# Problems on $\beta\N$

Klaas Pieter Hart  Faculty EEMCS
TU Delft
Postbus 5031
2600 GA Delft
the Netherlands<EMAIL_ADDRESS>http://fa.ewi.tudelft.nl/~hart and  Jan
van Mill  KdV Institute for Mathematics
University of Amsterdam
P.O. Box 94248
1090 GE Amsterdam
The Netherlands<EMAIL_ADDRESS>https://staff.fnwi.uva.nl/j.vanmill

###### Abstract.

This is an update on, and expansion of, our paper Open problems on
$\beta\omega$ in the book Open Problems in Topology.

We invite comments, corrections, more problems, …

## Introduction

In 1990 we contributed a paper to the book Open problems in Topology,
[MR1078636], titled Open problems on $\beta\omega$.

Through the years some of these problems were solved, some were shown to be
related to other problems, and some are still unsolved. In the first years
after the publication of the book there were regular updates on the problems
in the journal Topology and its Applications; in 2004 these were collated and
extended in a comprehensive status report, [MR2023411], by Elliott Pearl.

The COVID-19 pandemic provided a good opportunity to go through our original
paper again and provide a new update of the status of the problems as well as
to collect and formulate new questions on the fascinating object that is
$\beta\N$.

Many of the comments below incorporate information from Elliott Pearl’s update
mentioned above, but there have, of course, been many developments in the
years since.

The numbering of the problems is different from that in the first paper
because we have moved some questions around and combined related questions
into more comprehensive problems. We have made no attempt to separate the
solved problems from the unsolved ones. We wanted to keep related problems
together and even though we consider a problem solved the reader may disagree
and be inspired to investigate variations or strengthenings of the answers.

We should also mention the book Open problems in Topology II [MR2367385],
edited by Elliot Pearl, that contains many more problems in topology, and in
particular a paper by Peter Nyikos, Čech-Stone remainders of discrete spaces,
that, as the title indicates, deals with problems on $\beta\kappa$ for
arbitrary infinite cardinals $\kappa$.

##  1\. Preliminaries

The main objects of study in this paper are the space $\beta\N$ and its
subspace $\N^{*}$.

For a quick overview of their properties we refer to Chapter D-18 of
[MR2049453]; the paper [MR776630] offers a more comprehensive treatment. Here
we collect some of the basic facts about our spaces in order to fix some
notation that will be used throughout the paper.

To begin: $\beta\N$ is the set of ultrafilters on the set $\N$ of natural
numbers, endowed with the topology generated by the base
$\\{\overline{A}:A\subseteq\N\\}$, where $\overline{A}$ denotes the set of
ultrafilters that contain $A$. The readily established equality
$\beta\N\setminus\overline{A}=\overline{\N\setminus A}$ confirms what the
notation $\overline{A}$ suggests: the set $\overline{A}$ is open _and_ closed,
and also equal to the closure of $A$ in $\beta\N$.

We identify an element $n$ of $\N$ with the ultrafilter $\\{A:n\in A\\}$ and
thus consider $\N$ to be a subset of $\beta\N$. The complement
$\beta\N\setminus\N$ is the set of _free_ ultrafilters on $\N$ and is denoted
$\N^{*}$; we extend this notation to all subsets of $\N$ and write
$A^{*}=\overline{A}\setminus A$ whenever $A\subseteq\N$.

A map $\varphi$ from $\N$ to itself induces a map $\beta\varphi$ from
$\beta\N$ to itself: $\beta\varphi(u)$ is the ultrafilter generated by
$\\{\varphi[A]:A\in u\\}$.

##  2\. Autohomeomorphisms

The autohomeomorphisms of $\beta\N$ correspond to the permutations of $\N$ and
are, as such, not very interesting topologically. The autohomeomorphisms of
$\N^{*}$ offer more challenges.

In what follows $\mathsf{Aut}$ denotes the autohomeomorphism group of
$\N^{*}$, and $\mathsf{Triv}$ denotes the subgroup of trivial
autohomeomorphisms. Here a _trivial_ autohomeomorphism is one with an ‘easy’
description: an autohomeomorphism $h$ of $\N^{*}$ is trivial if there are co-
finite subsets $A$ and $B$ of $\N$ and a bijection $\varphi:A\to B$ such that
$h=\varphi^{*}$, where $\varphi^{*}$ denotes the restriction of $\beta\varphi$
to $\N^{*}$.

1.  Can $\mathsf{Triv}$ be a proper normal subgroup of $\mathsf{Aut}$, and if _yes_ what is (or can be) the structure of the factor group $\mathsf{Aut}/\mathsf{Triv}$; and if _no_ what is (or can be) $[\mathsf{Triv}:\mathsf{Aut}]$?

A very concrete first step would be to investigate what can one say about an
autohomeomorphism $h$ that satisfies $h^{-1}\circ\mathsf{Triv}\circ
h=\mathsf{Triv}$.

A related question: what is the minimum number of autohomeomorphisms necessary
to add to $\mathsf{Triv}$ to get a generating set for $\mathsf{Aut}$?

Of course that number is $0$ when $\mathsf{Aut}=\mathsf{Triv}$, but can it be
non-zero and finite?

If $h\in\mathsf{Aut}$ then $I(h)$ denotes the family of subsets of $\omega$ on
which $h$ is trivial, that is, $A\in I(h)$ iff there is a function
$h^{\prime}:A\to\omega$ such that $h(B^{*})=h^{\prime}[B]^{*}$ whenever
$B\subseteq A$.

If $I(h)$ contains an infinite set then $h$ is _somewhere trivial_ , otherwise
$h$ is totally non-trivial.

The ideal $I(h)$ determines an open set $O_{h}$: the union
$\bigcup\\{A^{*}:A\in I(h)\\}$; its complement $F_{h}$ is closed and could be
said to be the set of points of $\N^{*}$ where $h$ is truly non-trivial.

2. Does the existence of a (totally) non-trivial automorphism imply that $\mathsf{Aut}$ is simple?

This question asks more than the opposite of question 2; a yes answer here
would imply a no answer there, but a negative answer there could go together
with a negative answer here.

3. Is it consistent with $\mathsf{MA}+\neg\mathsf{CH}$ that a totally non-trivial automorphism exists?

The answer yes. This was established by Shelah and Steprāns in [MR1896046].
Consistency is the best one can hope for: in [MR935111] Shelah and Steprāns
proved that $\mathsf{PFA}$ implies all autohomeomorphisms of $\N^{*}$ are
trivial; they also indicated how the implicit large cardinal assumption can be
avoided and use $\diamond$ on $\omega_{2}$ to capture and eliminate any
potential non-trivial autohomeomorphisms in a countable support iteration of
length $\omega_{2}$. Though not stated explicitly by the authors it is clear
that one can modify the iteration so as to obtain a model that satisfies
$\mathsf{MA}_{\aleph_{1}}$ as well. In [MR1202874] Veličković showed that the
conjunction of $\mathsf{MA}_{\aleph_{1}}$ and $\mathsf{OCA}$ implies that all
autohomeomorphisms are trivial.

4. Is it consistent to have a non-trivial automorphism, while for every $h\in\mathsf{Aut}$ the ideal $I(h)$ is the intersection of finitely many prime ideals?

In topological terms: can one have non-trivial autohomeomorphisms but only
very mild ones; the set of points where an autohomeomorphism is truly non-
trivial is always finite.

5. Is every ideal $I(h)$ a $P$-ideal?

This was asked explicitly in [MR935111]*Question 2 in case every
autohomeomorphism is somewhere trivial, after it was shown that $\mathsf{PFA}$
implies a yes answer. However, as mentioned above, $\mathsf{PFA}$ implies that
all autohomeomorphisms are trivial, so that $I(h)$ is, in fact, always an
improper ideal.

Of course this question only makes sense in case $I(h)$ is not equal to the
ideal of finite sets. Also, if _every_ autohomeomorphism is somewhere trivial
then every $I(h)$ is a tall ideal and hence the set of points of non-
triviality is nowhere dense.

Without the additional condition that every autohomeomorphism is somewhere
trivial the answer is consistently negative. The Continuum Hypothesis lets one
construct an autohomeomorphism $h$ that is trivial, in fact the identity, on
the members of a partition $\mathcal{A}$ of $\omega$ into infinite sets, and
so that there is a point $u$ on the boundary of
$\bigcup\\{A^{*}:A\in\mathcal{A}\\}$ such that $h$ is not trivial on each
neighbourhood of $u$. This implies there is no $B\in I(h)$ such that
$A\subseteq B$ for all $A\in\mathcal{A}$.

6. If every automorphism is somewhere trivial, is then every automorphism trivial?

This is undecidable.

Shelah proved the consistency of “all autohomeomorphisms are trivial”, see
[MR1623206]. Shelah and Steprāns proved the consistency with
$\mathsf{MA}_{\aleph_{1}}$ of “every autohomeomorphism is somewhere trivial,
yet there is a non-trivial autohomeomorphism” in [MR1271551]; as noted above
they proved in [MR1896046] that $\mathsf{MA}$ does not imply that all
autohomeomorphisms are somewhere trivial.

Given a cardinal $\kappa$ call an autohomeomorphism $h$ _weakly
$\kappa$-trivial_ if the set $\\{p:p\not\mathrel{\equiv_{\mathbf{RK}}}h(p)\\}$
has cardinality less than $\kappa$. Here $p\mathrel{\equiv_{\mathbf{RK}}}q$
means that $p$ and $q$ have the same type, i.e., $q=\pi^{*}(p)$ for some
permutation $\pi$ of $\N$.

7. For what cardinals $\kappa$ is it consistent to have that all autohomeomorphisms are weakly $\kappa$-trivial?

Since a trivial autohomeomorphism is weakly $1$-trivial we see that $\kappa=1$
is a possibility. And of course the candidates are less than or equal to
$2^{\mathfrak{c}}$.

8. If $h$ is weakly $1$-trivial is $h$ then trivial?

This is a uniformization question: if for every $p\in\N^{*}$ there is a
permutation $\pi_{p}$ such that $h(p)=\pi_{p}^{*}(p)$ is there then one
(almost) permutation $\pi$ of $\N^{*}$ such that $h(p)=\pi^{*}(p)$ for all
$p$?

9. ($\mathsf{MA}+\lnot\mathsf{CH}$) if $p$ and $q$ are $P_{\mathfrak{c}}$-points is there an $h$ in $\mathsf{Aut}$ such that $h(p)=q$?

This is undecidable.

Shelah and Steprāns proved the consistency of $\mathsf{MA}+\lnot\mathsf{CH}$
with “all autohomeomorphisms are trivial” in [MR935111]; in this model there
are $\mathfrak{c}$ many autohomeomorphisms and $2^{\mathfrak{c}}$ many
$P_{\mathfrak{c}}$-points.

Steprāns proved the consistency of a positive answer in [MR1239060].

In the investigations into the previous question the following equivalence
relation was used: $p\equiv q$ means that there are two partitions
$\\{A_{n}:n\in\omega\\}$ and $\\{B_{n}:n\in\omega\\}$ of $\N$ into finite sets
such that

$(\forall P\in p)(\exists Q\in q)(\forall n)\bigl{(}\mathopen{|}P\cap
A_{n}\mathclose{|}=\mathopen{|}Q\cap B_{n}\mathclose{|}\bigr{)}$

The following question was left open.

10. Is $\equiv$ different from $\mathrel{\equiv_{\mathbf{RK}}}$ in $\mathsf{ZFC}$?

11. Are the autohomeomorphisms of $\N^{*}$ induced by the shift map $\sigma:n\mapsto n+1$ and by its inverse conjugate?

See [MR1035463, MR4138425]. Under $\mathsf{CH}$ the autohomeomorphism group of
$\N^{*}$ is simple, yet it has the maximum possible number of conjugacy
classes: $2^{\mathfrak{c}}$. This suggests questions about the number and
nature of conjugacy classes of this group, in $\mathsf{ZFC}$ or under various
familiar extra set-theoretical assumptions, see also [HartvanMill2022].

12. Does $\N^{*}$ have a universal autohomeomorphism?

This is a question with many possible variations. The definition of
universality that we adopt here is as follows: $f:\N^{*}\to\N^{*}$ is
universal if for every closed subspace $F$ of $\N^{*}$ and every
autohomeomorphism $g$ of $F$ there is an embedding $e:F\to\N^{*}$ such that
$g=e^{-1}\circ(f\restr e[F])\circ e$.

One can ask whether there is a universal autohomeomorphism at all, whether the
shift $\sigma$ is universal (for autohomeomorphisms without fixed points),
whether there is a universal autohomeomorphism just for autohomeomorphisms
without fixed points.

The authors have shown that there is a universal autohomeomorphism of $\N^{*}$
under $\mathsf{CH}$ and that there is no trivial universal autohomeomorphism.
See [MR4398473].

We also note that $\N$ has a universal permutation: take a permutation of $\N$
that has infinitely many $n$-cycles, for every $n$, and infinitely many
infinite cycles (copies of $\Z$ with the shift). Every other permutation of
$\N$ can be embedded into this one.

##  3\. Subspaces

13. For what $p$ are $\N^{*}\setminus\\{p\\}$ and $\beta\N\setminus\\{p\\}$ non-normal?

Originally this question had the word ‘equivalently’ after the ‘and’ (in
parentheses). Since $\N^{*}\setminus\\{p\\}$ is closed in
$\beta\N\setminus\\{p\\}$ there is an implication between the non-normality of
these spaces but we do not know whether that implication is reversible. Thus
this question may actually be two separate ones.

Under $\mathsf{CH}$ the answer is, in both cases, “for every point”, see
[MR234422, MR321012, MR292035, MR861501]. There are some results for some
special types of points, see, e.g., Błaszczyk and Szymański
[BlaszczykSzymanski1980], Gryzlov [MR760274], and Logunev [MR4405820], but a
general answer is wanting.

14. Is it consistent that there is a non-butterfly point in $\N^{*}$?

We call $p$ a butterfly point if there are disjoint sets $A$ and $B$ such that
$p$ is the only common accumulation point of $A$ and $B$, that is: $A^{d}\cap
B^{d}=\\{p\\}$.

The points used by Błaszczyk and Szymański [BlaszczykSzymanski1980] in their
(partial) answer to the previous question are easy-to-describe butterfly
points. Let $X=\\{x_{n}:n\in\omega\\}$ be a discrete subset of $\N^{*}$, let
$A=\operatorname{cl}X\setminus X$ and take $p\in A$. Let
$\\{B_{n}:n\in\omega\\}$ be a partition of $\N$ such that $B_{n}\in x_{n}$ for
all $n$ and let $q$ be the ultrafilter
$\\{Q:p\in\operatorname{cl}\\{x_{n}:n\in Q\\}\\}$. The set $B=\bigcap_{Q\in
q}\bigl{(}\bigcup\\{B_{n}:n\in Q\\}\bigr{)}^{*}$ is closed and
$\\{q\\}=A^{d}\cap B^{d}$. Thus butterfly points exist.

By contrast, in [MR1058805] Bešlagić and Van Douwen showed that it is
consistent with all consistent cardinal arithmetic that all points of $\N^{*}$
are butterfly points.

15. Is it consistent that $\N^{*}\setminus\\{p\\}$ is $C^{*}$-embedded in $\N^{*}$ for some but not all $p\in\N^{*}$?

The answer is yes: in [MR1434375] Alan Dow showed that in the Miller model
$\N^{*}\setminus\\{p\\}$ is $C^{*}$-embedded iff $p$ is not a $P$-point. There
are $P$-points in the Miller model: every ground-model $P$-point generates a
$P$-point in the extension.

16. What spaces can be embedded in $\beta\omega$?

This is a very general question and a definitive answer looks out of reach for
now, even for closed subspaces.

The Continuum Hypothesis implies that the closed subspaces of $\beta\omega$
are exactly the compact zero-dimensional $F$-spaces; in fact, these are also
exactly the closed $P$-sets in $\N^{*}$. The implication does not reverse: in
[MR1152978] it is shown that every compact zero-dimensional $F$-space is a
(closed) subspace of $\N^{*}$ in any model obtained by adding $\aleph_{2}$
many Cohen reals to a model of $\mathsf{CH}$.

Dow and Vermeer proved in [MR1137221] that it is consistent that the
$\sigma$-algebra of Borel sets of the unit interval is not the quotient of any
complete Boolean algebra. By Stone duality, this yields a compact basically
disconnected space, hence a compact zero-dimensional $F$-space, of weight
$\mathfrak{c}$ that cannot be embedded into any extremally disconnected space,
in particular not into $\beta\N$.

Some $\mathsf{ZFC}$ results are available. For instance: if $X$ is a compact
space of countable cellularity that is a continuous image of $\N^{*}$ then its
projective cover $E(X)$ can be embedded in $\N^{*}$ as a $\mathfrak{c}$-OK set
(a weakening of the notion of a $P$-set). This was proved by van Mill in
[MR637426] and applies to all separable compact extremally disconnected spaces
as well as to the projective covers of Suslin lines and of Bell’s ccc non-
separable remainder [MR624458].

Van Douwen proved in unpublished work that every $P$-space of weight
$\mathfrak{c}$ (or less) can be embedded into $\beta\N$. In fact he proved
that for every infinite cardinal $\kappa$ every $P$-space of weight
$2^{\kappa}$ can be embedded in $\beta\kappa$. The argument was sketched and
extended in [MR674103] and we summarize it here for the reader’s convenience.

Let $X$ be a $P$-space of weight $2^{\kappa}$ and embed it into the Cantor
cube $C=2^{2^{\kappa}}$ of weight $2^{\kappa}$. Next consider the projective
cover $\pi:E(C)\to C$ of this cube. The Cantor cube is a group under
coordinatewise addition modulo $2$, so for every $p\in C$ the map
$\lambda_{p}:x\mapsto x+p$ is a homeomorphism; this homeomorphism lifts to a
homeomorphism $\Lambda_{p}:E(C)\to E(C)$ with the property that
$\pi\circ\Lambda_{p}=\lambda_{p}\circ\pi$. Now take one point $u_{0}\in E(C)$
that maps to the neutral element $0$ of $C$ and consider the subspace
$X^{\prime}=\\{\Lambda_{p}(u_{0}):p\in X\\}$ of $E(C)$. Using the fact that
regular open sets in $C$ are, up to permutation of the coordinates, of the
form $U\times 2^{I}$ where $U$ is regular open in the Cantor set $2^{\omega}$
and $I=2^{\kappa}\setminus\omega$, one shows that $\pi$ is actually a
homeomorphism from $X^{\prime}$ to $X$. Finally then, as $\pi$ is irreducible
and $C$ has density $\kappa$, the density of $E(C)$ is equal to $\kappa$ as
well. Therefore there is a continuous surjection $f:\beta\kappa\to E(C)$ and
one can take a closed subset $F$ of $\beta\kappa$ such that $f\restr F$ is
irreducible and onto. As $E(C)$ is extremally disconnected this restriction is
a homeomorphism and we find our copy of $X$ in $F$.

The extension in [MR674103] delivers more but at a cost: one embeds $\beta X$
in a suitable Cantor cube, possibly of a larger weight than that of $X$
itself. What this delivers is that the copy of $X$ in $\beta\lambda$ (where
$\lambda$ may be larger than the $\kappa$ above) is $C^{*}$-embedded.

Thus we get the general statement that every $P$-space can be $C^{*}$-embedded
in a compact extremally disconnected space.

This argument also shows that $2^{\aleph_{0}}=2^{\aleph_{1}}$ implies that
$\beta\omega_{1}$ embeds into $\beta\N$. For $\beta\omega_{1}$ embeds into the
Cantor cube $2^{2^{\omega_{1}}}$, which under our assumption is the same as
$2^{\mathfrak{c}}$. The latter is a continuous image of $\beta\N$ and an
irreducible preimage of $\beta\omega_{1}$ will be homeomorphic to
$\beta\omega_{1}$. If $2^{\aleph_{0}}<2^{\aleph_{1}}$ then $\beta\omega_{1}$
can not be embedded into $\N^{*}$ because its weight, which is
$2^{\aleph_{1}}$, is larger than that of $\N^{*}$.

17. Describe the closed $P$-sets of $\N^{*}$.

This has a quite definitive answer under $\mathsf{CH}$: every compact zero-
dimensional $F$-space of weight $\mathfrak{c}$ can be embedded in $\N^{*}$ as
a $P$-set. What we are looking for are properties that can be established in
$\mathsf{ZFC}$, or provably can not. For example: one cannot prove in
$\mathsf{ZFC}$ that there is a $P$-set homeomorphic to $\N^{*}$ itself, see
[MR976360], or that there is a $P$-set that satisfies the ccc, see
[MR1253914].

One can ask if cellularity less than $\mathfrak{c}$ is at all possible.

There are various nowhere dense closed $P$-sets that one can write down
explicitly. To give two familiar examples, among many, we consider the density
ideal $\mathcal{I}_{d}$ and the summable ideal $\mathcal{I}_{\Sigma}$. The
first is defined by

$I\in\mathcal{I}_{d}\text{ \quad iff \quad}\lim_{n\to\infty}\frac{1}{n}|A\cap
n|=0$

and the second as

$I\in\mathcal{I}_{\Sigma}\text{ \quad iff \quad}\sum_{n\in A}\frac{1}{n}\text{
converges.}$

These ideals have been studied widely but we would like to know: what are the
topological properties of the nowhere dense closed $P$-sets

$F_{d}=\N^{*}\setminus\bigcup\\{A^{*}:A\in\mathcal{I}_{d}\\}\text{ \quad and
\quad}F_{\Sigma}=\N^{*}\setminus\bigcup\\{A^{*}:A\in\mathcal{I}_{\Sigma}\\}$

Rudin established in [MR0216451] that $F_{d}$ contains no $P$-points and even
that no countable set of $P$-points accumulates at a point of $F_{d}$. Indeed
let $u$ be a $P$-point and observe first that for every $n$ there is an
$i_{n}<n$ such that $U_{n}=\\{m\in\N:m\equiv i_{n}\pmod{n}\\}$ belongs to $u$.
Because $u$ is a $P$-point there is then a $U\in u$ such that
$U\subseteq^{*}U_{n}$ for all $n$. But this implies $U\in\mathcal{I}_{d}$, so
$u\notin F_{d}$. Because $F_{d}$ is a $P$-set this implies that no countable
set of $P$-points has accumulation points in $F_{d}$.

There are certain similarities between the two sets and $\N^{*}$ itself.
Consider the map $f:\N\to\N$, defined by $f(n)=k$ iff $k!<n\leq(k+1)!$. It is
an elementary exercise to show that

$\limsup_{n\to\infty}\frac{1}{n}|f^{\leftarrow}[X]\cap n|=1\text{ \quad en
\quad}\sum_{n\in f^{\leftarrow}[X]}\frac{1}{n}=\infty$

whenever $X$ is an infinite subset of $\N$. This implies that $\beta f$ maps
both $F_{d}$ and $F_{\Sigma}$ onto $\N^{*}$ and it allows for the lifting of
many combinatorial structures on $\N^{*}$ to these sets. It is clear that the
restriction of $\beta f$ to $\N^{*}$ is an open map onto $\N^{*}$ itself,
whether its restrictions to $F_{d}$ and $F_{\Sigma}$ are open as well is less
clear.

18. Which compact zero-dimensional $F$-spaces admit an open map onto $\N^{*}$?

This question is related to Van Douwen’s paper [MR1062775], where open maps
are used to transfer information from $\N^{*}$ to other remainders. As a
special case one can investigate whether the sets $F_{d}$ and $F_{\Sigma}$
from the Question 3 admit open maps onto $\N^{*}$ (if the map $\beta f$ given
there does not already give open maps).

19. Is there a nowhere dense copy of $\N^{*}$ in $\N^{*}$ that is a $\mathfrak{c}$-OK-set in $\N^{*}$?

Alan Dow showed in [MR3209343] that there a nowhere dense copy of $\N^{*}$
that is not of the form $\operatorname{cl}{D}\setminus D$ for some countable
and discrete subset $D$ of $\N^{*}$. This was later improved by Dow and van
Mill in [DowVanMill23] to a nowhere dense copy that is a weak $P$-set. In
light of the comments for question 3 the present question asks for the best
that we can get in $\mathsf{ZFC}$. Most likely the answer to this question
will require a new idea as the constructions in the papers cited above produce
sets that are definitely not $\mathfrak{c}$-OK in $\N^{*}$.

20. Is every subspace of $\N^{*}$ strongly zero-dimensional?

It is clear that every subspace is zero-dimensional and that closed subspaces
are even strongly zero-dimensional, but for general subspaces this question is
quite open. Until recently it was not even known whether there was an example
of a zero-dimensional $F$-space that is not strongly zero-dimensional, see
[MR4384168].

If the answer is negative then a secondary question suggests itself
immediately: is there an upper bound to the covering dimension of subspaces of
$\N^{*}$?

21. Is every nowhere dense subset of $\N^{*}$ a $\mathfrak{c}$-set?

In general a set $A$ is called a $\kappa$-set if there is a pairwise disjoint
family $\mathcal{O}$ of open sets of cardinality $\kappa$ and such that
$A\subseteq\bigcap\\{\operatorname{cl}O:O\in\mathcal{O}\\}$.

That the answer is positive is called by some “The $\mathfrak{c}$-set
conjecture”. In [MR1056181] Simon proved that this question is the same as “Is
there a maximal nowhere dense subset in $\N^{*}$?”. The questions are the same
in that the answer “no” to one is equivalent to the answer “yes” to the other:
Every nowhere dense set in $\N^{*}$ is a $\mathfrak{c}$-set if and only if
every nowhere dense set in $\N^{*}$ is a nowhere dense subset of another
nowhere dense set (this is the order that we are considering).

There is a purely combinatorial reformulation of this question, denoted
$\operatorname{RPC}(\omega)$ in [MR991597]: if $\mathcal{A}$ is an infinite
maximal almost disjoint family then $\mathcal{I}^{+}(\mathcal{A})$ has an
almost disjoint refinement. Here, $\mathcal{I}^{+}(\mathcal{A})$ is the family
of sets not in the ideal $\mathcal{I}(\mathcal{A})$ generated by $\mathcal{A}$
and the finite sets and an almost disjoint refinement is an almost disjoint
family $\mathcal{B}$ with a map $X\mapsto B_{X}$ from
$\mathcal{I}^{+}(\mathcal{A})$ to $\mathcal{B}$ such that
$B_{X}\subseteq^{*}X$ for all $X$.

Finally, we should mention that the answer is positive for one-point sets: all
points of $\N^{*}$ are $\mathfrak{c}$-points, see [MR567994].

22. Does there exist a completely separable maximal almost disjoint family?

This question is related to Question 3 because by [MR991597]*Theorem 4.19 a
positive answer to that question implies the existence of an abundance of
completely separable maximal almost disjoint families; where a maximal almost
disjoint family $\mathcal{A}$ is _completely separable_ if it is itself an
almost disjoint refinement of $\mathcal{I}^{+}(\mathcal{A})$.

Whether completely separable maximal almost disjoint families exist is a
problem first raised by Erdős and Shelah in [MR319770].

Currently the best result is due to Shelah who showed in [MR2894445] that the
answer is positive if $\mathfrak{c}<\aleph_{\omega}$ and that a negative
solution would imply consistency of the existence of large cardinals.

It is not (yet) clear whether this question and Question 3 are equivalent.
Thus far _constructions_ of completely separable maximal almost disjoint
families (in some model or another) could always be adapted to prove
$\operatorname{RPC}(\omega)$, but there is currently no proof of
$\operatorname{RPC}(\omega)$ from the _mere existence_ of such a family.

23. Describe the retracts of $\beta\N$ and $\N^{*}$, as well as their _absolute_ retracts.

A retract of $\beta\N$ is necessarily a closed separable extremally
disconnected subspace. It is known that a compact separable extremally
disconnected can be embedded as a retract of $\beta\N$. If $X$ is such a space
then there is a continuous surjection $f:\beta\N\to X$ and if $K$ is such that
$f\restr K$ is irreducible then $f\restr K$ is a homeomorphism to $X$ and
$(f\restr K)^{-1}\circ f$ is a retraction of $\beta\N$ onto $K$.

Shapiro [MR810825] and Simon [MR0869226] have shown independently and by quite
different means that not every closed separable subset of $\beta\N$ is a
retract. This gives rise to the notion of an absolute retract of $\beta\N$: a
(sub)space that is a retract irrespective of how it is embedded.

Bella, Błaszczyk and Szymański proved in [MR1295157] that if $X$ is compact,
extremally disconnected, without isolated points and of $\pi$-weight
$\aleph_{1}$ or less then $X$ is an absolute retract for extremally
disconnected spaces iff $X$ is the absolute of one of the following three
spaces: the Cantor set, the Cantor cube $\vphantom{2}{}^{\omega_{1}}2$, or the
sum of these two spaces. This shows that under $\mathsf{CH}$ there are very
few absolute retracts of $\beta\N$.

We have less information about the retracts of $\N^{*}$, absolute or not. Of
course if a subset of $\N^{*}$ is a retract of $\beta\N$ then it is a retract
of $\N^{*}$ as well. We do not know whether the converse is true, for
separable closed subsets of course.

We do know that non-trivial zero-sets are not retracts. Such a set is of the
form $Z=\N^{*}\setminus\bigcup_{n\in\omega}A_{n}^{*}$, where the $A_{n}$ are
infinite and pairwise disjoint subsets of $\N$. We write
$C=\bigcup_{n\in\omega}A_{n}^{*}$. Now the closure of $C$ is a $P$-set in
$\N^{*}$, it is the union of $C$ and the boundary of $Z$, and if we take one
point $u_{n}\in A_{n}^{*}$ for each $n$ then
$K=\operatorname{cl}\\{x_{n}:n\in\omega\\}$ is a copy of $\beta\N$ and
$K^{*}=K\setminus\\{x_{n}:n\in\omega\\}$ is a $P$-set in the boundary of $Z$
and hence in $Z$. If we now take assume $r:\N^{*}\to Z$ is a retraction then
$r\restr K^{*}$ is the identity and for all but finitely many $n$ we must have
$r(x_{n})\in K^{*}$. But this would imply that $K^{*}$ is separable, a
contradiction.

In addition the closure of a non-trivial (not itself closed) cozero-set may,
under $\mathsf{CH}$ ([MR0248057]), or may not, in the $\aleph_{2}$ Cohen model
([MR0813288]*Theorem 4.5), be a retract of $\N^{*}$.

##  4\. Individual Ultrafilters

24. Is there a model in which there are no $P$-points and no $Q$-points?

The Continuum Hypothesis implies that both kinds of points exist. If
$\mathfrak{c}=\aleph_{2}$ then at least one kind exists; this depends on the
value of $\mathfrak{d}$. If $\mathfrak{d}=\mathfrak{c}$ then $P$-points exist,
in fact Ketonen showed in [MR433387] that then every filter of cardinality
less than $\mathfrak{c}$ can be extended to a $P$-point. In the present case,
if $\mathfrak{d}<\mathfrak{c}$ then $\mathfrak{d}=\aleph_{1}$ and then the
result of Coplakova and Vojtáš from [MR863903] applies to show that there are
$Q$-points; this relies on the fact that the Novák number of $\N^{*}$ is at
least $\aleph_{2}$, see [MR0600576].

The current methods for creating models without $P$-points involve iterations
with countable supports and these invariably produce models where
$\mathfrak{c}=\aleph_{2}$, and hence these will contain $Q$-points. A recent
exception is [MR3990958], where models without $P$-points and arbitrarily
large continuum are constructed. However $\mathfrak{d}=\aleph_{1}$ in these
models, hence these contain $Q$-points as well.

25. Is there a model in which there is a rapid ultrafilter but in which there is no $Q$-point?

In [MR4246814] it was shown that the existence of a countable non-discrete
extremally disconnected group implies the existence of rapid ultrafilters.

26. What are the possible compactifications of spaces of the form $\N\cup\\{p\\}$ for $p\in\N^{*}$?

Of course for every $p$ we have $\beta(\N\cup\\{p\\})=\beta\N$. There are
points where this phenomenon persists: Dow and Zhou showed that is
$f:\beta\N\to{}^{\mathfrak{c}}2$ is continuous and onto and $K\subset\N^{*}$
is a closed set such that $f\restr K$ is irreducible and onto then for _every_
point in $K$ _every_ compactification of $\N\cup\\{p\\}$ contains a copy of
$\beta\N$, see [MR1676677].

Other examples of spaces of the form $\N\cup\\{x\\}$, where $x$ is the only
non-isolated point, for which every compactification contains $\beta\N$ were
constructed by Van Douwen and Przymusiński in [MR532957].

The case of scattered compactifications has received considerable interest.

In [MR107849] Semadeni asked whether $\N\cup\\{p\\}$ always has a scattered
compactification.

In [MR263030] Ryll-Nardzewski and Telgarsky proved that the answer is yes if
$p$ is a $P$-point and the Continuum Hypothesis holds; the compactification is
a version of the compactification $\gamma\N$ of Franklin-Rajagopalan from
[MR283742], where $\gamma\N\setminus\N$ is a copy of the ordinal
$\omega_{1}+1$ and $p$ corresponds to the point $\omega_{1}$.

In [MR410671] Jayachandran and Rajagopalan constructed a scattered
compactification of $\N\cup\\{p\\}$, where $p$ is a $P$-point limit of a
sequence of $P$-points.

Solomon, Telgarski, and Malykhin, in [MR448298], [MR461441], and [MR478101],
respectively, exhibited points $p$ in $\N^{*}$ such that $\N\cup\\{p\\}$ has
no scattered compactification.

Malykhin’s paper and the paper [MR461442] by Telgarsky contain investigations
of the structure of the (complementary) sets $S$ and $\mathit{NS}$ of points
for which $\N\cup\\{p\\}$ does and does not have a scattered compactification
respectively. The set $\mathit{NS}$ is quite rich: it contains the closures of
all of its countable subsets and it is upward closed in the Rudin-Frolík
order.

This richness foreshadowed a later result of Malykhin’s from
[Malykhin:betaNandnotCHa, Malykhin:betaNandnotCHb]: in the Cohen model it is
the case that for _every_ point $p\in\N^{*}$ _every_ compactification of
$\N\cup\\{p\\}$ contains a copy of $\beta\N$; in particular
$\mathrm{NS}=\N^{*}$ in this model.

27. Is there $p\in\Q_{d}^{*}$ such that $\mathcal{B}=\\{A\in p:A$ is closed and nowhere dense in $\Q$ and without isolated points$\\}$ is a base for $p$?

To eliminate possible confusion: we wrote $p\in\Q_{d}^{*}$ to emphasize that
we are asking for an ultrafilter on the countable _set of rationals_ (with the
discrete topology), and $\Q$ in the description of $\mathcal{B}$ to emphasize
that we want a base for the ultrafilter that is closely connected to the
topological structure of the _space of rationals_.

One could ask the question in the opposite direction: is there a point $x$ in
$\beta\Q\setminus\Q$ (the _space_ of rationals) that, when considered as an
ultrafilter of closed sets has a base consisting of closed nowhere dense
copies of $\Q$ and that also generates a real ultrafilter on the set $\Q$.

A third way of looking at this question is to consider
$\beta\operatorname{Id}:\beta\Q_{d}\to\beta\Q$, where $\operatorname{Id}$ is
the identity map and look for points in $\beta\Q\setminus\Q$ with one-point
preimages. Such points are easily found in the closure of $\N$ for example,
but we want a point whose elements are topologically as rich as possible.

These ultrafilters were dubbed ‘gruff ultrafilters’ by Van Douwen. This
question is still open but there are many consistent positive answers:

  * •

Van Douwen [MR1192307]: from $\mathsf{MA}_{\mathrm{countable}}$,

  * •

Coplakova and Hart [MR1676672]: from $\mathfrak{b}=\mathfrak{c}$,

  * •

Ciesielski and Pawlikowski [MR1997781]: from a version of the Covering
Property Axiom (hence in the Sacks model),

  * •

Millán [MR2182932]: from the same assumption a $Q$-point with this property,

  * •

Fernández-Bretón and Hrušák [MR3539743]: from a parametrized
$\diamond$-principle, from $\mathfrak{d}=\mathfrak{c}$, and in the random real
model; a correction in [MR3712981] points out that in the third case one needs
to add $\aleph_{1}$ many Cohen reals first

28. Is there a $p\in\N^{*}$ such that whenever $\langle x_{n}:n\in\omega\rangle$ is a sequence in $\Q$ there is an $A\in p$ such that $\\{\,x_{n}:n\in A\,\\}$ is nowhere dense?

Such ultrafilters are called _nowhere dense_. A $P$-point is nowhere dense: it
will have a member $A$ such that $\\{x_{n}:n\in A\\}$ converges to a point or
is closed and discrete. On the other hand, in [MR1690694] Shelah showed that
it is consistent that there are no nowhere dense ultrafilters. In [MR1833478]
it is shown that a nowhere dense ultrafilter exists iff there is a
$\sigma$-centered partial order that does _not_ add a Cohen real.

Research into this type of problem was initiated by Baumgartner in
[MR1335140]: the general situation involves a set $S$ and a notion of
smallness on $S$, usually expressed in terms of ideals. One then calls an
ultrafilter $u$ on $\N$ small if for every map $f:\N\to S$ there is a member
of $u$ whose image under $f$ is small.

29. Is there an ultrafilter $u$ such that for every map $f:\N\to\N$ there is a member $U$ of $u$ such that $f[U]$ has density zero?

This is a special case of the general problem mentioned in the comments above.
We mention it here because it is related to some special cases of problem 5,
which deals with permutations, rather than arbitrary maps.

30. Is there in $\mathsf{ZFC}$ an ultrafilter that is Sacks-indestructible?

This question is inspired by the many proofs that ultrafilters of small
character may exist. Sacks forcing preserves selective ultrafilters,
$P$-points and many ultrafilters constructed from these. Those ultrafilters
need not exist of course, so the question becomes if there are ultrafilters
that are preserved by this partial order.

##  5\. Dynamics, Algebra, and Number Theory

31. Is there a point in $\N^{*}$ that is not an element of any maximal orbit closure?

In this problem we consider the integers $\Z$ rather than $\N$ and the shift
map $\sigma$, defined by $\sigma(n)=n+1$. The orbit of $u\in\N^{*}$ is the set
$\\{\sigma(u):n\in\Z\\}$ and its closure $C_{u}$ is the _orbit closure_ of
$u$.

32. Is there an infinite strictly increasing sequence of orbit closures?

This problem is related to the previous problem: if there is no increasing
sequence of orbit closures then the family of orbit closures is well-founded
under reverse inclusion and every point is in some maximal orbit closure. A
negative answer to this question, and hence to Question 5, was given recently
by Zelenyuk in [MR4456336].

33. Is there a $p\in\N^{*}$ such that for every pair of commuting continuous maps $f,g:\vphantom{2}^{\omega}2\to\vphantom{2}^{\omega}2$ there is an $x\in\vphantom{2}^{\omega}2$ such that $p\streepje\\!\lim f^{n}(x)=p\streepje\\!\lim g^{n}(x)=x$?

This question is related in two ways to Birkhoff’s multiple recurrence
theorem, which states that commuting continuous self-maps of the Cantor set
have common recurrent points. Using ultrafilters one can state this theorem
as: for every pair of commuting continuous maps
$f,g:\vphantom{2}^{\omega}2\to\vphantom{2}^{\omega}2$ there are $p\in\N^{*}$
and $x\in\vphantom{2}^{\omega}2$ such that $p\streepje\\!\lim
f^{n}(x)=p\streepje\\!\lim g^{n}(x)=x$.

So the first connection to our question is clear: is there one single
ultrafilter that works for all pairs.

The second connection is the question whether the theorem holds for the Cantor
cube $\vphantom{2}{}^{\mathfrak{c}}2$?

If it does then the answer to our question is positive. To see this note first
that there are $\mathfrak{c}$ many pairs of commuting self-maps of
$\vphantom{2}{}^{\omega}2$, enumerated these as
$\\{\langle{f_{\alpha}},{g_{\alpha}}\rangle:\alpha<\mathfrak{c}\\}$. These
determine one pair $\langle{f},{g}\rangle$ of commuting self maps of
$\vphantom{2}{}^{\mathfrak{c}}2$: write $\vphantom{2}{}^{\mathfrak{c}}2$ as
$\vphantom{2}{}^{\mathfrak{c}\times\omega}2$, and let
$f=\prod_{\alpha<\mathfrak{c}}f_{\alpha}$ and
$g=\prod_{\alpha<\mathfrak{c}}g_{\alpha}$. The maps $f$ and $g$ commute and if
$x\in\vphantom{2}^{\mathfrak{c}\times\omega}2$ is a common recurrent point
then $p\streepje\\!\lim f^{n}(x)=p\streepje\\!\lim g^{n}(x)=x$ for some
$p\in\N^{*}$. But then also $p\streepje\\!\lim
f_{\alpha}^{n}(x_{\alpha})=p\streepje\\!\lim
g_{\alpha}^{n}(x_{\alpha})=x_{\alpha}$ for all $\alpha$.

34. For what nowhere dense sets $A\subseteq\N^{*}$ do we have $\bigcup_{\pi\in S_{\N}}\pi^{*}[A]\neq\N^{*}$?

Here $S_{\N}$ denotes the permutation group of $\N$.

It is consistent to assume that this happens for all nowhere dense sets. In
[MR0600576] Balcar, Pelant and Simon studied $\mathfrak{n}$, the Novák number
of $\N^{*}$, defined as the smallest number of nowhere dense sets needed to
cover $\N^{*}$. The inequality $\mathfrak{c}<\mathfrak{n}$ is consistent and
yields the consistency of “for all nowhere dense sets”; it follows from
$\mathsf{CH}$ (because $\mathfrak{n}\geq\aleph_{2}$), but is also consistent
with other values of $\mathfrak{c}$.

The inequality $\mathfrak{n}\leq\mathfrak{c}$ is also consistent and that case
there is not such an easy way out and it becomes an interesting project to
investigate whether the permutations of individual nowhere sets do, or do not,
cover $\N^{*}$ in $\mathsf{ZFC}$.

Permuting a singleton will not yield a cover, as $|\N^{*}|=2^{\mathfrak{c}}$.

Less obvious is Gryzlov’s result from [MR782711] that the permutations of the
set $F_{d}$ from Question 3 do not form a cover. This was improved by Flašková
in [MR2337416]: the permutations of the larger set $F_{\Sigma}$ do not cover
$\N^{*}$ either.

There is another natural nowhere dense subset of $\N^{*}$ the permutations of
which may, or may not, cover $\N^{*}$. Identify $\N$ with $\N\times\N$ and for
$k\in\N$ and $f:\N\to\N$ write $U(f,k)=\\{\langle{m},{n}\rangle:m\geq k$ and
$n\geq f(m)\\}$. The set $B=\bigcap_{f,k}U(f,k)^{*}$ is nowhere dense and it
is well known then $\bigcup_{\pi\in S_{\N}}\pi[B]$ consists of all non
$P$-points of $\N^{*}$. Hence the permutations of $B$ cover $\N^{*}$ iff there
are no $P$-points.

35. For what nowhere dense sets $A\subseteq\N^{*}$ do we have $\bigcup\\{h[A]:h\in\mathsf{Aut}\\}\neq\N^{*}$?

This question is more difficult than the previous one.

For example, singleton sets still do not provide covers in $\mathsf{ZFC}$, but
the easy counting argument is replaced by the non-trivial fact that $\N^{*}$
is not homogeneous.

We have no information about the sets $F_{d}$ and $F_{\Sigma}$ in this
context, except for the general fact that under $\mathsf{CH}$ the space
$\N^{*}$ cannot be covered by nowhere dense $P$-sets, see [MR548097]. Also, in
[MR620204] it was shown that it is consistent that $\N^{*}$ can be covered by
nowhere dense $P$-sets, and the principle $\mathsf{NCF}$ (Near Coherence of
Filters) implies that $\N^{*}$ is even the union of a _chain_ of nowhere dense
$P$-sets, see [MR1261700], but the sets in these covers are unrelated to the
sets $F_{d}$ and $F_{\Sigma}$. It is also unclear whether any one of the
individual sets in these families will produce a cover when moved around by
the members of $\mathsf{Aut}$.

The answer for the set $B$ remains the same because the union
$\bigcup\\{h[B]:h\in\mathsf{Aut}\\}$ consists of all non-$P$-points.

##  6\. Other

36. Are $\omega_{0}^{*}$ and $\omega_{1}^{*}$ ever homeomorphic?

This is known as the Katowice Problem, or rather the last remaining case of
this problem. It was posed in full by Marian Turzański, when he was in
Katowice (hence the name of the problem). The general question is: if $\kappa$
and $\lambda$ are infinite cardinals, endowed with the discrete topology, and
the remainders $\kappa^{*}$ and $\lambda^{*}$ are homeomorphic must the
cardinals $\kappa$ and $\lambda$ be equal?

Since the weight of $\kappa^{*}$ is equal to $2^{\kappa}$ it is immediate that
the Generalized Continuum Hypothesis implies a yes answer. In joint work
Balcar and Frankiewicz established that the answer is actually positive
without any additional assumptions, _except possibly for the first two
infinite cardinals_. More precisely, see [MR0461444, MR511955]: If
$\langle{\kappa},{\lambda}\rangle\neq\langle{\aleph_{0}},{\aleph_{1}}\rangle$
and $\kappa<\lambda$ then the remainders $\kappa^{*}$ and $\lambda^{*}$ are
not homeomorphic.

The paper [MR3563083] contains a list of the current known of consequences of
$\omega_{0}^{*}$ and $\omega_{1}^{*}$ being homeomorphic; all but one of these
can be made to hold in a single model of $\mathsf{ZFC}$.

By Stone-duality the Katowice problem can be formulated algebraically: are the
quotient (Boolean) algebras $\mathcal{P}(\omega_{0})/\mathrm{fin}$ and
$\mathcal{P}(\omega_{1})/\mathrm{fin}$ ever isomorphic? In this form the
question even makes sense in $\mathsf{ZF}$: in models without non-trivial
ultrafilters the spaces $\omega_{0}^{*}$ and $\omega_{1}^{*}$ are empty (and
so trivially homeomorphic) but the structures of the algebras may still
differ.

37. Is there consistently an uncountable cardinal $\kappa$ such that $\omega^{*}$ and $U(\kappa)$ are homeomorphic?

This problem is part of the uniform version of the Katowice problem, Question
6. The full question asks whether for distinct infinite cardinals $\kappa$ and
$\lambda$ spaces $U(\kappa)$ and $U(\lambda)$ of _uniform_ ultrafilters can be
homeomorphic, or algebraically whether the quotient algebras
$\mathcal{P}(\kappa)/[\kappa]^{<\kappa}$ and
$\mathcal{P}(\lambda)/[\lambda]^{<\lambda}$ can be isomorphic. This is
Question 47 in [MR588216], where we also find the information that, in
general, the algebra $\mathcal{P}(\kappa)/[\kappa]^{<\kappa}$ has cardinality
$2^{\kappa}$ and is $\mu$-complete for $\mu<\operatorname{cf}\kappa$ but not
$\operatorname{cf}\kappa$-complete. Therefore we can concentrate on cases
where $2^{\kappa}=2^{\lambda}$ and
$\operatorname{cf}\kappa=\operatorname{cf}\lambda$.

In [MR1103989] Van Douwen investigated the statements $S_{n}$:

$\hbox{if $\kappa\neq\aleph_{n}$ then $\mathcal{P}(\kappa)/[\kappa]^{<\kappa}$
and $\mathcal{P}(\omega_{n})/[\omega_{n}]^{<\aleph_{n}}$ are not isomorphic}.$

Thus, our question is whether it is consistent that $S_{0}$ is false. Van
Douwen showed that there is at most one $n$ for which $S_{n}$ is false, but
the proof offers no information on the location of that $n$ (if any) as it
simply establishes the implication “if $m<n$ and $S_{m}$ is false then $S_{n}$
holds”.

38. What is the structure of the sequences $\langle n((\N^{*})^{n}):n\in\N\rangle$ and $\langle wn((\N^{*})^{n}):n\in\N\rangle$?

Here $n$ and $wn$ denote the Novaḱ and weak Novák numbers, defined as the
minimum cardinality of a family of nowhere dense sets that covers the space,
or has a dense union, respectively.

It is clear that if $N$ is nowhere dense in a space $X$ then $N\times Y$ is
nowhere dense in the product $X\times Y$. This shows that, in general,
$n(X\times Y)\leq\min\\{n(X),n(Y)\\}$ and likewise for $wn$. It follows that
both sequences in our question are non-increasing and hence must become
constant eventually.

One could ask when they do become constant. For $wn$ this is undetermined: in
[MR1641157] Shelah and Spinas showed that for every $n$ there is a model in
which $wn((\N^{*})^{n})>wn((\N^{*})^{n+1})$. In particular
$wn(\N^{*})>wn(\N^{*}\times\N^{*})$ is possible, in [MR1751223] the latter
inequality was shown to hold in the Mathias model.

For the Novák numbers of the finite powers nothing is known as yet.

39. What is the status of the statement that all Parovichenko spaces are co-absolute (with $\N^{*}$)?

This question is related to Parovichenko’s theorem from [MR0150732], which
states that under $\mathsf{CH}$ all Parovichenko spaces are homeomorphic to
$\N^{*}$. Of course Parovichenko spaces were named after this theorem was
proved: they are compact, zero-dimensional $F$-spaces of weight $\mathfrak{c}$
without isolated points in which every non-empty $G_{\delta}$-set has non-
empty interior. For the nonce we say that a space is of _Parovichenko type_ if
it satisfies the conditions above, except for possibly the weight restriction.

In [MR612009] Broverman and Weiss proved that under $\mathsf{CH}$ all spaces
of Parovichenko type of $\pi$-weight $\mathfrak{c}$ are co-absolute (with
$\N^{*}$). They also established that if $\mathsf{CH}$ fails and
$\mathfrak{c}=2^{<\mathfrak{c}}$ then there is a Parovichenko space that is
not co-absolute with $\N^{*}$. They also proved that $\omega_{0}^{*}$ and
$\omega_{1}^{*}$ are co-absolute or, in algebraic terms that the Boolean
algebras $\mathcal{P}(\omega_{0})/\mathrm{fin}$ and
$\mathcal{P}(\omega_{1})/\mathrm{fin}$ have isomorphic completions, which
shows that completions do not have a direct effect on Question 6.

In [MR648079] Williams also established the $\pi$-weight result and showed
that $\N^{*}$ is co-absolute with a linearly ordered space.

In [MR676966] Van Mill and Williams improved the negative result of Broverman
and Weiss: if our statement holds then not only do we have
$\mathfrak{c}<2^{<\mathfrak{c}}$, but even $\mathfrak{c}<2^{\aleph_{1}}$.

In [MR759135] Dow proved that the equality
$\operatorname{cf}\mathfrak{c}=\aleph_{1}$ already implies that all
Parovichenko spaces are co-absolute.

The definition of the absolute as the Stone space of the Boolean algebra of
regular open sets makes sense for any compact space, so one may also seek co-
absolutes of $\N^{*}$ among spaces that are not zero-dimensional. In
[MR234422] Comfort and Negrepontis showed that under $\mathsf{CH}$ if $X$ is
locally compact and $\sigma$-compact, but not compact, and if
$\bigl{|}C(X)\bigr{|}=\mathfrak{c}$ then the set of $P$-points in $X^{*}$ is
homeomorphic to the $G_{\delta}$-modification of the ordered space
$2^{\omega_{1}}$; Parovichenko had already established this fact for $\N^{*}$
in [MR0150732]. This implies that for such spaces the remainders share a
homeomorphic dense subspace and hence that all such remainders are co-absolute
with $\N^{*}$, still under $\mathsf{CH}$ of course. So, for example, under
$\mathsf{CH}$ the spaces $\N^{*}$ and $\HH^{*}$ are co-absolute.

In [MR1619290] Dow showed that in the Mathias model $\N^{*}$ and $\HH^{*}$ are
not co-absolute.

40. Let $X$ be a compact space that can be mapped onto $\N^{*}$. Is $X$ non-homogeneous?

Since $\N^{*}$ maps onto $\beta\N$, a space as in the question will also map
onto $\beta\N$. If the weight of $X$ is at most $\mathfrak{c}$ then Theorem
4.1 (c) of [MR644652] applies and we find that $X$ is indeed non-homogeneous.

41. Is it consistent that every compact space contains either a convergent sequence or a copy of $\beta\N$?

Efimov asked in [MR0253290] whether every compact space contains either a
convergent sequence or a copy of $\beta\N$ and a counterexample is now called
a _Efimov space_. In [hart:efimov] one finds a survey of the status of the
problem in 2007; it lists various consistent Efimov spaces, which explains why
the present formulation asks for a consistency result. We mention here some of
the additional results that have been obtained in the meantime.

To begin there is a positive answer in [MR3164725] to Question 1 from
[hart:efimov]: Martin’s Axiom, or even the equality
$\mathfrak{b}=\mathfrak{c}$, implies that there is a Efimov space.

In addition there has been progress on two related questions due to Juhász and
Hušek. The latter asked whether every compact Hausdorff space contains either
a convergent $\omega$-sequence or a convergent $\omega_{1}$-sequence; Juhász’
question is stronger: must a compact Hausdorff space that does not contain a
convergent $\omega_{1}$-sequence be first-countable? A counterexample to
Hušek’s question would be a Efimov space because $\beta\N$ contains a
convergent $\omega_{1}$-sequence. In [MR1707489] one finds a result that
provides many models in which Juhász’ question, and hence that of Hušek’s,
have a positive answer. One of these models satisfies
$\mathfrak{b}=\mathfrak{c}$, hence Efimov’s question is strictly stronger than
that of Hušek’s.

42. Is there a locally connected continuum such that every proper subcontinuum contains a copy of $\beta\N$?

There are various continua that have the property that every proper
subcontinuum contains a copy of $\beta\N$: the remainders
$\beta\R^{n}\setminus\R^{n}$ all have this property for example. The reason is
that they are $F$-spaces, hence the closure of every countable relatively
discrete subset is a copy of $\beta\N$. However, these remainders are not
locally connected; indeed if a space $X$ is not pseudocompact then one can use
an unbounded continuous function to exhibit points in $X^{*}$ at which neither
$\beta X$ nor $X^{*}$ is locally connected, see [MR96195].

In [MR1934264] we find a construction, from $\mathsf{CH}$, of a locally
connected continuum without non-trivial convergent sequences. This
construction, an inverse limit in which all potential convergent sequences are
destroyed, can be modified with some extra bookkeeping to yield a locally
connected continuum in which every infinite subset contains a countable
discrete subset whose closure is homeomorphic to $\beta\N$, still under
$\mathsf{CH}$ of course.

This leaves the question for a $\mathsf{ZFC}$-example open but also suggest
some further variations. The example has the property that _some_ countable
relatively discrete subsets have $\beta\N$ as their closures. One can ask
whether one can ensure this for _all_ countable relatively discrete subsets,
or whether one can even make all countable subsets $C^{*}$-embedded. The
reason for this is that a compact $F$-space cannot be locally connected, hence
we would like to know how close to an $F$-space a locally connected continuum
can be.

We would also like to know whether there is a natural example that answers our
question; natural in the sense that one can simply write it down, as in
“$\beta\N$ is a compact space without convergent sequences” and “$\HH^{*}$ is
a continuum in which every proper subcontinuum contains a copy of $\beta\N$”.

43. Is there an extremally disconnected normal locally compact space that is not paracompact?

The ordinal space $\omega_{1}$ is locally compact and normal, but not
paracompact. There are, however, various additional assumptions that when
added to local compactness and normality will ensure paracompactness. Extremal
disconnectedness may or may not be such an assumption: Kunen and Parsons
showed in [MR540504] that if $\kappa$ is weakly compact then
$\beta\kappa\setminus U(\kappa)$ is normal and locally compact but not
paracompact. As weak compactness is a large cardinal property the answer to
this question can go many ways: a consistent counterexample, a real
counterexample, or even an equiconsistency result involving a large cardinal.

The weaker property of basic disconnectedness does not work, as shown by Van
Douwen’s example in [MR546947]. In this paper Van Douwen attributes the
present question to Grant Woods.

44. Is every compact hereditarily paracompact space of weight at most $\mathfrak{c}$ a continuous image of $\N^{*}$? Is every hereditarily c.c.c. compact space a continuous image of $\N^{*}$?

These questions are part of the general problem of identifying the continuous
images of $\N^{*}$. Przymusiński proved in [MR671232] that all perfectly
normal compact spaces are continuous images of $\N^{*}$. One can therefore
look for weakenings of perfect normality that still make the space an image of
$\N^{*}$. The present two properties are such weakenings and they have not
been ruled out yet.

Another weakening, first-countability, was ruled out by Bell in [MR1058795]:
the $\aleph_{2}$-Cohen model contains a first-countable compact space that is
not a continuous image of $\N^{*}$; this space is also hereditarily
metacompact. In the same paper Bell showed that the compact ordered space
$2^{\omega_{1}}$ (with the lexicographic order) is an image of $\N^{*}$.
Theorems 15 and 17 in Chapter 1 of [MR0220252] imply that every compact
ordered space that is first-countable is a continuous image of the latter
space, hence also of $\N^{*}$.

In connection with the latter result we note that it is consistent with the
negation of $\mathsf{CH}$ that all linear orders of cardinality $\mathfrak{c}$
are embeddable into the Boolean algebra $\mathcal{P}(\N)/\mathrm{fin}$, see
[MR567675]. By a combination of the Stone and Wallman dualities this implies
that it is consistent with $\neg\mathsf{CH}$ that every compact ordered space
of weight $\mathfrak{c}$ is a continuous image of $\N^{*}$.

This was later generalized in [MR1095691] to the consistency of Martin’s Axiom
for $\sigma$-linked partial orders, the negation of $\mathsf{CH}$, and the
statement that all compact spaces of weight $\mathfrak{c}$ are continuous
images of $\N^{*}$.

In both cases the proof constructs an embedding of a universal linear order or
a universal Boolean algebra of cardinality $\mathfrak{c}$ into
$\mathcal{P}(\N)/\mathrm{fin}$. This raises the question whether there is a
universal compact space of weight $\mathfrak{c}$; one that maps onto all such
spaces. The answer is negative, see [MR1707489]*Section 6.

45. Is every compact space of weight at most $\aleph_{1}$ a $1$-soft remainder of $\omega$?

A compactification $\gamma\N$ of $\N$ is $1$-soft if for every subset $A$ of
$\N$ with $\operatorname{cl}A\cap\operatorname{cl}(\N\setminus
A)\neq\emptyset$ there is an autohomeomorphism $h$ of $\gamma\N$ that is the
identity on $\gamma\N\setminus\N$ and is such that $\\{n\in A:h(n)\notin A\\}$
is infinite.

See Question 351527 on MathOverFlow, [flow351527], and also the papers
[MR4142223] and [MR4266612] for related information.

46. Is there a universal compact space of weight $\aleph_{1}$?

We mean universal in the mapping-onto sense; the dual question has the well-
known answer $[0,1]^{\omega_{1}}$ and Parovichenko’s theorem suggests that the
answer might be positive. The answer is negative in the $\aleph_{2}$-Cohen
model but a good reference is hard to find. There are references to the
result, [MR1676311, MR739914, MR1057268, MR1056364], but no concrete proof.

However, the argument in [MR1707489]*Section 6 can readily be adapted to
provide an accessible proof. We apply Stone duality and show that in the model
there is no Boolean algebra of cardinality $\aleph_{1}$ in which every Boolean
algebra of that cardinality can be embedded. Let
$\operatorname{Fn}(\omega_{2}\times\omega_{0},2)$ denote the Cohen partial
order and let $G$ be a generic filter.

The main steps are: we can assume that the Boolean algebra is determined by a
partial order $\prec$ on a subset of $\omega_{1}$. By the ccc of
$\operatorname{Fn}(\omega_{2}\times\omega_{0},2)$ the order $\prec$ is a
member of $V[G\restr\alpha]$ for some $\alpha<\omega_{2}$. Take the next
$\aleph_{1}$ many Cohen reals $\langle c_{\beta}:\beta<\omega_{1}\rangle$,
defined by $c_{\beta}(n)=\bigcup G(\alpha+\beta,n)$. The union, $T$, of the
binary tree $2^{<\omega}$ and the set $\\{c_{\beta}:\beta<\omega_{1}\\}$ is a
partially ordered set which, when turned upside-down generates a Boolean
algebra $B$. Assume $\varphi:T\to\omega_{1}$ is the restriction of an
embedding of $B$ into $\langle{\omega_{1}},{\prec}\rangle$. There is a
countable subset $C$ of $\omega_{2}$ such that the restriction of $\varphi$ to
$2^{<\omega}$ belongs to $V[G\restr(\alpha\cup C)]$. Now take
$\beta\in\omega_{1}$ such that $\alpha+\beta\notin C$. Then $c_{\beta}$ does
not belong to $V[G\restr(\alpha\cup C)]$, yet it can be defined from the
elements $\gamma=\varphi(c_{\beta})$ and $\varphi\restr 2^{<\omega}$ by the
formula $\bigcup\\{s:\gamma\prec\varphi(s)\\}$.

47. Investigate ultrafilters as topological spaces.

This is a very general question, so let us discuss some specific ones that may
be investigated. An ultrafilter can be viewed as a subspace of the Cantor set
$\vphantom{2}{}^{\omega}2$, if one identifies a subset of $\omega$ with its
characteristic function.

Of course this makes ultrafilters separable metric spaces, and hence
relatively well-behaved. But not too well-behaved: free ultrafilters are non-
measurable and do not have the property of Baire.

To begin one can repeat many of the investigations into the Rudin-Keisler
order using more general kinds of maps. We know
$p\mathrel{\leq_{\mathbf{RK}}}q$ means that there is a map $\varphi:\N\to\N$
such that $\beta\varphi(q)=p$. The map $\varphi$ determines a continuous map
from $\vphantom{2}{}^{\omega}2$ to itself, so the following definition
suggests itself at once: say $p\leq_{c}q$ if there is a continuous map
$f:\vphantom{2}^{\omega}2\to\vphantom{2}^{\omega}2$ such that $f[q]=p$.

One can ask whether $p\leq_{c}q$ and $q\leq_{c}p$ together imply that
$p\equiv_{c}q$, which means that there is a homeomorphism of
$\vphantom{2}{}^{\omega}2$ that maps $p$ to $q$. The structure of the partial
order $\leq_{c}$, minimal elements, incomparable elements, etc., would warrant
investigation as well.

There is no reason to stop there of course: one can ask the same questions
about Borel maps of any specific order, or of maps of arbitrary Baire classes.

One need not work with maps on $\vphantom{2}{}^{\omega}2$, though that may
make life easier, one can investigate what it means for two ultrafilters to be
homeomorphic, or what it means that one is a continuous image of the other.
The methods of [MR1277880] may be of use in determining the possible sizes of
sets of ultrafilters that are incomparable in this sense.

We note that an ultrafilter can be homeomorphic to at most $\mathfrak{c}$ many
other ultrafilters: if $f:p\to q$ is a homeomorphism then Lavrentieff’s
theorem implies that $f$ can be extended to a homeomorphism of
$G_{\delta}$-subsets of $\vphantom{2}{}^{\omega}2$, and the number of such
homeomorphisms is equal to $\mathfrak{c}$.

The paper [MR2879361] contains many results on the topology of ultrafilters.

48. Is it consistent that all free ultrafilters have the same Tukey type?

Isbell [MR201316] raised the question of the number of Tukey types of
ultrafilters on $\N$ and gave the obvious bounds $2$ (trivial or not) and
$2^{\mathfrak{c}}$. Tukey types of free ultrafilters were investigated by
Dobrinen and Todorčević in [MR3069290] who gave a combinatorial
characterization of ultrafilters that are Tukey-equivalent to the partial
order of finite subsets of $\mathfrak{c}$: the ultrafilter $\mathcal{U}$
should contain a subfamily $\mathcal{X}$ of cardinality $\mathfrak{c}$ such
that for every infinite subfamily $\mathcal{Y}$ of $\mathcal{X}$ the
intersection $\bigcap\mathcal{Y}$ does not belong to $\mathcal{U}$.

Such ultrafilters exist see [MR201316]*Theorem 5.4; they are the ultrafilters
of character $\mathfrak{c}$ constructed from an independent family of
cardinality $\mathfrak{c}$, see also [MR1454].

In [MR3990958]*Announcement 9 Chodounský and Guzmán announce a result that
comes close to the statement that all free ultrafilters have this property.

49. Is the space of minimal prime ideals of $C(\N^{*})$ _not_ basically disconnected?

For a commutative ring $R$ we let $mR$ denote the set of minimal prime ideals
endowed with the hull-kernel topology. In [MR0144921, MR194880] Henriksen and
Jerison studied this space and asked whether $mC(\N^{*})$ is basically
disconnected.

In the papers [MR958091] and [MR1057626] various conditions were found that
imply $mC(\N^{*})$ is _not_ basically disconnected. For example, $\mathsf{MA}$
implies that $mC(\N^{*})$ is not even an $F$-space ([MR958091]). In
[MR1057626] it was shown that the equality
$\operatorname{cf}[\mathfrak{d}]^{\aleph_{0}}=\mathfrak{d}$ suffices to show
that $mC(\N^{*})$ is not basically disconnected. Failure of this equality
entails the existence of inner models with measurable cardinals. The actual
consequence, called Mel, of this equality that was used in the proof
identifies $\N$ with $\Q$ and asks for a $P$-filter $\mathcal{F}$ on $\Q$, and
two countable disjoint dense subsets $A$ and $B$ of $\R\setminus\Q$ such that
the closure in $\R$ of every member of $\mathcal{F}$ meets both $A$ and $B$.

Thus, to show that $mC(\N^{*})$ is not basically disconnected it suffices to
show that Mel holds, or the following stronger, but possibly more manageable,
statement: the ideal of nowhere dense subsets of $\Q$ can be extended to a
$P$-ideal.

50. Is there a c.c.c. forcing extension of $L$ in which there are no $P$-points?

The consistency of the nonexistence of $P$-points was proven by Shelah, see
[MR728877] and also [MR1623206]*VI §4.

After this there have been various attempts to (dis)prove the existence of
$P$-points in various standard models. Quite often the outcome was that ground
model $P$-points remained ultrafilters and $P$-points in the extension.

A notable exception is the Silver model: in [MR3990958] we find a proof that
iterating Silver forcing $\omega_{2}$ times with countable supports produces a
model without $P$-points; the same holds for the countable support product of
arbitrarily many copies of the partial order. This establishes the consistency
of the nonexistence of $P$-points with arbitrarily large values of
$\mathfrak{c}$.

A question that is still open is whether $P$-points exist in the random real
model. If not then this would answer the present question positively. If there
are $P$-points in this model then our question gains interest as it is as yet
unknown whether c.c.c. forcing can be used to kill $P$-points.

51. What is the relationship between ultrafilters of small character (less than $\mathfrak{c}$) and $P$-points?

One of the first ultrafilters of small character can be found in
[MR597342]*Exercise VII.A10; it is a simple $P_{\aleph_{1}}$-point constructed
by iterated forcing over a model of $\neg\mathsf{CH}$. There are many more
examples of ultrafilters of small character but their constructions seem to
involve $P$-points in some form or another. A common method is to start with a
model of $\mathsf{CH}$ and enlarge the continuum while preserving some
ultrafilters; these will then have character $\aleph_{1}$, which is smaller
than $\mathfrak{c}$. Almost always these ‘indestructible’ ultrafilters are
$P$-points (or stronger) and remain $P$-points in the extension. There are a
few exceptions, see [MR987317] for instance, but there the ultrafilters are
built using $P$-points and these are preserved as well.

52. We let $\mathrm{Sp}_{\chi}$ denote the set of characters of ultrafilters on $\N$, the _character spectrum_ of $\N$. The general question is what one can say about this set.

We know that $\mathfrak{c}\in\mathrm{Sp}_{\chi}$, and that
$\mathrm{Sp}_{\chi}=\\{\mathfrak{c}\\}$ is possible.

In [MR2365799] Shelah showed the consistency of there being three cardinals
$\kappa$, $\lambda$, and $\mu$ such that $\kappa<\lambda<\mu$, and
$\kappa,\mu\in\mathrm{Sp}_{\chi}$ and $\lambda\notin\mathrm{Sp}_{\chi}$. The
construction uses a c.c.c. forcing over a ground model in which the three
cardinals are regular, $\lambda$ is measurable, and there is another
measurable cardinal below $\kappa$. In [MR2847327] he extended this result by
showing how to build, given two disjoint sets $\Theta_{1}$ and $\Theta_{2}$ of
regular cardinals, a cardinal-preserving partial order that forces
$\Theta_{1}$ to be a subset of $\mathrm{Sp}_{\chi}$ and $\Theta_{2}$ to be
disjoint from it; the construction requires $\Theta_{2}$ to consist of
measurable cardinals. The same paper also contains models in which
$\\{n:\aleph_{n}\in\mathrm{Sp}_{\chi}\\}$ can be any subset of $\N$, starting
from infinitely many compact cardinals. This answers a question from
[MR1686797], namely whether if there are ultrafilters of character
$\aleph_{1}$ and $\aleph_{3}$ there must be one of character $\aleph_{2}$, but
at the cost of large cardinals.

This leaves open the question whether the conjunction of
$\aleph_{1},\aleph_{3}\in\mathrm{Sp}_{\chi}$ and
$\aleph_{2}\notin\mathrm{Sp}_{\chi}$ can be proven consistent from the
consistency of just $\mathsf{ZFC}$. To be very specific we ask whether there
is an ultrafilter of character $\aleph_{2}$ in the model(s) of
[MR597342]*Exercise VII.A10, where one starts with a model of
$\mathfrak{c}=\aleph_{3}$, and in the side-by-side Sacks model where
$\mathfrak{c}=\aleph_{3}$.

53. Is there consistently a point in $\N^{*}$ whose $\pi$-character has countable cofinality?

The paper [MR1686797] contains a wealth of material on $\pi$-characters of
ultrafilters, including a model with an ultrafilter of $\pi$-character
$\aleph_{\omega}$.

Unlike the results on the character spectrum the results on the
$\pi$-character spectrum do not require large cardinals.

54. Is it consistent that $t(p,\N^{*})<\chi(p)$ for some $p\in\N^{*}$?

There are plenty of compact spaces with points where the tightness is smaller
than the character; the one-point compactification of the any uncountable
discrete space will do: the tightness at the point at infinity is countable,
the character of the point is not.

Let us remark that no point of $\N^{*}$ has countable tightness: certainly at
$P$-points the tightness is uncountable; if $p$ is not a $P$-point then it
lies on the boundary of a zero-set $C$ and in the closure of its interior, but
the closure of every countable subset of that interior is a subset of that
interior. This implies that $t(p,\N^{*})=\chi(p)=\mathfrak{c}$ if
$\mathsf{CH}$ holds, hence the question for a consistency result.

As an aside we mention that there are consistent examples of regular
extremally disconnected spaces of countable tightness: in [MR0458392] and
[MR461464] one finds constructions of extremally disconnected $S$-spaces. The
constructions use $\clubsuit$ and that some extra assumption is necessary is
shown in [MR615971]: there are no extremally disconnected $S$-spaces if
${\mathsf{MA}+\neg\mathsf{CH}}$ holds. Both [MR461464] and [MR615971] contain
constructions of extremally disconnected $S$-spaces in $\beta\N$.

55. If $C(\omega+1,\C)$ admits an incomplete norm then does $C(\beta\N,\C)$ admit one too?

This question is related to a conjecture/question of Kaplansky’s about algebra
norms on the spaces $C(X,\C)$, with $X$ compact. The question is whether every
algebra norm is equivalent to the sup-norm $\|{\cdot}\|_{\infty}$. The answer
is positive if the norm is complete, hence the question became whether every
algebra norm on $C(X,\C)$ is complete.

The book [MR942216] surveys the solution to this problem: under $\mathsf{CH}$
every $C(X,\C)$ carries an incomplete algebra norm (Dales and Esterlé) and it
is consistent that every algebra norm on every $C(X,\C)$ is complete (Solovay
and Woodin).

The present question comes from the results that if $C(\beta\N,\C)$ admits an
incomplete norm then so does every $C(X,\C)$, and if _some_ $C(X,\C)$ carries
an incomplete norm then so does $C({\omega+1},\C)$. In short it asks whether
all compact spaces are equivalent for Kaplansky’s conjecture.

The question can be translated into terms of individual ultrafilters and this
leads to some interesting subquestions. A seminorm on an algebra is a function
that satisfies all conditions of an algebra norm except for the condition that
non-zero elements should have non-zero norm. An algebra is semi-normable if it
carries a non-trivial seminorm.

For a point $p$ of $\beta\N$ we let $A_{p}$ denote the quotient algebra
$M_{p}/I_{p}$, where $M_{p}=\\{f\in C(\beta\N,\C):f(p)=0\\}$, and
$I_{p}=\\{f\in C(\beta\N,\C):(\exists P\in p)(f\restr P=0)\\}$. We also let
$c_{0}$ be the subalgebra of $C(\beta\N,\C)$ of functions that vanish on
$\N^{*}$ and we let $c_{0}/p$ denote the quotient algebra $c_{0}/(c_{0}\cap
I_{p})$.

Theorem 2.21 in [MR942216] shows why we should be interested in these
algebras: The algebra $C(\beta\N,C)$ admits an incomplete norm iff for some
$p$ the algebra $A_{p}$ is seminormable, and $C({\omega+1},\C)$ admits an
incomplete norm iff for some $q$ the algebra $c_{0}/q$ is seminormable.

We see that _if_ there is a $p$ such that $A_{p}$ is seminormable _then_ there
is a $q$ such that $c_{0}/q$ is seminormable. The present question ask whether
this implication can be reversed.

Further questions regarding these algebras suggest themselves: is it the case
that the seminormability of $A_{p}$ implies that of $c_{0}/p$? In other words
can we get $q=p$ in the previous paragraph?

Also, what is the answer to the stronger version of our question: if $c_{0}/p$
is seminormable is $A_{p}$ seminormable too?

We recommend [MR942216]*Chapters 1, 2 and 3 for more detailed information on
this question.

56. ($\mathsf{MA}+\lnot\mathsf{CH}$) Are there $\Gap$ and $p$ ($P$-point, selective) such that $p\subseteq I_{\Gap}^{+}$?

Here $\Gap$ denotes a Hausdorff-gap in ${}^{\omega}\omega$ and $I_{\Gap}$ is
the ideal of sets over which $\Gap$ is filled.

S. Kamo [MR1138202] proved that if $V$ is obtained from a model of
$\mathsf{CH}$ by adding Cohen reals then in $V$ an ideal is a gap-ideal iff it
is $\leq\aleph_{1}$-generated. Also, $\mathsf{CH}$ implies that any nontrivial
ideal is a gap-ideal.

The commentary in [MR2023411] mentions a further preprint by Kamo, [kamo1993],
where it is shown that, under $\mathsf{MA}+\lnot\mathsf{CH}$, for every
Hausdorff gap $\Gap$ there are both selective ultrafilters and non-$P$-points
consisting of positive sets (with respect to the gap-ideal $I_{\Gap}$). Also
under $\mathsf{MA}+\lnot\mathsf{CH}$ there is a selective
non-$P_{\aleph_{2}}$-point that meets every gap-ideal.

Unfortunately we were unable to locate this preprint and verify these
statements.

##  7\. Orders

57. Is there for every $p\in\N^{*}$ a $q\in\N^{*}$ such that $p$ and $q$ are $\mathrel{\leq_{\mathbf{RK}}}$-incomparable?

This question has a long history; it is as old as the Rudin-Keisler order
itself. In [MR314619] Kunen constructed two points that are
$\mathrel{\leq_{\mathbf{RK}}}$-incomparable. In [MR540490] Shelah and Rudin
proved that there is a set of $2^{\mathfrak{c}}$ incomparable points. In
[MR0863940] Simon proved that these points may be taken to be $\aleph_{1}$-OK.
In [MR1277880] Dow showed that that there are many more situations where such
sets may be constructed.

However, none of these results shed light on the present question. Some
partial results are available: in [MR931732] Hindman proved: if $p$ is such
that $\chi(r)=\mathfrak{c}$ whenever $r\mathrel{\leq_{\mathbf{RK}}}p$ then
there is a point that is incomparable with $p$, so the answer to the present
question is positive if all ultrafilters have character $\mathfrak{c}$.
Furthermore if $\mathfrak{c}$ is singular and $\chi(p)=\mathfrak{c}$ then
again there is a point that is incomparable with $p$. The latter result was
extended by Butkovičová in [MR1045131]: if $\kappa<\mathfrak{c}$ is such that
$\mathfrak{c}<2^{\kappa}$ then for every ultrafilter of character
$\mathfrak{c}$ there are $2^{\kappa}$ many ultrafilters incomparable with it.
Note that these results all impose conditions on individual ultrafilters in
order to find an incomparable point; only the condition “all ultrafilters have
character $\mathfrak{c}$” answers this question directly.

In [MR1623206]*XVIII §4 Shelah proved that it is consistent that up to
permutation there is one $P$-point.

We recall the definition of the Rudin-Frolík order: we say
$p\mathrel{\leq_{\mathbf{RF}}}q$ if there is an embedding
$f:\beta\N\to\beta\N$ such that $f(p)=q$. This is a preorder that induces a
partial order on the types of ultrafilters. To see this note that
$p\mathrel{\leq_{\mathbf{RF}}}q$ implies $p\mathrel{\leq_{\mathbf{RK}}}q$:
given $f$ take a partition $\\{A_{n}:n\in\N\\}$ of $N$ such that $A_{n}\in
f(n)$ for all $n$. The map $g=\bigcup_{n}(A_{n}\times\\{n\\})$ satisfies
$p=g(q)$ and shows $p\mathrel{\leq_{\mathbf{RK}}}q$.

As usual $p\mathrel{<_{\mathbf{RF}}}q$ will mean
$p\mathrel{\leq_{\mathbf{RF}}}q$ plus not-$q\mathrel{\leq_{\mathbf{RF}}}p$,
and this is readily seen to be equivalent to there being an embedding
$f:\beta\N\to\N^{*}$ such that $f(p)=q$.

The Rudin-Frolík order is tree-like: if $p,q\mathrel{\leq_{\mathbf{RF}}}r$
then $p\mathrel{\leq_{\mathbf{RF}}}q$ or $q\mathrel{\leq_{\mathbf{RF}}}p$. And
due to the relation with $\mathrel{\leq_{\mathbf{RK}}}$ we see at once that
$\\{p:p\mathrel{\leq_{\mathbf{RK}}}q\\}$ always has cardinality at most
$\mathfrak{c}$.

In many papers on the Rudin-Frolík order Frolík’s original notation is
employed where one writes $X=f[\N]$, and $q=\Sigma(X,p)$ as well as
$p=\Omega(X,q)$.

58. For what cardinals $\kappa$ is there a strictly decreasing chain of copies of $\beta\N$ in $\N^{*}$ with a one-point intersection?

This question is related to decreasing chains in
$\mathrel{\leq_{\mathbf{RF}}}$. A decreasing sequence of copies of $\beta\N$
determines and is determined by a sequence $\langle
X_{\alpha}:\alpha\in\delta\rangle$ of countable discrete subsets of $\N^{*}$
with the property that
$X_{\alpha}\subseteq\operatorname{cl}{X_{\beta}}\setminus X_{\beta}$ whenever
$\beta\in\alpha$. Take a point $p$ in the intersection of the sequence; then
$\langle\Omega(X_{\alpha},p):\alpha\in\delta\rangle$ is a decreasing
$\mathrel{\leq_{\mathbf{RF}}}$-chain.

To ensure that this chain does not have a lower bound one should make sure
that $p$ is not an accumulation point of a countable discrete subset of the
intersection. Having a one-point intersection is certainly sufficient for
this. In [MR863908] Van Douwen showed that it is possible to have a chain of
length $\mathfrak{c}$ with a one-point intersection. In [MR633575] and
[MR1007490] we find constructions of decreasing
$\mathrel{\leq_{\mathbf{RF}}}$-chains of type $\omega$ and of type $\mu$ for
uncountable $\mu<\mathfrak{c}$ respectively. The latter two constructions
provide a point in the intersection of a suitable chain of copies of $\beta\N$
that is not an accumulation point of a countable discrete subset of that
intersection.

We want to know when in these cases the intersection can be made to be a one-
point set.

59. If $\kappa\leq\mathfrak{c}$ has uncountable cofinality and if $\left\langle X_{\alpha}:\alpha<\kappa\right\rangle$ is a strictly decreasing sequence of copies of $\beta\N$ with intersection $K$, is there a point $p$ in $K$ that is not an accumulation point of any countable discrete subset of $K$?

This is related to Question 7: the chains of copies of $\beta\N$ in the
positive results were chosen with care. We want to know if that care is
necessary.

60. What are the possible lengths of unbounded $\mathbf{RF}$-chains?

Since every point has at most $\mathfrak{c}$ predecessors every chain has
cardinality at most $\mathfrak{c}^{+}$. In [MR633575] we find a point with
exactly $\aleph_{0}$ many predecessors, with the order type of the set of
negative integers.

Every unbounded chain will have cardinality at least $\mathfrak{c}$ (this
follows from results in [MR277371]*Theorem 2.9 or [MR698391]), so the
cardinality of an unbounded chain is equal to either $\mathfrak{c}$ or
$\mathfrak{c}^{+}$. In [MR730151] and [MR817833] Butkovičová constructed
unbounded chains of order-type $\mathfrak{c}^{+}$ and $\omega_{1}$
respectively.

What other order-type are possible? Can one prove that a chain or order-type
$\mathfrak{c}$ (or its cofinality) exists, irrespective of $\mathsf{CH}$?

61. Is every finite partial order embeddable in the Rudin-Keisler order?

See MathOverFlow https://mathoverflow.net/questions/375365. To get a positive
answer it suffices to embed every finite power set into this order. It is
relatively easy to adapt the construction of two incomparable ultrafilters to
yield an embedding of the power set of $\\{0,1\\}$ (see [ruitje]), but an
embedding of the power set of $\\{0,1,2\\}$ already poses unexpected
difficulties.

The analogous question for the Rudin-Frolík order has an easier answer. This
order is tree-like in that the predecessors of a point are linearly ordered,
and every point has $2^{\mathfrak{c}}$ many successors. This implies that
every finite rooted tree, and only those among the finite partial orders, can
be embedded into this order.

##  8\. Uncountable Cardinals

62.  Is there consistently an uncountable cardinal $\kappa$ with $p\in U(\kappa)$ such that $\chi(p)<2^{\kappa}$?

It is well known that if $\kappa$ is an infinite cardinal then there are
$2^{2^{\kappa}}$ many uniform ultrafilters on $\kappa$ with character equal to
$2^{\kappa}$, see [MR1454].

It is also well known, and referred to in other questions, that it is
consistent that there are ultrafilters on $\N$ of character less than
$\mathfrak{c}$.

Of course the Generalized Continuum Hypothesis implies that every uniform
ultrafilter on every $\kappa$ has character $2^{\kappa}$, but we are not aware
of any consistency result the other way for uncountable cardinals.

We formulate two special cases of our question:

  * •

Is it consistent to have a uniform ultrafilter on $\omega_{1}$ of character
$\aleph_{2}$ (with $\aleph_{2}<2^{\aleph_{1}}$ of course)?

  * •

Is it consistent to have a measurable cardinal $\kappa$ with a $p\in
U(\kappa)$ such that $\chi(p)<2^{\kappa}$?

The first question simply looks at the smallest possible case and the second
question asks, implicitly, if having a uniform ultrafilter of small character
is actually a large-cardinal property of $\aleph_{0}$.

There has been recent activity in this area; the paper [MR3832086] deals with
the character spectrum of uncountable cardinals of countable cofinality, and
in [MR4081063] one finds models with $\mathfrak{u}_{\kappa}<2^{\kappa}$ for
$\kappa=\mathfrak{c}$ and for $\kappa=\aleph_{\omega+1}$. These results use
large cardinals in the ground model: the spectrum result uses a supercompact
and many measurables; the results for $\mathfrak{c}$ and $\aleph_{\omega+1}$
use a measurable and supercompact cardinal respectively.

63. Is it consistent to have cardinals $\kappa<\lambda$ with points $p\in U(\kappa)$ and $q\in U(\lambda)$ such that $\chi(p)>\chi(q)$?

This is a follow-up question to Question 8: if uniform ultrafilters of small
character are at all possible, how much variation can we achieve among various
cardinals?

64. If $\kappa$ is regular and uncountable, $\mathcal{F}$ is a countably complete uniform filter on $\kappa$ then what is the cardinality of the closed set $U_{\mathcal{F}}=\\{u\in U(\kappa):\mathcal{F}\subseteq u\\}$?

In case $\kappa$ is measurable one can use a measure ultrafilter to create
filters $\mathcal{F}$ such that $U_{\mathcal{F}}$ is finite or a copy of
$\beta\lambda$, for any $\lambda<\kappa$.

For other cardinals the set $U_{\mathcal{F}}$ will always be at least infinite
and given the nature of $\beta\kappa$ the cardinality will be closely related
to numbers of the form $2^{2^{\lambda}}$ for $\lambda\leq\kappa$.

For the closed unbounded filter the answer is $2^{2^{\kappa}}$: using a family
of $\kappa$ many pairwise disjoint stationary sets and an independent family
on $\kappa$ of cardinality $2^{\kappa}$ one can produce a map from
$U_{\mathcal{F}}$ onto the Cantor cube of weight $2^{\kappa}$.

65. Assume that $\kappa$ is regular, that $\kappa\subseteq X\subseteq\beta\kappa$ is such that $[X]_{<\kappa}=X$ and $\beta_{X}\kappa=X$. Now if $Y$ is a closed subspace of a power of $X$, is then also $X$ a closed subspace of a power of $Y$?

Some notation: $[X]_{<\kappa}$ denotes
$\bigcup\\{\operatorname{cl}B:B\in[X]^{<\kappa}\\}$, and if $\kappa\subseteq
X\subseteq\beta\kappa$ then $\beta_{X}\kappa$ is the maximal subset of
$\beta\kappa$ such that every function from $\kappa$ to $X$ has a continuous
extension from $\beta_{X}\kappa$ to $X$.

66. Are there $\kappa$ and $p\in U(\kappa)$ such that $|\R_{p}|>|\R_{p}/\equiv|=\mathfrak{c}$?

Here $\R_{p}$ denotes the ultrapower of $\R$ by the ultrafilter $p$. The
relation $\equiv$ is that of Archimedean equivalence: $a\equiv b$ means that
there is an $n\in\N$ such that both
$\mathopen{|}a\mathclose{|}<\mathopen{|}nb\mathclose{|}$ and
$\mathopen{|}b\mathclose{|}<\mathopen{|}na\mathclose{|}$.

67. Is there a $C^{*}$-embedded bi-Bernstein set in $U(\omega_{1})$?

68. Are there open sets $G_{1}$ and $G_{2}$ in $U(\omega_{1})$ such that $\operatorname{cl}{G_{1}}\cap\operatorname{cl}{G_{2}}$ consists of exactly one point?

## References


Acknowledgments: The final part of this work was completed when the authors
participated in a Research in Teams (RIT) project “Modular Forms and Theta
Correspondence for Exceptional Groups” hosted by the Erwin Schrodinger
Institute, University of Vienna. We thank the Erwin Schrodinger Institute for
its generous support and wonderful working environment, as well as our other
RIT members Nadya Gurevich and Aaron Pollack for helpful discussions. We would
like to thank B. Bekka and M. Tadić for their help with the Fell topology and
G. Chenevier for his help with the unacceptability of $\mathrm{Spin}_{7}$.
Finally, we are grateful to Benedict Gross, whose inspirational ideas on the
exceptional theta correspondence have been sustaining our work for the past 25
years. W.T. Gan is partially supported by a Singapore government MOE Tier 1
grant R-146-000-320-114. G. Savin is partially supported by a National Science
Foundation grant DMS-1901745.

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