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.

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    $\begingroup$ The perfect timing for this question would have been two months ago, when X-Files returned for a mini-season. You're too late! $\endgroup$ – Asaf Karagila Apr 5 '16 at 22:01
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    $\begingroup$ @AsafKaragila: Actually, I was waiting for your comment. I was wondering whether people would think I'm just late for April's fools day. With the answer below, the name ufo can be recycled for future cardinals. :) $\endgroup$ – Boaz Tsaban Apr 6 '16 at 5:48
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    $\begingroup$ Me specifically? I also never cared much for April Fools. But The X-Files? :-) $\endgroup$ – Asaf Karagila Apr 6 '16 at 6:08
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    $\begingroup$ In the same spirit, in French, my mother tongue, a large cardinal is translated as "grand cardinal". I have no idea how tall Mazarin or Richelieu were though. $\endgroup$ – Sylvain JULIEN Apr 7 '16 at 20:01
  • $\begingroup$ @SylvainJULIEN: This reminds me of the second paragraph in my microblog. $\endgroup$ – Boaz Tsaban Apr 7 '16 at 22:23

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.

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    $\begingroup$ Welcome to math overflow! Nice to see you here! $\endgroup$ – Goldstern Apr 8 '16 at 22:40
  • $\begingroup$ Much appreciated! $\endgroup$ – Boaz Tsaban Apr 9 '16 at 19:40

It is the case that $\mathfrak{ufo} = \aleph_1$, by considering Shelah's construction of ultrafilters over $\lambda$ with $lcf(\omega, \mathcal{D})= \kappa$ for $\aleph_0 < \kappa \; (\leq 2^{\lambda})$. So for instance, we can let $\kappa = \aleph_1$, and then the construction gives an increasing sequence of $\aleph_1$ many filters on $\lambda$ whose union is an ultrafilter on $\lambda$; the cofinality of the construction is $\aleph_1$. See Classification Theory Ch. VI, namely Theorem 3.12 (page 357/366) and especially Lemma 3.18 (page 360).

We take a large family of independent functions with range $\omega$ and arrange these into $\aleph_1$ many rows. Then we can find an increasing sequence of filters that are maximal with respect to the remaining rows being independent. This give you an ultrafilter ornament as each step fails to be an ultrafilter due to the remaining independent functions, and the union is an ultrafilter. ($\mathcal{G}^*$ is empty in the parlance of Lemma 3.18).

Note this cannot give us a countable sequence of filters as we truly need that the cofinality of the number of rows is uncountable for Lemma 3.18. Thus we do get $\mathfrak{ufo} = \aleph_1$.

A (Slightly) More Detailed Proof

Definition: For a filter $\mathcal{F}$, a set of functions, $\mathcal{G} = \{ f_{\alpha}: \lambda \rightarrow \omega \mid \alpha < \kappa \}$ is independent mod $\mathcal{F}$ if for any finite set of distinct $\{f_1, ... , f_n\} \subset \mathcal{G}$, any finite set $i_1, ... , i_n \in \omega$, and any $F \in \mathcal{F}$: $$\{ j \in \omega \mid f_k(j) = i_k \textrm{for all k} \} \cap F \neq \emptyset. $$

Then it is an old result that there is a family in $\omega^\omega$, of cardinality $2^{\aleph_0}$, that is independent mod $\{ \omega \}$. (For instance, see Kunen Theorem 3.4).

Really, the linked proof must be modified to account for only having a range of $\omega$ for general $\lambda$, but this is easily accomplished by letting the $r$ in the referenced Engelking-Karłowicz proof have range $\omega$. However, it is simple enough to show that there is a family of at least size $\aleph_1$ by a diagonal argument on the finite sets of functions/indices (with respect to the Frechet (cofinite) filter so that we don't end up generating a principle filter). Then we can (with Zorn's) extend this to a maximal family of independent functions and work with that.

Then we can arrange this family into $\aleph_1$ many disjoint pieces; for $\beta < \omega_1$ let $\mathcal{G}_{\beta}$ be the first $\beta$ many pieces. The first claim is then that for each $\beta < \omega_1$, we can build a filter $\mathcal{D}_{\beta}$ using $\mathcal{G}_{\beta}$ that is as large as possible with the constraint that $\mathcal{G} - \mathcal{G}_{\beta}$ is independent mod $\mathcal{D}_{\beta}$. This is done by adding sets of the form $A_n = \{ j \mid f_{\alpha}(j) \geq n \}$. Note our independence property for the functions gives us that each of these can be added without causing an empty intersection somewhere. Then we maximize our filter while keeping the remaining functions still independent (Claim 3.15(3) page 358).

Then finally the second claim is that the union of all of these is maximal with respect to the empty family of functions, i.e. it is an ultrafilter. This is the key Lemma 3.18 with $\mathcal{G}^* = \emptyset$. Part i) just states that we are generating each filter from the family of independent functions in that row; and ii) and iii) say that we are maximizing each of our $\aleph_1$ many steps. Then the $\mathcal{D}^*$ is our desired ultrafilter.

To show that the union is an ultrafilter, it suffices to show that any function from $\lambda$ to $\omega$ is not independent mod the union. This is because if we only had a filter we could build a function that is independent. So consider any such function $f$. If it was in our original $\mathcal{G}$, then it will not be independent mod the union as it is already taken care of at some step. If it was not in $\mathcal{G}$, then it must be dependent, due to the maximality of our family $\mathcal{G}$. So some $i \in \omega$ has $f^{-1}(i)$ with empty intersection with finitely many terms from $\mathcal{G}$. Then removing that index from our function gives us a new function where we can do the same procedure. We do this countably many times (less than cofinality of $\aleph_1$) to get some point where the entire function has already been taken care of in our construction. This is the contrapositive of the statement that a collection that is not an ultrafilter has some countable partition of the underlying set that has some element intersecting the complement of every set in the filter.

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  • $\begingroup$ I added a bit more exposition on the idea. I don't have time at the moment to really expound on the core of the proof, that this union is indeed an ultrafilter. Though intuitively, we "use up" all of our independent functions as we construct along each row. To test that this filter is an ultrafilter, we must show that any other function is not independent. So since the range of any other function is less than the cofinality of our construction, there is some point that we already took care of it. $\endgroup$ – David Casey Apr 5 '16 at 21:56
  • $\begingroup$ Again, I added a bit more on how the proof works. $\endgroup$ – David Casey Apr 6 '16 at 8:34
  • $\begingroup$ $\lambda$ is the cardinality of the set the our ultrafilter is over. So interestingly, we get that this value is $\aleph_1$ for any size set, including $\omega$. $\endgroup$ – David Casey Apr 7 '16 at 10:38
  • $\begingroup$ I see. I edited your answer accordingly, to reduce confusion. $\endgroup$ – Boaz Tsaban Apr 7 '16 at 19:34
  • $\begingroup$ I would appreciate more details on the "Really, ..." section. Maybe there are subtleties there, or just the full proof may be so long as to merit a journal publication for you. $\endgroup$ – Boaz Tsaban Apr 7 '16 at 19:37

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