# Limits of rearranged sequences along ultrafilters

Suppose that a bounded sequence of real numbers $s_i$ ($i\in\omega$) has a limit $\alpha$ along some ultrafilter $\mu_1\in \beta{\Bbb N}\setminus{\Bbb N}$. Then given another ultrafilter $\mu_2\in \beta{\Bbb N}\setminus{\Bbb N}$, surely there exists some rearrangement $s_{r(i)}$ of $s_i$ that has the same limit $\alpha$.

One can easily extend this simple observation to a countable family of sequences.

Now given $s_{i;j}$ ( $i,j\in \omega$; values bounded for each fixed $j$) with limits $\alpha_j$ along a fixed $\mu_1\in \beta{\Bbb N}\setminus{\Bbb N}$, and given another ultrafilter $\mu_2\in \beta{\Bbb N}\setminus{\Bbb N}$, there exists a simultaneous rearrangement $s_{r(i);j}$ having the same limits $\alpha_j$ along $\mu_2$.

All this fails if we pass to size $c=2^\omega$ families of sequences. Indeed $s_{i;j}$ could then enumerate all bounded sequences. But all the limits $\alpha_j$ together would determine $\mu_1$. Taking limits of a simultaneous rearrangement of all the sequences amounts, equivalently, to taking limits of the original sequences along an ultrafilter $\mu_2'$ in the orbit of $\mu_2$ under the action of the symmetric group of $\Bbb N$ extended to $\beta\Bbb N$. Equality of all those limits thus forces $\mu_1=\mu_2'$, and that places $\mu_1$ and $\mu_2$ in the same orbit of the symmetric group action, a severe restriction on $\mu_2$.

Question: If CH fails, what happens for a size $\omega_1$ family of sequences?

• I don't mean to be too nitpicky, but your first statement is not true if one of the $\mu_i$'s is principal. – Paul McKenney Apr 6 '16 at 13:03
• @Paul McKenney. You're correct, of course. Fixed. – David Feldman Apr 6 '16 at 14:22
• It seems likely to be a cardinal characteristic. – Joel David Hamkins Apr 7 '16 at 0:07

If $\mathfrak{p} > \kappa$, then you can get such a $\pi$ for every family of $\kappa$-many bounded sequences.

To see this, suppose $\mathfrak{p} > \kappa$, $\mu_1$ and $\mu_2$ are nonprincipal ultrafilters, and $s^\xi\in\mathbb{R}^\mathbb{N}$ ($\xi < \kappa$) is a family of bounded sequences. Let $\lambda^\xi$ be the $\mu_1$-limit of $s^\xi$ and let $$X_n^\xi = \left\{m\in\mathbb{N}\; :\; |s_m^\xi - \lambda^\xi| < \frac{1}{n}\right\}$$ Then each $X_n^\xi$ is a member of $\mu_1$, and hence every intersection of finitely-many of them is infinite. Since $\mathfrak{p} > \kappa$, there is then an infinite set $P$ which is contained (mod-finite) in all of them. Let $\pi$ be any permutation such that $\pi''(P)\in \mu_2$. Then if $t^\xi$ is the rearrangement of $s^\xi$ by $\pi$, we have $t^\xi_m = s^\xi_{\pi^{-1}(m)}$, so

$$Y^\xi_n = \left\{ m\in\mathbb{N}\; :\; |t_m^\xi - \lambda^\xi| < \frac{1}{n}\right\} = \pi''(X^\xi_n) \supseteq^* \pi''(P)$$

and hence all of these $Y^\xi_n$'s are in $\mu_2$, which means that the $\mu_2$-limit of each $t^\xi$ is still $\lambda^\xi$.

It would be nice to improve this to a larger cardinal characteristic than $\mathfrak{p}$.

EDIT: I think $\mathfrak{u}$ (the minimal size of a base for a nonprincipal ultrafilter) is an upper bound.

Let $\mu_1$ be a nonprincipal ultrafilter with a base $\mathcal{B}$ of minimal size. Consider the family of sequences $\chi_A$ ($A\in\mathcal{B}$), where $\chi_A$ is the characteristic function of $A$. The limit of each such $\chi_A$ along $\mu_1$ is $1$. If $\pi$ is a permutation, then the sequence we get by rearranging $\chi_A$ using $\pi$ is exactly $\chi_{\pi^{-1}(A)}$. So, given $\mu_2$, we'd like to find a $\pi$ such that for each $A\in\mathcal{B}$, $\pi^{-1}(A)\in \mu_2$. Clearly this implies that $\pi^{-1}(A)\in \mu_2$ for all $A\in \mu_1$. But this implies that $\mu_2$ is exactly

$$\{\pi^{-1}(A) \; : \; A\in\mu_1\}$$

(If there were some $B\in\mu_2$ not in the above set, then consider $\pi''(B)$. This is not in $\mu_1$ by assumption, hence $\pi''(\mathbb{N}\setminus B) \in \mu_1$, hence $\mathbb{N}\setminus B\in \mu_2$, a contradiction.)

So in this case we can only find such a $\pi$ when $\mu_2$ is in the orbit of $\mu_1$ under $S_\mathbb{N}$, and of course this orbit is not all of $\beta\mathbb{N}\setminus \mathbb{N}$. (There are only continuum-many points in the orbit of $\mu_1$ but $2^{\mathfrak{c}}$-many ultrafilters total.)

As Joel David Hamkins commented, this is likely to be a cardinal characteristic. Let me define (I hope temporarily) the relevant characteristic $\mathfrak{dfpmk}$ (named after the OP and the author of the accepted answer) to be the smallest cardinal $\kappa$ such that, for some non-principal ultrafilters $\mathcal U$ and $\mathcal V$ on $\omega$ and some family of $\kappa$ sequences $(s_n^\xi)_{n\in\omega}$ (indexed by $\xi\in\kappa$), the sequences all converge to limits $\lambda^\xi$ with respect to $\mathcal U$ but no simultaneous rearrangements $(s_{r(n)}^\xi)_{n\in\omega}$ all converge to the corresponding $\lambda^\xi$ with respect to $\mathcal V$. So what Paul McKenney proved is that $\mathfrak p\leq\mathfrak{dfpmk}\leq\mathfrak u$.

Of course, what Joel probably intended is that $\mathfrak{dfpmk}$ might be a known cardinal characteristic. I can't prove anything like that, but I can improve in some models the upper bound $\mathfrak u$ in Paul's answer. Namely, if there exists a P-point, then $\mathfrak{dfpmk}\leq\mathfrak d$. (This is an improvement only if there is a P-point and $\mathfrak d<\mathfrak u$; there are models where that happens, for example, the random real model.)

Before giving the proof, let me indicate an alternative way to view $\mathfrak{dfpmk}$, which will be useful (at least for me) in the proof. To avoid excess verbiage, assume that all filters in the following extend the cofinite filter on $\omega$, so in particular all ultrafilters are non-principal.) I claim that $\mathfrak{dfpmk}$ is the smallest number of generators needed for a filter $\mathcal F$ on $\omega$ such that, for some ultrafilter $\mathcal U$, no isomorphic copy of $\mathcal U$ extends $\mathcal F$.

To prove the equivalence, assume first that $\kappa<\mathfrak{dfpmk}$, let $\mathcal F$ be any filter with a basis $\mathcal B$ of size $\kappa$, and let $\mathcal U$ be any ultrafilter. Consider the characteristic functions of the sets in $\mathcal B$ and consider any ultrafilter $\mathcal V$ extending $\mathcal F$. These characteristic functions all converge to 1 with respect to $\mathcal V$, and there are only $\kappa<\mathfrak{dfpmk}$ of them, so some simultaneous rearrangements of them, say by a permutation $r$, all converge to 1 with respect to $\mathcal U$. But that means that the original (not rearranged) characteristic functions converge to 1 with respect to $r^{-1}(\mathcal U)$. That means $r^{-1}(\mathcal U)$ contains all the sets in $\mathcal B$ and therefore extends $\mathcal F$.

For the converse, suppose $\kappa$ is smaller than my proposed alternative view of $\mathfrak{dfpmk}$, and let $\kappa$ sequences $(s_n^\xi)$ converge to limits $\lambda^\xi$ with respect to an ultrafilter $\mathcal U$.This convergence means that $\mathcal U$ contains all the sets $A_{\xi,k}=\{n\in\omega: |s_n^\xi-\lambda^\xi|<2^{-k}\}$ for $\xi<\kappa$ and $k\in\omega$. These $\kappa$ sets $A_{\xi,k}$ generate a filter $\mathcal F$, and, by choice of $\kappa$, every ultrafilter $\mathcal V$ has an isomorphic copy $r(\mathcal V)$ extending $\mathcal F$. Then the sequences $(s_n^\xi)$ converge to $\lambda^\xi$ with respect to $r(\mathcal V)$, and therefore the simultaneous rearrangements $(s_{r^{-1}(n)}^\xi)$ converge to $\lambda^\xi$ with respect to $\mathcal V$.

This completes the proof of the equivalence between the two views of $\mathfrak{dfpmk}$. Now to prove my claim about $\mathfrak d$, consider the filter $\mathcal F$ on the "plane" $\omega\times\omega$ consisting of those sets $X$ such that, for all but finitely many $x\in\omega$, the "column" at abscissa $x$ has all but finitely many of its points in $X$. That is, for all but finitely many $x$, for all but finitely many $y$, $(x,y)\in X$. (This filter is often called the tensor square or the Fubini square of the cofinite filter on $\omega$.) This filter is generated by $\mathfrak d$ sets, namely the sets $\{(x,y):x>n\}$ for each $n\in\omega$ and the sets $\{(x,y):y>f(x)\}$ for each $f$ in some dominating family of cardinality $\mathfrak d$.

If an ultrafilter $\mathcal U$ extends $\mathcal F$, then $\mathcal U$ cannot be a P-point, because any set on which the projection to the first factor, $(x,y)\mapsto x$, is finite-to-one or constant is the complement of a set if $\mathcal F$ and therefore cannot be in $\mathcal U$. Furthermore, since the property of being a P-point is preserved by isomorphism, no P-point can be isomorphic to an extension of $\mathcal F$. So, as long as there is a P-point, the filter $\mathcal F$ witnesses, in the alternative view of $\mathfrak{dfpmk}$ above, that $\mathfrak{dfpmk}\leq\mathfrak d$.

A very similar argument shows, under the weaker hypothesis that there exists a nowhere dense ultrafilter, that $\mathfrak{dfpmk}\leq\mathfrak{cof}(B)$. (Here "nowhere dense" is in the sense of Baumgartner's $I$-ultrafilters; it means that the image of $\mathcal U$ under any map $\omega\to\mathbb Q$ contains a nowhere dense set. $\mathcal{cof}(B)$ is the cofinality number for Baire category, the minimum number of sets needed to generate the ideal of meager subsets of $\mathbb R$.)