An *ultrafilter ornament* is a chain of free filters on $\mathbb{N}$ that are not ultrafilters, whose union is an ultrafilter.


Let $\mathfrak{ufo}$ be the minimal cardinality of 
an ultrafilter ornament. 


I arrived at this definition back in 2008, while teaching Ramsey theory at the Weizmann Institute of Science, based on a nice book by I. Protasov.
Topologically, the cardinal number $\mathfrak{ufo}$ is the minimal length of a transfinite convergent sequence in $\beta(\mathbb{N})$, but you may proceed with the combinatorial definition if you prefer.


**Basic facts.** 


1. $\aleph\_1\le\mathfrak{ufo}\le\mathfrak{c}$.
2. $\mathfrak{ufo}$ is a regular cardinal number.
3. $\mathfrak{ufo}\le\mathfrak{u}$, indeed $\mathfrak{ufo}\le\operatorname{cof}(\kappa)$ for each cardinality $\kappa$ of a basis for an ultrafilter.


Item 1 follows from the topological interpretation, but can also be proved combinatorially. Blass supplied me with one such proof. 
This is a cute exercise (or see below). Items 2-3 are immediate.
Item 3 was pointed out to me by Blass. 


**Problem.** Can the cardinal number $\mathfrak{ufo}$ be identified as a classic combinatorial cardinal characteristic of the continuum?


(For me, $\aleph\_1$ is definitely a (potential) positive answer.)


Here is Blass's proof. I thank him for the permission to include it here.
I made tiny changes. If you find errors, this must be my fault. :)


Suppose, toward a contradiction, that some ultrafilter $U$ is the union of a countable, strictly increasing sequence of filters $F\_n$. For each $n$, pick a set $A\_n\in F\_{n+1}\setminus F\_n$. By intersecting each $A\_n$ with all the earlier $A\_m$'s, we may assume that the $A\_n$ sequence is decreasing (with respect to set-inclusion). Let $D\_n$ be the set-difference $A\_n\setminus A\_{n+1}$; so the sets $D\_n$ are pairwise disjoint. Let $X$ be the union of the sets $D\_n$ for the even $n$, and let Y be the union the remaining sets $D\_n$. Then 
$$
\mathbb{N}=X\cup Y\cup (\mathbb{N}\setminus A\_1)\cup \bigcap\_n A\_n.
$$
The last of these isn't in $U$, because, being a subset of $A\_n$, it can't be in $F\_n$ for any $n$. And $\mathbb{N}\setminus A\_1$ isn't in $U$, because $A\_1$ is. So either $X$ or $Y$ is in $U$; without loss of generality, suppose $X$ is in $U$. Then $X$ is in $F\_n$ for some $n$, and we may assume that $n$ is odd (just add 1 to $n$ if necessary). So $F\_n$ contains $X$ and $A\_n$ but not $A\_{n+1}$. But $An\setminus A\_{n+1}=D\_n$ is disjoint from $X$ (because $n$ is odd), which means that $A\_{n+1}$ is a superset of the intersection of $X$ and $A\_n$. The preceding two sentences contradict the fact that $F\_n$ is a filter.



$\mathfrak{ufo}$: An unidentified combinatorial cardinal characteristic of the continuum?
This invariant is known for Boolean algebras in general as pseudo-altitude. That it is $\omega\_1$ is proved in van Douwen's chapter in the Boolean algebra handbook 11.1 and 12.7, in the more general form that this holds for any weakly countably complete BA.


