# New Hindman spaces

Rafał Filipów<EMAIL_ADDRESS>http://mat.ug.edu.pl/ rfilipow ,
Krzysztof Kowitz<EMAIL_ADDRESS>,  Adam Kwela
<EMAIL_ADDRESS>http://kwela.strony.ug.edu.pl/ and  Jacek Tryba
<EMAIL_ADDRESS>
###### Abstract.

We introduce a method that allows to turn topological questions about Hindman
spaces into purely combinatorial questions about the Katětov order of ideals
on $\mathbb{N}$. We also provide two applications of the method.

  1. (1)

We characterize $F_{\sigma}$ ideals $\mathcal{I}$ for which there is a Hindman
space which is not an $\mathcal{I}$-space under the continuum hypothesis. This
reduces a topological question of Albin L. Jones about consistency of
existence of a Hindman space which is not van der Waerden to the question
whether the ideal of all non AP-sets is not below the ideal of all non IP-sets
in the Katětov order.

  2. (2)

Under the continuum hypothesis, we construct a Hindman space which is not an
$\mathcal{I}_{1/n}$-space. This answers a question posed by Jana Flašková at
the 22nd Summer Conference on Topology and its Applications.

###### Key words and phrases:

Hindman space, van der Waerden space, sequentially compact space, Mrowka
space, ideal, filter, summable ideal, almost disjoint family, IP-set, AP-set,
IP-convergence, ideal convergence

###### 2010 Mathematics Subject Classification:

Primary: 54A20, 05A17, 03E35; Secondary: 03E50, 05C55, 11P99

In our paper we consider topological spaces which satisfy some properties
connected with well known combinatorial theorems, in particular van der
Waerden’s arithmetical progressions theorem and Hindman’s finite sums theorem.
Research into this topic was initiated by Kojman and later continued by Shelah
and others in [16, 15, 17, 13, 8, 4, 6].

A topological space $X$ is called _van der Waerden_ [16] if for every sequence
$\langle x_{n}\rangle_{n\in\mathbb{N}}$ in $X$ there exists a convergent
subsequence $\langle x_{n}\rangle_{n\in A}$ with $A$ being an AP-set (i.e. $A$
contains arithmetic progressions of arbitrary finite length).

A nonempty family $\mathcal{I}\subseteq\mathcal{P}(\mathbb{N})$ of subsets of
$\mathbb{N}$ is an _ideal on $\mathbb{N}$_ if it is closed under taking
subsets and finite unions of its elements, $\mathbb{N}\notin\mathcal{I}$ and
$\mathcal{I}$ contains all finite subsets of $\mathbb{N}$. It is easy to see
that the family $\mathcal{I}_{1/n}=\\{A\subseteq\mathbb{N}:\sum_{n\in
A}\frac{1}{n+1}<\infty\\}$ is an ideal on $\mathbb{N}$, and it follows from
van der Waerden’s theorem [20] that the family
$\mathcal{W}=\\{A\subseteq\mathbb{N}:\text{$A$ is not an AP-set}\\}$ is an
ideal on $\mathbb{N}$. An ideal on $\mathbb{N}$ is called _$F_{\sigma}$_ if it
is an $F_{\sigma}$ subset of $\mathcal{P}(\mathbb{N})$ with a topology induced
from the Cantor space $\\{0,1\\}^{\mathbb{N}}$ (recall: a set is $F_{\sigma}$
if it is a countable union of closed sets). It is not difficult to check that
both $\mathcal{I}_{1/n}$ and $\mathcal{W}$ are $F_{\sigma}$.

A topological space $X$ is called an _$\mathcal{I}$ -space_ [8] if for every
sequence $\langle x_{n}\rangle_{n\in\mathbb{N}}$ in $X$ there exists a
converging subsequence $\langle x_{n}\rangle_{n\in A}$ with
$A\notin\mathcal{I}$. In particular, van der Waerden spaces coincide with
$\mathcal{W}$-spaces.

A set $A\subseteq\mathbb{N}$ is an _IP-set_ if there exists an infinite set
$D\subseteq\mathbb{N}$ such that $\operatorname{FS}(D)\subseteq A$ where
$\operatorname{FS}(D)$ denotes the set of all finite non-empty sums of
distinct elements of $D$. The family
$\mathcal{H}=\\{A\subseteq\mathbb{N}:\text{$A$ is not an IP-set}\\}$ is an
ideal on $\mathbb{N}$ (it follows from Hindman’s theorem [11]).

In [15], Kojman proved that only finite spaces are $\mathcal{H}$-spaces, and
to obtain a meaningful topological counterpart of Hindman’s finite sums
theorem he utilized the notion of IP-convergence introduced by Furstenberg and
Weiss in [9].

An _IP-sequence_ in $X$ is a sequence indexed by $\operatorname{FS}(D)$ for
some infinite $D\subseteq\mathbb{N}$. An IP-sequence $\langle
x_{n}\rangle_{n\in\operatorname{FS}(D)}$ in a topological space $X$ _IP-
converges_ [9] to a point $x\in X$ if for every neighborhood $U$ of the point
$x$ there exists $m\in\mathbb{N}$ so that
$\\{x_{n}:n\in\operatorname{FS}(D\setminus\\{0,1,\dots,m\\})\\}\subseteq U$
(point $x$ is called the _IP-limit_ of the sequence).

A topological space $X$ is called _Hindman_ [15] if for every sequence
$\langle x_{n}\rangle_{n\in\mathbb{N}}$ in $X$ there exists an infinite set
$D\subseteq\mathbb{N}$ such that the subsequence $\langle
x_{n}\rangle_{n\in\operatorname{FS}(D)}$ IP-converges to some $x\in X$.

In [14], Katětov introduced the following order to study convergence in
topological spaces. For ideals $\mathcal{I},\mathcal{J}$ we say that
$\mathcal{I}$ is below $\mathcal{J}$ in the _Katětov order_ if there is a
function $f:\mathbb{N}\to\mathbb{N}$ such that $f^{-1}[A]\in\mathcal{J}$ for
every $A\in\mathcal{I}$. We denote it by $\mathcal{I}\leq_{K}\mathcal{J}$.
When $\mathcal{I}$ is not below $\mathcal{J}$ in Katětov order, we denote it
by $\mathcal{I}\not\leq_{K}\mathcal{J}$.

In 2003, Kojman and Shelah [17] constructed under the continuum hypothesis a
van der Waerden space (i.e. $\mathcal{W}$-space) which is not Hindman. Later
Flašková [8] extended their result by constructing an $\mathcal{I}$-space
which is not Hindman for every $F_{\sigma}$ ideal $\mathcal{I}$ under the
assumption of Martin’s axiom. In particular, she constructed
$\mathcal{I}_{1/n}$-space which is not Hindman and posed a question [7]
whether it is consistent that there is a Hindman space which is not an
$\mathcal{I}_{1/n}$-space. In Section 3 we show that the answer to her
question is positive under the continuum hypothesis. This follows from our
main result: in Section 2 we show that if an ideal $\mathcal{I}$ is not below
the ideal $\mathcal{H}$ in the Katětov order then there exists a Hindman space
which is not an $\mathcal{I}$-space.

The spaces constructed by Kojman, Shelah and Flašková are examples of Mrówka
spaces. In Section 1 we show that Mrówka spaces are not Hindman, therefore the
spaces we construct in Section 2 are not Mrówka spaces.

In [13] Jones posed a question whether there is a Hindman space which is not
van der Waerden. In fact, we show that, in the realm of $F_{\sigma}$ ideals,
the consistency of existence of a Hindman space which is not an
$\mathcal{I}$-space is equivalent to $\mathcal{I}\not\leq_{K}\mathcal{H}$.
Thus, the topological question of Jones is reduced to a purely combinatorial
question whether $\mathcal{W}\not\leq_{K}\mathcal{H}$.

##  1\. Mrówka spaces are not Hindman

In the sequel we use $\omega$ to denote the set of all natural numbers, and we
identify $n\in\omega$ with the set $\\{0,1,\dots,n-1\\}$ (in particular
$A\setminus n=A\setminus\\{0,1,\dots,n-1\\}$).

We say that sets $A,B$ are _almost disjoint_ if $A\cap B$ is finite.

Let $\mathcal{A}$ be an infinite pairwise almost disjoint family of infinite
subsets of $\omega$. Define a topological space $\Psi(\mathcal{A})$ as
follows: the underlying set of $\Psi(\mathcal{A})$ is $\omega\cup\mathcal{A}$,
the points of $\omega$ are isolated and a basic neighborhood of
$A\in\mathcal{A}$ has the form $\\{A\\}\cup(A\setminus F)$, with $F$ finite.
(The space $\Psi(\mathcal{A})$ was introduced in [19].)

Let $\Phi(\mathcal{A})=\Psi(\mathcal{A})\cup\\{\infty\\}$ be the one-point
compactification of $\Psi(\mathcal{A})$.

If $\mathcal{A}$ is a _mad family on $\omega$_ (i.e. infinite maximal pairwise
almost disjoint family of infinite subsets of $\omega$), then the space
$\Phi(\mathcal{A})$ is called a _Mrówka space_.

In [15], Kojman proved that if $\mathcal{A}\subseteq\mathcal{H}$, then the
Mrówka space $\Phi(\mathcal{A})$ is not Hindman, and using this observation
various spaces were shown to be not Hindman (see e.g. [17, 13, 8]). Below we
show that a Mrówka space is not Hindman even without assuming that
$\mathcal{A}\subseteq\mathcal{H}$.

######  Proposition 1.1.

No Mrówka space is Hindman.

###### Proof.

Fix a Mrówka space $\Phi(\mathcal{A})$ defined by a mad family $\mathcal{A}$.

Let $P_{k}=\\{n2^{k+1}+2^{k}:n\in\omega\\}$ for each $k\in\omega$ and notice
that for any infinite $D\subseteq\omega$ we have:

  1. (1)

$\operatorname{FS}(D)\cap P_{k}$ is infinite for some $k\in\omega$,

  2. (2)

$\operatorname{FS}(D)\setminus P_{k}$ is infinite for each $k\in\omega$.

Indeed, we have two cases: $D\cap P_{k}$ is infinite for some $k$ or $D\cap
P_{k}$ is finite for each $k$.

In the first case, take $k_{0}$ such that $D\cap P_{k_{0}}$ is infinite. Then
$\operatorname{FS}(D)\cap P_{k_{0}}$ is infinite as it contains $D\cap
P_{k_{0}}$. Moreover for $k\neq k_{0}$ we have $\operatorname{FS}(D)\setminus
P_{k}$ is infinite as $P_{k}\cap P_{k_{0}}=\emptyset$ and $D\cap
P_{k_{0}}\subseteq\operatorname{FS}(D)$, and $\operatorname{FS}(D)\setminus
P_{k_{0}}$ is infinite as $a+b\in\operatorname{FS}(D)\setminus P_{k_{0}}$ for
any $a,b\in D\cap P_{k_{0}}$.

In the second case, take $k_{0}$ such that $D\cap P_{k_{0}}\neq\emptyset$.
Then $\operatorname{FS}(D)\cap P_{k_{0}}$ is infinite as
$a+b\in\operatorname{FS}(D)\cap P_{k_{0}}$ for any $a\in D\cap P_{k_{0}}$ and
$b\in D\cap P_{l}$ with $l>k_{0}$. Moreover, $\operatorname{FS}(D)\setminus
P_{k}$ is infinite as $D\subseteq\operatorname{FS}(D)$ and $D\setminus P_{k}$
is infinite for any $k$.

Let $A_{1},A_{2},\dots$ be a sequence of distinct sets from $\mathcal{A}$. Let
$f:\omega\to\Phi(\mathcal{A})$ be a sequence such that $f\restriction
P_{k}:P_{k}\to A_{k}\setminus\bigcup_{i<k}A_{i}$ is a bijection for each
$k\in\omega$. We claim that $f$ has no IP-convergent IP-subsequence.

We take any infinite $D\subseteq\omega$ and show that
$f\restriction\operatorname{FS}(D)$ is not IP-convergent to any
$x\in\Phi(\mathcal{A})$.

If $x\in\omega$, then $f\restriction\operatorname{FS}(D)$ is not IP-convergent
to $x$ as $\omega$ is discrete in $\Phi(\mathcal{A})$ and $f$ is one-to-one.

Suppose that $x=\infty$. Take $A\in\mathcal{A}$ such that $f[D]\cap A$ is
infinite (there is one such $A$ since $\mathcal{A}$ is a mad family) and
notice that for each $n\in\omega$ the set $f[D\setminus n]\cap A$ is also
infinite. Then $U=\Phi(\mathcal{A})\setminus(\\{A\\}\cup A)$ is an open
neighborhood of $\infty$ such that $f[\operatorname{FS}(D\setminus
n)]\setminus U$ is infinite for each $n$. Hence
$f\restriction\operatorname{FS}(D)$ is not IP-convergent to $\infty$.

Suppose that $x\in\mathcal{A}$. We have two cases: $x\neq A_{k}$ for any
$k\in\omega$ or $x=A_{k_{0}}$ for some $k_{0}\in\omega$.

In the first case, $U=\\{x\\}\cup x$ is an open neighborhood of $x$ such that
$f[\operatorname{FS}(D\setminus n)]\setminus U$ is infinite for each $n$ and
therefore $f\restriction\operatorname{FS}(D)$ is not IP-convergent to $x$.

In the second case,
$U=\\{A_{k_{0}}\\}\cup(A_{k_{0}}\setminus\bigcup_{i<k_{0}}(A_{i}\cap
A_{k_{0}}))$ is an open neighborhood of $x$ such that
$f[\operatorname{FS}(D\setminus n)]\setminus U$ is infinite for each $n$ and
therefore $f\restriction\operatorname{FS}(D)$ is not IP-convergent to $x$. ∎

##  2\. New Hindman spaces

In this section we prove our main result that under the continuum hypothesis
if $\mathcal{I}\not\leq_{K}\mathcal{H}$ then there is a Hindman space which is
not an $\mathcal{I}$-space.

First we prove some lemmas we are going to use in the proof of the main
theorem.

######  Proposition 2.1.

Let $X$ be a topological space. The following conditions are equivalent.

  1. (1)

Limits of convergent sequences in $X$ are unique.

  2. (2)

IP-limits of IP-convergent IP-sequences in $X$ are unique.

###### Proof.

$(1)\implies(2)$ Follows from the fact that if $\langle
x_{n}\rangle_{n\in\operatorname{FS}(D)}$ is an IP-sequence IP-convergent to
$x$ then $\langle x_{n}\rangle_{n\in D}$ is a sequence convergent to $x$.

$(2)\implies(1)$ Suppose that there is a sequence $x_{n}\in X$ with two
distinct limits $x,y\in X$. Let $P_{k}=\\{n2^{k+1}+2^{k}:n\in\omega\\}$ for
each $k\in\omega$. Now we define a sequence $y_{n}\in X$ by $y_{n}=x_{k}\iff
n\in P_{k}$. Let $E=\\{2^{k}:k\in\omega\\}$. Then it is not difficult to see
that $\langle y_{n}\rangle_{n\in\operatorname{FS}(E)}$ is IP-convergent to
both $x$ and $y$. ∎

An infinite set $D\subseteq\omega$ is _sparse_ if for every
$x\in\operatorname{FS}(D)$ there is a unique nonempty finite set $F\subseteq
D$ with $x=\sum_{i\in F}i$. A sparse set $D$ is _very sparse_ if additionally
for each $F,F^{\prime}\in[D]^{<\omega}$, if $F\cap F^{\prime}\neq\emptyset$
then $\sum_{i\in F}i+\sum_{i\in F^{\prime}}i\notin\operatorname{FS}(D)$. (The
term a “sparse set” was coined by Kojman in [15], however the sparsity of sets
was used earlier as this notion is crucial in Hindman’s original proof [11].
The same notion has been recently renamed “apartness” in some other contexts
[2].)

######  Lemma 2.2 (Essentially [10, Lemma 2.3]).

For every infinite $D$ there is an infinite very sparse $D^{\prime}\subseteq
D$.

###### Proof.

Let $D$ be infinite. Inductively pick $d_{i}\in D$ in such a way that
$d_{n}>2\sum_{i<n}d_{i}$. Define $D^{\prime}=\\{d_{i}:i\in\omega\\}$.
Obviously, $D^{\prime}\subseteq D$. Moreover, $D^{\prime}$ is sparse. Indeed,
if $x\in\operatorname{FS}(D)$ and $F,F^{\prime}\in[D]^{<\omega}$ are such that
$x=\sum_{i\in F}i=\sum_{i\in F^{\prime}}i$, then $\sum_{i\in F\setminus
F^{\prime}}i=\sum_{i\in F^{\prime}\setminus F}i$. Assume to the contrary that
$F\neq F^{\prime}$. Since $(F\setminus F^{\prime})\cap(F^{\prime}\setminus
F)=\emptyset$, without loss of generality we may assume that
$d_{n}=\max(F\setminus F^{\prime})>\max(F^{\prime}\setminus F)$. Observe that
$\sum_{i\in F^{\prime}\setminus F}i\leq\sum_{i<n}d_{i}<d_{n}\leq\sum_{i\in
F\setminus F^{\prime}}i$. Thus, $F=F^{\prime}$.

Finally, we show inductively that $D^{\prime}$ is very sparse. For the first
inductive step note that if $F,F^{\prime}\subseteq\\{d_{0}\\}$ and $F\cap
F^{\prime}\neq\emptyset$ then $F=F^{\prime}=\\{d_{0}\\}$ and $\sum_{i\in
F}i+\sum_{i\in F^{\prime}}i=2d_{0}<d_{1}$. Thus, $\sum_{i\in F}i+\sum_{i\in
F^{\prime}}\notin\operatorname{FS}(D^{\prime})$. Assume now that for some
$n\in\omega$, if $F,F^{\prime}\subseteq\\{d_{i}:i<n\\}$ and $F\cap
F^{\prime}\neq\emptyset$ then $\sum_{i\in F}i+\sum_{i\in
F^{\prime}}\notin\operatorname{FS}(D^{\prime})$. Fix
$F,F^{\prime}\subseteq\\{d_{i}:i\leq n\\}$ such that $F\cap
F^{\prime}\neq\emptyset$.

There are two possibilities: either $\max(F\cup F^{\prime})<d_{n}$ and then
$\sum_{i\in F}i+\sum_{i\in F^{\prime}}i\notin\operatorname{FS}(D^{\prime})$ by
the inductive assumption, or $\max(F\cup F^{\prime})=d_{n}$. In the latter
case note that $\sum_{i\in F}i+\sum_{i\in F^{\prime}}i\leq 2\sum_{i\leq
n}d_{i}<d_{n+1}$. Moreover, if $d_{n}\in F\cap F^{\prime}$ then $\sum_{i\in
F}i+\sum_{i\in F^{\prime}}i\geq 2d_{n}>\sum_{i\leq n}d_{i}$ and thus
$\sum_{i\in F}i+\sum_{i\in F^{\prime}}i\notin\operatorname{FS}(D^{\prime})$.
On the other hand, if $d_{n}\in F\setminus F^{\prime}$, then suppose to the
contrary that $\sum_{i\in F}i+\sum_{i\in
F^{\prime}}i\in\operatorname{FS}(D^{\prime})$. In this case, since $\sum_{i\in
F}i+\sum_{i\in F^{\prime}}i<d_{n+1}$, $\sum_{i\in F}i+\sum_{i\in
F^{\prime}}i\in FS(\\{d_{i}:i\leq n\\})=\\{d_{n}\\}\cup
FS(\\{d_{i}:i<n\\})\cup(d_{n}+FS(\\{d_{i}:i<n\\}))$. As
$d_{n}>2\sum_{i<n}d_{i}$ and $d_{n}\in F\setminus F^{\prime}$, we get that
$\sum_{i\in F}i+\sum_{i\in F^{\prime}}i\in d_{n}+FS(\\{d_{i}:i<n\\})$ i.e.
$\sum_{i\in F\setminus\\{d_{n}\\}}i+\sum_{i\in
F^{\prime}}i\in\operatorname{FS}(\\{d_{i}:i<n\\})$. But this contradicts the
inductive assumption. Thus, $D^{\prime}$ is very sparse. ∎

######  Lemma 2.3.

Let $(D_{i})\subseteq[\omega]^{\omega}$ be such that $D_{0}$ is very sparse
and
$\operatorname{FS}(D_{0})\supseteq\operatorname{FS}(D_{1})\supseteq\ldots$.
Then there is an infinite $E$ such that
$\operatorname{FS}(E)\subseteq\operatorname{FS}(D_{0})$ and for each
$n\in\omega$ there is $k\in\omega$ with $\operatorname{FS}(E\setminus
k)\subseteq\operatorname{FS}(D_{n})$.

###### Proof.

Since $D_{0}$ is very sparse, for each $x\in\operatorname{FS}(D_{0})$ there is
a unique $F(x)\in[D_{0}]^{<\omega}$ with $x=\sum_{i\in F(x)}i$. Let $e_{0}\in
D_{0}$,
$e_{1}\in\operatorname{FS}(D_{1})\setminus\\{x\in\operatorname{FS}(D_{0}):F(x)\cap
F(e_{0})\neq\emptyset\\}$ and
$e_{n}\in\operatorname{FS}(D_{n})\setminus\\{x\in\operatorname{FS}(D_{0}):(\exists
i<n)\,F(x)\cap F(e_{i})\neq\emptyset\\}$ for all $n>1$. This is possible as
$\\{x\in\operatorname{FS}(D_{0}):(\exists i<n)\,F(x)\cap
F(e_{i})\neq\emptyset\\}\in\mathcal{H}\restriction\operatorname{FS}(D_{0})$
for each $n\in\omega$.

Define $E=\\{e_{i}:i\in\omega\\}$. Then $E$ is infinite. We need to check that
$E$ is as needed.

Let $x\in\operatorname{FS}(E)$. Then there is a finite $F\subseteq\omega$ such
that $x=\sum_{i\in F}e_{i}$. As $F(e_{i})$ are pairwise disjoint, we get that
$x\in\operatorname{FS}(D_{0})$.

Fix $n\in\omega$ and let $x\in\operatorname{FS}(E\setminus e_{n})$. Then there
is a finite $F\subseteq\omega\setminus n$ such that $x=\sum_{i\in F}e_{i}$.
Recall that each $e_{i}$ for $i\geq n$ is in $\operatorname{FS}(D_{n})$. Thus,
for each $i\in F$ we can find finite $T_{i}\subseteq D_{n}$ with
$e_{i}=\sum_{d\in T_{i}}d$. Observe that $F(d)\cap F(d^{\prime})=\emptyset$
for all $d,d^{\prime}\in D_{n}$ as otherwise we would get
$d+d^{\prime}\in\operatorname{FS}(D_{n})\setminus\operatorname{FS}(D_{0})$
which contradicts $\operatorname{FS}(D_{0})\supseteq\operatorname{FS}(D_{n})$.
We claim that the sets $T_{i}$ are pairwise disjoint. Indeed, if $d\in
T_{i}\cap T_{j}\subseteq D_{n}$ for some $i,j\in F$ then
$F(d)\subseteq\bigcup_{d^{\prime}\in T_{i}}F(d^{\prime})=F(e_{i})$ and
$F(d)\subseteq\bigcup_{d^{\prime}\in T_{j}}F(d^{\prime})=F(e_{j})$ (as $F(x)$
are unique). This leads to $F(e_{i})\cap F(e_{j})\neq\emptyset$ which implies
$e_{i}+e_{j}\notin\operatorname{FS}(D_{0})$ and contradicts the fact that
$F(e_{i})$ are pairwise disjoint. Since the sets $T_{i}\in[D_{n}]^{<\omega}$
are pairwise disjoint, $x=\sum_{i\in F}\sum_{d\in
T_{i}}d\in\operatorname{FS}(D_{n})$. ∎

######  Lemma 2.4.

Let $D\subseteq\omega$ be infinite and $A_{n}\in\mathcal{H}$ for each
$n\in\omega$. Then there is an infinite $D^{\prime}$ such that
$\operatorname{FS}(D^{\prime})\subseteq\operatorname{FS}(D)$ and for each
$n\in\omega$ there is $k\in\omega$ with $\operatorname{FS}(D^{\prime}\setminus
k)\cap A_{n}=\emptyset$.

###### Proof.

By Lemma 2.2, without loss of generality we may assume that $D$ is very
sparse. Since $A_{0}\in\mathcal{H}$, the set $\operatorname{FS}(D)\setminus
A_{0}$ contains $\operatorname{FS}(D_{0})$ for some infinite
$D_{0}\subseteq\omega$. Similarly, the set $\operatorname{FS}(D_{0})\setminus
A_{1}$ contains $\operatorname{FS}(D_{1})$ for some infinite
$D_{1}\subseteq\omega$. Thus, we can inductively define a sequence of sets
$D_{n}\in[\omega]^{\omega}$ such that
$\operatorname{FS}(D)\supseteq\operatorname{FS}(D_{0})\supseteq\operatorname{FS}(D_{1})\supseteq\ldots$
and $\operatorname{FS}(D_{i})\cap A_{i}=\emptyset$ for all $i\in\omega$.
Applying Lemma 2.3 we get an infinite $E$ such that for each $n\in\omega$
there is $k\in\omega$ with $\operatorname{FS}(E\setminus
k)\subseteq\operatorname{FS}(D_{n})$. Let $k_{0}$ be such that
$\operatorname{FS}(E\setminus k_{0})\subseteq\operatorname{FS}(D_{0})$ and
define $D^{\prime}=E\setminus k_{0}$. Then
$\operatorname{FS}(D^{\prime})=\operatorname{FS}(E\setminus
k_{0})\subseteq\operatorname{FS}(D_{0})\subseteq\operatorname{FS}(D)$.
Moreover, given $n\in\omega$, as there is $k\in\omega$ with
$\operatorname{FS}(E\setminus k)\subseteq\operatorname{FS}(D_{n})$, we have
that $\operatorname{FS}(D^{\prime}\setminus
k)\subseteq\operatorname{FS}(E\setminus k)\subseteq\operatorname{FS}(D_{n})$,
so $\operatorname{FS}(D^{\prime}\setminus k)\cap A_{n}=\emptyset$. ∎

Finally, we are ready for the main result of this section. To some extent, the
idea of the proof of the following theorem shares some similarity with Kojman-
Shelah’s construction of van der Waerden space which is not Hindman ([17]).

######  Theorem 2.5.

Assume the continuum hypothesis. If $\mathcal{I}\not\leq_{K}\mathcal{H}$, then
there is a $T_{1}$ separable Hindman space which is not an
$\mathcal{I}$-space. Moreover, limits of convergent sequences and IP-limits of
IP-convergent IP-sequences in this space are unique.

###### Proof.

Using the continuum hypothesis, we only need countable ordinals to fix a list
$\langle(H_{\alpha},f_{\alpha}):\alpha<\omega_{1}\rangle$ of all sets
$H_{\alpha}\notin\mathcal{H}$ and functions $f_{\alpha}:H_{\alpha}\to\omega$
satisfying $f_{\alpha}^{-1}[\\{n\\}]\in\mathcal{H}$ for all $n\in\omega$.

We will construct a sequence $\langle D_{\alpha}:\alpha<\omega_{1}\rangle$ of
infinite subsets of $\omega$ such that for every $\alpha<\omega_{1}$ we have
$\operatorname{FS}(D_{\alpha})\subseteq H_{\alpha}$ and
$f_{\alpha}[\operatorname{FS}(D_{\alpha})]\in\mathcal{I}$ and one of the
following conditions holds:

(W1)
$\forall{\beta<\alpha}\,\forall{i\in\omega}\,\exists{k\in\omega}\,\left(f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus
k)]\cap(i\cup f_{\beta}[\operatorname{FS}(D_{\beta}\setminus
k)])=\emptyset\right)$

or

(W2)
$\exists{\beta<\alpha}\,\forall{k\in\omega}\,\exists{n\in\omega}\,\,\left(f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus
n)]\subseteq f_{\beta}[\operatorname{FS}(D_{\beta}\setminus k)]\setminus
k\right).$

Suppose that $\alpha<\omega_{1}$ and that $D_{\beta}$ have been chosen for all
$\beta<\alpha$ . Since $\mathcal{H}$ is homogeneous (see [18, Example 2.6])
and $\mathcal{I}\not\leq_{K}\mathcal{H}$, there is $C\in\mathcal{I}$ such that
$f^{-1}_{\alpha}[C]\notin\mathcal{H}\restriction H_{\alpha}$. Find infinite
$D\subseteq\omega$ with $\operatorname{FS}(D)\subseteq H_{\alpha}$ and
$f_{\alpha}[\operatorname{FS}(D)]\subseteq C$. By Lemma 2.2 we can assume that
$D$ is very sparse.

We have 2 cases:

(P1)
$\forall{D^{\prime}\in[\omega]^{\omega}}\,\forall{\beta<\alpha}\,\left(\operatorname{FS}(D^{\prime})\subseteq\operatorname{FS}(D)\implies\exists{k\in\omega}\,\left(\operatorname{FS}(D^{\prime})\setminus
f_{\alpha}^{-1}[f_{\beta}[\operatorname{FS}(D_{\beta}\setminus
k)]]\notin\mathcal{H}\right)\right)$

or

(P2)
$\exists{D^{\prime}\in[\omega]^{\omega}}\,\exists{\beta<\alpha}\,\left(\operatorname{FS}(D^{\prime})\subseteq\operatorname{FS}(D)\land\forall{k\in\omega}\,\left(\operatorname{FS}(D^{\prime})\setminus
f_{\alpha}^{-1}[f_{\beta}[\operatorname{FS}(D_{\beta}\setminus
k)]]\in\mathcal{H}\right)\right).$

In the first case, let $\alpha\times\omega=\\{(\beta_{n},i_{n}):n<\omega\\}$,
taking into account that $\alpha$ is countable. Using condition (P1)
repeatedly and the fact that $f_{\alpha}^{-1}[i]\in\mathcal{H}$ for every
$i\in\omega$, one can easily construct a sequence $\langle
E_{n}:n\in\omega\rangle$ of infinite subsets of $\omega$ such that

  1. (1)

$\operatorname{FS}(E_{0})\subseteq\operatorname{FS}(D)$,

  2. (2)

$\forall{n\in\omega}\,(\operatorname{FS}(E_{n+1})\subseteq\operatorname{FS}(E_{n}))$,

  3. (3)

$\forall{n\in\omega}\,\exists{k\in\omega}\left(\operatorname{FS}(E_{n})\cap
f_{\alpha}^{-1}[i_{n}\cup
f_{\beta_{n}}[\operatorname{FS}(E_{\beta_{n}}\setminus k)]]=\emptyset\right)$.

Now using Lemma 2.3 we find an infinite set $E^{\prime}\subseteq\omega$ such
that $\operatorname{FS}(E^{\prime})\subseteq\operatorname{FS}(D)$ and for
every $n\in\omega$ there is $l\in\omega$ with
$\operatorname{FS}(E^{\prime}\setminus l)\subseteq\operatorname{FS}(E_{n})$.

It is not difficult to see that $D_{\alpha}=E^{\prime}$ satisfies
$\operatorname{FS}(D_{\alpha})\subseteq H_{\alpha}$ and (W1).

Consider the second case. Let $D^{\prime}\in[\omega]^{\omega}$ and
$\beta<\alpha$ be such that
$\operatorname{FS}(D^{\prime})\subseteq\operatorname{FS}(D)$ and
$\operatorname{FS}(D^{\prime})\setminus
f_{\alpha}^{-1}[f_{\beta}[\operatorname{FS}(D_{\beta}\setminus
k)]]\in\mathcal{H}$ for each $k\in\omega$. Since
$f_{\alpha}^{-1}[i]\in\mathcal{H}$ for every $i\in\omega$, we also have
$\operatorname{FS}(D^{\prime})\setminus
f_{\alpha}^{-1}[f_{\beta}[\operatorname{FS}(D_{\beta}\setminus k)]\setminus
k]\in\mathcal{H}$ for each $k\in\omega$. Using Lemma 2.4, we find an infinite
set $D^{\prime\prime}\subseteq\omega$ such that
$\operatorname{FS}(D^{\prime\prime})\subseteq\operatorname{FS}(D^{\prime})$
and for every $k\in\omega$ there is $n\in\omega$ with
$\operatorname{FS}(D^{\prime\prime}\setminus
n)\cap(\operatorname{FS}(D^{\prime})\setminus
f_{\alpha}^{-1}[f_{\beta}[\operatorname{FS}(D_{\beta}\setminus k)]\setminus
k])=\emptyset$.

It is not difficult to see that $D_{\alpha}=D^{\prime\prime}$ satisfies
$\operatorname{FS}(D_{\alpha})\subseteq H_{\alpha}$ and (W2).

The construction of sets $D_{\alpha}$ is finished.

We are ready to define the required space. Let
$T=\\{\alpha<\omega_{1}:\text{$D_{\alpha}$ satisfies \eqref{eq:W1}}\\}$ and

$X=\omega\cup\\{\operatorname{FS}(D_{\alpha}):\alpha\in T\\}\cup\\{\infty\\}.$

For every $x\in X$ we define the family
$\mathcal{B}(x)\subseteq\mathcal{P}(X)$ as follows:

  * •

$\mathcal{B}(n)=\left\\{\\{n\\}\right\\}$ for $n\in\omega$,

  * •

$\mathcal{B}(\operatorname{FS}(D_{\alpha}))=\left\\{\\{\operatorname{FS}(D_{\alpha})\\}\cup
f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus k)]\setminus
k:k\in\omega\right\\}$ for $\alpha\in T$,

  * •

$\mathcal{B}(\infty)=\left\\{\\{\infty\\}\cup\bigcup_{\alpha\in T\setminus
F}U_{\alpha}:F\in[T]^{<\omega}\land
U_{\alpha}\in\mathcal{B}(FS(D_{\alpha}))\text{ for $\alpha\in T\setminus
F$}\right\\}$.

It is not difficult to check that the family
$\mathcal{N}=\\{\mathcal{B}(x):x\in X\\}$ satisfies:

  * •

for each $x\in X$ we have $\mathcal{B}(x)\neq\emptyset$ and if
$U\in\mathcal{B}(x)$ then $x\in U$;

  * •

for each $x\in X$, if $U,U^{\prime}\in\mathcal{B}(x)$ then there is
$V\in\mathcal{B}(x)$ with $V\subseteq U\cap U^{\prime}$;

  * •

if $x,y\in X$ are such that $y\in U$ for some $U\in\mathcal{B}(x)$ then there
is $V\in\mathcal{B}(y)$ with $V\subseteq U$.

Hence $\mathcal{N}$ is a neighborhood system (see e.g. [3, Proposition
1.2.3]). We claim that $X$ with the topology generated by $\mathcal{N}$ is a
topological space we are looking for.

First, observe that $X$ is separable as $\omega$ is a countable dense subset
of $X$.

Second, it is not difficult to see that $X$ is $T_{1}$. In fact
$X\setminus\\{\infty\\}$ is $T_{2}$ (i.e. Hausdorff), however it is unlikely
to be able to separate points $x=\infty$ and
$y=\operatorname{FS}(D_{\alpha})$.

Now we show that $X$ is a Hindman space. Fix any $f:\omega\to X$. If there is
$x\in X$ with $f^{-1}[\\{x\\}]\notin\mathcal{H}$ then find
$D\in[\omega]^{\omega}$ with $FS(D)\subseteq f^{-1}[\\{x\\}]$ and observe that
$\langle f(n)\rangle_{n\in FS(D)}$ IP-converges to $x$. Thus, we can assume
that $f^{-1}[\\{x\\}]\in\mathcal{H}$ for all $x\in X$. There are two possible
cases: $f^{-1}[X\setminus\omega]\notin\mathcal{H}$ or
$f^{-1}[\omega]\notin\mathcal{H}$.

If $f^{-1}[X\setminus\omega]\notin\mathcal{H}$ then find
$D\in[\omega]^{\omega}$ with $FS(D)\subseteq f^{-1}[X\setminus\omega]$. As
$f^{-1}[\\{x\\}]\in\mathcal{H}$ for all $x\in X$ and
$f^{-1}[\\{x\\}]\neq\emptyset$ only for countably many $x\in X$, using Lemma
2.4 we can find $D^{\prime}\in[\omega]^{\omega}$ with $FS(D^{\prime})\subseteq
FS(D)$ and such that for each $x\in X\setminus\omega$ there is $k\in\omega$
with $FS(D^{\prime}\setminus k)\cap f^{-1}[\\{x\\}]=\emptyset$. Since for each
$U\in\mathcal{B}(\infty)$ there are only finitely many $\alpha\in T$ with
$\operatorname{FS}(D_{\alpha})\notin U$, $\langle f(n)\rangle_{n\in
FS(D^{\prime})}$ IP-converges to $\infty$.

If $f^{-1}[\omega]\notin\mathcal{H}$ then there is $\alpha<\omega_{1}$ with
$(f^{-1}[\omega],f_{\upharpoonright f^{-1}[\omega]})=(H_{\alpha},f_{\alpha})$.
We have two subcases: $\alpha\in T$ and $\alpha\notin T$.

Assume $\alpha\in T$. Using Lemma 2.4, we find an infinite set
$D^{\prime}\subseteq\omega$ such that
$\operatorname{FS}(D^{\prime})\subseteq\operatorname{FS}(D_{\alpha})$ and for
each $k\in\omega$ there is $n\in\omega$ with
$\operatorname{FS}(D^{\prime}\setminus n)\cap f^{-1}[k]=\emptyset$. Since each
$U\in\mathcal{B}(\operatorname{FS}(D_{\alpha}))$ is of the form
$\\{\operatorname{FS}(D_{\alpha})\\}\cup f[FS(D_{\alpha}\setminus k)]\setminus
k$ for some $k\in\omega$, the subsequence $\langle f(n)\rangle_{n\in
FS(D^{\prime})}$ IP-converges to $\operatorname{FS}(D_{\alpha})$.

Assume $\alpha\notin T$. Then there is $\beta<\alpha$ such that for each
$k\in\omega$ there is $n\in\omega$ with $f_{\alpha}[FS(D_{\alpha}\setminus
n)]\subseteq f_{\beta}[FS(D_{\beta}\setminus k)]\setminus k$. If we take the
smallest $\beta<\alpha$ with the above property, then $\beta\in T$ and
$\langle f(n)\rangle_{n\in FS(D_{\alpha})}$ IP-converges to
$\operatorname{FS}(D_{\beta})\in X$.

Now we check that $X$ is not an $\mathcal{I}$-space. Define $f:\omega\to X$ by
$f(n)=n$ and fix any $B\notin\mathcal{I}$. We claim that $\langle
f(n)\rangle_{n\in B}$ does not converge. Clearly, $\langle f(n)\rangle_{n\in
B}$ cannot converge to any $x\in\omega$. Moreover, it cannot converge to any
$\operatorname{FS}(D_{\alpha})$ as $U=\\{\operatorname{FS}(D_{\alpha})\\}\cup
f_{\alpha}[\operatorname{FS}(D_{\alpha})]$ is an open neighborhood of
$\operatorname{FS}(D_{\alpha})$ and $f[B]\setminus U=B\setminus U$ is infinite
(since $B\notin\mathcal{I}$ and
$f_{\alpha}[\operatorname{FS}(D_{\alpha})]\in\mathcal{I}$).

We will show that $\langle f(n)\rangle_{n\in B}$ cannot converge to $\infty$.
Since $B$ is infinite, there is $\alpha<\omega_{1}$ with $H_{\alpha}=\omega$
and $f_{\alpha}:H_{\alpha}\to B$ being the increasing enumeration of $B$. In
particular, $f_{\alpha}[H_{\alpha}]=B$. We have 2 cases: $\alpha\in T$ or
$\alpha\notin T$.

Suppose first that $\alpha\in T$. Then
$f_{\alpha}[\operatorname{FS}(D_{\alpha})]\subseteq B$. Since
$f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus k)]$ is infinite for every
$k$, we can inductively take $c_{k}\in
f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus k)]\setminus\\{c_{i}:i<k\\}$
to obtain an infinite set $C=\\{c_{k}:k\in\omega\\}$ such that $C\subseteq B$
and $C\setminus f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus k)]$ is
finite for all $k\in\omega$. Now observe that for each $\beta\in
T\setminus\\{\alpha\\}$ there is
$U_{\beta}\in\mathcal{B}(\operatorname{FS}(D_{\beta}))$ with $U_{\beta}\cap
C=\emptyset$. Indeed, fix $\beta\in T\setminus\\{\alpha\\}$. By condition (W1)
there is $k\in\omega$ such that
$f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus k)]\cap
f_{\beta}[\operatorname{FS}(D_{\beta}\setminus k)]=\emptyset$. Let
$k_{\beta}=\max(k,\max(C\setminus
f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus k)]))$. Then
$U_{\beta}=\\{\operatorname{FS}(D_{\beta})\\}\cup
f_{\beta}[\operatorname{FS}(D_{\beta}\setminus k_{\beta})]\setminus k_{\beta}$
is in $\mathcal{B}(\operatorname{FS}(D_{\beta}))$ and $U_{\beta}\cap
C=\emptyset$. Therefore, $U=\\{\infty\\}\cup\bigcup_{\beta\in
T\setminus\\{\alpha\\}}U_{\beta}$ is an open neighborhood of $\infty$ which is
disjoint with $C$. Since $C\in[B]^{\omega}$, this shows that $\langle
f(n)\rangle_{n\in B}$ cannot converge to $\infty$.

Suppose now that $\alpha\notin T$. Then there is $\beta<\alpha$ such that for
each $k\in\omega$ there is $n\in\omega$ with
$f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus n)]\subseteq
f_{\beta}[FS(D_{\beta}\setminus k)]$. If we take the smallest $\beta<\alpha$
with the above property, then $\beta\in T$. Then similarly as in the first
case, we find an infinite $C\subseteq B$ with $C\setminus
f_{\beta}[FS(D_{\beta}\setminus k)]$ finite for all $k\in\omega$, and
consequently we can find an open neighborhood $U$ of $\infty$ disjoint with
$C$.

To finish the entire proof we have to check that limits of convergent
sequences and IP-limits of IP-convergent IP-sequences in $X$ are unique. By
Proposition 2.1 we only have to show the uniqueness of limits.

Fix a convergent sequence $x_{n}\in X$. Since $X\setminus\\{\infty\\}$ is
$T_{2}$ (and limits in $T_{2}$ space are unique) and $\omega$ is discrete in
$X$ (so sequences with limits in $\omega$ have unique limits) we only need to
check that if $\langle x_{n}\rangle$ converges to some
$\operatorname{FS}(D_{\alpha})$ then it cannot converge to $\infty$. So
suppose that $\langle x_{n}\rangle$ converges to $A_{\alpha}$.

If $x_{n}=FS(D_{\alpha})$ for infinitely many $n$ then
$U=\\{\infty\\}\cup\bigcup_{\beta\in
T\setminus\\{\alpha\\}}\\{\operatorname{FS}(D_{\beta})\\}\cup
f_{\beta}[\operatorname{FS}(D_{\beta})]$ is an open neighborhood of $\infty$
which omits infinitely many elements of $\langle x_{n}\rangle$. Hence $\langle
x_{n}\rangle$ cannot converge to $\infty$.

Suppose now that $x_{n}\neq FS(D_{\alpha})$ for all but finitely many $n$.
Then the set $C=\omega\cap\\{x_{n}:n\in\omega\\}$ is infinite. Since $\langle
x_{n}\rangle$ converges to $FS(D_{\alpha})$, $C\setminus
f_{\alpha}[\operatorname{FS}(D_{\alpha}\setminus k)]$ is finite for any
$k\in\omega$. Then, using condition (W1), for each $\beta\in
T\setminus\\{\alpha\\}$ we can find $k_{\beta}\in\omega$ with
$C\cap(f_{\beta}[\operatorname{FS}(D_{\beta}\setminus k_{\beta})]\setminus
k_{\beta})=\emptyset$. Therefore, $U=\\{\infty\\}\cup\bigcup_{\beta\in
T\setminus\\{\alpha\\}}\\{\operatorname{FS}(D_{\beta})\\}\cup
f_{\beta}[\operatorname{FS}(D_{\beta}\setminus k_{\beta})]\setminus k_{\beta}$
is an open neighborhood of $\infty$ which is disjoint with $C$. Hence $\langle
x_{n}\rangle$ cannot converge to $\infty$. ∎

At this point a natural conjecture would be that if
$\mathcal{I}\leq_{K}\mathcal{H}$ then every Hindman space is an
$\mathcal{I}$-space. We were able to prove it only under one additional
assumption on the ideal $\mathcal{I}$.

An ideal $\mathcal{I}$ is called _P +-ideal_ if for every decreasing sequence
$A_{1},A_{2},\dots\notin\mathcal{I}$ there is $A\notin\mathcal{I}$ such that
$A\setminus A_{n}$ is finite for each $n\in\omega$ ([12, p. 32]).

######  Proposition 2.6.

If $\mathcal{I}\leq_{K}\mathcal{H}$ and $\mathcal{I}$ is a P+-ideal, then
every Hindman space is an $\mathcal{I}$-space. In particularly, if
$\mathcal{I}$ is an $F_{\sigma}$ ideal and $\mathcal{I}\leq_{K}\mathcal{H}$
then every Hindman space is an $\mathcal{I}$-space.

###### Proof.

The “in particular” part follows from the first part as it is well known that
each $F_{\sigma}$ ideal is a P+-ideal (see e.g. [5, Proposition 5.1]).

Let $X$ be a Hindman space and fix a sequence $\langle
x_{n}\rangle_{n\in\omega}$ in $X$. Let $f:\omega\to\omega$ be a witness for
$\mathcal{I}\leq_{K}\mathcal{H}$ and define $y_{n}=x_{f(n)}$ for all
$n\in\omega$. Then there is an infinite $D\subseteq\omega$ and $x\in X$ such
that for each open neighborhood $U$ of the point $x$ there is $m\in\omega$
with $\\{y_{n}:n\in\operatorname{FS}(D\setminus m)\\}\subseteq U$. We have
$f[\operatorname{FS}(D\setminus n)]\notin\mathcal{I}$ for each $n\in\omega$.
Consider the decreasing sequence of $\mathcal{I}$-positive sets given by
$A_{i}=f[\operatorname{FS}(D\setminus i)]$ for all $i\in\omega$. Since
$\mathcal{I}$ is a P+-ideal, there is $B\notin\mathcal{I}$ with $B\setminus
A_{i}$ finite for all $i\in\omega$. We claim that $\langle x_{n}\rangle_{n\in
B}$ converges. Indeed, let $U$ be an open neighborhood of $x$. Then there is
$k\in\omega$ with $\\{x_{n}:n\in f[\operatorname{FS}(D\setminus
k)]\\}=\\{y_{n}:n\in\operatorname{FS}(D\setminus k)\\}\subseteq U$. Thus,
$\\{n\in B:x_{n}\notin U\\}\subseteq B\setminus f[FS(D\setminus k)]=B\setminus
A_{k}$ and the latter set is finite. ∎

######  Example 2.7.

The family $\mathcal{F}=\\{A\subseteq\omega:\exists k\in\omega\,\forall
D\in[\omega]^{k}\,(\operatorname{FS}(D)\setminus A\neq\emptyset)\\}$ is an
ideal on $\mathbb{N}$ (it follows from a version of Folkman’s theorem see e.g.
[1, a remark before Theorem 1.3 at page 43]). (This ideal was introduced in
[8], where the author proved that consistently there is an $\mathcal{F}$-space
which is not an $\mathcal{I}_{1/n}$-space. This ideal was also considered in
[18], where the authors posed a question whether it is homogeneous.) It is not
difficult to show that $\mathcal{F}$ is $F_{\sigma}$. It is obvious that
$\mathcal{F}\subseteq\mathcal{H}$, so $\mathcal{F}\leq_{K}\mathcal{H}$ and
each Hindman space is an $\mathcal{F}$-space.

Combining Theorem 2.5 and Proposition 2.6 we get the following result.

######  Corollary 2.8.

If $\mathcal{I}$ is an $F_{\sigma}$ ideal on $\omega$, then the following
conditions are equivalent.

  1. (1)

$\mathcal{I}\not\leq_{K}\mathcal{H}$.

  2. (2)

The continuum hypothesis implies that there is a Hindman space which is not an
$\mathcal{I}$-space.

  3. (3)

It is consistent that there is a Hindman space which is not an
$\mathcal{I}$-space.

In [13] Jones posed a question whether it is consistent that there is a
Hindman space which is not van der Waerden. Since the ideal $\mathcal{W}$ is
$F_{\sigma}$ and $\mathcal{W}$-spaces coincide with van der Waerden spaces,
the question of Jones is reduced to a purely combinatorial question whether
$\mathcal{W}\not\leq_{K}\mathcal{H}$.

##  3\. A Hindman space which is not an $\mathcal{I}_{1/n}$-space

From the previous section we know that there are $F_{\sigma}$ ideals below
$\mathcal{H}$ in the Katětov order. At this point one may ask about existence
of $F_{\sigma}$ ideals that are not below $\mathcal{H}$ in the Katětov order.
In this section we show that $\mathcal{I}_{1/n}$ is an example of such ideal.
By the main theorem from the previous section, this answers in positive a
question posed by Jana Flašková at the 22nd Summer Conference on Topology and
its Applications [7] about existence of a Hindman space which is not
$\mathcal{I}_{1/n}$-space.

######  Lemma 3.1.

If $D\in[\omega]^{\omega}$ is very sparse and $E\in[\omega]^{\omega}$ is such
that $\operatorname{FS}(E)\subseteq\operatorname{FS}(D)$, then for every
$d\in\omega$ there is $e\in\omega$ such that $\operatorname{FS}(E\setminus
e)\subseteq\operatorname{FS}(D\setminus d)$.

###### Proof.

Suppose that there is $d\in\omega$ such that for every $e\in\omega$ we have
$\operatorname{FS}(E\setminus e)\setminus\operatorname{FS}(D\setminus
d)\neq\emptyset$. Then we pick inductively numbers $x_{n}$ such that
$x_{n}\in\operatorname{FS}(E\setminus\\{0,1,\dots,x_{n-1}\\})\setminus\operatorname{FS}(D\setminus
d)$. Then $x_{n}+x_{k}\in\operatorname{FS}(E)$ for any $n\neq k$. Since
$x_{n}\in\operatorname{FS}(E)\subseteq\operatorname{FS}(D)$ and $D$ is sparse,
there is a unique set $F_{n}\subseteq D$ with $x_{n}=\sum F_{n}$. Since
$x_{n}\notin\operatorname{FS}(D\setminus d)$, we have $F_{n}\cap
d\neq\emptyset$. Thus, there is $n\neq k$ with $F_{n}\cap F_{k}\neq\emptyset$.
But $D$ is very sparse, so $x_{n}+x_{k}\notin\operatorname{FS}(D)$, a
contradiction. ∎

######  Theorem 3.2.

  1. (1)

$\mathcal{I}_{1/n}\not\leq_{K}\mathcal{H}$ i.e. there is no function
$f:\omega\to\omega$ such that $f^{-1}[A]\in\mathcal{H}$ for every
$A\in\mathcal{I}_{1/n}$.

  2. (2)

Assume the continuum hypothesis. There is a Hindman space which is not
$\mathcal{I}_{1/n}$-space.

###### Proof.

Since (2) follows from (1) and Theorem 2.5, below we show only (1).

Let $f:\omega\to\omega$ be any function. If there is $n\in\omega$ with
$f^{-1}[\\{n\\}]\notin\mathcal{H}$ then we are done, so we can assume that
$f^{-1}[\\{n\\}]\in\mathcal{H}$ for all $n\in\omega$.

Let $a_{0},a_{1},\ldots\in\omega$ be such that

$\sum_{n\in\omega}\frac{2^{n+1}}{a_{n}+1}<\infty.$

Since $f^{-1}[\\{0,1,\ldots,a_{n}\\}]\in\mathcal{H}$ for all $n\in\omega$,
there is a very sparse infinite $D\subseteq\omega$ such that for each
$n\in\omega$ there is $k\in\omega$ with $FS(D\setminus k)\cap
f^{-1}[\\{0,1,\ldots,a_{n}\\}]=\emptyset$ (by Lemmas 2.2 and 2.4). By
shrinking and re-enumerating $D$ if necessary, we can assume that
$\operatorname{FS}(D\setminus d_{n})\cap
f^{-1}[\\{0,1,\ldots,a_{n}\\}]=\emptyset$, where $\langle
d_{i}:i\in\omega\rangle$ is the increasing enumeration of $D$.

We will inductively pick $x_{n}\in\omega$, $D_{n}\in[\omega]^{\omega}$ and
auxiliary functions $g_{n}$ such that for each $n\in\omega$:

  * •

$x_{n}<x_{n+1}$, $x_{n}\in D_{n}$ and $\operatorname{FS}(D_{n}\setminus
x_{n})\cap f^{-1}[\\{0,1,\ldots,a_{n}\\}]=\emptyset$;

  * •

$\operatorname{FS}(D)\supseteq\operatorname{FS}(D_{n})\supseteq\operatorname{FS}(D_{n+1})$,

  * •

$g_{n}:\operatorname{FS}(\\{x_{i}:i<n\\})\to\\{0,1\\}$,

and for each $y\in\operatorname{FS}(\\{x_{i}:i<n\\})$:

  * •

if $g_{n}(y)=1$ then $(y+\operatorname{FS}(D_{n}))\cap
f^{-1}[\\{0,1,\ldots,a_{n}\\}]=\emptyset$ (in particular, $f(y+x_{n})>a_{n}$),

  * •

if $g_{n}(y)=0$ and either $g_{n-1}(y)=0$ or
$y\in(x_{n-1}+(\operatorname{FS}(\\{x_{i}:i<n-1\\})\cap
g^{-1}_{n-1}[\\{0\\}]))$ then $f\restriction(y+\operatorname{FS}(D_{n}))$ is a
constant function whose value is some $c\in
f[\operatorname{FS}(\\{x_{i}:i<n\\})]$,

  * •

if $g_{n}(y)=0$ and either $g_{n-1}(y)=1$ or
$y\in(x_{n-1}+(\operatorname{FS}(\\{x_{i}:i<n-1\\})\cap
g^{-1}_{n-1}[\\{1\\}]))$ or $y=x_{n-1}$ then $f\restriction(y+FS(D_{n}))$ is a
constant function whose value is some $c>a_{n-1}$.

(The initial step for $n=0$) We start with $D_{0}=D$, $g_{0}=\emptyset$ and
any $x_{0}\in D_{0}$. Since $d_{0}\leq x_{0}$, we have
$\operatorname{FS}(D_{0}\setminus x_{0})\cap
f^{-1}[\\{0,1,\ldots,a_{0}\\}]=\emptyset$.

(The initial step for $n=1$) We have 2 cases.

_Case 1._ There is $c\in\omega$ such that $f^{-1}[\\{c\\}]$ contains
$x_{0}+\operatorname{FS}(C)$ for some infinite $C$ with
$\operatorname{FS}(C)\subseteq\operatorname{FS}(D_{0}\setminus\\{0,1,\dots,x_{0}\\})$.
We define $D_{1}=C$, $g_{1}(x_{0})=0$ and, using Lemma 3.1, pick $x_{1}\in
D_{1}$ such that $\operatorname{FS}(D_{1}\setminus
x_{1})\subseteq\operatorname{FS}(D\setminus d_{1})$.

Since $\operatorname{FS}(D\setminus d_{1})\cap
f^{-1}[\\{0,1,\ldots,a_{1}\\}]=\emptyset$, $\operatorname{FS}(D_{1}\setminus
x_{1})\cap f^{-1}[\\{0,1,\ldots,a_{1}\\}]=\emptyset$.

Since $x_{0}+\operatorname{FS}(D_{1})\subseteq
x_{0}+\operatorname{FS}(D_{0}\setminus\\{0,1,\dots,x_{0}\\})\subseteq\operatorname{FS}(D_{0}\setminus
x_{0})$ and $\operatorname{FS}(D_{0}\setminus x_{0})\cap
f^{-1}[\\{0,1,\ldots,a_{0}\\}]=\emptyset$, we obtain $c>a_{0}$.

_Case 2._ For each $c\in\omega$ the set $f^{-1}[\\{c\\}]$ does not contain any
$x_{0}+\operatorname{FS}(C)$ for infinite $C$ with
$\operatorname{FS}(C)\subseteq\operatorname{FS}(D_{0}\setminus\\{0,1,\dots,x_{0}\\})$,
so using Lemma 2.4 we can find an infinite $C$ with
$\operatorname{FS}(C)\subseteq\operatorname{FS}(D_{0}\setminus\\{0,1,\dots,x_{0}\\})$
such that for each $m\in\omega$ there is $k_{m}\in\omega$ with
$(x_{0}+\operatorname{FS}(C\setminus k_{m}))\cap
f^{-1}[\\{0,1,\ldots,a_{m}\\}]=\emptyset$. Put $D_{1}=C\setminus k_{1}$,
$g_{1}(x_{0})=1$ and, using Lemma 3.1, pick $x_{1}\in D_{1}$ such that
$\operatorname{FS}(D_{1}\setminus x_{1})\subseteq\operatorname{FS}(D\setminus
d_{1})$.

Since $\operatorname{FS}(D\setminus d_{1})\cap
f^{-1}[\\{0,1,\ldots,a_{1}\\}]=\emptyset$, $\operatorname{FS}(D_{1}\setminus
x_{1})\cap f^{-1}[\\{0,1,\ldots,a_{1}\\}]=\emptyset$

Since
$x_{1}\in\operatorname{FS}(D_{1})\subseteq\operatorname{FS}(D_{0}\setminus\\{0,1,\dots,x_{0}\\})$,
we obtain $x_{0}<x_{1}$.

Moreover, $(x_{0}+\operatorname{FS}(D_{1}))\cap
f^{-1}[\\{0,1,\ldots,a_{1}\\}]=(x_{0}+\operatorname{FS}(C\setminus k_{1}))\cap
f^{-1}[\\{0,1,\ldots,a_{1}\\}]=\emptyset$.

(The inductive step) Suppose now that $x_{i}$, $g_{i}$ and $D_{i}$ for all
$i<n$ are already defined. Let

$\begin{split}\\{t_{0},\ldots,t_{k}\\}&=\operatorname{FS}(\\{x_{i}:i<n\\})\setminus\\\
&\left(g_{n-1}^{-1}[\\{0\\}]\cup(x_{n-1}+(\operatorname{FS}(\\{x_{i}:i<n-1\\})\cap
g_{n-1}^{-1}[\\{0\\}]))\right)\end{split}$

and define $g_{n}(y)=0$ for each $y\in
g_{n-1}^{-1}[\\{0\\}]\cup(x_{n-1}+(\operatorname{FS}(\\{x_{i}:i<n-1\\})\cap
g_{n-1}^{-1}[\\{0\\}]))$.

We will inductively construct auxiliary sets
$E_{-1},E_{0},\ldots,E_{k}\in[\omega]^{\omega}$ such that
$\operatorname{FS}(D_{n-1}\setminus\\{0,1,\dots,x_{n-1}\\})\supseteq\operatorname{FS}(E_{-1})\supseteq\dots\supseteq\operatorname{FS}(E_{k})$.

We start with $E_{-1}=D_{n-1}\setminus\\{0,1,\dots,x_{n-1}\\}$. Suppose now
that $i\leq k$ and $E_{j}$ for all $j<i$ are defined. We have 2 cases.

_Case 1._ There is $c\in\omega$ such that $f^{-1}[\\{c\\}]$ contains
$t_{i}+\operatorname{FS}(C)$ for some infinite $C$ with
$\operatorname{FS}(C)\subseteq\operatorname{FS}(E_{i-1})$. We define $E_{i}=C$
and $g_{n}(t_{i})=0$.

Note that $c>a_{n-1}$ in this case. Indeed, we have 3 cases:

  1. (1)

$t_{i}=x_{n-1}$,

  2. (2)

$t_{i}\in x_{n-1}+(\operatorname{FS}(\\{x_{j}:j<n-1\\})\cap
g_{n-1}^{-1}[\\{1\\}])$,

  3. (3)

$t_{i}\in\operatorname{FS}(\\{x_{j}:j<n-1\\})$.

In the first case, $t_{i}+\operatorname{FS}(E_{i})\subseteq
t_{i}+\operatorname{FS}(D_{n-1}\setminus\\{0,1,\dots,x_{n-1}\\})$ and
$\operatorname{FS}(D_{n-1}\setminus x_{n-1})\cap
f^{-1}[\\{0,\ldots,a_{n-1}\\}]=\emptyset$, so $c>a_{n-1}$.

In the second case, $t_{i}=y+x_{n-1}$ for some
$y\in\operatorname{FS}(\\{x_{j}:j<n-1\\})$ with $g_{n-1}(y)=1$. Then
$t_{i}+\operatorname{FS}(E_{i})\subseteq
t_{i}+\operatorname{FS}(D_{n-1}\setminus\\{0,1,\dots,x_{n-1}\\})=y+x_{n-1}+\operatorname{FS}(D_{n-1}\setminus\\{0,1,\dots,x_{n-1}\\})\subseteq
y+\operatorname{FS}(D_{n-1}\setminus x_{n-1})$. Since $g_{n-1}(y)=1$,
$(y+\operatorname{FS}(D_{n-1}))\cap
f^{-1}[\\{0,1,\ldots,a_{n-1}\\}]=\emptyset$. Thus $c>a_{n-1}$.

In the third case, $g_{n-1}(t_{i})=1$, so
$(t_{i}+\operatorname{FS}(D_{n-1}))\cap
f^{-1}[\\{0,1,\ldots,a_{n-1}\\}]=\emptyset$. Since
$t_{i}+\operatorname{FS}(E_{i})\subseteq t_{i}+\operatorname{FS}(D_{n-1})$, we
obtain $c>a_{n-1}$.

_Case 2._ For each $c\in\omega$ the set $f^{-1}[\\{c\\}]$ does not contain any
$t_{i}+\operatorname{FS}(C)$ for infinite $C$ with
$\operatorname{FS}(C)\subseteq\operatorname{FS}(E_{i-1})$, so using Lemma 2.4
we can find an infinite $C$ with
$\operatorname{FS}(C)\subseteq\operatorname{FS}(E_{i-1})$ such that for each
$m\in\omega$ there is $k_{m}\in\omega$ with
$(t_{i}+\operatorname{FS}(C\setminus k_{m}))\cap
f^{-1}[\\{0,1,\ldots,a_{m}\\}]=\emptyset$. Put $E_{i}=C\setminus k_{n}$ and
$g_{n}(t_{i})=1$.

Once all $E_{i}$ for $i\leq k$ are constructed, put $D_{n}=E_{k}$ and, using
Lemma 3.1, pick $x_{n}\in D_{n}$ such that $\operatorname{FS}(D_{n}\setminus
x_{n})\subseteq\operatorname{FS}(D\setminus d_{n})$.

This finishes the inductive construction of the sequences $\langle
x_{n}\rangle$, $\langle D_{n}\rangle$ and $\langle g_{n}\rangle$.

Define $X=\\{x_{i}:i\in\omega\\}$. Obviously,
$\operatorname{FS}(X)\notin\mathcal{H}$. To end the proof we need to check
that $f[\operatorname{FS}(X)]\in\mathcal{I}_{1/n}$ i.e. we have to show that

$\sum_{i\in f[\operatorname{FS}(X)]}\frac{1}{i+1}=\sum_{i\in
f[\operatorname{FS}(X_{0})]}\frac{1}{i+1}+\sum_{n=1}^{\infty}\left(\sum_{i\in
f[\operatorname{FS}(X_{n})]\setminus
f[\operatorname{FS}(X_{n-1})]}\frac{1}{i+1}\right)<\infty,$

where $X_{n}=\\{x_{0},x_{1},\ldots,x_{n}\\}$ for every $n\in\omega$.

Let $n\geq 1$. If $i\in f[\operatorname{FS}(X_{n})]\setminus
f[\operatorname{FS}(X_{n-1})]$, then either $i=f(x_{n})$ or $i=f(y+x_{n})$ for
some $y\in\operatorname{FS}(X_{n-1})$. It follows from the properties of the
sequence $\langle x_{n}\rangle$ that $f(x_{n})>a_{n}$ (hence
$1/(i+1)<1/(a_{n}+1)$) and for the value $f(y+x_{n})$ we have three cases.

  * •

If $g_{n}(y)=1$ then $f(y+x_{n})>a_{n}$, so $1/(i+1)<1/(a_{n}+1)$.

  * •

If $g_{n}(y)=0$ and either $g_{n-1}(y)=0$ or
$y\in(x_{n-1}+(\operatorname{FS}(\\{x_{i}:i<n-1\\})\cap
g^{-1}_{n-1}[\\{0\\}]))$, then $f\restriction(y+\operatorname{FS}(D_{n}))$ is
a constant function whose value is some $c\in
f[\operatorname{FS}(\\{x_{i}:i<n\\})]$. Note that this case is not possible
here because $i\notin f[\operatorname{FS}(X_{n-1})]$.

  * •

If $g_{n}(y)=0$ and either $g_{n-1}(y)=1$ or
$y\in(x_{n-1}+(\operatorname{FS}(\\{x_{i}:i<n-1\\})\cap
g^{-1}_{n-1}[\\{1\\}]))$ or $y=x_{n-1}$, then $f\restriction(y+FS(D_{n}))$ is
a constant function whose value is some $c>a_{n-1}$, so
$1/(i+1)<1/(a_{n-1}+1)$.

Thus,

$\begin{split}\sum_{i\in f[\operatorname{FS}(X_{n})]\setminus
f[\operatorname{FS}(X_{n-1})]}\frac{1}{i+1}&\leq\frac{1}{a_{n}+1}+\frac{|\operatorname{FS}(X_{n-1})|}{a_{n-1}+1}\leq\frac{2^{n}}{a_{n-1}+1}.\end{split}$

All in all,

$\sum_{i\in f[\operatorname{FS}(X)]}\frac{1}{i+1}\leq\sum_{i\in
f[\operatorname{FS}(X_{0})]}\frac{1}{i+1}+\sum_{n=1}^{\infty}\frac{2^{n}}{a_{n-1}+1}=\frac{1}{f(x_{0})+1}+\sum_{n=1}^{\infty}\frac{2^{n}}{a_{n-1}+1}<\infty.$

This finishes the proof. ∎

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