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.**

- $\aleph_1\le\mathfrak{ufo}\le\mathfrak{c}$.
- $\mathfrak{ufo}$ is a regular cardinal number.
- $\mathfrak{ufo}\le\mathfrak{u}$, 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.