# Invariant Ideal Axiom

Michael Hrušák  Centro de Ciencias Matemáticas, Universidad Nacional Autónoma
de México, Campus Morelia, Morelia, Michoacán, 58089, México
<EMAIL_ADDRESS>http://www.matmor.unam.mx/ michael and  Alexander
Shibakov  Department of Mathematics, Tennessee Tech. University, 110
University Drive, Cookeville, Tennessee 38505, USA<EMAIL_ADDRESS>http://www.math.tntech.edu/alex

(Date: March 2021)

###### Abstract.

We introduce and prove the consistency of a new set theoretic axiom we call
the _Invariant Ideal Axiom_. The axiom enables us to provide (consistently) a
full topological classification of countable sequential groups, as well as
fully characterize the behavior of their finite products.

We also construct examples that demonstrate the optimality of the conditions
in IIA, and list a number of open questions.

###### 2010 Mathematics Subject Classification:

22A05, 03C20, 03E05, 03E35, 54H11

The research of the first author was supported by PAPIIT grants IN100317 and
IN104220, and CONACyT grant A1-S-16164.

## Introduction

In this paper we introduce a new set theoretic principle we call the
_Invariant Ideal Axiom_ (IIA for short) and prove its consistency with the
usual axioms of ZFC. As the main application of IIA we show that it implies
that all countable sequential groups are either metrizable or $k_{\omega}$
and, in particular, every countable sequential group has a definable (in fact
$F_{\sigma\delta}$) topology, thus concluding the project initiated in the
1970’s of determining the structural theory of countable Fréchet and
sequential groups (see [6, 8, 23, 34, 42, 43, 46, 47, 49, 84] for some early
papers on the subject). This line of research is considered central in
topological algebra (see [11, 21, 38, 62, 64, 65, 67, 79, 82, 81], and several
others) with some of the early questions answered only recently (see [19, 39,
66, 68]). For a more comprehensive overview of the field, including the
history, the fundamental results, and the open problems the reader may wish to
consult several excellent surveys available on the subject, as well as a
number of books on topological algebra: [9, 24, 40, 45, 61].

The axiom is fully accessible to the mathematicians working in topology or
algebra and does not require any knowledge of modern set theory. Aside from
giving the ultimate structural result for countable sequential groups, the
axiom has a profound impact on product properties of sequential groups.

Our hope and expectation is that the axiom IIA provides both a canonical
environment, and a test model for future study of convergence in topological
algebra.

To better relate our results to the existing body of research one may recall
that arguments in analysis and topology often depend on establishing the
extent of various special classes of topological spaces. Fusing algebraic and
topological properties proved to be among the most fruitful techniques.
Classical examples of such results are the implication $T_{1}\Rightarrow
T_{3{1\over 2}}$ in general topological groups (or even $T_{3}\Rightarrow
T_{3{1\over 2}}$ in _para_ topological groups, see [10]) and the _Birkhoff-
Kakutani theorem_ (see [9] for these and other facts about topological groups)
on the metrizability of first countable $T_{1}$ topological groups.
Metrizability theorems in particular drew a lot of attention, stimulating a
search for the weakest set of conditions that jointly imply that a given
topology is generated by a metric.

In the class of topological groups, compactness and countable tightness
together imply metrizability (see [9]) so it is natural to look for a similar
yet less restrictive set of conditions that may yield the same result.
Reasoning along these lines led V. Malykhin to ask about the existence of
countable (equivalently, separable) Fréchet non metrizable groups (see [6]).

Malykhin’s problem generated a large body of research that illustrates another
important quality shared by several results in this area. Namely, the effect
of a particular set of restrictions is greatly influenced by set theory. As a
case in point, Malykhin’s problem has an affirmative answer in a variety of
set theoretic universes, including models of MA. The conclusive result,
establishing the independence of the answer to Malykhin’s problem of the
axioms of ZFC was obtained by the first author and U. A. Ramos-García in [39]
using a forcing construction. The same paper also contains a construction of a
countable Fréchet non metrizable group under a very weak set theoretic
assumption $\diamondsuit(2=)$.

Malykhin asked (see [58]) a related question about the class of countable
sequential abelian groups (see below for all the appropriate definitions).
This question was fully solved in [58] by E. Zelenyuk and V. Protasov who
established (in ZFC) the existence of countable sequential group topologies
that are not Fréchet on any infinite countable abelian group. The existence of
such topology on _nonabelian_ countable groups (specifically, the free group)
was well-known (see [55]).

The investigation into the class of sequential groups prompted P. Niykos (see
[46]) to look at their _sequential order_ , which can be roughly thought of as
the ordinal measure of the complexity of the closure operator in sequential
spaces. The existence of sequential groups (of any size) of sequential order
strictly between 1 and $\omega_{1}$ turned out to be independent of the axioms
of ZFC, as well (see [65], [68], and [71]).

A thorough review of existing ZFC constructions of sequential non Fréchet
groups (see [40], [76], [58], [21]) reveals a structure common to all such
examples. Their topology is determined by a countable family of (countably)
compact subspaces (i.e. is $k_{\omega}$, see [32] and the definition below).
Perhaps the most widely known family of $k_{\omega}$ spaces is the class of
countable CW-complexes (see [16]). Various results in algebraic topology (such
as the homotopy equivalence for filtered spaces theorem of J. Milnor, see
[15]) heavily depend on the $k_{\omega}$ property. The class of $k_{\omega}$
spaces is well behaved, in particular, it is finitely productive, and every
countable $k_{\omega}$ space is sequential and analytic (see below for the
definitions and further discussion).

In [69], answering a question of S. Todorčević and C. Uzcátegui, the second
author showed that at least in the definable case (more specifically, in the
class of countable _analytic_ groups), the only sequential examples of
countable groups are $k_{\omega}$ or metrizable. This naturally brought about
the question (posed in [68]) whether it is consistent with ZFC that all
countable sequential groups are either metrizable or $k_{\omega}$
(equivalently, whether all countable sequential groups are analytic).

The main tool introduced in this paper, _the Invariant Ideal Axiom_ , or IIA,
is used to answer this question in the affirmative. As important corollaries,
we show that IIA generates a complete classification of sequential group
topologies on countable groups, as well as allows for a transparent
description of products of such groups.

To support our claim of the optimal nature of IIA in the study of convergence
in countable groups, we show that its natural generalization fails to stay
consistent. We also construct an example demonstrating the differences between
the case of countable sequential groups and their separable counterparts. We
conclude by listing a few open questions we believe will lead to greater
insight about this field of research.

##  1\. Preliminaries

All topological spaces and groups considered are $T_{1}$ and completely
regular. To see more about topological groups consult [6, 9, 22, 23, 24, 61,
77].

Recall that a topological space $X$ is _Fréchet_ if for any
$x\in\overline{A}\subseteq X$ there is a sequence $S\subseteq A$ such that
$S\to x$. A space $X$ is _sequential_ if for every $A\subseteq X$ which is not
closed there is a $C\subseteq A$ such that $C\to x\not\in A$. The term
‘Fréchet space’ appears to have been coined by Arkhangel’skii in [3], while
the term ‘sequential’ appears for the first time in Franklin’s [30], where the
following notion is defined:

Given $A\subseteq X$, define the _sequential closure_ of $A$ as

$[A]^{\prime}={\\{\,x\in X:C\to x\text{ for some }C\subseteq A\,\\}}\text{,
and then recursively}$ $[A]_{0}=A\text{ and
}[A]_{\alpha}=\cup{\\{\,[[A]_{\beta}]^{\prime}:\beta<\alpha\,\\}}\text{ for
}\alpha\leq\omega_{1}.$

Then $X$ is sequential if and only if $\overline{A}=[A]_{\omega_{1}}$ for
every $A\subseteq X$, and the _sequential order_ of $X$ is defined as

$\mathop{\mathfrak{so}}(X)=\min{\\{\,\alpha\leq\omega_{1}:[A]_{\alpha}=\overline{A}\text{
for every }A\subseteq X\,\\}}.$

Fréchet spaces are easily seen to be exactly those sequential spaces $X$ for
which $\mathop{\mathfrak{so}}(X)\leq 1$. The following definition plays a
central role in our investigation.

######  Definition 1.

A topological space $X$ is called a _$k_{\omega}$ -space_ (_$c_{\omega}$
-space_) if there exists a countable family ${\mathcal{K}}$ of (countably)
compact subspaces of $X$ such that a $U\subseteq X$ is open in $X$ if and only
if $U\cap K$ is relatively open in $K$ for every $K\in{\mathcal{K}}$.

Countable $k_{\omega}$ spaces are always sequential and the class of
$k_{\omega}$-spaces is productive. Such spaces are definable objects and have
$F_{\sigma\delta}$ topologies.

What follows is a short discussion of _test spaces_ :

  * •

Arens space ([2]): $S_{2}=[\omega]^{\leq 2}$ where $U\subseteq S_{2}$ is open
if and only if for every $s\in U$ such that $|s|<2$ the set
${{\\{\,s\cup\\{n\\}\in S_{2}:s\cup\\{n\\}\not\in U\,\\}}}$ is finite,

  * •

sequential fan ([1]): the quotient $S(\omega)=S_{2}/[\omega]^{\leq 1}$, and
finally

  * •

convergent sequence of discrete sets ([25]):
$D(\omega)=\omega\times\omega\cup\\{(\omega,\omega)\\}\subseteq(\omega+1)^{2}$
in the natural product topology.

The sequential fan $S(\omega)$ and the convergent sequence of discrete sets
$D(\omega)$ are both Fréchet spaces, $D(\omega)$ is metrizable while
$S(\omega)$ has character $\mathfrak{d}$. The space $S_{2}$ is sequential, and
$\mathop{\mathfrak{so}}(S_{2})=2$. Both $S(\omega)$ and $S_{2}$ are
$k_{\omega}$-spaces while $D(\omega)$ is not.

######  Proposition 2.

  1. (1)

(Y. Tanaka [75]) A sequential space contains a copy of $S(\omega)$ if and only
if it contains a closed copy of $S(\omega)$.

  2. (2)

(Y. Tanaka [75]) A countable sequential topological group is Fréchet if and
only if it does not contain a closed copy of $S(\omega)$.

  3. (3)

(T. Banakh and L. Zdomskyĭ [11]) If a topological group ${\mathbb{G}}$
contains closed copies of $S(\omega)$ and $D(\omega)$, it also contains a
subset $D$ such that $D$ is not closed in ${\mathbb{G}}$ and is almost
disjoint from every convergent sequence in ${\mathbb{G}}$.

Recall that the _Cantor-Bendixson_ derivative $A^{\prime}$ of a topological
space $A$ is defined by $A^{\prime}=A\setminus{\\{\,x:x\text{ is an isolated
point of $A$}\,\\}}$. The Cantor-Bendixson derivative can be
iterated—recursively define $(A)_{\alpha}$ for any ordinal $\alpha$ by putting
$(A)_{0}=A$, $(A)_{\alpha+1}=(A)_{\alpha}^{\prime}$ and
$(A)^{\lambda}=\bigcap_{\alpha<\lambda}(A)_{\alpha}$ for $\lambda$ limit.

The _full Cantor-Bendixson derivative_ (also called the _Cantor-Bendixson_ (or
_perfect_) _kernel_) of a space $A$ is $(A)_{\alpha}$ where $\alpha$ is an
ordinal such that $(A)_{\alpha}=(A)_{\beta}$ for any $\beta\geq\alpha$.

A topological space $A$ is _scattered_ if every subset of $A$ contains an
isolated (in the subset) point, equivalently if its full Cantor-Bendixson
derivative is empty. Every scattered space is thus naturally stratified into
levels, $x\in A$ belonging to the _$\alpha$ -th level_ (denoted by
$\mathop{\rm scl}(x,A)=\alpha$) if and only if $\alpha$ is the unique ordinal
such that $x\in(A)_{\alpha}\setminus(A)_{\alpha+1}$. The _height_ of $A$
($\mathop{\rm scl}(A)$) is the smallest ordinal $\alpha$ such that
$(A)_{\alpha}=\varnothing$.

Throughout the paper, $\mathop{\bf csc}(X)$ denotes the ideal generated by
closed scattered subsets of $X$, $\mathop{\bf nwd}(X)$ is the ideal of nowhere
dense subsets of $X$, and $\mathop{\bf cpt}(X)$ stands for the ideal generated
by all the compact subsets of $X$.

##  2\. The Invariant ideal Axiom

###  2.1. Introducing IIA

Analyzing the proofs of

######  Theorem 3 ([39]).

It is consistent that every countable Fréchet group is metrizable.

######  Theorem 4 ([68]).

It is consistent that every separable sequential group is either metrizable or
has sequential order $\omega_{1}$.

we have isolated the Invariant Ideal Axiom IIA which we shall present next.

First let us introduce the relevant notation. Recall that an _ideal_ is a
family $\mathcal{I}\subseteq\mathcal{P}({\mathbb{G}})$ closed under taking
subsets and finite unions, and it is _invariant_ if both $g\cdot I=\\{g\cdot
h:h\in I\\}$ and $I\cdot g=\\{h\cdot g:h\in I\\}$, as well as
$I^{-1}=\\{h^{-1}:h\in I\\}$ are in $\mathcal{I}$ for every $I\in\mathcal{I}$
and $g\in{\mathbb{G}}$. We shall assume throughout the paper that all ideals
contain all finite subsets of ${\mathbb{G}}$. Recall also that
$\mathcal{I}^{+}=\mathcal{P}({\mathbb{G}})\setminus\mathcal{I}$. Given a point
$x$ in a topological space (or a topological group) we denote by

$\mathcal{I}_{x}=\\{A\subseteq X:x\not\in\overline{A}\\}$

the dual ideal to the filter of neighborhoods of $x$. An ideal ${\mathcal{I}}$
on a set $X$ is _$\omega$ -hitting_ if for every countable family
$\mathcal{Y}$ of infinite subsets of $X$ there is an $I\in\mathcal{I}$ such
that $Y\cap I$ is infinite for every $Y\in\mathcal{Y}$.

We call an ideal $\mathcal{I}$ _tame_ if for every $Y\in\mathcal{I}^{+}$ every
$f:Y\to\omega$ and every $\omega$-hitting ideal $\mathcal{J}$ on $\omega$
there is a $J\in\mathcal{J}$ such that $f^{-1}[J]\not\in\mathcal{I}$, i.e. if
no ideal _Katětov-below_ a restriction of $\mathcal{I}$ to a positive set is
$\omega$-hitting (see e.g. [37] for more on Katětov order and $\omega$-hitting
ideals). Finally, we call an ideal $\mathcal{I}$ of subsets of a topological
group ${\mathbb{G}}$ _weakly closed_ if for every set $A\subseteq{\mathbb{G}}$
and every sequence $C\subseteq{\mathbb{G}}$ convergent to $1_{\mathbb{G}}$,

$A\in\mathcal{I}\text{ if and only if }A\cup\\{x:C\cdot
x\subseteq^{*}A\\}\in\mathcal{I}.$

It is immediate from the definition that every ideal generated by
(sequentially) closed subsets of ${\mathbb{G}}$ is weakly closed, in
particular the ideals $\mathop{\bf csc}({\mathbb{G}})$, $\mathop{\bf
nwd}({\mathbb{G}})$ and $\mathop{\bf cpt}({\mathbb{G}})$ are all invariant
weakly closed ideals in any topological group.

We call a subset $Y$ of a topological space $X$ _entangled_ if
$\mathcal{I}_{x}\restriction Y$ is $\omega$-hitting for every $x\in X$. We
shall call a topological space $X$ _groomed_ if it does not contain a dense
entangled set.

The class of groomed spaces includes all non-discrete Fréchet and sequential
spaces, as well as all non-discrete _subsequential_ spaces (i.e. subspaces of
sequential spaces).

######  Lemma 5.

Every non-discrete subsequential space is groomed.

###### Proof.

Let $X$ be a dense subspace of a sequential space $Y$. Let $D\subseteq X$ be
dense and let $x\in X$ be a point which is not isolated. As $Y$ is sequential,
there are countably many disjoint infinite subsets $\\{C_{n}:n\in\omega\\}$ of
$D$ such that

  1. (1)

Each $C_{n}$ converges to some point $x_{n}\in Y$ (not necessarily distinct),
and

  2. (2)

For every neighborhood $U$ of $x$ there are infinitely many $n\in\omega$ such
that $x_{n}\in U$.

This is easily proved by induction on the sequential order of $x$ in $D$:

If there is a sequence $C\subseteq D\setminus\\{x\\}$ convergent to $x$ let
$\\{C_{n}:n\in\omega\\}$ be any collection of infinite pairwise disjoint
subsets of $C$.

For the inductive step, assume that $x\in Y$ is the limit of a convergent
sequence $\\{x_{n}:n\in\omega\\}$ such that for each $x_{n}$ (by the inductive
hypothesis) exist pairwise disjoint sequences $\\{C^{n}_{m}:m\in\omega\\}$ of
elements of $D$ convergent each to a point $x^{n}_{m}$ such that every open
set $U$ containing $x_{n}$ contains infinitely many of the $x^{n}_{m}$. Let
$\\{D_{m}^{n}:n,m\in\omega\\}$ be a disjoint refinement of
$\\{C_{m}^{n}:n,m\in\omega\\}$. Then it is a collection of pairwise disjoint
sequences convergent in $Y$ and every neighborhood $U$ of $x$ will contain all
but finitely many of the $\\{x_{n}:n\in\omega\\}$, and, consequently,
infinitely many of the $\\{x_{m}^{n}:n,m\in\omega\\}$.

Then, however, $D$ is not entangled, as $x\in\overline{Z}$ for every $Z$ such
that $|Z\cap C_{n}|=\omega$ for every $n\in\omega$. ∎

Examples of spaces which are not groomed are discrete spaces, $\omega^{*}$ and
$2^{\mathfrak{c}}$.

We are now ready to introduce the Invariant Ideal Axiom:

IIA: For every countable groomed topological group ${\mathbb{G}}$ and every
tame, weakly closed invariant ideal $\mathcal{I}\subseteq 2^{\mathbb{G}}$ one
of the following holds:

  1. (1)

there is a countable $\mathcal{S}\subseteq\mathcal{I}$ such that for every
infinite sequence $C$ convergent in ${\mathbb{G}}$ there is an
$I\in\mathcal{S}$ such that $C\cap I$ is infinite,

  2. (2)

there is a countable $\mathcal{H}\subseteq\mathcal{I}^{+}$ such that for every
non-empty open $U\subseteq{\mathbb{G}}$ there is an $H\in\mathcal{H}$ such
that $H\setminus U\in\mathcal{I}$.

We refer to the $\mathcal{S}$ from the first alternative as a _sequence
capturing_ set, and the set $\mathcal{H}$ from the second alternative as an
_almost $\pi$-network_.

To see the relevance of the axiom let us deduce the solution to the Malykhin
problem from it. We first recall the following simple lemma from the
literature (we include the short proofs for the sake of completeness):

######  Lemma 6.

Let $X$ be a countable Fréchet space without isolated points.

  1. (1)

([12]) If $x\in X$ and $\mathcal{X}$ is a countable collection of nowhere
dense subsets of $X$ then there is a $C\subseteq X$ convergent to $x$ such
that $X\cap N$ is finite for every $N\in\mathcal{X}$.

  2. (2)

([39]) The ideal $\mathop{\bf nwd}(X)$ of nowhere dense subsets of $X$ is
tame.

###### Proof.

Ad (1): Enumerate $\mathcal{X}$ as $\langle M_{n}\colon n\in\omega\rangle$. As
$X$ is Fréchet without isolated points, there is a one-to-one sequence
$\langle x_{n}\colon n\in\omega\rangle\subseteq X\setminus\\{x\\}$ converging
to $x$. For each $n\in\omega$ the set
$X_{n}=X\setminus\left(\\{x\\}\cup\bigcup_{i<n}M_{i}\right)$ is dense in $X$,
so using the Fréchet property again, there is for each $n\in\omega$ a sequence
$\langle y_{i}^{n}\colon i\in\omega\rangle\subseteq X_{n}$ converging to
$x_{n}$. Then $x\in\overline{\\{y_{i}^{n}\colon i,n\in\omega\\}}$, hence there
is a sequence $C\subseteq\\{y_{i}^{n}\colon i,n\in\omega\\}$ converging to
$x$. Now, $C\cap M_{n}\subseteq\\{y_{i}^{m}\colon i\in\omega,\ m\leq n\\}$ and
as each sequence $\langle y_{i}^{n}\colon i\in\omega\rangle\subseteq X_{n}$
converges to $x_{n}\neq x$, $C\cap M_{n}$ is finite for every $n\in\omega$.

To see (2), let $Y\in\mathsf{nwd}^{+}(X)$, let $f:Y\to\omega$, and let an
$\omega$-hitting ideal $\mathcal{J}$ on $\omega$ be given. Put
$Z=\text{Int}(\overline{Y})\cap Y$. Then either

  1. (a)

there is an $n\in\omega$ such that $f^{-1}(n)\in\mathsf{nwd}^{+}(X)$, or

  2. (b)

for every $x\in Z$ there is a sequence $C_{x}\subseteq Z\setminus\\{x\\}$
converging to $x$ such that ${f}|_{C_{x}}$ is finite-to-one.

If there is an $n\in\omega$ such that $f^{-1}(n)\in\mathsf{nwd}^{+}(X)$, let
$J=\\{n\\}\in\mathcal{J}$. If, on the other hand,
$f^{-1}(n)\in\mathsf{nwd}(X)$ for all $n\in\omega$, apply (1) to $Z$ and every
$x\in Z$ with $\mathcal{X}=\\{f^{-1}(n)\cap Z:n\in\omega\\}$ to get
$\\{C_{x}:x\in Z\\}$ as in (b). The family$\\{f[C_{x}]:x\in Z\\}$ is then a
countable collection of infinite subsets of $\omega$, hence there is a
$J\in\mathcal{J}$ such that $J\cap f[C_{x}]$ is infinite for every $x\in Z$.
Then $f^{-1}[J]$ is dense in $Z$, hence in either case $J$ is an element of
$\mathcal{J}$ such that $f^{-1}[J]\in\mathsf{nwd}^{+}(X)$. ∎

######  Theorem 7.

Assuming IIA, every separable Fréchet group is metrizable.

###### Proof.

Let $\mathbb{H}$ be a separable Fréchet group and let
${\mathbb{G}}\subseteq\mathbb{H}$ be a dense countable subgroup. Apply IIA to
${\mathbb{G}}$ and $\mathop{\bf nwd}({\mathbb{G}})$. Alternative (1) fails by
Lemma 6 so there is a countable family $\mathcal{X}$ of somewhere dense
subsets of ${\mathbb{G}}$, hence also somewhere dense in $\mathbb{H}$ such
that every open set contains $\bmod\mathop{\bf nwd}$ an element of
$\mathcal{X}$. Then

$\\{\mathrm{int}(\overline{X}):X\in\mathcal{X}\\},$

where the interior and closure are taken in $\mathbb{H}$, form a countable
$\pi$-base, and as $\pi$-weight and weight coincide in topological groups, the
group $\mathbb{H}$ is second countable hence metrizable. ∎

######  Corollary 8.

Assuming IIA, $\mathfrak{p}=\omega_{1}$ and $\mathfrak{b}>\omega_{1}$.

###### Proof.

It is well known that if either $\mathfrak{p}>\omega_{1}$ or
$\mathfrak{b}=\omega_{1}$ then there is a separable non-metrizable Fréchet
group see e.g. [46, 47, 56]. ∎

The next remark we would like to make is that the assumption that the group is
groomed cannot be dropped from the statement of IIA, as is shown in the
proposition below.

######  Proposition 9.

There is a countable topological group ${\mathbb{G}}$ and a tame, weakly
closed, invariant ideal $\mathcal{I}\subseteq\mathcal{P}(\omega)$ such that
IIA fails for ${\mathbb{G}}$ and $\mathcal{I}$.

###### Proof.

Without loss of generality we can assume IIA as its failure provides an
example. So, in particular, we can assume that $\mathfrak{b}>\omega_{1}$.

Let $\\{A_{\alpha},B_{\alpha}:\alpha<\omega_{1}\\}\subseteq[\omega]^{\omega}$
be a _Hausdorff gap_ , i.e.

  1. (1)

$A_{\alpha}\subseteq^{*}A_{\beta}\subseteq^{*}B_{\beta}\subseteq^{*}B_{\alpha}$
for $\alpha<\beta<\omega_{1}$, and

  2. (2)

there is no $X$ such that $A_{\alpha}\subseteq^{*}X\subseteq^{*}B_{\alpha}$
for every $\alpha<\omega_{1}$.

Topologize the group ${\mathbb{G}}=[\omega]^{<\omega}$ by declaring the sets
(in fact, subgroups) $[F]^{<\omega}$ open neighborhoods of $\varnothing$,
where $F$ is such that there is an $\alpha<\omega_{1}$ with
$B_{\alpha}\subseteq^{*}F$, and let

$\mathcal{I}=\\{A\subseteq{\mathbb{G}}:\ \forall\alpha<\omega_{1}\ \bigcup
A\subseteq^{*}B_{\alpha}\\}.$

Now, the fact that $\mathcal{I}$ is tame follows easily by noting that no
restriction of $\mathcal{I}$ to a positive set is tall, and $\mathcal{I}$ is
inviariant as $\bigcup a\triangle I=^{*}\bigcup I$ for every $I\in\mathcal{I}$
and $a\in{\mathbb{G}}$. The fact that $\mathcal{I}$ is weakly closed follows
immediately from the fact that every set of the form $[C]^{<\omega}$ is closed
in the topology of ${\mathbb{G}}$, and every set in $\mathcal{I}$ is contained
in en element of $\mathcal{I}$ of this form ($C\in\mathcal{I}$ if and unly if
$[\bigcup C]^{<\omega}\in\mathcal{I}$).

To see that the alternative (1) of IIA fails for $\mathcal{I}$ note first that
$C\to 0$ if and only if $C$ is a point-finite family of finite sets and
$C\in\mathcal{I}$. The fact that there cannot be a countable family of
elements of $\mathcal{I}$ intersecting infinitely every convergent sequence
follows directly from the fact that we started with a Hausdorff gap (hence
there cannot be a single such element of $\mathcal{I}$) and the fact that
$\mathfrak{b}>\omega_{1}$, hence $\\{B_{\alpha}:\alpha<\omega_{1}\\}$ cannot
form the upper half of an $(\omega,\omega_{1})$-gap by the Theorem of
Rothberger.

Alternative (2) fails as $X\in\mathcal{I}^{+}$ iff $\bigcup X\setminus
B_{\alpha}$ is infinite for some $\alpha<\omega_{1}$, and having countably
many such $X$, there is an $\alpha$ which is a witness for all of them, hence
none of them is mod $\mathcal{I}$ contained in $[B_{\alpha}]^{<\omega}$. ∎

We shall return to further discussion of consequences of IIA later on, but
first we shall see that the axiom is consistent.

###  2.2. Consistency of IIA

All the tools to prove consistency of the _Invariant Ideal Axiom_ have been
presented in [39] and [19], though the models constructed there are not models
of IIA. We shall first recall all the relevant lemmata from the above
mentioned papers. Those we can quote directly we do not prove. Those that
require some (very) minor changes we do prove, though in all cases the changes
are mere technicalities.

Recall first that the Laver-Mathias-Prikry forcing
${\mathbb{L}}_{\mathcal{F}}$ associated to a filter $\mathcal{F}$ on $\omega$,
is defined as the set of those trees $T\subseteq\omega^{<\omega}$ with stem
$s_{T}$ such that for all $s\in T$ extending $s_{T}$ the set
$\mbox{succ}_{T}(s)=\\{n\in\omega\colon s^{\frown}n\in T\\}$ belongs to
$\mathcal{F}$. The set ${\mathbb{L}}_{\mathcal{F}}$ is ordered by inclusion.

The forcing ${\mathbb{L}}_{\mathcal{F}}$ is $\sigma$-centered, and adds
generically a dominating real $\dot{\ell}_{\mathcal{F}}\colon\omega\to\omega$
(The function $\dot{\ell}_{\mathcal{F}}$ is the unique branch through
$\omega^{<\omega}$ that belongs to all trees in the generic filter, and it
eventually dominates all ground model reals). Its range
$\dot{A}_{gen}=\text{ran}(\dot{\ell}_{\mathcal{F}})$ _separates_ the filter
$\mathcal{F}$ (that is, the set $\dot{A}_{gen}$ is almost contained in all
members of $\mathcal{F}$ and has infinite intersection with every
$\mathcal{F}$-positive set).

Names for reals in forcings of the type ${\mathbb{L}}_{\mathcal{F}}$ can be
analyzed using _ranks_ as introduced by Baumgartner and Dordal in [13], and
further developed by Brendle [17, 18]. Given a formula $\varphi$ in the
forcing language and $s\in\omega^{<\omega}$, we say that $s$ favors $\varphi$
if there is no condition $T\in{\mathbb{L}}_{\mathcal{F}}$ with stem $s$ such
that $T\Vdash``\neg\varphi"$, or equivalently, every condition
$T\in{\mathbb{L}}_{\mathcal{F}}$ with stem $s$ has an extension $T^{\prime}$
such that $T^{\prime}\Vdash``\varphi"$.

Recall also, that a forcing notion $\mathbb{P}$ strongly preserves
$\omega$-hitting if for every sequence $\langle\dot{A}_{n}\colon
n\in\omega\rangle$ of $\mathbb{P}$-names for infinite subsets of $\omega$
there is a sequence $\langle B_{n}\colon n\in\omega\rangle$ of infinite
subsets of $\omega$ such that for any $B\in[\omega]^{\omega}$, if $B\cap
B_{n}$ is infinite for all $n$ then $\Vdash_{\mathbb{P}}``B\cap\dot{A}_{n}$ is
infinite for all $n"$.

In our terminology, one of the lemmas of [19] becomes:

######  Lemma 10 ([19]).

Let $\mathcal{I}$ be an ideal on $\omega$ and let
$\mathcal{F}=\mathcal{I}^{*}$ be the dual filter. Then the following are
equivalent:

  1. (1)

$\mathcal{I}$ is tame,

  2. (2)

${\mathbb{L}}_{\mathcal{F}}$ strongly preserves $\omega$-hitting.

  3. (3)

${\mathbb{L}}_{\mathcal{F}}$ preserves $\omega$-hitting.

and the standard preservation under finite support iteration argument gives

######  Lemma 11 ([19]).

Finite support iteration of ccc forcings strongly preserving $\omega$-hitting
strongly preserves $\omega$-hitting.

Recall also, that given an ideal $\mathcal{I}$, a forcing notion $\mathbb{P}$,
and a $\mathbb{P}$-name $\dot{A}$ for a subset of $\omega$ we say that
$\mathbb{P}$ seals the ideal $\mathcal{I}$ via $\dot{A}$ if
$\Vdash_{\mathbb{P}}``\dot{A}\in\mathcal{I}^{+}\wedge\,\mathcal{I}\restriction\dot{A}$
is $\omega$-hitting”.

Following [39] we say that an ideal $\mathcal{I}$ is $\omega$-hitting mod
filter $\mathcal{F}$ if $\mathcal{I}\cap\mathcal{F}=\varnothing$ and for every
countable family $\mathcal{H}\subset\mathcal{F}^{+}$ there is an
$I\in\mathcal{I}$ such that $H\cap I\in\mathcal{F}^{+}$ for all
$H\in\mathcal{H}$.

######  Lemma 12 ([39]).

The forcing ${\mathbb{L}}_{\mathcal{F}}$ seals an ideal $\mathcal{I}$ via
$\dot{A}_{gen}$ if and only if $\mathcal{I}$ is $\omega$-hitting mod
$\mathcal{F}$.

The following corollary is the main tool for the consistency of IIA.

######  Corollary 13 ([39]).

Let ${\mathbb{G}}$ be a countable topological group and
$\mathcal{I}\subseteq\mathcal{P}(\omega)$ be a tame invariant ideal such that
both alternatives of IIA fail. Then:

  1. (1)

${\mathbb{L}}_{\mathcal{I}^{*}}$ forces ${\mathbb{G}}$ is not groomed, and

  2. (2)

${\mathbb{L}}_{\mathcal{I}^{*}}$ strongly preserves $\omega$-hitting.

###### Proof.

To see (1), one essentially only needs to translate from one language to
another; to force that ${\mathbb{G}}$ is not groomed, i.e. contains a dense
entangled set, means (in the language of sealing) adding a dense subset
$\dot{A}$ of ${\mathbb{G}}$ such that ${\mathbb{L}}_{\mathcal{I}^{*}}$ seals
the ideal $\mathcal{I}_{g}$for every $g\in{\mathbb{G}}$. Now, by Lemma 12, it
is enough to show that the ideal $\mathcal{I}_{g}$ is $\omega$-hitting mod
$\mathcal{I}^{*}$, and hence, ${\mathbb{L}}_{\mathcal{I}^{*}}$ seals the ideal
$\mathcal{I}_{g}$ via $\dot{A}_{gen}$ for every $g\in{\mathbb{G}}$, which is
simply the negation of the existence of a countable almost $\pi$-network of
${\mathcal{I}}$-positive sets (alternative (2) of IIA). The failure of
capturing of convergent sequences (alternative (1) of IIA) guarantees that
every element of $\mathcal{I}^{*}$ is dense in ${\mathbb{G}}$, hence also
${\mathbb{L}}_{\mathcal{I}^{*}}$ forces $\dot{A}_{gen}$ to be dense in
${\mathbb{G}}$.

(2) follows directly from Lemma 10. ∎

The last part of the forcing argument which deals with the preservation of
dense entangled sets, really uses algebra.

Let $({\mathbb{G}},\cdot)$ be an abstract group and let
$A\subseteq{\mathbb{G}}\setminus\\{{1_{{\mathbb{G}}}}\\}$. A subset $Y$ of
${\mathbb{G}}$ is called $A$-large if for every $a\in A$ and
$b\in{\mathbb{G}}$, either $b\in Y$ or $a\cdot b^{-1}\in Y$. By
$A$-$\mathsf{large}$ we will denote the collection of all subsets of
${\mathbb{G}}$ which are $A$-large. A family $\mathcal{C}$ of subsets of
${\mathbb{G}}$ is $\omega$-hitting w.r.t. $A$ if given $\langle Y_{n}\colon
n\in\omega\rangle\subset A$-$\mathsf{large}$ there is a $C\in\mathcal{C}$ such
that $C\cap Y_{n}$ is infinite for all $n$. Finally, We say that a relation
$R\subseteq{\mathbb{G}}\times{\mathbb{G}}$ is large if for every
$a,b\in{\mathbb{G}}$, either $\langle a,b\rangle\in R$ or $\langle a,a\cdot
b^{-1}\rangle\in R$.

######  Lemma 14 ([39]).

Let ${\mathbb{G}}$ be a countable group and let $\mathcal{I}$ be a weakly
closed invariant ideal on ${\mathbb{G}}$ for which (1) of the IIA fails, and
let $\langle R_{n}\colon n\in\omega\rangle$ be a sequence of large relations.
Then there is a sequence $C$ convergent to $1_{\mathbb{G}}$ such that
$R_{n}^{-1}[C\setminus F]\in\mathcal{I}^{+}$ for every $n\in\omega$ and
$F\in[{\mathbb{G}}]^{<\omega}$.

###### Proof.

For every $n\in\omega$ let $B_{n}=\\{b\in{\mathbb{G}}\colon
R_{n}^{-1}(b)\in\mathcal{I}\\}$ and put

$\mathcal{S}=\\{R_{n}^{-1}(b)\cdot b^{-1}\colon b\in
B_{n},n\in\omega\\}\cup\\{B_{n}\colon B_{n}\in\mathcal{I}\\}.$

As (1) of IIA fails, there is a sequence $C$ converging to $1_{\mathbb{G}}$
such that $C\cap S$ is finite for every $S\in\mathcal{S}$. We claim that
$R_{n}^{-1}[C\setminus F]\in\mathcal{I}^{+}$ for every $n\in\omega$ and
$F\in[{\mathbb{G}}]^{<\omega}$. To see this, let $n\in\omega$ and
$F\in[{\mathbb{G}}]^{<\omega}$ be given. Consider two cases.

Case 1. $B_{n}\in\mathcal{I}$.

Then there is an $b\in C\setminus F$ such that
$R_{n}^{-1}(b)\in\mathcal{I}^{+}$.

Case 2. $B_{n}\in\mathcal{I}^{+}$.

Fix $b\in B_{n}$. Then $a\notin R_{n}^{-1}(b)\cdot b^{-1}$ (or, equivalently,
$a\cdot b\notin R_{n}^{-1}(b)$) for all but finitely many $a\in C$. Since
$R_{n}$ is a large relation, $\langle a\cdot b,a\rangle\in R_{n}$ for all but
finitely many $a\in C$. In particular, $\\{a\cdot b\colon a\in
C\\}\subseteq^{*}R_{n}^{-1}[C\setminus F]$ and $\\{a\cdot b\colon a\in C\\}$
converges to $b$. Thus,

$B_{n}\subseteq R_{n}^{-1}[C\setminus F]\cup\\{b\in{\mathbb{G}}:C\cdot
b\subseteq^{*}R_{n}^{-1}[C\setminus F]\\},$

hence, as $\mathcal{I}$ is weakly closed, also $R_{n}^{-1}[C\setminus
F]\in\mathcal{I}^{+}$. ∎

######  Lemma 15 ([39]).

Let ${\mathbb{G}}$ be a countable topological group and $\mathcal{I}$ an
invariant ideal on ${\mathbb{G}}$ for which (1) of the IIA fails. Then

$\Vdash_{{\mathbb{L}}_{\mathcal{I}^{*}}}``\mathcal{C}\text{ is
}\omega\text{-hitting w.r.t. }\dot{A}_{gen}\text{'',}$

where $\mathcal{C}=\mathcal{I}_{1_{{\mathbb{G}}}}^{\perp}$ is the ideal
consisting of sequences converging to $1_{{\mathbb{G}}}$.

###### Proof.

Aiming for a contradiction, assume that there are a sequence
$\langle\dot{B}_{n}\colon n\in\omega\rangle$ of
${\mathbb{L}}_{\mathcal{I}^{*}}$-names and a condition
$T^{*}\in{\mathbb{L}}_{\mathcal{I}^{*}}$ such that $T^{*}\Vdash``\forall
n\in\omega\,(\dot{B}_{n}\in\dot{A}_{gen}$-$\mathsf{large}$)”, and for every
$C\in\mathcal{C}$ there are a condition
$T_{C}\in{\mathbb{L}}_{\mathcal{I}^{*}}$ with $T_{C}\leqslant T^{*}$, a
natural number $n_{C}$, and a finite subset $F_{C}$ of ${\mathbb{G}}$ such
that

$T_{C}\Vdash``C\cap\dot{B}_{n_{C}}\subseteq F_{C}".\qquad(\star)$

For each $s\in T^{*}$ with $s\supseteq s_{T^{*}}$ and each natural number $n$,
put

$R_{s,n}=\\{\langle a,b\rangle\colon a\in\text{succ}_{T^{*}}(s)\Rightarrow
s^{\frown}a\text{ favors }b\in\dot{B}_{n}\\}.$

Claim. The relation $R_{s,n}$ is large.

Let $a$ and $b$ be two elements of ${\mathbb{G}}$. Assume that $\langle
a,b\rangle\notin R_{s,n}$. Assuming $a\in\text{succ}_{T^{*}}(s)$ we have to
show that $\langle a,a\cdot b^{-1}\rangle\in R_{s,n}$. There is a condition
$T^{\prime}\leq T^{*}$ with $s_{T^{\prime}}=s^{\frown}a$ such that
$T^{\prime}\Vdash``b\notin\dot{B}_{n}"$. Then
$T^{\prime}\Vdash``a\in\dot{A}_{gen}$ and $b\notin\dot{B}_{n}$”, but also
$T^{\prime}\Vdash``\dot{B}_{n}\in\dot{A}_{gen}$-$\mathsf{large}$” so
$T^{\prime}\Vdash``a\cdot b^{-1}\in\dot{B}_{n}"$. This finishes the proof of
the claim.

By Lemma 14, there is a $C\in\mathcal{C}$ such that $R_{s,n}^{-1}[C\setminus
F]\in\mathcal{I}^{+}$ for every $s\in T^{*}$ with $s\supseteq s_{T^{*}}$,
$n\in\omega$ and $F\in[{\mathbb{G}}]^{<\omega}$. In particular,
$R_{s_{C},n_{C}}^{-1}[C\setminus F_{C}]\in\mathcal{I}^{+}$, where
$s_{C}=s_{T_{C}}$. Pick an $a\in\text{succ}_{T_{C}}(s_{C})\cap
R_{s_{C},n_{C}}^{-1}[C\setminus F_{C}]$. Then, there is a $b\in C\setminus
F_{C}$ such that $s_{C}^{\frown}a$ favors $b\in\dot{B}_{n_{C}}$, and hence
there is a condition $T\leqslant T_{C}$ whose stem extends $s_{C}^{\frown}a$
such that $T\Vdash``b\in\dot{B}_{n_{C}}"$, a contradiction to the initial
assumption $(\star)$. ∎

We say that a forcing notion $\mathbb{P}$ strongly preserves $\omega$-hitting
w.r.t. $A$ if for every $\mathbb{P}$-name $\dot{Y}$ for an $A$-large subset of
a group ${\mathbb{G}}$ there is a sequence $\langle Y_{n}\colon
n\in\omega\rangle\subset A$-$\mathsf{large}$ such that for any
$C\subseteq{\mathbb{G}}$, if $C\cap Y_{n}$ is infinite for all $n$ then
$\Vdash_{\mathbb{P}}``C\cap\dot{Y}$ is infinite”. Clearly, every forcing
notion that strongly preserves $\omega$-hitting w.r.t. $A$ preserves
$\omega$-hitting w.r.t. $A$.

######  Lemma 16 ([39]).

Let $\mathbb{P}$ be a $\sigma$-centered forcing notion. Then $\mathbb{P}$
strongly preserves $\omega$-hitting w.r.t. $A$.

######  Lemma 17 ([39]).

Finite support iteration of ccc forcings strongly preserving $\omega$-hitting
w.r.t. $A$ strongly preserves $\omega$-hitting w.r.t. $A$.

Now we are in position to state and prove the main theorem of this section:

######  Theorem 18.

The Invariant Ideal Axiom IIA together with the Martin’s Axiom
$\mathsf{MA}(\sigma$-centered strongly $\omega$-hitting preserving$)$ is
consistent with ZFC.

###### Proof.

Assume that the ground model $\mathbf{V}$ satisfies CH, split the set
$S_{1}^{2}$—the stationary subset of $\omega_{2}$ consisting of ordinals of
cofinality $\omega_{1}$—into two disjoint stationary sets $S_{0}$ and $S_{1}$,
and suppose $\langle A_{\alpha}\colon\alpha\in S_{1}^{2}\rangle$ witnesses
that both $\diamondsuit(S_{0})$, and $\diamondsuit(S_{1})$ hold111Given a
stationary subset $S$ of $\omega_{2}$, the principle $\diamondsuit(S)$ asserts
the existence of a sequence $\langle A_{\alpha}:\alpha\in S\rangle$ of subsets
of $\omega_{2}$ such that for any $A\subset\omega_{2}$ the set $\\{\alpha\in
S:A_{\alpha}=A\cap\alpha\\}$ is stationary..

Construct a finite support iteration
$\mathbb{P}_{\omega_{2}}=\langle\mathbb{P}_{\alpha},\dot{{\mathbb{Q}}}_{\alpha}:\alpha<\omega_{2}\rangle$
so that at a stage $\alpha\in S_{0}$, if $A_{\alpha}$ codes a
$\mathbb{P}_{\alpha}$-name for a $\sigma$-centered forcing
$\hat{{\mathbb{Q}}}$ which strongly preserves $\omega$-hitting families, then
let ${\mathbb{Q}}_{\alpha}=\hat{{\mathbb{Q}}}$, otherwise let
$\dot{\mathbb{Q}}_{\alpha}$ be a $\mathbb{P}_{\alpha}$-name for
${\mathbb{L}}_{\mathop{\bf nwd}^{*}({\mathbb{Q}})}$ where ${\mathbb{Q}}$ are
the rational numbers; at a stage $\alpha\in S_{1}$, if $A_{\alpha}$ codes a
group operation $\circ$ on $\omega$, a $\mathbb{P}_{\alpha}$-name for a
regular group topology $\tau$ with no isolated points on $(\omega,\circ)$ and
a $\circ$-invariant tame ideal such that neither (1) nor (2) of the IIA hold,
we let $\dot{\mathbb{Q}}_{\alpha}$ be a $\mathbb{P}_{\alpha}$-name for
${\mathbb{L}}_{\mathcal{I}^{*}}$. If $\alpha$ is not of this form, let
$\dot{\mathbb{Q}}_{\alpha}$ be again $\mathbb{P}_{\alpha}$-name for
${\mathbb{L}}_{\mathop{\bf nwd}^{*}({\mathbb{Q}})}$ where ${\mathbb{Q}}$ are
the rational numbers. Let $G_{\omega_{2}}$ be a
$\mathbb{P}_{\omega_{2}}$-generic over $\mathbf{V}$. A standard argument shows
that MA for $\sigma$-centered partial orders strongly preserving
$\omega$-hitting families holds in ${\bf V}[G_{\omega_{2}}]$.

We shall show that, in ${\bf V}[G_{\omega_{2}}]$, IIA holds.

Aiming toward a contradiction, assume that in $\mathbf{V}[G_{\omega_{2}}]$
there is a countable groomed group ${\mathbb{G}}$ with a group topology $\tau$
and a tame invariant ideal $\mathcal{I}$ on ${\mathbb{G}}$ satisfying neither
(1) nor (2) of IIA.

Now, by a standard closing off argument, there is a set $E\subset S_{1}^{2}$
which is a club relative to $S_{1}^{2}$ such that for all $\alpha\in E$,

  1. (1)

${\bf V}[G_{\alpha}]\models\tau_{\alpha}$ is groomed where
$\tau_{\alpha}=\tau\cap{\bf V}[G_{\alpha}]$,

  2. (2)

${\bf V}[G_{\alpha}]\models\mathcal{I}_{\alpha}=\mathcal{I}\cap{\bf
V}[G_{\alpha}]$ is a ${\mathbb{G}}$-invariant tame ideal satisfying neither
(1) nor (2) of IIA, and

  3. (3)

every sequence in ${\bf V}[G_{\alpha}]$ which is $\tau_{\alpha}$-convergent is
forced to be $\tau$-convergent in ${\bf V}[G_{\omega_{2}}]$.

Therefore, at some stage $\alpha\in S_{1}$, we would have added a set
$A_{gen}$ such that ${\bf V}[G_{\alpha+1}]\models A_{gen}$ is a dense
entangled subset of ${\mathbb{G}}$, i.e. the ideal
${\mathcal{I}_{g}(\tau_{\alpha})}|_{A_{gen}}$ is $\omega$-hitting for every
$g\in{\mathbb{G}}$ (Proposition 13 (1)).

We claim that $A_{gen}$ is also dense in ${\bf V}[G_{\omega_{2}}]$, i.e.
$A_{gen}\in\mathcal{I}_{g}^{+}(\tau)$ for every $g\in{\mathbb{G}}$. As
${\mathbb{G}}$ is a group it suffices to show this at ${0_{{\mathbb{G}}}}$. If
it were not true, in ${\bf V}[G_{\omega_{2}}]$, there is a $\tau$-open
neighborhood $U$ of $0$ disjoint from $A_{gen}$ such that $U\cdot U\cap
A_{gen}=\varnothing$. Then $Y={\mathbb{G}}\setminus U$ is $A_{gen}$-large. By
Lemma 15, in ${\bf V}[G_{\alpha+1}]$ the ideal
$\mathcal{I}_{0}^{\perp}(\tau_{\alpha})=\mathcal{I}^{\perp}_{0}(\tau)\cap\mathbf{V}[G_{\alpha}]$
is $\omega$-hitting w.r.t. $A_{gen}$, and by Lemmata 16 and 17, it follows
that the ideal $\mathcal{I}_{0}^{\perp}(\tau_{\alpha})$ is also
$\omega$-hitting w.r.t. $A_{gen}$ in ${\bf V}[G_{\omega_{2}}]$. In particular,
there is a $C\in\mathcal{I}_{0}^{\perp}(\tau_{\alpha})$ such that $C\cap Y$ is
infinite, _i.e._ , $C\in{\bf V}[G_{\alpha}]$ is a sequence
$\tau_{\alpha}$-converging to $0$, which intersects $Y$ infinitely often.
However, by the item 2,
$\mathcal{I}_{0}^{\perp}(\tau_{\alpha})\subset\mathcal{I}_{0}^{\perp}(\tau)$,
therefore $C$ is also $\tau$-converging to $0$. This however leads to a
contradiction as both $Y\cap U=\varnothing$ and $C\setminus U$ is finite.

By Proposition 13 (2) and Lemma 11, in ${\bf V}[G_{\omega_{2}}]$ the ideal
${\mathcal{I}_{g}(\tau_{\alpha})}|_{A_{gen}}$ is $\omega$-hitting for every
$g\in{\mathbb{G}}$ contradicting that ${\mathbb{G}}$ was groomed in ${\bf
V}[G_{\omega_{2}}]$. ∎

##  3\. Countable sequential groups under IIA

In this section we prove the main result of the paper, which confirms a
conjecture of the second author [68] by proving a common extension of Theorems
3 and 4 that provides the ultimate (consistent) classification for the
topologies of countable sequential topological groups, namely:

######  Theorem 19.

Assuming IIA, every countable sequential group is either metrizable or
$k_{\omega}$.

Countable $k_{\omega}$ groups are completely classified by their compact
scatteredness rank defined as the supremum of the Cantor-Bendixson index of
their compact subspaces by the theorem of Zelenyuk:

######  Theorem 20 (E. Zelenyuk [84]).

Countable $k_{\omega}$ groups of the same compact scatteredness rank are
homeomorphic.

To give a more concrete feel for how strong Theorem 19 actually is, let us
introduce the following notation: Given an indecomposable ordinal222An ordinal
number $\alpha$ is _indecomposable_ if it cannot be written as an ordinal sum
of two strictly smaller ordinals, equivalently, there is a $\beta\leq\alpha$
such that $\alpha=\omega^{\beta}$, where $\omega^{\beta}$ denotes ordinal
exponentiation. $\alpha<\omega_{1}$ let $\mathcal{K}_{\alpha}$ be a fixed
countable family of compact subsets of the rationals ${\mathbb{Q}}$ closed
under translations, inverse and algebraic sums such that
$\alpha=\sup\\{\mathop{\rm rank}_{CB}(K):K\in\mathcal{K}_{\alpha}\\}$, and let

$\tau_{\alpha}=\\{U\subseteq{\mathbb{Q}}:\ \forall
K\in\mathcal{K}_{\alpha}:U\cap K\text{ is open in }K\\}.$

Then $\tau_{\alpha}$ is a $k_{\omega}$ sequential group topology on
${\mathbb{Q}}$ and we denote
${\mathbb{Q}}_{\alpha}=(\mathbb{Q},\tau_{\alpha})$. Note, in particular, that
if $\alpha=0$ then $\tau_{0}$ is the discrete topology on ${\mathbb{Q}}$, and
that the usual topology on ${\mathbb{Q}}$ is similarly determined by taking
into account _all_ of its compact subsets, so it makes sense to denote it as
${\mathbb{Q}}_{\omega_{1}}$. Hence Theorem 19 can be reformulated as:

######  Theorem 21.

Assuming IIA, for every infinite countable sequential group ${\mathbb{G}}$
there is exactly one $\alpha\leq\omega_{1}$ such that ${\mathbb{G}}$ is
homeomorphic to ${\mathbb{Q}}_{\omega^{\alpha}}$.

Note that in the argument above we may have started with an arbitrary
countable _topologizable_ (i.e. admitting a nondiscrete Hausdorff group
topology) group ${\mathbb{G}}$ instead of ${\mathbb{Q}}$ by possibly choosing
a coarser first countable topology on ${\mathbb{G}}$ first. Thus _every_
countable topologizable group admits every possible $k_{\omega}$ group
topology showing that in a model of IIA the algebraic structure of the group
has almost no influence on the kind of sequential topology the group can
admit. Indeed, in such models the number of nonisomorphic topologizable
countable groups ($\mathfrak{c}$) is greater than the number of
nonhomeomorphic sequential group topologies ($\omega_{1}$).

We shall prove Theorem 19 in a sequence of lemmata.

######  Lemma 22.

Let ${\mathbb{G}}$ be a countable nondiscrete sequential group. Suppose
${\mathcal{P}}\subseteq\mathop{\bf csc}({\mathbb{G}})$ is a countable family
such that for every $S\to{1_{{\mathbb{G}}}}$ there exists a
$P\in{\mathcal{P}}$ such that $|S\cap P|=\omega$. Let
${\mathcal{D}}\subseteq[{\mathbb{G}}]^{\omega}$ be a countable family of
closed discrete subsets of ${\mathbb{G}}$. Then for every $g\in{\mathbb{G}}$
there exists an open $U\ni g$ such that $U\cap D$ is finite for every
$D\in{\mathcal{D}}$.

###### Proof.

Let ${\mathcal{D}}={\\{\,D_{n}:n\in\omega\,\\}}\subseteq 2^{\mathbb{G}}$ be a
collection of closed discrete subsets of ${\mathbb{G}}$. For brevity, call a
point $g\in{\mathbb{G}}$ a vD-point of ${\mathcal{D}}$ if for every open $U\ni
g$ there is a $D\in{\mathcal{D}}$ with the property $|U\cap D|=\omega$. The
statement of the lemma is equivalent to claiming that there are no vD-points
of ${\mathcal{D}}$. Suppose $g\in{\mathbb{G}}$ is a vD-point of
${\mathcal{D}}$. By translating each $D\in{\mathcal{D}}$ if necessary, we may
assume that $g={1_{{\mathbb{G}}}}$.

Let ${\mathcal{P}}={\\{\,P_{n}:n\in\omega\,\\}}$ be a collection of closed
scattered subsets of ${\mathbb{G}}$ such that for any $S\to{1_{{\mathbb{G}}}}$
there is a $P\in{\mathcal{P}}$ such that $|S\cap P|=\omega$. By requiring
${\mathcal{P}}$ to be closed under finite unions we may assume that
$S\subseteq^{*}P$.

Pick a family ${\\{\,O_{n}:n\in\omega\,\\}}$ of open neighborhoods of
${1_{{\mathbb{G}}}}$ such that $\overline{O_{n+1}}\subseteq O_{n}$ and
$\bigcap_{n\in\omega}O_{n}=\\{{1_{{\mathbb{G}}}}\\}$. Put
$P=\bigcup_{n\in\omega}P_{n}\cap\overline{O_{n}}$. One may verify that $P$ is
closed, scattered, and for any $S\to{1_{{\mathbb{G}}}}$, $S\subseteq^{*}P$.
Changing $P$ if necessary, we may further require that $\alpha_{P}=\mathop{\rm
scl}({1_{{\mathbb{G}}}},P)$ is the smallest one among all $P$ with such
properties.

Let $P^{\prime}={\\{\,p\in P\setminus\\{{1_{{\mathbb{G}}}}\\}:\mathop{\rm
scl}(p,P)\geq\alpha_{P}\,\\}}$. Note that
${1_{{\mathbb{G}}}}\not\in\overline{P^{\prime}}$ and we may therefore assume
(by taking an appropriate subset of $P$, if necessary) that $\mathop{\rm
scl}({1_{{\mathbb{G}}}},P)=\mathop{\rm scl}(P)>\mathop{\rm scl}(p,P)$ for any
$p\in P$ such that $p\not={1_{{\mathbb{G}}}}$. Now

  1. (1)

$P$ is a closed scattered subset of ${\mathbb{G}}$ such that $S\subseteq^{*}P$
for every $S\to{1_{{\mathbb{G}}}}$; moreover, $\alpha_{P}=\mathop{\rm
scl}(P)=\mathop{\rm scl}({1_{{\mathbb{G}}}},P)>\mathop{\rm scl}(p,P)$ for any
$p\in P\setminus\\{{1_{{\mathbb{G}}}}\\}$, and $\alpha_{P}$ is the smallest
possible.

Let $g\in{\mathbb{G}}$ and consider the family
${\mathcal{D}}_{g}={\\{\,D_{n}\cap(g\cdot P):n\in\omega\,\\}}$. Suppose $g$ is
not a vD-point of ${\mathcal{D}}_{g}$ for any $g\in{\mathbb{G}}$. For each
$g_{i}\in{\mathbb{G}}$ pick an open neighborhood $U_{i}\ni g_{i}$ such that
$U_{i}\cap(g_{i}\cdot P)\cap D_{n}=F_{i}^{n}$ is finite for every
$n\in\omega$. Put $D_{n}^{\prime}=D_{n}\setminus\bigcup_{i\leq n}F_{i}^{n}$.
Then ${1_{{\mathbb{G}}}}$ is a vD-point of
${\mathcal{D}}^{\prime}={\\{\,D_{n}^{\prime}:n\in\omega\,\\}}$, therefore,
${1_{{\mathbb{G}}}}\in\overline{\bigcup{\mathcal{D}}^{\prime}}$.

Let $S\subseteq\bigcup{\mathcal{D}}^{\prime}$ be an infinite sequence such
that $S\to g_{i}$ for some $g_{i}\in{\mathbb{G}}$. Then, by (1)
$S\subseteq^{*}U_{i}\cap(g_{i}\cdot P)$. Since every
$D\in{\mathcal{D}}^{\prime}$ is closed discrete, we may assume that $S=\langle
s_{n}:n\in\omega\rangle$ is such that $s_{n}\in D_{m(n)}^{\prime}$ for some
$m(n)\geq n$. Let $n>i$ be such that $s_{n}\in U_{i}\cap(g_{i}\cdot P)$. Then
$s_{n}\in F_{i}^{m(n)}$, $i<n<m(n)$ contradicting $s_{n}\in
D_{m(n)}^{\prime}=D_{m(n)}\setminus\bigcup_{i\leq m(n)}F_{i}^{m(n)}$. Thus no
such $S$ exists making $\bigcup{\mathcal{D}}^{\prime}$ almost disjoint from
every convergent sequence in ${\mathbb{G}}$ contradicting
${1_{{\mathbb{G}}}}\in\overline{\bigcup{\mathcal{D}}^{\prime}}$ and the
sequentiality of ${\mathbb{G}}$.

We may therefore assume that $D\subseteq P$ for every $D\in{\mathcal{D}}$ and
some $g\in{\mathbb{G}}$ is a vD-point of ${\mathcal{D}}$. Note that $g\in P$,
since $P$ is closed. Let $p\in P$ be a vD-point of ${\mathcal{D}}$ such that
$\mathop{\rm scl}(p,P)$ is the smallest. By picking a neighborhood $U\ni p$
relatively open in $P$ such that $\mathop{\rm scl}(x,P)<\mathop{\rm scl}(p,P)$
for any $x\in\overline{U}\setminus\\{p\\}$ and restricting ${\mathcal{D}}$ to
$U$, if necessary, we may assume that $p$ is the only vD-point of
${\mathcal{D}}$. Using an argument similar to the one in the previous
paragraph, by possibly removing a finite subset from each $D\in{\mathcal{D}}$
we may assume that ${\mathbb{D}}=\bigcup{\mathcal{D}}\cup\\{p\\}$ is closed,
$p$ is the only nonisolated point of $\mathbb{D}$, and $p$ is a vD-point of
${\mathcal{D}}$.

Consider the translation ${\mathcal{D}}^{\prime\prime}={\\{\,D\cdot
p^{-1}:D\in{\mathcal{D}}\,\\}}$ of ${\mathcal{D}}$. Suppose
${1_{{\mathbb{G}}}}\in\overline{\bigcup{\mathcal{D}}^{\prime\prime}\setminus
P}$. By the closedness of $\mathbb{D}$ and property (1) the set
$\bigcup{\mathcal{D}}^{\prime\prime}\setminus P$ contains no infinite
converging sequence, contradicting the sequentiality of ${\mathbb{G}}$.

  1. (1)

There exists a countable family $\mathcal{D}$ of closed discrete subsets of
${\mathbb{G}}$ such that $\bigcup{\mathcal{D}}\subseteq P$,
${1_{{\mathbb{G}}}}$ is the only nonisolated point of
$\bigcup{\mathcal{D}}\cup\\{{1_{{\mathbb{G}}}}\\}$, which is closed in
${\mathbb{G}}$, and ${1_{{\mathbb{G}}}}$ is a vD-point of ${\mathcal{D}}$.

Suppose there exists an $S\to{1_{{\mathbb{G}}}}$ such that $(S\cdot
p)\setminus P$ is infinite for every $p\in
P\setminus\\{{1_{{\mathbb{G}}}}\\}$. We may then pick an infinite sequence
$S^{\prime}\subseteq S\cdot S$ so that $S^{\prime}\to{1_{{\mathbb{G}}}}$ and
$S^{\prime}\subseteq{\mathbb{G}}\setminus P$, contradicting (1). Thus for
every sequence $S\to{1_{{\mathbb{G}}}}$ there exists a $p\in
P\setminus\\{{1_{{\mathbb{G}}}}\\}$ such that $(S\cdot p)\subseteq^{*}P$.

Suppose $A\subseteq{\mathbb{G}}$ is such that for some ordinal $\beta$,
$\mathop{\rm scl}(a,P)\leq\beta<\alpha_{P}$ for every $a\in A\cap P$. Then
there exists a sequence $S\to{1_{{\mathbb{G}}}}$ such that
$S\setminus\bigcup_{a\in F}(P\cdot a^{-1})$ is infinite for every
$F\in[A]^{<\omega}$.

Indeed, suppose no such $S$ exists and let $A={\\{\,p_{n}:n\in\omega\,\\}}$
list all the points in $A$. For each $n\in\omega$ find a neighborhood
$U_{n}\ni p_{n}$ relatively open in $P$ so that $\mathop{\rm
scl}(\overline{U_{n}}\cap P)\leq\beta$ if $p_{n}\in P$ and put
$P_{n}^{\prime}=(\overline{U_{n}}\cap P)\cdot p_{n}^{-1}$. If $p_{n}\not\in
P$, put $P_{n}^{\prime}=\varnothing$. Note that
${\mathcal{P}}^{\prime}={\\{\,P_{n}^{\prime}:n\in\omega\,\\}}$ is a collection
of closed scattered subsets of ${\mathbb{G}}$ with the property that
$\mathop{\rm scl}(P_{n})\leq\beta$ for every $n\in\omega$, and for every
$S\to{1_{{\mathbb{G}}}}$ there exists an $F\in[\omega]^{<\omega}$ such that
$S\subseteq^{*}\bigcup_{n\in F}P_{n}^{\prime}$.

Repeating the construction used to build $P$ out of $P_{n}$ at the beginning
of this argument we may construct a closed scattered
$P^{\prime}\subseteq{\mathbb{G}}$ such that $\mathop{\rm
scl}(P^{\prime})\leq\beta$ and $S\subseteq^{*}P^{\prime}$ for every
$S\to{1_{{\mathbb{G}}}}$ contradicting the minimality of $\alpha_{P}$ in (1).

Let $P\setminus\\{{1_{{\mathbb{G}}}}\\}={\\{\,p_{n}:n\in\omega\,\\}}$ list all
the points in $P$ other than ${1_{{\mathbb{G}}}}$. For each $n\in\omega$ pick
an open $O_{n}\ni{1_{{\mathbb{G}}}}$ such that $\beta=\mathop{\rm
scl}(p_{n},P)=\mathop{\rm scl}(\overline{O_{n}\cdot p_{n}}\cap P)<\alpha_{P}$,
$\overline{O_{n+1}}\subseteq O_{n}$, and
$\bigcap_{n\in\omega}O_{n}=\\{{1_{{\mathbb{G}}}}\\}$.

Restricting ${\mathcal{D}}$ to $O_{0}$ if necessary, assume that
$\bigcup{\mathcal{D}}\subseteq O_{0}$. By induction, pick disjoint closed
discrete $D_{n}^{\prime}\subseteq\cup{\mathcal{D}}$ so that
$D_{n}^{\prime}\subseteq O_{n}$ and each $D_{n}$ is covered by finitely many
$D_{n}^{\prime}$. To see that this is possible, put $D_{n}^{\prime}=(D_{n}\cap
O_{n})\cup((O_{n}\setminus
O_{n+1})\cap(\bigcup{\mathcal{D}}\setminus\bigcup_{i<n}D_{i}^{\prime}))$ and
observe that the intersection inside the second pair of parentheses is a
closed and discrete subspace of ${\mathbb{G}}$, since ${1_{{\mathbb{G}}}}$ is
the only nonisolated point of $\mathbb{D}$. Put
${\mathcal{D}}^{\prime}={\\{\,D_{n}^{\prime}:n\in\omega\,\\}}$. Note that
${1_{{\mathbb{G}}}}$ is a vD-point of ${\mathcal{D}}^{\prime}$. To simplify
notation, we will assume that $D_{n}\subseteq O_{n}$ in what follows.

Let $n\in\omega$. By the choice of $O_{n}$, $D_{n}\subseteq O_{n}\subseteq
O_{i}$, so $D_{n}\cdot p_{i}\subseteq O_{i}\cdot p_{i}$, whenever $i\leq n$.
Thus there is a $\beta<\alpha_{P}$ such that $\mathop{\rm scl}(a,P)\leq\beta$
for every $a\in A_{n}=\bigcup_{i\leq n}D_{n}\cdot p_{i}$. Find a sequence
$S_{n}\to{1_{{\mathbb{G}}}}$ such that $S_{n}\setminus\bigcup_{a\in F}(P\cdot
a^{-1})$ is infinite for every $F\in[A_{n}]^{<\omega}$. Let
$D_{n}={\\{\,d_{i}:i\in\omega\,\\}}$ and $S_{n}={\\{\,s_{i}:i\in\omega\,\\}}$
be 1-1 listings of $D_{n}$ and $S_{n}$. For each $i\in\omega$ pick an $m(i)>n$
so that $m(i)$ is strictly increasing, $s_{m(i)}\cdot d_{i}\cdot p_{j}\not\in
P$ for every $i\in\omega$ and $j\leq n$, and $e_{i}^{n}=s_{m(i)}\cdot d_{i}\in
O_{n}$. Note that the latter is possible, since $S\to{1_{{\mathbb{G}}}}$ and
$d_{i}\in D_{n}\subseteq O_{n}$. Put $B_{n}={\\{\,e_{i}^{n}:i\in\omega\,\\}}$,
$B=\bigcup_{n\in\omega}B_{n}$ and note that each $B_{n}$ is a closed and
discrete subspace of ${\mathbb{G}}$.

Now, ${1_{{\mathbb{G}}}}\in\overline{B}$. Indeed, let $U\ni{1_{{\mathbb{G}}}}$
be any open neighborhood of ${1_{{\mathbb{G}}}}$. Find an open
$V\ni{1_{{\mathbb{G}}}}$ such that $V\cdot V\subseteq U$. Then $V\cap D_{n}$
is infinite for some $n\in\omega$, since ${1_{{\mathbb{G}}}}$ is a vD-point of
${\mathcal{D}}$. Also, $S_{n}\subseteq^{*}V$. Thus for some large enough
$i\in\omega$ both $d_{i}\in V$ and $s_{m(i)}\in V$ showing that $e_{i}^{n}\in
U$.

Let $C\subseteq B$ be an infinite sequence such that $C\to g$ for some
$g\in{\mathbb{G}}$. Since each $B_{n}$ is closed and discrete, we may assume
that $C={\\{\,e_{i(k)}^{n(k)}:k\in\omega\,\\}}$ where $n(k)$ is strictly
increasing. Since $e_{i}^{n}\in O_{n}$, $C\to{1_{{\mathbb{G}}}}$. Thus, there
exists a $j\in\omega$ such that $C\cdot p_{j}\subseteq^{*}P$. Pick a
$k\in\omega$ large enough so that $n(k)>j$ and $e_{i(k)}^{n(k)}\cdot p_{j}\in
P$. At the same time $e_{i(k)}^{n(k)}\cdot p_{j}=s_{m(i(k))}\cdot
d_{i(k)}\cdot p_{j}\not\in P$ by the choice of $s_{m(i)}$, a contradiction. ∎

######  Lemma 23.

Let ${\mathbb{G}}$ be a countable nondiscrete sequential group. Suppose
${\mathcal{P}}\subseteq\mathop{\bf nwd}({\mathbb{G}})$ is a countable family
such that for every $S\to{1_{{\mathbb{G}}}}$ there exists a
$P\in{\mathcal{P}}$ such that $|S\cap P|=\omega$. Then ${\mathbb{G}}$ does not
have a countable $\pi$-network at ${1_{{\mathbb{G}}}}$ that consists of dense
in themselves sets.

###### Proof.

Note that ${\mathbb{G}}$ is not Fréchet by Lemma 6, and thus does not contain
a closed subspace homeomorphic to ${\mathbb{D}}(\omega)$ by Proposition 2.

Let ${\mathcal{D}}={\\{\,D_{n}:n\in\omega\,\\}}$ be a $\pi$-network at
${1_{{\mathbb{G}}}}$ such that each $D\in{\mathcal{D}}$ is dense in itself. By
translating each element of ${\mathcal{D}}$ if necessary, we may assume that
${1_{{\mathbb{G}}}}\in D$ for every $D\in{\mathcal{D}}$.

Fix open $O_{n}\ni{1_{{\mathbb{G}}}}$ so that $\overline{O_{n+1}}\subseteq
O_{n}$ and $\bigcap_{n\in\omega}O_{n}=\\{{1_{{\mathbb{G}}}}\\}$.

Let ${\mathcal{P}}={\\{\,P_{n}:n\in\omega\,\\}}\subseteq\mathop{\bf
nwd}({\mathbb{G}})$ be such that for every $S\to{1_{{\mathbb{G}}}}$ there
exists a $P\in{\mathcal{P}}$ such that $|S\cap P|=\omega$. Just as in the
proof of Lemma 22 we may construct a $P\in\mathop{\bf nwd}({\mathbb{G}})$ such
that for every $S\to{1_{{\mathbb{G}}}}$, $S\subseteq^{*}P$. By taking the
closure of $P$ if necessary, we may assume that $P$ is closed.

Let $g\in{\mathbb{G}}$. Define
$d(g)=\mathop{\mathfrak{so}}(g,{\mathbb{G}}\setminus P)$. Let
$\alpha_{P}=d({1_{{\mathbb{G}}}})$. Note that $\alpha_{P}>1$ by the choice of
$P$. We proceed to prove the following claim by induction on $\alpha$.

  1. (1)

Let $p\in P$ and $d(p)=\alpha$ for some $\alpha<\omega_{1}$. There exists a
$T\subseteq{\mathbb{G}}\setminus P$ and a neighborhood assignment
$W:T\to\tau({\mathbb{G}})$ such that the following properties hold:

    1. (a)

$p\in[T]_{\alpha}$, if $p^{\prime}\in\overline{T}$ then
$d(p^{\prime})=\mathop{\mathfrak{so}}(p^{\prime},T)$;

    2. (b)

$g\in W(g)\setminus P$ for every $g\in T$, the $W(g)$ are disjoint; if
$s_{i}\in W(g_{i})\setminus P$ is such that $s_{i}\to g$ for some
$g\in{\mathbb{G}}$ and all $g_{i}$ are distinct then $g_{i}\to g$;

Let $d(p)=\alpha$ for some $p\in P$. If $\alpha=1$ there exists an infinite
sequence of $g_{i}\in{\mathbb{G}}\setminus P$ such that $g_{i}\to p$. Thinning
out the sequence and reindexing, if necessary, pick disjoint open $W(g_{i})\ni
g_{i}$ so that $W(g_{i})\subseteq O_{i}\cdot p$. Put
$T={\\{\,g_{i}:i\in\omega\,\\}}$. Properties (1a) and (1b) are easy to check.

Thus we can assume $\alpha>1$. Let $p^{n}\to p$ and $\alpha_{n}<\alpha$ be
such that $p^{n}\in O_{n}\cdot p$ and $p^{n}\in[{\mathbb{G}}\setminus
P]_{\alpha_{n}}$ for every $n\in\omega$. Since
$d(p^{n})\leq\alpha_{n}<\alpha$, by the induction hypothesis there exist
$T_{n}\subseteq{\mathbb{G}}\setminus P$ and $W_{n}:T_{n}\to\tau({\mathbb{G}})$
that satisfy (1). Pick a sequence of open disjoint $V_{n}\ni p^{n}$ such that
$V_{n}\subseteq O_{n}\cdot p$ after thinning out an reindexing if necessary.
By passing to subsets and reindexing again, if necessary, we may assume that
the $\overline{T_{n}}$ are disjoint, $T_{n}\subseteq O_{n}\cdot p\cap V_{n}$,
and $W_{n}(g)\subseteq O_{n}\cdot p\cap V_{n}$ for every $n\in\omega$ and
$g\in T_{n}$. Let $T=\bigcup_{n\in\omega}T_{n}$ and define
$W:T\to\tau({\mathbb{G}})$ by $W(g)=W_{n}(g)$ whenever $g\in T_{n}$.

By the choice of $T_{n}$ and $O_{n}$,
$\overline{T}=\\{p\\}\cup\bigcup_{n\in\omega}\overline{T_{n}}$. If
$p^{\prime}\in\overline{T_{n}}$ then
$d(p^{\prime})=\mathop{\mathfrak{so}}(p^{\prime},T_{n})=\mathop{\mathfrak{so}}(p^{\prime},T)$
by the inductive hypothesis and the choice of $T_{n}$. Since
$d(p)=(\sup_{n}\alpha_{n})+1$ and $d(p^{n})\leq\alpha_{n}$,
$d(p)=\mathop{\mathfrak{so}}(p,T)$.

Let $s_{i}\in W(g_{i})\setminus P$ for some $g_{i}\in T$ be such that
$s_{i}\to g$. By thinning out and reindexing we may assume that either
$g_{i}\in T_{n}$ for some fixed $n\in\omega$ or $g_{i}\in T_{n(i)}\subseteq
O_{n(i)}\cdot p$ for some strictly increasing $n(i)$. In the first case
$g_{i}\to g\in P$ by the choice of $T_{n}$. Otherwise $s_{i}\in
W_{n(i)}(g_{i})\subseteq O_{n(i)}\cdot p$ so $s_{i}\to p$ by the choice of
$O_{n}$ contradicting $s_{i}\in{\mathbb{G}}\setminus P$ and $d(p)=\alpha>1$.

Pick $T$ and $W$ that satisfy (1) for $p={1_{{\mathbb{G}}}}$ and let
$T={\\{\,t_{n}:n\in\omega\,\\}}$ be a 1-1 enumeration of the points of $T$.
Pick $U_{n}\subseteq(W(t_{n})\setminus P)\cdot t_{n}^{-1}$ so that
$U_{n+1}\subseteq U_{n}$,
$\bigcap_{n\in\omega}U_{n}=\\{{1_{{\mathbb{G}}}}\\}$, and
${1_{{\mathbb{G}}}}\in\overline{{\\{\,t_{k}:U_{k}\cdot t_{k}\subseteq
O_{n}\,\\}}}$ for every $n\in\omega$. Note that each $U_{n}\cdot
t_{n}\subseteq{\mathbb{G}}\setminus P$. Let $k\in\omega$ and show that

  1. (1)

there exists a closed discrete subset $E_{k}\subseteq D_{k}$ such that
$E_{k}={\\{\,e_{n}^{k}:n\in\omega\,\\}}$ and $e_{n}^{k}\in U_{n}$ for every
$n\in\omega$;

If no such $E_{k}$ exists, the $U_{n}\cap D_{k}$ form a countable base of
neighborhoods of ${1_{{\mathbb{G}}}}$ in $D_{k}$. Since ${1_{{\mathbb{G}}}}\in
D_{k}$, and each $D_{k}$ is dense in itself, this implies the existence of a
closed copy of $\mathbb{D}(\omega)$ in ${\mathbb{G}}$ contradicting the
sequentiality of ${\mathbb{G}}$ as noted at the beginning of this proof.

Consider the set $D^{k}={\\{\,d^{n}:d^{n}=e_{n}^{k}\cdot t_{n},U_{n}\cdot
t_{n}\subseteq O_{k},n\in\omega\,\\}}$. Then $D^{k}\subseteq O_{k}\setminus P$
for every $k\in\omega$ and $d^{n}\in W(t_{n})$ for every $n\in\omega$ by
$e_{n}^{k}\in U_{n}$ and the choice of $U_{n}$. Suppose $d^{n(i)}\to d$ for
some $d\in{\mathbb{G}}$. By (1b) $t_{n(i)}\to d$ so
$e_{n(i)}^{k}=d^{n(i)}\cdot t_{n(i)}^{-1}\to{1_{{\mathbb{G}}}}$ contradicting
the choice of $e_{n}^{k}$. Thus each $D^{k}\subseteq O_{k}$ is closed and
discrete in ${\mathbb{G}}$.

Suppose $S\to g$ for some $g\in{\mathbb{G}}$ is an infinite sequence such that
$S\subseteq\bigcup_{k\in\omega}D^{k}$. Since each $D^{k}$ is closed discrete,
we may assume that $S={\\{\,s_{n}:n\in\omega\,\\}}$ where $s_{n}\in D^{k(n)}$
for some strictly increasing $k(n)$. Then $s_{n}\in O_{k(n)}$ and
$S\to{1_{{\mathbb{G}}}}$ contradicting $S\subseteq{\mathbb{G}}\setminus P$ and
the choice of $P$.

Let $U\ni{1_{{\mathbb{G}}}}$ be open and find an open $V\ni{1_{{\mathbb{G}}}}$
such that $V\cdot V\subseteq U$. Let $k\in\omega$ be such that $D_{k}\subseteq
V$, and let $t_{n}$ be such that $U_{n}\cdot t_{n}\subseteq O_{k}$ and
$t_{n}\in V$. Then $d^{n}=e_{n}^{k}\cdot t_{n}\in D^{k}\cap V\cdot V$. Thus
${1_{{\mathbb{G}}}}\in\overline{\bigcup_{k\in\omega}D^{k}}$ contradicting the
sequentiality of ${\mathbb{G}}$. ∎

######  Lemma 24.

Let $X$ be a sequential space, $S\subseteq X$, and $x\in\overline{S}$ for some
$x\in X$. Let ${\mathcal{I}}\subseteq 2^{X}$ be a cover of $S$. Then either
there exists an $I\in{\mathcal{I}}$ such that $x\in\overline{S\cap I}$ or
there is a countable ${\mathcal{I}}^{*}\subseteq[{\mathcal{I}}]^{\omega}$ such
that whenever ${\mathcal{I}}^{\prime}\subseteq{\mathcal{I}}$ is such that
${\mathcal{I}}^{\prime}\cap{\mathcal{I}}^{\prime\prime}$ is infinite for every
${\mathcal{I}}^{\prime\prime}\in{\mathcal{I}}^{*}$,
$x\in\overline{S\cap\bigcup{\mathcal{I}}^{\prime}}$.

###### Proof.

The proof proceeds by induction on $\mathop{\mathfrak{so}}(x,S)$. The case
$\mathop{\mathfrak{so}}(x,S)=0$ is trivial so assume
$\mathop{\mathfrak{so}}(x,S)=\alpha+1$ and the lemma is proved for all
successor $\beta\leq\alpha$. Pick a sequence $T\subseteq X$ such that $T\to x$
and $\mathop{\mathfrak{so}}(y,S)=\beta_{y}\leq\alpha$ for every $y\in T$ and
consider the two alternatives that follow from the inductive hypothesis.

First, suppose the set $T^{\prime}={\\{\,y:y\in\overline{I_{y}\cap S}\text{
for some }I_{y}\in{\mathcal{I}}\,\\}}$ is infinite. If the family
${\mathcal{I}}^{\prime}={\\{\,I_{y}:y\in T^{\prime}\,\\}}$ is finite then
there is an $I\in{\mathcal{I}}^{\prime}$ such that $x\in\overline{S\cap I}$.
Otherwise put ${\mathcal{I}}^{*}=\\{{\mathcal{I}}^{\prime}\\}$.

Alternatively, assume for every $y\in T$ there is a countable
${\mathcal{I}}^{*}_{y}\subseteq[{\mathcal{I}}]^{\omega}$ such that
$y\in\overline{S\cap\bigcup{\mathcal{I}}^{\prime}}$ for any
${\mathcal{I}}^{\prime}$ such that
${\mathcal{I}}^{\prime}\cap{\mathcal{I}}^{\prime\prime}$ is infinite for every
${\mathcal{I}}^{\prime\prime}\in{\mathcal{I}}^{*}_{y}$. Put
${\mathcal{I}}^{*}=\bigcup_{y\in T}{\mathcal{I}}^{*}_{y}$. ∎

######  Lemma 25.

Let $X$ be a countable sequential space, ${\mathcal{I}}\subseteq 2^{X}$ be an
ideal with the following properties: ${\mathcal{I}}$ contains all singletons,
$\overline{I}\in{\mathcal{I}}$ for every $I\in{\mathcal{I}}$, and whenever
$A\in{\mathcal{I}}^{+}$ there is a $Y\subseteq A$, $Y\in{\mathcal{I}}^{+}$
such that $\overline{Y\setminus I}=\overline{Y}$ for any $I\in{\mathcal{I}}$.
Then ${\mathcal{I}}$ is tame.

###### Proof.

Suppose ${\mathcal{I}}$ is not tame and let $A\in{\mathcal{I}}^{+}$,
$f:A\to\omega$ witness this. Using the property of ${\mathcal{I}}$ from the
statement of the lemma, find $Y\subseteq A$, $Y\in{\mathcal{I}}^{+}$ such that
$\overline{Y\setminus I}=\overline{Y}$ for any $I\in{\mathcal{I}}$. Let $y\in
Y$. Since ${\mathcal{I}}$ contains $\\{y\\}$ and $X$ is sequential, there
exists a sequence $T\subseteq\overline{Y}\setminus\\{y\\}$ such that $T\to y$.
Let $T=\langle y_{i}:i\in\omega\rangle$, $I_{i}=f^{-1}(i)\in{\mathcal{I}}$,
and for every $i\in\omega$ pick a subset $S_{i}\subseteq
Y\setminus\bigcup_{j<i}I_{j}$ such that $y_{i}\in\overline{S_{i}}\not\ni y$.

Note that ${\mathcal{I}}_{f}={\\{\,I_{i}:i\in\omega\,\\}}$ is a cover of
$S=\bigcup_{i\in\omega}S_{i}$ and $y\in\overline{S}$. Since $I_{i}\cap
S\subseteq\bigcup_{j\leq i}S_{i}$ and $y\not\in\overline{\bigcup_{j\leq
i}S_{i}}$, the first alternative of Lemma 24 fails so there exists a countable
${\mathcal{I}}^{*}_{y}\subseteq[{\mathcal{I}}_{f}]^{\omega}$ such that
$y\in\overline{S\cap\bigcup{\mathcal{I}}^{\prime}}$ for any
${\mathcal{I}}^{\prime}\subseteq{\mathcal{I}}$ with the property that
${\mathcal{I}}^{\prime}\cap{\mathcal{I}}^{\prime\prime}$ is infinite for every
${\mathcal{I}}^{\prime\prime}\in{\mathcal{I}}^{*}_{y}$.

Let
${\mathcal{J}}={\\{\,f[\bigcup{\mathcal{I}}^{\prime\prime}]:{\mathcal{I}}^{\prime\prime}\in{\mathcal{I}}^{*}_{y},y\in
Y\,\\}}\subseteq 2^{\omega}$. Since $f[{{\mathcal{I}}}|_{A}]$ is
$\omega$-hitting, there exists a $J\in f[{{\mathcal{I}}}|_{A}]$ such that
$J\cap J^{\prime}$ is infinite for every $J^{\prime}\in{\mathcal{J}}$. Let
${\mathcal{I}}^{\prime}={\\{\,f^{-1}(n):n\in J\,\\}}$. Then
${\mathcal{I}}^{\prime}\cap{\mathcal{I}}^{\prime\prime}$ is infinite for every
${\mathcal{I}}^{\prime\prime}\in{\mathcal{I}}^{*}_{y}$ so
$y\in\overline{f^{-1}[J]}$ for every $y\in Y$. Thus
$Y\subseteq\overline{f^{-1}[J]}\in{\mathcal{I}}$ contradicting
$Y\in{\mathcal{I}}^{+}$. ∎

######  Lemma 26.

Let ${\mathbb{G}}$ be a countable nondiscrete sequential group. Then each of
$\mathop{\bf nwd}({\mathbb{G}})$, $\mathop{\bf cpt}({\mathbb{G}})$, and
$\mathop{\bf csc}({\mathbb{G}})$ is tame.

###### Proof.

For $\mathop{\bf cpt}({\mathbb{G}})$ the statement can be proved directly. For
$\mathop{\bf nwd}({\mathbb{G}})$ and $\mathop{\bf csc}({\mathbb{G}})$ it is
sufficient to establish the properties listed in Lemma 25. In the case of
$\mathop{\bf nwd}({\mathbb{G}})$ one may pick $Y=\mathop{\rm
Int}(\overline{A})\cap A$, while for $\mathop{\bf csc}({\mathbb{G}})$ the
choice of the full Cantor-Bendixson derivative of $A$ as $Y$ satisfies the
conditions of Lemma 25. ∎

######  Lemma 27.

Let ${\mathbb{G}}$ be a countable, sequential non metrizable, non $k_{\omega}$
group. Then one of the $\mathop{\bf nwd}({\mathbb{G}})$, $\mathop{\bf
cpt}({\mathbb{G}})$, or $\mathop{\bf csc}({\mathbb{G}})$ is a tame invariant
ideal that satisfies neither 1 nor 2 of the IIA.

###### Proof.

That each of the ideals is tame follows from Lemma 26. The invariance is
trivial.

Since $\mathop{\bf nwd}({\mathbb{G}})$ never satisfies 2 for a nonmetrizable
${\mathbb{G}}$ (see [39], Proposition 5.2), we may assume that $\mathop{\bf
nwd}({\mathbb{G}})$ satisfies 1. Suppose $\mathop{\bf csc}({\mathbb{G}})$
satisfies 2. Then there exists a countable ${\mathcal{D}}\subseteq\mathop{\bf
csc}^{+}({\mathbb{G}})$ such that for any nonempty open
$U\subseteq{\mathbb{G}}$ there exists a $D\in{\mathcal{D}}$ with the property
$D\setminus U\in\mathop{\bf csc}({\mathbb{G}})$. By replacing each $D$ with a
full Cantor-Bendixson derivative of itself we may assume that each $D$ is
dense in itself. Applying Lemma 23 we arrive at a contradiction. Thus
$\mathop{\bf csc}({\mathbb{G}})$ does not satisfy 2.

Suppose $\mathop{\bf csc}({\mathbb{G}})$ satisfies 1. If $\mathop{\bf
cpt}({\mathbb{G}})$ satisfies 2 there exists a countable family
${\mathcal{D}}$ of closed, noncompact subsets of ${\mathbb{G}}$ such that for
any open $U\subseteq{\mathbb{G}}$ there exists a $D\in{\mathcal{D}}$ such that
$D\setminus U$ is compact. By picking an infinite closed discrete subset in
each $D\in{\mathcal{D}}$ and applying Lemma 22 we arrive at a contradiction.
Thus either $\mathop{\bf csc}({\mathbb{G}})$ does not satisfy 1 or
$\mathop{\bf cpt}({\mathbb{G}})$ does not satisfy 2.

Since ${\mathbb{G}}$ is not $k_{\omega}$, $\mathop{\bf cpt}({\mathbb{G}})$
cannot satisfy 1. ∎

This concludes the proof of Theorem 19. The result has the following
corollary, which illuminates the behavior of sequential groups under taking
products (part (1) is an obvious corollary of Theorem 7 and has been included
for completeness):

######  Corollary 28.

Assume IIA.

  1. (1)

The product of at most countably many separable Fréchet groups is Fréchet, and

  2. (2)

The product of finitely many countable sequential groups which are either
discrete or not Fréchet is sequential.

###### Proof.

It suffices to note that

  1. (1)

${\mathbb{Q}}_{\omega^{\alpha}}\times{\mathbb{Q}}_{\omega^{\beta}}\simeq{\mathbb{Q}}_{\omega^{\beta}}$
if $\alpha<\beta<\omega_{1}$,

  2. (2)

${\mathbb{Q}}_{\omega^{\alpha}}\times{\mathbb{Q}}_{\omega_{1}}$ is not
sequential if $0<\alpha<\omega_{1}$,

  3. (3)

${\mathbb{Q}}_{0}\times{\mathbb{Q}}_{0}\simeq{\mathbb{Q}}_{0}$, and

  4. (4)

${\mathbb{Q}}_{0}\times{\mathbb{Q}}_{\omega_{1}}\simeq{\mathbb{Q}}_{\omega_{1}}\times{\mathbb{Q}}_{\omega_{1}}\simeq{\mathbb{Q}}_{\omega_{1}}$.

∎

We do not know at the moment whether it is consistent that the product of two
sequential groups which are not Fréchet is sequential (independently of their
cardinality).

##  4\. Examples, concluding remarks, and open questions

The example below can probably be constructed using the techniques of [76] but
we chose to provide a direct proof. An appeal to [76] would require a proof of
the normality of finite powers of $\gamma{\mathbb{N}}$ spaces, as well as an
adaptation of the free topological group arguments from [76] to the free
boolean group construction used here.

The nontrivial case of the proof below assumes ${\bf t}=\omega_{1}$, however,
the statement of the example is meant to emphasize the fact that one of the
two ‘pathologies’ exists in every model of ZFC: either there is a separable
nonmetrizable Fréchet group or a (possibly uncountable) sequential group that
is not $k_{\omega}$. The authors do not know whether any separable locally
compact first countable countably compact non compact space may be used in
place of $\gamma{\mathbb{N}}$.

######  Example 29.

If there is no separable nonmetrizable Fréchet group then there exists a
separable sequential $c_{\omega}$ group ${\mathbb{G}}$ that is not
$k_{\omega}$.

###### Proof.

Since ${\bf t}>\omega_{1}$ implies the existence of a separable nonmetrizable
Fréchet group (see, for example [38]) we may assume that ${\bf t}=\omega_{1}$.

Let $X=D\cup\omega_{1}$ be a countably compact $\gamma{\mathbb{N}}$ space (see
[48], Example 2.2) where $D$ is the set of isolated points, disjoint from
$\omega_{1}$, which has the usual topology. Let $X\cup\\{\infty\\}$ be the
one-point compactification of $X$, which is also a subspace of some boolean
group $H$ that is (algebraically) generated by $X\cup\\{\infty\\}$. We shall
assume that $X\cup\\{\infty\\}$ is linearly independent over $H$ (in
particular this means $0_{H}\not\in X\cup\\{\infty\\}$). The free boolean
group over $X\cup\\{\infty\\}$ (see [73]) would have all the desired
properties (in fact it is not difficult to show that any group satisfying the
properties above is naturally isomorphic to the free boolean group over
$X\cup\\{\infty\\}$). Below we use the convention that the elements of such a
group are finite sets of elements of $X\cup\\{\infty\\}$ with the symmetric
difference as the group operation.

Let ${\mathbb{G}}$ be the subgroup of $H$ generated by $X$ and let
$A\subseteq{\mathbb{G}}$ be such that
${0_{{\mathbb{G}}}}\in\overline{A}\setminus A$ (here the closure is taken in
the topology induced by $H$). We must show that there exists a sequence
$S\subseteq A$ such that $S\to x\in{\mathbb{G}}\setminus A$. Suppose no such
sequence exists.

Let $n\in\omega$ be the smallest number such that
${0_{{\mathbb{G}}}}\in\overline{A\cap\sum^{n}X}$. Note that such an $n$ exists
by the definition of the topology on $H$. We will assume that
$A\subseteq\sum^{n}X$. By the minimality of $n$ (truncating $A$ if necessary)
we may assume that $|a|=n$ for every $a\in A$. Write an arbitrary $a\in A$ as
$a=d+w$ where $d\in{\langle{D}\rangle}$ and $w\in{\langle{\omega_{1}}\rangle}$
and put $\delta(a)=|d|$. Note that for every $k\leq n$ the set
$A_{k}={\\{\,a\in A:\delta(a)\leq k\,\\}}$ is a sequentially closed subset of
$A$.

Let $k\leq n$ be the smallest such that
${0_{{\mathbb{G}}}}\in\overline{A_{k}}$. Replacing $A$ with $A_{k}$ and using
the minimality of $k$ we may assume (again truncating $A$ if necessary) that
$\delta(a)=k$ for every $a\in A$. Let $D_{A}=\bigcup{\\{\,t\in D:t\in a\in
A\,\\}}$ and suppose $D_{A}$ is infinite. Note that $\sum^{n}X$ is
sequentially compact. Using this and the property of $D_{A}$ we may pick a
convergent sequence $S\subseteq A$ such that $S\to x$ for some $x\in\sum^{n}X$
and for each $s\in S$ there is a $t_{s}\in s\cap D$ such that
${\\{\,t_{s}:s\in S\,\\}}\to\alpha\in\omega_{1}$. Then $\delta(x)<k$ so
$x\not\in A$.

Thus we may assume that $D_{A}$ is finite. Note that this implies that $D_{A}$
is empty (otherwise ${0_{{\mathbb{G}}}}\not\in\overline{A}$). Therefore
$A\in\sum^{n}\omega_{1}$. Define $l\leq n$ to be the largest with the
following property: for any $\alpha<\omega_{1}$ there exists an $a\in A$ such
that $|a\setminus\alpha|\geq l$. Note that we may assume that $l\geq 1$,
otherwise $A$ is countable with a metrizable closure.

Suppose $l\geq 2$. Recursively pick a sequence $a_{i}\in A$ such that for some
distinct $\alpha_{i},\beta_{i}\in a_{i}$,
$\alpha_{i+1},\beta_{i+1}>\max\\{\alpha_{i},\beta_{i}\\}$. By passing to a
subsequence if necessary we may assume that $a_{i}\to a\in\sum^{n}\omega_{1}$.
Since $\alpha_{i},\beta_{i}\to\gamma$ for some $\gamma\in\omega_{1}$, $|a|<n$,
showing that $a\not\in A$. We may thus assume that $l=1$.

This implies the existence of an $\alpha\in\omega_{1}$ such that every $a\in
A$ can be written as $a=\\{\beta_{a}\\}+b_{a}$ where $\beta_{a}>\alpha+1$ and
$b_{a}\in[\alpha+1]^{n-1}$. If $n>1$, the set $A^{\prime}={\\{\,b_{a}:a\in
A\,\\}}$ is a sequentially closed (and therefore compact) subset of $H$ such
that ${0_{{\mathbb{G}}}}\not\in A^{\prime}$ (otherwise
$A\cap\omega_{1}\not=\varnothing$ contradicting the choice of $A$ and $n>1$).
Now
$U=H\setminus(A^{\prime}+({\\{\,\beta:\beta>\alpha+1\,\\}}\cup\\{\infty\\}))$
is an open neighborhood of ${0_{{\mathbb{G}}}}$ such that $U\cap
A=\varnothing$ contradicting the choice of $A$. Hence $n=1$ implying
${0_{{\mathbb{G}}}}\not\in\overline{A}$, a contradiction. ∎

The statement of IIA given at the beginning of this paper may appear somewhat
technical in that it lists several restrictions on both the space (groomed),
as well as the ideal (tame, invariant, weakly closed). This complexity may be
significantly reduced in most applications, however. Most natural ideals
(including all used in this paper) in sequential spaces are generated by their
(sequentially) closed members while tameness can be replaced by the
topological condition defined in Lemma 25, namely the existence of a ‘kernel’
in each positive set. One may prefer a weaker version of IIA that states that
for every invariant ideal generated by sequentially closed sets for which the
conditions in Lemma 25 are satisfied, one of the two alternatives in the
statement of IIA holds.

Limiting the class of spaces may also make applications of IIA more
transparent. Call the following satement the _Unrestricted Ideal Axiom_ or
UIA:

UIA: For every space $X$ in some class $\mathcal{P}$ and every ideal
$\mathcal{I}\subseteq 2^{X}$ one of the following holds for every $x\in X$:

  1. (1)

there is a countable $\mathcal{S}\subseteq\mathcal{I}$ such that for every
infinite sequence $C$ convergent to $x$ in $X$ there is an $I\in\mathcal{S}$
such that $C\cap I$ is infinite,

  2. (2)

there is a countable $\mathcal{H}\subseteq\mathcal{I}^{+}$ such that for every
non-empty open $U\subseteq X$, $x\in U$ there is an $H\in\mathcal{H}$ such
that $H\setminus U\in\mathcal{I}$.

It is not dificult to see that any countable space that is either $k_{\omega}$
or first countable satisfies UIA. Theorem 19 shows IIA implies UIA holds for
the class of all countable sequential groups (note that there are no
restrictions on the ideal whatsoever, not even invariance). The authors do not
know at the moment if UIA for all _groomed_ groups is implied by IIA or even
whether it is consistent. There are countable Fréchet spaces for which UIA
fails in ZFC with the ideal of the nowhere dense subsets as the witness (see
[27]).

To shed some light on the topology of groomed spaces the following more
detailed treatment of the concept of a vD-point from Lemma 22 may be helpful.

######  Definition 30.

Let $X$ be a topological space. Let ${\mathcal{D}}$ be a countable family of
infinite closed discrete subspaces of $X$. We call ${\mathcal{D}}$ a _(strict)
vD-network_ at $x\in X$ if for every open $U\ni x$ there is a
$D\in{\mathcal{D}}$ such that $D\cap U$ is inifinite ($D\subseteq^{*}U$).

If ${\mathcal{D}}$ is a (strict) vD-network at $x$ we will refer to the space
$\bigcup{\mathcal{D}}\cup\\{x\\}$ as a _(strict) vD-subspace of $X$_ and the
point $x$ as a _(strict) vD-point of ${\mathcal{D}}$ in $X$_.

Now the lemma below offers a topological descripton of groomed spaces. We omit
an elementary proof.

######  Lemma 31.

A countable topological space $X$ is groomed if and only if for every dense
$D\subseteq X$ there exists a point $x\in X$ such that there either exists an
infinite sequence $S\subseteq D$ such that $S\to x$ or a strict vD-network
${\mathcal{D}}\subseteq 2^{D}$ at $x$.

As indicated by Corollary 28 the classes of (countable) Fréchet and sequential
non Fréchet groups are both finitely productive assuming IIA holds. It appears
the following question is open, including the intriguing possibility of the
negative answer in ZFC.

######  Question 1.

Does there exist a (separable, or even countable) sequential non Fréchet group
with a non sequential square?

The answer to the next question is known to be independent of ZFC for
separable groups, while the non separable case in ZFC remains open.

######  Question 2.

Does there exist a Fréchet group with a non Fréchet square?

Example 29 shows that the dichotomy of Theorem 19 does not hold for general
sequential groups even in the separable case. However, the more general
question below appears to be open.

######  Question 3.

Does there exist a separable sequential group that is neither $c_{\omega}$ nor
metrizable?

Finally, while the class of $k_{\omega}$ spaces is finitely productive, it is
not clear if the same is true about sequential $c_{\omega}$ spaces.

######  Question 4.

Are sequential $c_{\omega}$-spaces preserved by finite products?

## Acknowledgements

The authors would like to thank the anonymous referees for their careful
reading of the paper and many helpful suggestions that significantly improved
the exposition. In particular, one of the referees noted that our proof of the
consistency of the IIA implicitly used a closedness-like assumption for the
sets that generate the ideal. This assumption was made explicit in the
definition of the weakly closed ideal.

The second author (AS) would also like to thank the first author and Centro de
Ciencias Matemáticas in Morelia, México for their support and hospitality
during his visit in October of 2019 which served as a starting point for the
research in this paper.

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