0,1]$. Since $\mathcal{U}\neq\mathcal{V}$, there is an $A\in\mathcal{U}$ such that $\mathbb{N}\setminus A\in\mathcal{V}$. In particular, the indicator function $1_A$ on $A$ is continuous, since $\mathbb{N}$ is discrete. But for every $B\in\mathcal{U}\cap\mathcal{V}$ we have $1_A(B)=\{0,1\}$, so $1_A(\mathcal{U}\cap\mathcal{V})$ is not Cauchy. This is a contradiction.
Secondly, every filter is the intersection of all finer ultrafilters. The previous step gives that therefore only ultrafilters can be Cauchy.
