# Katětov order between Hindman, Ramsey, van der Waerden and summable ideals

Rafał Filipów  Institute of Mathematics
Faculty of Mathematics, Physics and Informatics
University of Gdańsk
ul. Wita Stwosza 57
80-308 Gdańsk
Poland<EMAIL_ADDRESS>http://mat.ug.edu.pl/~rfilipow ,  Krzysztof
Kowitz  Institute of Mathematics
Faculty of Mathematics
Physics and Informatics
University of Gdańsk
ul. Wita Stwosza 57
80-308 Gdańsk
Poland<EMAIL_ADDRESS>and  Adam Kwela  Institute of
Mathematics
Faculty of Mathematics
Physics and Informatics
University of Gdańsk
ul. Wita Stwosza 57
80-308 Gdańsk
Poland<EMAIL_ADDRESS>https://mat.ug.edu.pl/~akwela

###### Abstract.

A family $\mathcal{I}$ of subsets of a set $X$ is an _ideal on $X$_ if it is
closed under taking subsets and finite unions of its elements. An ideal
$\mathcal{I}$ on $X$ is below an ideal $\mathcal{J}$ on $Y$ in the _Katětov
order_ if there is a function $f:Y\to X$ such that $f^{-1}[A]\in\mathcal{J}$
for every $A\in\mathcal{I}$. We show that the Hindman ideal, the Ramsey ideal
and the summable ideal are pairwise incomparable in the Katětov order, where

  * •

the _Ramsey ideal_ consists of those sets of pairs of natural numbers which do
not contain a set of all pairs of any infinite set (equivalently do not
contain, in a sense, any infinite complete subgraph),

  * •

the _Hindman ideal_ consists of those sets of natural numbers which do not
contain any infinite set together with all finite sums of its members
(equivalently do not contain IP-sets that are considered in Ergodic Ramsey
theory),

  * •

the _summable ideal_ consists of those sets of natural numbers such that the
series of the reciprocals of its members is convergent.

Moreover, we show that in the Katětov order the above mentioned ideals are not
below the _van der Waerden ideal_ that consists of those sets of natural
numbers which do not contain arithmetic progressions of arbitrary finite
length.

###### Key words and phrases:

Katetov order, ideal, filter, Ramsey’s theorem for coloring graphs, Hindman’s
finite sums theorem, van der Waerden’s arithmetical progressions theorem

###### 2020 Mathematics Subject Classification:

Primary: 03E05, 05D10.

##  1\. Introduction

The Katětov order is an efficient tool for studying ideals over countable sets
[19, 20, 21, 22, 35, 37]. Originally, the Katětov order (introduced by Katětov
[24] in 1968) was used to study convergence in topological spaces, and our
interest in Katětov order between the Hindman, Ramsey, van der Waerden and
summable ideals stems from the study of sequentially compact spaces defined
as, in a sense, topological counterparts of well-known combinatorial theorems:
Ramsey’s theorem for coloring graphs, Hindman’s finite sums theorem and van
der Waerden’s arithmetical progressions theorem [3, 4, 25, 26, 27, 28]. It is
known [11, 30] that an existence of a sequentially compact space which
distinguishes the above mentioned classes of spaces is reducible to a question
whether particular ideals are incomaparable in the Katětov order.

Beside our primary interest in the Katětov order described above we mention
one more strength of this order. Using the Katětov order, we can classify non-
definable objects (like ultrafilters or maximal almost disjoint families)
using Borel ideals [20]. For instance, an ultrafilter $\mathcal{U}$ is a
P-point if and only if the dual ideal $\mathcal{U}^{*}$ is not Katětov above
$\mathrm{Fin}^{2}$ (equivalently $\mathcal{U}$ is a
$\mathrm{Fin}^{2}$-ultrafilter as defined by Baumgartner [2]). It is known
[10] that an existence of an ultrafilter which distinguishes between some
classes of ultrafilters is reducible to a question whether particular ideals
are incomaparable in the Katětov order.

Below we describe the results obtain in this paper and introduce a necessary
notions and notations.

We write $\omega$ to denote the set of all natural numbers (with zero).

We write $[A]^{2}$ to denote the set of all unordered pairs of elements of
$A$, $[A]^{<\omega}$ to denote the family of all finite subsets of $A$ and
$[A]^{\omega}$ to denote the family of all infinite countable subsets of $A$.

A family $\mathcal{I}\subseteq\mathcal{P}(X)$ of subsets of a set $X$ is an
_ideal on $X$_ if it is closed under taking subsets and finite unions of its
elements, $X\notin\mathcal{I}$ and $\mathcal{I}$ contains all finite subsets
of $X$. By $\mathrm{Fin}(X)$ we denote the family of all finite subsets of $X$
and we write $\mathrm{Fin}$ instead of $\mathrm{Fin}(\omega)$.

For an ideal $\mathcal{I}$ on $X$, we write $\mathcal{I}^{+}=\\{A\subseteq
X:A\notin\mathcal{I}\\}$ and call it the _coideal of $\mathcal{I}$_, and we
write $\mathcal{I}^{*}=\\{X\setminus A:A\in\mathcal{I}\\}$ and call it the
_filter dual to $\mathcal{I}$_. It is easy to see that
$\mathcal{I}\restriction A=\\{A\cap B:B\in\mathcal{I}\\}$ is an ideal on $A$
if and only if $A\in\mathcal{I}^{+}$.

For a set $B\subseteq\omega$, we write $FS(B)$ to denote the set of all finite
(nonempty) sums of distinct elements of $B$ i.e.
$\operatorname{FS}(B)=\\{\sum_{n\in
F}n:F\in[B]^{<\omega}\setminus\\{\emptyset\\}\\}$.

In this paper we are interested in the following four ideals:

  * •

the _Ramsey ideal_

$\mathcal{R}=\left\\{A\subseteq[\omega]^{2}:\forall
B\in[\omega]^{\omega}\,([B]^{2}\not\subseteq A)\right\\},$

  * •

the _Hindman ideal_

$\mathcal{H}=\left\\{A\subseteq\omega:\forall
B\in[\omega]^{\omega}\,\operatorname{FS}(B)\not\subseteq A\right\\},$

  * •

the _van der Waerden ideal_

$\begin{split}\mathcal{W}=\\{A\subseteq\omega:&\text{ $A$ does not contain
arithmetic progressions}\\\ &\text{ of arbitrary finite
length}\\},\end{split}$

  * •

the _summable ideal_

$\mathcal{I}_{1/n}=\left\\{A\subseteq\omega:\sum_{n\in
A}\frac{1}{n+1}<\infty\right\\}.$

The Ramsey ideal was introduced by Meza-Alcántara and Hrušák [22] (the authors
noted that if we identify a set $A\subseteq[\omega]^{2}$ with a graph
$G_{A}=(\omega,A)$, the ideal $\mathcal{R}$ can be seen as an ideal consisting
of graphs without infinite complete subgraphs). Both the Hindman and van der
Waerden ideals were introduced by Flaškova [14, p. 109]. The summable ideal is
a particular instance of the so-called _summable ideals_ which seem to be
“ancient” compare to previously mentioned ideals as they were introduced in
1972 by Mathias [31, Example 3, p.206].

We say that an ideal $\mathcal{I}$ on $X$ is below an ideal $\mathcal{J}$ on
$Y$ in the _Katětov order_ [24] if there is a function $f:Y\to X$ such that
$f^{-1}[A]\in\mathcal{J}$ for every $A\in\mathcal{I}$ (equivalently,
$f[B]\notin\mathcal{I}$ for all $B\notin\mathcal{J}$). Note that the Katětov
order has been extensively examined (even in its own right) for many years so
far [1, 2, 5, 6, 7, 16, 17, 18, 19, 20, 21, 22, 34, 35, 36, 37, 39].

The aim of this paper is to prove the following

######  Theorem 1.1.

  1. (1)

The ideals $\mathcal{R}$, $\mathcal{H}$ and $\mathcal{I}_{1/n}$ are pairwise
incomparable in the Katětov order.

  2. (2)

The ideals $\mathcal{R}$, $\mathcal{H}$ and $\mathcal{I}_{1/n}$ are not below
the ideal $\mathcal{W}$ in the Katětov order.

As far as we are concerned, the remaining three questions about these ideals
are still open.

######  Question 1.2.

Is the ideal $\mathcal{W}$ below the ideal $\mathcal{R}$ ($\mathcal{H}$,
$\mathcal{I}_{1/n}$, resp.) in the Katětov order?

Note that in the case of the summable ideal, Question 1.2 is a weakening of
the famous Erdős-Turán conjecture which says that
$\mathcal{W}\subseteq\mathcal{I}_{1/n}$.

##  2\. Preliminaries

An ideal $\mathcal{I}$ on $X$ is _tall_ [32, Definition 0.6] if for every
infinite set $A\subseteq X$ there exists an infinite set $B\subseteq A$ such
that $B\in\mathcal{I}$. It is not difficult to see that $\mathcal{I}$ is not
tall $\iff$ $\mathcal{I}\leq_{K}\mathcal{J}$ for every ideal $\mathcal{J}$
$\iff$ $\mathcal{I}\leq_{K}\mathrm{Fin}$ $\iff$ $\mathcal{I}\restriction
A=\mathrm{Fin}(A)$ for some $A\in\mathcal{I}^{+}$. It is easy to show the
following

######  Proposition 2.1.

The ideals $\mathcal{H}$, $\mathcal{R}$, $\mathcal{W}$ and $\mathcal{I}_{1/n}$
are tall.

Ideals $\mathcal{I}$ and $\mathcal{J}$ on $X$ and $Y$, respectively are
_isomorphic_ (in short: $\mathcal{I}\approx\mathcal{J}$) if there exists a
bijection $\phi:X\to Y$ such that $A\in\mathcal{I}\iff\phi[A]\in\mathcal{J}$
for each $A\subseteq X$. An ideal $\mathcal{I}$ is _homogeneous_ [29,
Definition 1.3] if the ideals $\mathcal{I}$ and $\mathcal{I}\restriction A$
are isomorphic for every $A\in\mathcal{I}^{+}$.

######  Proposition 2.2 ([29, Examples 2.5 and 2.6]).

The ideals $\mathcal{H}$, $\mathcal{R}$ and $\mathcal{W}$ are homogeneous.

By identifying subsets of $X$ with their characteristic functions, we equip
$\mathcal{P}(X)$ with the topology of the space $2^{X}$ (the product topology
of countably many copies of the discrete topological space $\\{0,1\\}$) and
therefore we can assign topological notions to ideals on $X$. In particular,
an ideal $\mathcal{I}$ is _Borel_ (_$F_{\sigma}$_ , resp.) if $\mathcal{I}$ is
a Borel ($F_{\sigma}$, resp.) subset of $2^{X}$.

If $A\subseteq\omega$ and $n\in\omega$, we write $A+n=\\{a+n:a\in A\\}$ and
$A-n=\\{a-n:a\in A,a\geq n\\}$.

A set $D\subseteq\omega$ is _sparse_ [25, p. 1598] if for each
$x\in\operatorname{FS}(D)$ there exists the unique set $\alpha\subseteq D$
such that $x=\sum_{n\in\alpha}n$. This unique set will be denoted by
$\alpha_{D}(x)$. For instance, the set $E=\\{2^{n}:n\in\omega\\}$ is sparse,
and in the sequel, we write $\alpha(x)$ instead of $\alpha_{E}(x)$.

A set $D\subseteq\omega$ is _very sparse_ [12, p. 894] if it is sparse and

$\forall
x,y\in\operatorname{FS}(D)\,\left(\alpha_{D}(x)\cap\alpha_{D}(y)\neq\emptyset\implies
x+y\notin\operatorname{FS}(D)\right).$

In the sequel, we will use the following

######  Lemma 2.3 ([12, Lemma 2.2]).

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

##  3\. Summable and van der Waerden ideals are not above Hindman and Ramsey
ideals

To show that $F_{\sigma}$ ideals are not above $\mathcal{H}$ nor $\mathcal{R}$
in the Katětov order one can use the following ideal on $\omega^{2}$
introduced by Katětov [23, Definition 5.1]:

$\mathrm{Fin}^{2}=\left\\{C\subseteq\omega^{2}:\\{n\in\omega:\\{k\in\omega:(n,k)\in
C\\}\notin\mathrm{Fin}\\}\in\mathrm{Fin}\right\\}.$

The following lemma and proposition can be found in [11], but we decided to
include proofs here for the sake of completeness.

######  Lemma 3.1 ([11, Proposition 7.2]).

  * (a)

$\mathrm{Fin}^{2}\leq_{K}\mathcal{H}$.

  * (b)

$\mathrm{Fin}^{2}\leq_{K}\mathcal{R}$.

###### Proof.

(a): Let $A_{k}=\\{2^{k}(2n+1):n\in\omega\\}$ for each $k\in\omega$. Let
$f:\omega\to\omega^{2}$ be any injective function such that
$f[A_{k}]\subseteq\\{k\\}\times\omega$ for all $k\in\omega$. In [12, item (2)
in the proof of Proposition 1.1], the authors showed that
$A_{k}\in\mathcal{H}$ for every $k\in\omega$ (so
$f^{-1}[\\{k\\}\times\omega]\in\mathcal{H}$ for all $k\in\omega$), whereas in
[12, item (1) in the proof of Proposition 1.1] it is shown that for every
$B\notin\mathcal{H}$ there is $k\in\omega$ such that $B\cap A_{k}$ is infinite
(so $f^{-1}[C]\in\mathcal{H}$ whenever $C\subseteq\omega^{2}$ is such that
$C\cap(\\{k\\}\times\omega)$ is finite for all $k\in\omega$). Thus, the
function $f$ witnesses the fact that $\mathrm{Fin}^{2}\leq_{K}\mathcal{H}$.

(b): Let $A_{n}=\\{\\{k,i\\}:i>k\geq n\\}$ for every $n\in\omega$. Then
$A_{n}\notin\mathcal{R}$, $A_{0}=[\omega]^{2}$,
$\bigcap_{n\in\omega}A_{n}=\emptyset$ and $A_{n}\setminus
A_{n+1}=\\{\\{n,i\\}:i>n\\}\in\mathcal{R}$. Let $f:[\omega]^{2}\to\omega^{2}$
be any injective function such that $f[A_{n}\setminus
A_{n+1}]\subseteq\\{n\\}\times\omega$. Then
$f^{-1}[\\{n\\}\times\omega]\in\mathcal{R}$ for all $n\in\omega$. Suppose, for
sake of contradiction, that there is $C\subseteq\omega^{2}$ such that
$C\cap(\\{k\\}\times\omega)$ is finite for all $k\in\omega$ (so
$C\in\mathrm{Fin}^{2}$), but $B=f^{-1}[C]\notin\mathcal{R}$. Then
$B\subseteq^{*}A_{n}$ for every $n\in\omega$. Let $H=\\{h_{n}:n\in\omega\\}$
be an infinite set such that $[H]^{2}\subseteq B$ and $h_{n}<h_{n+1}$ for
every $n\in\omega$. Since $[H]^{2}\subseteq^{*}A_{h_{1}}$, there is a finite
set $F$ such that $[H]^{2}\setminus F\subseteq A_{h_{1}}$. Since $F$ is
finite, there is $k>0$ such that $\\{h_{0},h_{n}\\}\notin F$ for every $n\geq
k$. Then $\\{\\{h_{0},h_{n}\\}:n\geq k\\}\subseteq[H]^{2}\setminus F$ and
$\\{\\{h_{0},h_{n}\\}:n\geq k\\}\cap A_{h_{1}}=\emptyset$, a contradiction. ∎

######  Proposition 3.2 ([11, Theorem 7.7]).

  * (a)

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

  * (b)

$\mathcal{R}\not\leq_{K}\mathcal{W}$.

  * (c)

$\mathcal{H}\not\leq_{K}\mathcal{I}_{1/n}$.

  * (d)

$\mathcal{R}\not\leq_{K}\mathcal{I}_{1/n}$.

###### Proof.

(a): Suppose otherwise: $\mathcal{H}\leq_{K}\mathcal{W}$. Using Lemma 3.1 we
get that $\mathrm{Fin}^{2}\leq_{K}\mathcal{H}$, so
$\mathrm{Fin}^{2}\leq_{K}\mathcal{W}$. However, since $\mathcal{W}$ is
$F_{\sigma}$ (see [13, Example 4.12]),
$\mathrm{Fin}^{2}\not\leq_{K}\mathcal{W}$ (by [8, Theorems 7.5 and 9.1] and
[1, Example 4.1]). A contradiction.

The proofs of items (b), (c) and (d) are similar to the proof of item (a),
since $\mathrm{Fin}^{2}\leq_{K}\mathcal{R}$ (by Lemma 3.1) and
$\mathcal{I}_{1/n}$ is $F_{\sigma}$ (see [33, Example 1.5]). ∎

##  4\. Summable ideal is not below van der Waerden ideal

######  Proposition 4.1.

$\mathcal{I}_{1/n}\not\leq_{K}\mathcal{W}$.

###### Proof.

Suppose for sake of contradiction that there is a function
$\phi:\omega\to\omega$ such that $\phi^{-1}[B]\in\mathcal{W}$ for every
$B\in\mathcal{I}_{1/n}$. We construct a sequence $(F_{n}:n\in\omega)$ of
finite subsets of $\omega$ such that for every $n\in\omega$ we have

  1. (1)

$F_{n}$ is an arithmetic progression of length $n$,

  2. (2)

$\phi(x)\geq n2^{n}$ for every $x\in F_{n}$.

Suppose that $F_{i}$ are constructed for $i<n$. Since
$B=\\{i\in\omega:i<n2^{n}\\}$ is finite, $A=\phi^{-1}[B]\in\mathcal{W}$. Then
$\omega\setminus A\notin\mathcal{W}$, so there is an arithmetic progression
$F_{n}\subseteq\omega\setminus A$ of length $n$. This finishes the
construction of $F_{n}$.

Let $A=\bigcup\\{F_{n}:n\in\omega\\}$. Then $A\notin\mathcal{W}$, but

$\sum_{y\in\phi[A]}\frac{1}{y+1}\leq\sum_{n\in\omega}\left(\sum_{x\in
F_{n}}\frac{1}{\phi(x)+1}\right)\leq\sum_{n\in\omega}\left(\sum_{x\in
F_{n}}\frac{1}{n2^{n}+1}\right)=\sum_{n\in\omega}\frac{n}{n2^{n}+1}<\infty,$

so $\phi[A]\in\mathcal{I}_{1/n}$, a contradiction. ∎

##  5\. Summable ideal is not below Hindman ideal

######  Theorem 5.1.

$\mathcal{I}_{1/n}\not\leq_{K}\mathcal{H}$.

###### Proof.

This is proved in [12, Theorem 3.2], but below we provide a simpler proof.

Let $\phi:\omega\to\omega$ be an arbitrary function. We will show that $\phi$
is not a witness for $\mathcal{I}_{1/n}\leq_{K}\mathcal{H}$ i.e. we will find
an infinite set $D\subseteq\omega$ such that
$\phi[\operatorname{FS}(D)]\in\mathcal{I}_{1/n}$.

Using Canonical Hindman Theorem ([38, Theorem 2.1], see also [15, Theorem 5 at
p. 133]), there is an infinite set $C=\\{c_{n}:n\in\omega\\}\subseteq\omega$
such that $\max\alpha(c_{n})<\min\alpha(c_{n+1})$ for every $n\in\omega$ and
one of the following five cases holds:

  1. (1)

$\forall x,y\in\operatorname{FS}(C)(\phi(x)=\phi(y))$,

  2. (2)

$\forall
x,y\in\operatorname{FS}(C)(\phi(x)=\phi(y)\iff\min\alpha(x)=\min\alpha(y))$,

  3. (3)

$\forall
x,y\in\operatorname{FS}(C)(\phi(x)=\phi(y)\iff\max\alpha(x)=\max\alpha(y))$,

  4. (4)

$\forall
x,y\in\operatorname{FS}(C)(\phi(x)=\phi(y)\iff(\min\alpha(x)=\min\alpha(y)\text{
and }\max\alpha(x)=\max\alpha(y)))$,

  5. (5)

$\forall x,y\in\operatorname{FS}(C)(\phi(x)=\phi(y)\iff x=y)$.

_Case 1._ We take $D=C$ and see that the set $\phi[\operatorname{FS}(D)]$ has
only one element, so it belongs to $\mathcal{I}_{1/n}$.

_Case 2._ We construct a strictly increasing sequence $\\{k_{n}:n\in\omega\\}$
such that $\phi(c_{k_{n}})>2^{n}$ for every $n\in\omega$.

Suppose that $k_{i}$ are constructed for $i<n$. Since
$\max\alpha(c_{k})<\min\alpha(c_{k+1})$ for every $k\in\omega$,
$\min\alpha(c_{k})\neq\min\alpha(c_{l})$ for distinct $k,l\in\omega$.
Consequently, $\phi\restriction C$ is one-to-one, so we can find
$k_{n}>k_{n-1}$ such that $\phi(c_{k_{n}})>2^{n}$. That finishes the inductive
construction of $k_{n}$.

Let $D=\\{c_{k_{n}}:n\in\omega\\}$. If we show that
$\phi[\operatorname{FS}(D)]\in\mathcal{I}_{1/n}$, the proof of this case will
be finished. Using the properties of $c_{k_{n}}$’s we can see that
$\phi[c_{k_{n}}+\operatorname{FS}(\\{c_{k_{i}}:i>n\\})]=\\{\phi(c_{k_{n}})\\},$
for every $n\in\omega$, so

$\begin{split}\sum_{y\in\phi[\operatorname{FS}(D)]}\frac{1}{y+1}&=\sum_{n\in\omega}\left(\sum_{y\in\\{\phi(c_{k_{n}})\\}\cup\phi[c_{k_{n}}+\operatorname{FS}(\\{c_{k_{i}}:i>n\\})]}\frac{1}{y+1}\right)\\\
&=\sum_{n\in\omega}\frac{1}{\phi(c_{k_{n}})+1}\leq\sum_{n\in\omega}\frac{1}{2^{n}+1}<\infty.\end{split}$

_Case 3._ We construct a strictly increasing sequence $\\{k_{n}:n\in\omega\\}$
such that $\phi(c_{k_{n}})>2^{n}$ for every $n\in\omega$.

Suppose that $k_{i}$ are constructed for $i<n$. Since
$\max\alpha(c_{k})<\min\alpha(c_{k+1})$ for every $k\in\omega$,
$\max\alpha(c_{k})\neq\max\alpha(c_{l})$ for distinct $k,l\in\omega$.
Consequently, $\phi\restriction C$ is one-to-one, so we can find
$k_{n}>k_{n-1}$ such that $\phi(c_{k_{n}})>2^{n}$. That finishes the inductive
construction of $k_{n}$.

Let $D=\\{c_{k_{n}}:n\in\omega\\}$. If we show that
$\phi[\operatorname{FS}(D)]\in\mathcal{I}_{1/n}$, the proof of this case will
be finished. Using the properties of $c_{k_{n}}$’s we can see that
$\phi[c_{k_{n}}+\operatorname{FS}(\\{c_{k_{i}}:i<n\\})]=\\{\phi(c_{k_{n}})\\},$
for every $n\in\omega$, so

$\begin{split}\sum_{y\in\phi[\operatorname{FS}(D)]}\frac{1}{y+1}&=\sum_{n\in\omega}\left(\sum_{y\in\\{\phi(c_{k_{n}})\\}\cup\phi[c_{k_{n}}+\operatorname{FS}(\\{c_{k_{i}}:i<n\\})]}\frac{1}{y+1}\right)\\\
&=\sum_{n\in\omega}\frac{1}{\phi(c_{k_{n}})+1}\leq\sum_{n\in\omega}\frac{1}{2^{n}+1}<\infty.\end{split}$

_Case 4._ We construct a strictly increasing sequence $\\{k_{n}:n\in\omega\\}$
such that

$\forall n\in\omega\,\forall
i<n\,\left(\phi(c_{k_{n}})>n2^{n}\land\phi(c_{k_{n}}+c_{k_{i}})>n2^{n}\right).$

Suppose that $k_{i}$ are constructed for $i<n$. Since
$\max\alpha(c_{k})<\min\alpha(c_{k+1})$ for every $k\in\omega$, we obtain that
$\min\alpha(c_{k}+c_{k_{i}})\neq\min\alpha(c_{k}+c_{k_{j}})$ and
$\min\alpha(c_{k})\neq\min\alpha(c_{k}+c_{k_{i}})$ for every $k>k_{n-1}$ and
$i<j\leq n-1$. Consequently, the function
$\phi\restriction(\\{c_{k}+c_{k_{i}}:k>k_{n-1},i<n\\}\cup\\{c_{k}:k>k_{n-1}\\})$
is one-to-one, so using pigeonhole principle we can find $k_{n}>k_{n-1}$ such
that $\phi(c_{k_{n}})>n2^{n}$ and $\phi(c_{k_{n}}+c_{k_{i}})>n2^{n}$ for every
$i<n$. That finishes the inductive construction of $k_{n}$.

Let $D=\\{c_{k_{n}}:n\in\omega\\}$. If we show that
$\phi[\operatorname{FS}(D)]\in\mathcal{I}_{1/n}$, the proof of this case will
be finished. Using the properties of $c_{k_{n}}$’s we can see that
$\phi[c_{k_{m}}+\operatorname{FS}(\\{c_{k_{i}}:m<i<n\\})+c_{k_{n}}]=\\{\phi(c_{k_{m}}+c_{k_{n}})\\}$
for every $m<n$, $m,n\in\omega$, so

$\begin{split}\sum_{y\in\phi[\operatorname{FS}(D)]}\frac{1}{y+1}&=\sum_{n\in\omega}\frac{1}{\phi(c_{k_{n}})+1}\\\
&+\sum_{n\in\omega}\sum_{m<n}\left(\sum_{y\in\\{\phi(c_{k_{m}}+c_{k_{n}})\\}\cup\phi[c_{k_{m}}+\operatorname{FS}(\\{c_{k_{i}}:m<i<n\\})+c_{k_{n}}]}\frac{1}{y+1}\right)\\\
&=\sum_{n\in\omega}\frac{1}{\phi(c_{k_{n}})+1}+\sum_{n\in\omega}\sum_{m<n}\frac{1}{\phi(c_{k_{m}}+c_{k_{n}})+1}\\\
&\leq\sum_{n\in\omega}\frac{1}{n2^{n}+1}+\sum_{n\in\omega}\sum_{m<n}\frac{1}{n2^{n}+1}<\infty.\end{split}$

_Case 5._ We construct inductively a strictly increasing sequence
$\\{k_{n}:n\in\omega\\}$ such that

$\forall n\in\omega\,\forall
x\in\operatorname{FS}(\\{c_{k_{i}}:i<n\\})\,\left(\phi(c_{k_{n}})>2^{2n}\land\phi(c_{k_{n}}+x)>2^{2n}\right).$

Suppose that $k_{i}$ are constructed for $i<n$. Let $m\in\omega$ be such that
$m>2^{2n}$ and $m>\phi(x)$ for every
$x\in\operatorname{FS}(\\{c_{k_{i}}:i<n\\})$. Since
$\phi\restriction\operatorname{FS}(C)$ is one-to-one, the set
$F=\phi^{-1}[\\{0,1,\dots,m\\}]$ is finite. Let $k_{n}\in\omega$ be such that
$c_{k_{n}}>\max F$. Since $c_{k_{n}}>\max F$, we obtain that $c_{k_{n}}\notin
F$ and consequently $\phi(c_{k_{n}})>m>2^{2n}$. Similarly, for every
$x\in\operatorname{FS}(\\{c_{k_{i}}:i<n\\})$ we have
$c_{k_{n}}+x>c_{k_{n}}>\max F$, so $\phi(c_{k_{n}}+x)>m>2^{2n}$. That finishes
the inductive construction of $k_{n}$.

Let $D=\\{c_{k_{n}}:n\in\omega\\}$. If we show that
$\phi[\operatorname{FS}(D)]\in\mathcal{I}_{1/n}$, the proof of this case will
be finished. Using the properties of $c_{k_{n}}$’s we can see that:

$\begin{split}\sum_{y\in\phi[\operatorname{FS}(D)]}\frac{1}{y+1}&=\sum_{n\in\omega}\left(\frac{1}{\phi(c_{k_{n}})+1}+\sum_{x\in\operatorname{FS}(\\{c_{k_{i}}:i<n\\})}\frac{1}{\phi(c_{k_{n}}+x)+1}\right)\\\
&\leq\sum_{n\in\omega}\left(\frac{1}{2^{2n}+1}+\sum_{x\in\operatorname{FS}(\\{c_{k_{i}}:i<n\\})}\frac{1}{2^{2n}+1}\right)\\\
&\leq\sum_{n\in\omega}\left(\frac{1}{2^{2n}+1}+(2^{n}-1)\cdot\frac{1}{2^{2n}+1}\right)<\infty.\qed\end{split}$

##  6\. Summable ideal is not below Ramsey ideal

######  Theorem 6.1.

$\mathcal{I}_{1/n}\not\leq_{K}\mathcal{R}$.

###### Proof.

Let $\phi:[\omega]^{2}\to\omega$ be an arbitrary function. We will show that
$\phi$ is not a witness for $\mathcal{I}_{1/n}\leq_{K}\mathcal{R}$ i.e. we
will find an infinite set $H\subseteq\omega$ such that
$\phi[[H]^{2}]\in\mathcal{I}_{1/n}$. Using Canonical Ramsey Theorem ([9,
Theorem II], see also [15, Theorem 2 at p. 129]), there is an infinite set
$T\subseteq\omega$ such that one of the following four cases holds:

  1. (1)

$\forall x,y\in[T]^{2}(\phi(x)=\phi(y))$,

  2. (2)

$\forall x,y\in[T]^{2}(\phi(x)=\phi(y)\iff\min x=\min y)$,

  3. (3)

$\forall x,y\in[T]^{2}(\phi(x)=\phi(y)\iff\max x=\max y)$,

  4. (4)

$\forall x,y\in[T]^{2}(\phi(x)=\phi(y)\iff x=y)$.

_Case 1._ We take $H=T$ and see that the set $\phi[[H]^{2}]$ has only one
element, so it belongs to $\mathcal{I}_{1/n}$.

_Case 2._ In this case, for every $t\in T$ the restriction
$\phi\restriction\\{\\{t,s\\}:s\in T,s>t\\}$ is constant with distinct values
for distinct $t$. Thus, for every $t\in T$ there is $k_{t}$ such that
$\\{k_{t}\\}=\phi[\\{\\{t,s\\}:s\in T,s>t\\}]$.

Since $k_{t_{n}}$ are pairwise distinct, we can find a one-to-one sequence
$\\{t_{n}:n\in\omega\\}\subseteq T$ such that $k_{t_{n}}>2^{n}$ for every
$n\in\omega$.

Now, we take $H=\\{t_{n}:n\in\omega\\}$ and notice that

$\sum_{k\in\phi[[H]^{2}]}\frac{1}{k+1}=\sum_{n=0}^{\infty}\left(\sum_{k\in\phi[\\{\\{t_{n},t_{i}\\}:i>n\\}]}\frac{1}{k+1}\right)=\sum_{n=0}^{\infty}\frac{1}{k_{t_{n}}+1}\leq\sum_{n=0}^{\infty}\frac{1}{2^{n}}<\infty,$

so $\phi[[H]^{2}]\in\mathcal{I}_{1/n}$.

_Case 3._ In this case, for every $t\in T$ the restriction
$\phi\restriction\\{\\{s,t\\}:s\in T,s<t\\}$ is constant with distinct values
for distinct $t$. Thus, for every $t\in T$ there is $k_{t}$ such that
$\\{k_{t}\\}=\phi[\\{\\{t,s\\}:s\in T,s<t\\}]$.

Since $k_{t_{n}}$ are pairwise distinct, we can find a one-to-one sequence
$\\{t_{n}:n\in\omega\\}\subseteq T$ such that $k_{t_{n}}>2^{n}$ for every
$n\in\omega$.

Now, we take $H=\\{t_{n}:n\in\omega\\}$ and notice that

$\sum_{k\in\phi[[H]^{2}]}\frac{1}{k+1}=\sum_{n=0}^{\infty}\left(\sum_{k\in\phi[\\{\\{t_{i},t_{n}\\}:i<n\\}]}\frac{1}{k+1}\right)=\sum_{n=0}^{\infty}\frac{1}{k_{t_{n}}+1}\leq\sum_{n=0}^{\infty}\frac{1}{2^{n}}<\infty,$

so $\phi[[H]^{2}]\in\mathcal{I}_{1/n}$.

_Case 4._ We construct inductively a one-to-one sequence
$\\{t_{n}:n\in\omega\\}\subseteq T$ such that $\phi(\\{t_{i},t_{n}\\})>n\cdot
2^{n}$ for every $n\in\omega$ and every $i<n$.

Suppose that $t_{i}$ are constructed for $i<n$. Since there are only finitely
many numbers below $n\cdot 2^{n}$ and the function $\phi$ is one-to-one on
$[T]^{2}$ there is $t_{n}\in T\setminus\\{t_{i}:i<n\\}$ such that
$\phi(\\{t,t_{n}\\})>n\cdot 2^{n}$ for every $t\in T$. That finishes the
inductive construction of $t_{n}$.

Now, we take $H=\\{t_{n}:n\in\omega\\}$ and notice that

$\sum_{k\in\phi[[H]^{2}]}\frac{1}{k+1}=\sum_{n=0}^{\infty}\sum_{i<n}\frac{1}{\phi(\\{t_{i},t_{n}\\})+1}\leq\sum_{n=0}^{\infty}\sum_{i<n}\frac{1}{n\cdot
2^{n}+1}\leq\sum_{n=0}^{\infty}\frac{1}{2^{n}}<\infty,$

so $\phi[[H]^{2}]\in\mathcal{I}_{1/n}$. ∎

##  7\. Hindman ideal is not below Ramsey ideal

######  Lemma 7.1.

If $D$ is very sparse, then
$\\{x\in\operatorname{FS}(D):\alpha_{D}(x)\cap\alpha_{D}(y)\neq\emptyset\\}\in\mathcal{H}$
for every $y\in\operatorname{FS}(D)$.

###### Proof.

Let $(d_{n})_{n\in\omega}$ be the increasing enumeration of all elements of
$D$ and $\alpha_{D}(y)=\\{k_{0},\dots,k_{n}\\}$. Since

$\\{x\in\operatorname{FS}(D):\alpha_{D}(x)\cap\alpha_{D}(y)\neq\emptyset\\}=\bigcup_{i\leq
n}\\{x\in\operatorname{FS}(D):k_{i}\in\alpha_{D}(x)\\},$

we only need to show that
$\\{x\in\operatorname{FS}(D):k_{i}\in\alpha_{D}(x)\\}\in\mathcal{H}$ for every
$i\leq n$.

If $y,z\in\\{x\in\operatorname{FS}(D):k_{i}\in\alpha_{D}(x)\\}$, then
$k_{i}\in\alpha_{D}(y)\cap\alpha_{D}(z)\neq\emptyset$, so
$y+z\notin\operatorname{FS}(D)$ (since $D$ is very sparse). Thus, there is no
infinite (even two-element) set $C$ such that
$\operatorname{FS}(C)\subseteq\\{x\in\operatorname{FS}(D):k_{i}\in\alpha_{D}(x)\\}$.
∎

######  Theorem 7.2.

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

###### Proof.

Let $D\subseteq\omega$ be a very sparse set (which exists by Lemma 2.3). Since
the ideal $\mathcal{H}$ is homogeneous (see Proposition 2.2), it suffices to
show that
$\mathcal{H}\restriction\operatorname{FS}(D)\not\leq_{K}\mathcal{R}$.

Assume to the contrary that there exists
$f:[\omega]^{2}\to\operatorname{FS}(D)$ which witnesses
$\mathcal{H}\restriction\operatorname{FS}(D)\leq_{K}\mathcal{R}$.

We will recursively define infinite sets $B_{n}\subseteq\omega$ and pairwise
distinct elements $b_{n}\in\omega$ such that for all $n\in\omega$ the
following conditions are satisfied:

  1. (a)

$b_{n}\in B_{n}$, $b_{n+1}>b_{n}$,

  2. (b)

$B_{n+1}\subseteq B_{n}$, $B_{0}=\omega$,

  3. (c)

for each $y\in f\left[[\\{b_{i}:i<n\\}]^{2}\right]$ we have

$f[[B_{n}]^{2}]\cap\\{x\in\operatorname{FS}(D):\
\alpha_{D}(x)\cap\alpha_{D}(y)\neq\emptyset\\}=\emptyset,$

  4. (d)

for each $y\in f\left[[\\{b_{i}:i<n\\}]^{2}\right]$ and $i<n$ we have

$f\left[\\{\\{b_{i},b\\}:b\in B_{n}\\}\right]-y\in\mathcal{H}.$

Let $b_{0}=0$ and $B_{0}=\omega$. Then $b_{0}$ and $B_{0}$ are as required.
Assume that $b_{i}$ and $B_{i}$ have been constructed for $i<n$ and satisfy
items (b)–(d).

Since
$\\{x\in\operatorname{FS}(D):\alpha_{D}(x)\cap\alpha_{D}(y)\neq\emptyset\\}\in\mathcal{H}$
for every $y\in
f\left[[\\{b_{i}:i<n\\}]^{2}\right]\subseteq\operatorname{FS}(D)$ (by Lemma
7.1), $[B_{n-1}]^{2}\in\mathcal{R}^{+}$ and we assumed that $f$ witnesses
$\mathcal{H}\restriction\operatorname{FS}(D)\leq_{K}\mathcal{R}$, there exists
an infinite set $B\subseteq\omega$ such that

$[B]^{2}\subseteq[B_{n-1}]^{2}\setminus\bigcup_{y\in
f\left[[\\{b_{i}:i<n\\}]^{2}\right]}f^{-1}\left[\\{x\in\operatorname{FS}(D):\
\alpha_{D}(x)\cap\alpha_{D}(y)\neq\emptyset\\}\right].$

Observe that for each infinite set $E\subseteq\omega$ and $b,y\in\omega$ there
exists an infinite set $C\subseteq E$ such that $f[\\{\\{b,c\\}:c\in
C\\}]-y\in\mathcal{H}.$ Indeed, let $g:E\setminus\\{b\\}\to\omega$ be given by
$g(x)=f(\\{b,x\\})-y$. Since $\mathcal{H}$ is a tall ideal (Proposition 2.1),
$\mathcal{H}\not\leq_{K}\mathrm{Fin}(E\setminus\\{b\\})$. Thus, there is
$C\notin\mathrm{Fin}(E\setminus\\{b\\})$ such that $C\subseteq
E\setminus\\{b\\}$ and $g[C]=f[\\{\\{b,c\\}:c\in C\\}]-y\in\mathcal{H}$.

Now, using recursively the above observation we can find an infinite set
$C\subseteq B$ such that $f[\\{\\{b_{i},c\\}:c\in C\\}]-y\in\mathcal{H}$ for
every $i<n$ and $y\in f\left[[\\{b_{i}:i<n\\}]^{2}\right]$.

We put $B_{n}=C$ and pick any $b_{n}\in B_{n}$ with $b_{n}>b_{n-1}$.

The construction of the sequences $(B_{n})_{n\in\omega}$ and
$(b_{n})_{n\in\omega}$ is finished.

Let $B=\\{b_{n}:\ n\in\omega\\}$. Since $B$ is infinite,
$[B]^{2}\in\mathcal{R}^{+}$. Since we assumed that $f$ witnesses
$\mathcal{H}\restriction\operatorname{FS}(D)\leq_{K}\mathcal{R}$,
$f[[B]^{2}]\in\mathcal{H}^{+}\restriction\operatorname{FS}(D)$, and
consequently there exists an infinite set $C\subseteq\omega$ such that
$\operatorname{FS}(C)\subseteq f[[B]^{2}]$.

Pick any $c\in C$ and let $j,n\in\omega$ be such that $c=f(\\{b_{j},b_{n}\\})$
and $j<n$.

Since $X=\left[\\{b_{i}:i\leq n\\}\right]^{2}$ is finite,
$f[X]-c\in\mathcal{H}$.

Let $Y=\\{\\{b_{i},b_{k}\\}:i\leq n<k\\}\\}$. Since $\\{b_{k}:k>n\\}\subseteq
B_{n+1}$ and $B_{n+1}$ satisfies item (d) applied to $y=c$, we have
$f\left[Y\right]-c\in\mathcal{H}$.

Let $Z=[\\{b_{i}:i>n\\}]^{2}$. We claim that
$\operatorname{FS}(C\setminus\\{c\\})\cap\left(f\left[Z\right]-c\right)=\emptyset$.
Suppose to the contrary that there exists
$a\in\operatorname{FS}(C\setminus\\{c\\})\cap\left(f\left[Z\right]-c\right)$.
Then $a+c\in\operatorname{FS}(C)\cap f[Z]\subseteq\operatorname{FS}(D)\cap
f[[B_{n+1}]^{2}]$, so by item (c) applied to $y=c$,
$\alpha_{D}(c)\cap\alpha_{D}(a+c)=\emptyset$. On the other hand,
$a,c\in\operatorname{FS}(D)$, $D$ is very sparse and
$a+c\in\operatorname{FS}(D)$, so $\alpha_{D}(a)\cap\alpha_{D}(c)=\emptyset$.
Consequently, $\alpha_{D}(a+c)=\alpha_{D}(a)\cup\alpha_{D}(c)$, so
$\alpha_{D}(c)\cap\alpha_{D}(a+c)=\alpha_{D}(c)\neq\emptyset$, a
contradiction.

Since $[B]^{2}=X\cup Y\cup Z$ and
$\operatorname{FS}(C\setminus\\{c\\})\subseteq\operatorname{FS}(C)-c\subseteq
f[[B]^{2}]-c$, we have

$\begin{split}\operatorname{FS}(C\setminus\\{c\\})&\subseteq(f[X]-c)\cup(f[Y]-c)\cup((f[Z]-c)\cap\operatorname{FS}(C\setminus\\{c\\}))\\\
&=(f[X]-c)\cup(f[Y]-c)\cup\emptyset\in\mathcal{H},\end{split}$

a contradiction. ∎

##  8\. Ramsey ideal is not below Hindman ideal

######  Theorem 8.1.

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

###### Proof.

By $\Gamma$ we will denote the set
$\Gamma=\\{(z_{0},z_{1})\in\omega^{2}:z_{0}>z_{1}\\}$. In this proof we will
view $\mathcal{R}$ as an ideal on $\Gamma$ consisting of those
$A\subseteq\Gamma$ that do not contain any $B^{2}\cap\Gamma$, for infinite
$B\subseteq\omega$.

By Lemma 2.3, there is a very sparse $X\in[\omega]^{\omega}$. The ideal
$\mathcal{H}$ is homogeneous (see Proposition 2.2), so it suffices to show
that $\mathcal{R}\not\leq_{K}\mathcal{H}\restriction\operatorname{FS}(X)$. Fix
any $f:\operatorname{FS}(X)\to\Gamma$ and assume to the contrary that it
witnesses $\mathcal{R}\leq_{K}\mathcal{H}$. There are two possible cases.

######  Case 1.

There are $k\in\omega$ and very sparse $D\in[\omega]^{\omega}$,
$\operatorname{FS}(D)\subseteq\operatorname{FS}(X)$, such that for all $n>k$
and $x\in\operatorname{FS}(D)$ we have:
$(f^{-1}[(\omega\times\\{n\\})\cap\Gamma]\cap\\{y\in
FS(D):\alpha_{D}(x)\subseteq\alpha_{D}(y)\\})-x\in\mathcal{H}\restriction\operatorname{FS}(X).$

In this case we recursively pick
$\\{x_{n}:n\in\omega\\}\subseteq\operatorname{FS}(D)$ and
$\\{D_{n}:n\in\omega\cup\\{-1\\}\\}\subseteq[\omega]^{\omega}$ such that
$D_{-1}=D$ and for all $n\in\omega$ we have:

  * (a)

$x_{n}\in\operatorname{FS}(D_{n-1})\setminus\big{(}\\{x_{i}:i<n\\}\cup\bigcup_{i<n}\bigcup_{j<n}\\{y\in\operatorname{FS}(D_{j}):\alpha_{D_{j}}(y)\cap\alpha_{D_{j}}(x_{i})\neq\emptyset\\}\big{)}$
(here we put $\alpha_{D_{j}}(x_{i})=\emptyset$ whenever $x_{i}\notin D_{j}$);

  * (b)

$D_{n}$ is very sparse;

  * (c)

$\operatorname{FS}(\\{x_{0},\ldots,x_{n}\\})\subseteq\operatorname{FS}(D)$;

  * (d)

$\operatorname{FS}(D_{n})\subseteq\operatorname{FS}(D_{n-1})\subseteq\operatorname{FS}(D)$;

  * (e)

$\left(f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]-x\right)\cap\operatorname{FS}(D_{n})=\emptyset$
for every $x\in\operatorname{FS}(\\{x_{0},\ldots,x_{n}\\})$ and $1\leq i\leq
n+1$;

  * (f)

$f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]\cap\operatorname{FS}(D_{n})=\emptyset$
for all $1\leq i\leq n+1$.

The initial step of the construction is given by the requirement $D_{-1}=D$.
Suppose now that $x_{i}$ and $D_{i}$ for all $i<n$ are defined.

Find $x_{n}\in\operatorname{FS}(D_{n-1})$ such that
$\operatorname{FS}(\\{x_{0},\ldots,x_{n}\\})\subseteq\operatorname{FS}(D)$ and
$x_{n}\neq x_{i}$ for all $i<n$. This is possible since it suffices to pick
any point from the set

$\operatorname{FS}(D_{n-1})\setminus\bigcup_{i<n}\bigcup_{j<n}\\{y\in\operatorname{FS}(D_{j}):\alpha_{D_{j}}(y)\cap\alpha_{D_{j}}(x_{i})\neq\emptyset\\},$

which is nonempty as
$\operatorname{FS}(D_{n-1})\notin\mathcal{H}\restriction\operatorname{FS}(X)$
and

$\bigcup_{i<n}\bigcup_{j<n}\\{y\in\operatorname{FS}(D_{j}):\alpha_{D_{j}}(y)\cap\alpha_{D_{j}}(x_{i})\neq\emptyset\\}\in\mathcal{H}\restriction\operatorname{FS}(X)$

by Lemma 7.1 and item (b) for all $j<n$ (here we put
$\alpha_{D_{j}}(x_{i})=\emptyset$ whenever $x_{i}\notin D_{j}$).

Enumerate
$\operatorname{FS}(\\{x_{0},\ldots,x_{n}\\})=\\{c_{0},c_{1},\ldots,c_{2^{n+1}-2}\\}$.
We will define sets $E_{t}\in[\omega]^{\omega}$ for $-1\leq t\leq n$ such that
$E_{-1}=D_{n-1}$ and for all $0\leq t\leq n$:

  * •

$\operatorname{FS}(E_{t})\subseteq\operatorname{FS}(E_{t-1})\subseteq\operatorname{FS}(D_{n-1})$,

  * •

$\left(\bigcup_{1\leq i\leq
n+1}f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]-c_{l}\right)\cap\operatorname{FS}(E_{t})=\emptyset$
for every $0\leq l\leq 2^{n+1}-2$.

Such construction is possible. Indeed, since we are on Case 1 and each
$c_{l}\in\operatorname{FS}(\\{x_{0},\ldots,x_{n}\\})\subseteq\operatorname{FS}(D)$,
we know that:

$\begin{split}&\left(\bigcup_{1\leq i\leq
n+1}f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]-c_{l}\right)\cap\left(\\{y\in\operatorname{FS}(D):\alpha_{D}(c_{l})\subseteq\alpha_{D}(y)\\}-c_{l}\right)\\\
=&\left(\bigcup_{1\leq i\leq
n+1}f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]\cap\\{y\in\operatorname{FS}(D):\alpha_{D}(c_{l})\subseteq\alpha_{D}(y)\\}\right)-c_{l}\in\mathcal{H}\restriction\operatorname{FS}(D).\end{split}$

On the other hand, we get:

$\begin{split}&\operatorname{FS}(E_{t-1})\cap\left(\\{y\in\operatorname{FS}(D):\alpha_{D}(c_{l})\subseteq\alpha_{D}(y)\\}-c_{l}\right)\\\
\supseteq&\operatorname{FS}(E_{t-1})\cap\left(\operatorname{FS}(D)\setminus\\{y\in\operatorname{FS}(D):\alpha_{D}(c_{l})\cap\alpha_{D}(y)\neq\emptyset\\}\right)\\\
\supseteq&\operatorname{FS}(E_{t-1})\setminus\\{y\in\operatorname{FS}(D):\alpha_{D}(c_{l})\cap\alpha_{D}(y)\neq\emptyset\\}\notin\mathcal{H}\restriction\operatorname{FS}(D),\end{split}$

as
$\\{y\in\operatorname{FS}(D):\alpha_{D}(c_{l})\cap\alpha_{D}(y)\neq\emptyset\\}\in\mathcal{H}\restriction\operatorname{FS}(D)$
(by Lemma 7.1). Then

$\begin{split}&\operatorname{FS}(E_{t-1})\setminus\left(\bigcup_{1\leq i\leq
n+1}f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]-c_{l}\right)\\\
\supseteq&\left(\operatorname{FS}(E_{t-1})\cap\left(\\{y\in\operatorname{FS}(D):\alpha_{D}(c_{l})\subseteq\alpha_{D}(y)\\}-c_{l}\right)\right)\\\
&\setminus\left(\left(\bigcup_{1\leq i\leq
n+1}f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]-c_{l}\right)\right.\\\
&\cap\left.\left(\\{y\in\operatorname{FS}(D):\alpha_{D}(c_{l})\subseteq\alpha_{D}(y)\\}-c_{l}\right)\right)\notin\mathcal{H}\restriction\operatorname{FS}(D).\end{split}$

Thus, there is $E_{t}\in[\omega]^{\omega}$ as needed.

Once all $E_{t}$ are defined, observe that

$\bigcup_{1\leq i\leq n+1}(\omega\times\\{k+i\\})\cap\Gamma\in\mathcal{R}.$

Since we assumed that $f$ witnesses $\mathcal{R}\leq_{K}\mathcal{H}$,

$\operatorname{FS}(E_{n})\setminus\bigcup_{1\leq i\leq
n+1}f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]\notin\mathcal{H}.$

Hence, there is a very sparse $D_{n}\in[\omega]^{\omega}$ such that

$\operatorname{FS}(D_{n})\subseteq\operatorname{FS}(E_{n})\setminus\bigcup_{1\leq
i\leq n+1}f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]$

(by Lemma 2.3). Note that
$\operatorname{FS}(D_{n})\subseteq\operatorname{FS}(E_{n})\subseteq\operatorname{FS}(D_{n-1})\subseteq\operatorname{FS}(D)$
and

$\bigcup_{1\leq i\leq
n+1}\left(f^{-1}[(\omega\times\\{k+i\\})\cap\Gamma]-x\right)\cap\operatorname{FS}(D_{n})=\emptyset$

for all $x\in\operatorname{FS}(\\{x_{0},\ldots,x_{n}\\})$.

This finishes the construction of
$\\{x_{n}:n\in\omega\\}\subseteq\operatorname{FS}(D)$ and
$\\{D_{n}:n\in\omega\cup\\{-1\\}\\}\subseteq[\omega]^{\omega}$.

Define $B=\operatorname{FS}(\\{x_{n}:\ n\in\omega\\})$. Obviously,
$B\notin\mathcal{H}\restriction\operatorname{FS}(X)$ as
$\operatorname{FS}(\\{x_{0},\ldots,x_{n}\\})\subseteq\operatorname{FS}(D)\subseteq\operatorname{FS}(X)$
for all $n\in\omega$. We will show that
$f[B]\cap(\omega\times\\{n\\})\cap\Gamma$ is finite for all $n>k$. This will
finish the proof in this case as $\bigcup_{n\leq
k}(\omega\times\\{n\\})\cap\Gamma\in\mathcal{R}$ and any set finite on each
$(\omega\times\\{n\\})\cap\Gamma$ belongs to $\mathcal{R}$.

Assume that $f(x)\in(\omega\times\\{k+m+1\\})\cap\Gamma$ for some $m\in\omega$
and $x=x_{n_{0}}+\ldots+x_{n_{t}}\in B$, where $n_{0}<\ldots<n_{t}$. If
$n_{0}>m$, then $x\in\operatorname{FS}(\\{x_{n}:\
n>m\\})\subseteq\operatorname{FS}(D_{m})$ which contradicts
$f(x)\in(\omega\times\\{k+m+1\\})\cap\Gamma$ (by item (f)). If $n_{0}\leq m$
but $J=\\{j\leq t:n_{j}>m\\}\neq\emptyset$, then let $j=\min J$ and note that
$x\in\sum_{i<j}x_{n_{i}}+\operatorname{FS}(D_{m})$. As
$\sum_{i<j}x_{n_{i}}\in\operatorname{FS}(\\{x_{0},\ldots,x_{m}\\})$, item (e)
gives us a contradiction with $f(x)\in(\omega\times\\{k+m+1\\})\cap\Gamma$.
Hence, the only possibility is that $n_{j}\leq m$ for all $j\leq t$. Thus,
$f[B]\cap(\omega\times\\{k+m+1\\})\cap\Gamma\subseteq
f[\operatorname{FS}(\\{x_{0},\ldots,x_{m}\\})]$, which is a finite set.

######  Case 2.

For every $k\in\omega$ and very sparse $D\in[\omega]^{\omega}$,
$\operatorname{FS}(D)\subseteq\operatorname{FS}(X)$, there are $n>k$ and
$x\in\operatorname{FS}(D)$ such that:

$(f^{-1}[(\omega\times\\{n\\})\cap\Gamma]\cap\\{y\in\operatorname{FS}(D):\alpha_{D}(x)\subseteq\alpha_{D}(y)\\})-x\notin\mathcal{H}\restriction\operatorname{FS}(X).$

In this case we will pick $\\{n_{i}:i\in\omega\\}\subseteq\omega$,
$\\{j_{i}:i\in\omega\\}\subseteq\\{0,1\\}$,
$\\{x_{i}:i\in\omega\\}\subseteq\operatorname{FS}(D)$,
$\\{D_{i}:i\in\omega\cup\\{-1\\}\\}\subseteq[\omega]^{\omega}$,
$\\{k_{i}:i\in\omega\\}\subseteq\omega\cup\\{-1\\}$ and
$\\{F_{i}:i\in\omega\\}\subseteq\mathrm{Fin}$ such that $D_{-1}=X$ and for
each $i\in\omega$:

  * (a)

    * (a1)

$n_{i}>n_{i-1}$ (here we put $n_{-1}=-1$);

    * (a2)

$n_{i}>\min\\{a\in\omega:f[\operatorname{FS}(\\{x_{j}:j<i\\})]\subseteq\\{0,1,\ldots,a\\}^{2}\cap\Gamma\\}$;

  * (b)

    * (b1)

$\operatorname{FS}(D_{i})\subseteq\operatorname{FS}(D_{i-1})\subseteq\operatorname{FS}(X)$;

    * (b2)

$D_{i}$ is very sparse;

  * (c)

if $j_{i}=0$, then:

    * (c1)

$k_{i}=-1$;

    * (c2)

$F_{i}=\emptyset$;

    * (c3)

$x_{i}\in\operatorname{FS}(D_{i-1})\cap
f^{-1}[(\omega\times\\{n_{i}\\})\cap\Gamma]$;

    * (c4)

$x_{i}+\operatorname{FS}(D_{i})\subseteq
f^{-1}[(\omega\times\\{n_{i}\\})\cap\Gamma]$;

  * (d)

if $j_{i}=1$, then:

    * (d1)

$k_{i}\in\\{0\leq u<i:u\notin\bigcup_{q<i}F_{q},j_{u}=0\\}$;

    * (d2)

$F_{i}=\\{k_{i},k_{i}+1,\ldots,i-1\\}$;

    * (d3)

      * (d3a)

$x_{i}\in f^{-1}[\\{(n_{i},n_{k_{i}})\\}]$;

      * (d3b)

$x_{i}\in
x_{k_{i}}+(\\{0\\}\cup\operatorname{FS}(\\{x_{r}:k_{i}<r<i,r\notin\bigcup_{q<i}F_{q}\\}))+\operatorname{FS}(D_{i-1})$;

    * (d4)

$x_{i}+\operatorname{FS}(D_{i})\subseteq f^{-1}[\\{(n_{i},n_{k_{i}})\\}]$;

  * (e)

if $x=\sum_{b\leq a}x_{t_{b}}$ for some $0\leq t_{0}<\ldots<t_{a}<i$,
$t_{b}\notin\bigcup_{q\leq i}F_{q}$ (so $x\in\operatorname{FS}(\\{x_{t}:t<i,\
t\notin\bigcup_{q\leq i}F_{q}\\})$), then:

    * (e1)

$(x+x_{i}+\operatorname{FS}(D_{i}))\cap
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$;

    * (e2)

$(x+\operatorname{FS}(D_{i}))\cap f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$;

    * (e3)

$(x+x_{i})\cap f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$;

  * (f)

$\operatorname{FS}(D_{i})\cap\\{y\in\operatorname{FS}(D_{t}):\alpha_{D_{t}}(y)\cap\alpha_{D_{t}}(x_{u})\neq\emptyset\\}=\emptyset$
for all $-1\leq t<i$ and $0\leq u\leq i$ such that
$x_{u}\in\operatorname{FS}(D_{t})$;

  * (g)

    * (g1)

$\operatorname{FS}(\\{x_{t}:t\leq i,t\notin\bigcup_{q\leq
i}F_{q}\\})\subseteq\operatorname{FS}(X)$;

    * (g2)

$\sum_{b\leq a}x_{t_{b}}\in x_{t_{0}}+\operatorname{FS}(D_{t_{0}})$ for every
$a>0$, $0\leq t_{0}<\ldots<t_{a}\leq i$, $t_{b}\notin\bigcup_{q\leq i}F_{q}$;

At first step, since we are in Case 2, for $k=0$ and $D=X$ there are $n_{0}>k$
(note that (a) is satisfied) and $x^{\prime}_{0}\in\operatorname{FS}(X)$ such
that:
$(f^{-1}[(\omega\times\\{n_{0}\\})\cap\Gamma]\cap\\{y\in\operatorname{FS}(X):\alpha_{X}(x^{\prime}_{0})\subseteq\alpha_{X}(y)\\})-x^{\prime}_{0}\notin\mathcal{H}\restriction\operatorname{FS}(X).$
Hence, there is $D^{\prime}_{0}\in[\omega]^{\omega}$ such that:
$x^{\prime}_{0}+\operatorname{FS}(D^{\prime}_{0})\subseteq
f^{-1}[(\omega\times\\{n_{0}\\})\cap\Gamma]\cap\\{y\in\operatorname{FS}(X):\alpha_{X}(x^{\prime}_{0})\subseteq\alpha_{X}(y)\\}\subseteq\operatorname{FS}(X).$
Put $j_{0}=0$, $k_{0}=-1$ and $F_{0}=\emptyset$ (note that (c1) and (c2) are
satisfied). Moreover, define $x_{0}=x^{\prime}_{0}+\min(D^{\prime}_{0})$ (note
that (c3) and (g1) are satisfied, because $x_{0}\in
x^{\prime}_{0}+\operatorname{FS}(D^{\prime}_{0})\subseteq\operatorname{FS}(X)$
and $x^{\prime}_{0}+\operatorname{FS}(D^{\prime}_{0})\subseteq
f^{-1}[(\omega\times\\{n_{0}\\})\cap\Gamma]$) and using Lemma 2.3 find a very
sparse $D_{0}\in[\omega]^{\omega}$ such that

$\operatorname{FS}(D_{0})\subseteq\operatorname{FS}(D^{\prime}_{0}\setminus\\{\min(D^{\prime}_{0})\\})\setminus\\{y\in\operatorname{FS}(X):\alpha_{X}(y)\cap\alpha_{X}(x_{0})\neq\emptyset\\},$

which is possible as
$\\{y\in\operatorname{FS}(X):\alpha_{X}(y)\cap\alpha_{X}(x_{0})\neq\emptyset\\}\in\mathcal{H}\restriction\operatorname{FS}(X)$
by Lemma 7.1 (note that (c4), (f) and (b) are satisfied, because
$x_{0}+\operatorname{FS}(D_{0})\subseteq
x^{\prime}_{0}+\operatorname{FS}(D^{\prime}_{0})\subseteq
f^{-1}[(\omega\times\\{n_{0}\\})\cap\Gamma]$ and
$\operatorname{FS}(D_{0})\subseteq\operatorname{FS}(D^{\prime}_{0})\subseteq\\{y\in\operatorname{FS}(X):\alpha_{X}(x^{\prime}_{0})\subseteq\alpha_{X}(y)\\}-x^{\prime}_{0}\subseteq\operatorname{FS}(X)$).
In conditions (e) and (g2) there is nothing to check. Thus, all the
requirements are met.

At $i$th step, where $i>0$, since we are in Case 2, if
$k=\max\\{n_{i-1},\min\\{a\in\omega:f[\operatorname{FS}(\\{x_{j}:j<i\\})]\subseteq\\{0,1,\ldots,a\\}^{2}\cap\Gamma\\}\\}$
and $D=D_{i-1}$, then there are $n_{i}>k$ (so (a) is satisfied) and
$x^{\prime}_{i}\in\operatorname{FS}(D_{i-1})$ such that

$\left(f^{-1}[(\omega\times\\{n_{i}\\})\cap\Gamma]\cap\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(x_{i}^{\prime})\subseteq\alpha_{D_{i-1}}(y)\\}\right)-x^{\prime}_{i}\notin\mathcal{H}\restriction\operatorname{FS}(X).$

Hence, there is $D^{\prime}_{i}\in[\omega]^{\omega}$ such that:
$x^{\prime}_{i}+\operatorname{FS}(D^{\prime}_{i})\subseteq
f^{-1}[(\omega\times\\{n_{i}\\})\cap\Gamma]\cap\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(x_{i}^{\prime})\subseteq\alpha_{D_{i-1}}(y)\\}.$
In particular,
$\operatorname{FS}(D^{\prime}_{i})\subseteq\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(x_{i}^{\prime})\subseteq\alpha_{D_{i-1}}(y)\\}-x^{\prime}_{i}=\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(x_{i}^{\prime})\cap\alpha_{D_{i-1}}(y)=\emptyset\\}\subseteq\operatorname{FS}(D_{i-1})$.
There are two possibilities.

Assume first that there is $x=\sum_{b\leq a}x_{t_{b}}$ for some
$t_{0}<\ldots<t_{a}<i$, $t_{b}\notin\bigcup_{q<i}F_{q}$ such that either
$x+x^{\prime}_{i}+\operatorname{FS}(\bar{D}_{i})\subseteq
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ or $x+\operatorname{FS}(\bar{D}_{i})\subseteq
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ for some $\bar{D}_{i}\in[\omega]^{\omega}$
such that
$\operatorname{FS}(\bar{D}_{i})\subseteq\operatorname{FS}(D^{\prime}_{i})$.
Define $j_{i}=1$ and let $k_{i}$ be minimal such that there is (one or more)
$x$ as above with $k_{i}=t_{0}$.

Notice that $k_{i}\in\\{0\leq u<i:\ u\notin\bigcup_{q<i}F_{q}\\}$. We will
show that $j_{k_{i}}=0$ (i.e., (d1) is satisfied). Suppose that
$j_{k_{i}}=j_{t_{0}}=1$. Observe that
$x^{\prime}_{i}+\operatorname{FS}(\bar{D}_{i})\subseteq\operatorname{FS}(D_{i-1})$
(by
$\operatorname{FS}(\bar{D}_{i})\cap\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(y)\cap\alpha_{D_{i-1}}(x_{i}^{\prime})\neq\emptyset\\}=\emptyset$)
and consequently
$x^{\prime}_{i}+\operatorname{FS}(\bar{D}_{i})\subseteq\operatorname{FS}(D_{t_{0}})$
(by item (b1)). Then items (b1), (f) and (g2) give us:

  * •

$x+x^{\prime}_{i}+\operatorname{FS}(\bar{D}_{i})\subseteq
x_{t_{0}}+\operatorname{FS}(D_{t_{0}})$,

  * •

$x+\operatorname{FS}(\bar{D}_{i})\subseteq
x_{t_{0}}+\operatorname{FS}(D_{t_{0}})$.

Then from (d4) we have:

  * •

$f[x+x^{\prime}_{i}+\operatorname{FS}(\bar{D}_{i})]\subseteq\\{(n_{t_{0}},n_{k_{t_{0}}})\\}$,

  * •

$f[x+\operatorname{FS}(\bar{D}_{i})]\subseteq\\{(n_{t_{0}},n_{k_{t_{0}}})\\}$.

This contradicts $x+x^{\prime}_{i}+\operatorname{FS}(\bar{D}_{i})\subseteq
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ or $x+\operatorname{FS}(\bar{D}_{i})\subseteq
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$, because $n_{t_{0}}<n_{i}$ (by $t_{0}<i$ and
item (a1)).

Define $F_{i}=\\{k_{i},k_{i}+1,\ldots,i-1\\}$ (so (d2) is satisfied) and
$\bar{x}_{i}=x+x^{\prime}_{i}$ (or $\bar{x}_{i}=x$ if
$x+\operatorname{FS}(\bar{D}_{i})\subseteq f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$).

To define $D_{i}$ and $x_{i}$, note that by the choice of $k_{i}$, for each
$y=\sum_{b\leq a}x_{t_{b}}$, $t_{0}<\ldots<t_{a}<i$,
$t_{b}\notin\bigcup_{q\leq i}F_{q}$ (so in fact $t_{a}<k_{i}$) we know that
$(y+\bar{x}_{i}+\operatorname{FS}(E))\not\subseteq
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ and $(y+\operatorname{FS}(E))\not\subseteq
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ for every $E\in[\omega]^{\omega}$ such that
$\operatorname{FS}(E)\subseteq\operatorname{FS}(D^{\prime}_{i})$. In other
words,
$f^{-1}[\\{(n_{i},n_{t_{0}})\\}]-(y+\bar{x}_{i})\in\mathcal{H}\restriction\operatorname{FS}(D^{\prime}_{i})$
and
$f^{-1}[\\{(n_{i},n_{t_{0}})\\}]-y\in\mathcal{H}\restriction\operatorname{FS}(D^{\prime}_{i})$,
for every such $y$. Thus, we can find $\tilde{D}_{i}\in[\omega]^{\omega}$ such
that:

  * •

$\operatorname{FS}(\tilde{D}_{i})\subseteq\operatorname{FS}(\bar{D}_{i})$;

  * •

$(y+\bar{x}_{i}+\operatorname{FS}(\tilde{D}_{i}))\cap
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$ and
$(y+\operatorname{FS}(\tilde{D}_{i}))\cap
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$ for every $y=\sum_{b\leq
a}x_{t_{b}}$, where $t_{0}<\ldots<t_{a}<i$, $t_{b}\notin\bigcup_{q\leq
i}F_{q}$;

  * •

$\operatorname{FS}(\tilde{D}_{i})\cap\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(y)\cap\alpha_{D_{i-1}}(x_{i}^{\prime})\neq\emptyset\\}=\emptyset$;

(the last item is trivial, as
$\operatorname{FS}(\tilde{D}_{i})\subseteq\operatorname{FS}(\bar{D}_{i})\subseteq\operatorname{FS}(D^{\prime}_{i})$
and
$\operatorname{FS}(D^{\prime}_{i})\subseteq\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(x_{i}^{\prime})\cap\alpha_{D_{i-1}}(y)=\emptyset\\}$).

Define $x_{i}=\bar{x}_{i}+\min(\tilde{D}_{i})$ and let
$D_{i}\in[\omega]^{\omega}$ be very sparse such that
$\operatorname{FS}(D_{i})\subseteq\operatorname{FS}(\tilde{D}_{i}\setminus\\{\min(\tilde{D}_{i})\\})$
and $D_{i}$ satisfies item (f). It is possible using Lemma 2.3, as
$\\{y\in\operatorname{FS}(D_{t}):\alpha_{D_{t}}(y)\cap\alpha_{D_{t}}(x_{u})\neq\emptyset\\}\in\mathcal{H}\restriction\operatorname{FS}(X)$
by Lemma 7.1 and item (b2) for all $-1\leq t<i$. Then (b2) is satisfied.
Observe that other conditions are met:

  * (b1)

$\operatorname{FS}(D_{i})\subseteq\operatorname{FS}(\tilde{D}_{i})\subseteq\operatorname{FS}(\bar{D}_{i})\subseteq\operatorname{FS}(D^{\prime}_{i})\subseteq\operatorname{FS}(D_{i-1})\subseteq\operatorname{FS}(X)$;

  * (d3b)

if $\bar{x}_{i}=x+x^{\prime}_{i}$, then
$x_{i}=\bar{x}_{i}+\min(\tilde{D}_{i})=x_{k_{i}}+(x-x_{k_{i}})+x^{\prime}_{i}+\min(\tilde{D}_{i})\in
x_{k_{i}}+(\\{0\\}\cup\operatorname{FS}(\\{x_{r}:\
k_{i}<r<i,r\notin\bigcup_{q<i}F_{q}\\}))+\operatorname{FS}(D_{i-1})$ by the
fact that
$\operatorname{FS}(\tilde{D}_{i})\cap\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(y)\cap\alpha_{D_{i-1}}(x_{i}^{\prime})\neq\emptyset\\}=\emptyset$
(if $\bar{x}_{i}=x$ this is even easier to show);

  * (d3a)

if $\bar{x}_{i}=x+x^{\prime}_{i}$, then $x_{i}\in
x+x^{\prime}_{i}+\operatorname{FS}(\bar{D}_{i})\subseteq
f^{-1}[\\{(n_{i},n_{k_{i}})\\}]$ (if $\bar{x}_{i}=x$ this is also true);

  * (d4)

if $\bar{x}_{i}=x+x^{\prime}_{i}$, then
$x_{i}+\operatorname{FS}(D_{i})\subseteq
x+x^{\prime}_{i}+\operatorname{FS}(\bar{D}_{i})\subseteq
f^{-1}[\\{(n_{i},n_{k_{i}})\\}]$ (if $\bar{x}_{i}=x$ this is also true);

  * (e)

for (e3), if $y=\sum_{b\leq a}x_{t_{b}}$, $t_{0}<\ldots<t_{a}<i$,
$t_{b}\notin\bigcup_{q\leq i}F_{q}$, then note that
$y+x_{i}=y+\bar{x}_{i}+\min(\tilde{D}_{i})\in
y+\bar{x}_{i}+\operatorname{FS}(\tilde{D}_{i})$ and recall that
$(y+\bar{x}_{i}+\operatorname{FS}(\tilde{D}_{i}))\cap
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$, thus $y+x_{i}\notin
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ ((e1) and (e2) are similar);

  * (g1)

$\operatorname{FS}(\\{x_{t}:t\leq i,t\notin\bigcup_{q\leq
i}F_{q}\\})\subseteq\operatorname{FS}(\\{x_{t}:t<i,t\notin\bigcup_{q<i}F_{q}\\})\cup(x_{i}+\operatorname{FS}(\\{x_{t}:t<i,t\notin\bigcup_{q<i}F_{q}\\})\subseteq\operatorname{FS}(X)$
by items (f) and (g1) applied to $i-1$ and item (d3b) applied to $i$;

  * (g2)

if $a>0$, $t_{0}<\ldots<t_{a}\leq i$, $t_{b}\notin\bigcup_{q\leq i}F_{q}$ then
either $t_{a}<i$ and $\sum_{b\leq a}x_{t_{b}}\in
x_{t_{0}}+\operatorname{FS}(D_{t_{0}})$ (by (g2) applied to $i-1$) or
$t_{a}=i$ and $\sum_{b\leq a}x_{t_{b}}=\sum_{b<a}x_{t_{b}}+x_{i}\in
x_{t_{0}}+(\\{0\\}\cup\operatorname{FS}(\\{x_{r}:t_{0}<r<k_{i},r\notin\bigcup_{q<i}F_{q}\\}))+x_{k_{i}}+(\\{0\\}\cup\operatorname{FS}(\\{x_{r}:k_{i}<r<i,r\notin\bigcup_{q<i}F_{q}\\}))+\operatorname{FS}(D_{i-1})\subseteq
x_{t_{0}}+\operatorname{FS}(D_{t_{0}})$ by items (b1), (d3b), (f) and (g2) for
$i-1$.

Hence, all the requirements are met. This finishes the case of $j_{i}=1$.

Assume now that for all $x=\sum_{b\leq a}x_{t_{b}}$, $t_{0}<\ldots<t_{a}<i$,
$t_{b}\notin\bigcup_{q<i}F_{q}$ we have
$x+x^{\prime}_{i}+\operatorname{FS}(E)\not\subseteq
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ and $x+\operatorname{FS}(E)\not\subseteq
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ for all $E\in[\omega]^{\omega}$ such that
$\operatorname{FS}(E)\subseteq\operatorname{FS}(D^{\prime}_{i})$. Put
$j_{i}=0$, $k_{i}=-1$ and $F_{i}=\emptyset$ (note that (c1) and (c2) are
satisfied).

Similarly as above (in the construction of $\tilde{D}_{i}$), we can find
$\tilde{D}_{i}\in[\omega]^{\omega}$ such that:

  * •

$\operatorname{FS}(\tilde{D}_{i})\subseteq\operatorname{FS}(D^{\prime}_{i})$;

  * •

$(x+x^{\prime}_{i}+\operatorname{FS}(\tilde{D}_{i}))\cap
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$ and
$(x+\operatorname{FS}(\tilde{D}_{i}))\cap
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$ for every $x=\sum_{b\leq
a}x_{t_{b}}$, $t_{0}<\ldots<t_{a}<i$, $t_{b}\notin\bigcup_{q\leq i}F_{q}$;

  * •

$\operatorname{FS}(\tilde{D}_{i})\cap\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(y)\cap\alpha_{D_{i-1}}(x_{i}^{\prime})\neq\emptyset\\}=\emptyset$.

Define $x_{i}=x^{\prime}_{i}+\min(\tilde{D}_{i})$ and let
$D_{i}\in[\omega]^{\omega}$ be very sparse such that
$\operatorname{FS}(D_{i})\subseteq\operatorname{FS}(\tilde{D}_{i}\setminus\\{\min(\tilde{D}_{i})\\})$
and $D_{i}$ satisfies item (f) (which is possible by Lemmas 2.3 and 7.1 and
(b2) applied to all $-1\leq t<i$). Note that (b2) is satisfied. Observe that
other conditions are met:

  * (b1)

$\operatorname{FS}(D_{i})\subseteq\operatorname{FS}(\tilde{D}_{i})\subseteq\operatorname{FS}(D^{\prime}_{i})\subseteq\operatorname{FS}(D_{i-1})\subseteq\operatorname{FS}(X)$;

  * (c3)

$x_{i}\in\operatorname{FS}(D_{i-1})$ as
$\operatorname{FS}(\tilde{D}_{i})\cap\\{y\in\operatorname{FS}(D_{i-1}):\alpha_{D_{i-1}}(y)\cap\alpha_{D_{i-1}}(x_{i}^{\prime})\neq\emptyset\\}=\emptyset$,
$x_{i}\in x^{\prime}_{i}+\operatorname{FS}(D^{\prime}_{i})\subseteq
f^{-1}[(\omega\times\\{n_{i}\\})\cap\Gamma]$;

  * (c4)

$x_{i}+\operatorname{FS}(D_{i})\subseteq
x^{\prime}_{i}+\operatorname{FS}(D^{\prime}_{i})\subseteq
f^{-1}[(\omega\times\\{n_{i}\\})\cap\Gamma]$;

  * (e)

for (e3), if $x=\sum_{b\leq a}x_{t_{b}}$, $t_{0}<\ldots<t_{a}<i$,
$t_{b}\notin\bigcup_{q\leq i}F_{q}$, then note that
$x+x_{i}=x+x^{\prime}_{i}+\min(\tilde{D}_{i})\in
x+x^{\prime}_{i}+\operatorname{FS}(\tilde{D}_{i})$ and recall that
$(x+x^{\prime}_{i}+\operatorname{FS}(\tilde{D}_{i}))\cap
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]=\emptyset$, thus $x+x_{i}\notin
f^{-1}[\\{(n_{i},n_{t_{0}})\\}]$ ((e1) and (e2) are similar);

  * (g1)

by items (f) and (g1) applied to $i-1$ and item (c3) applied to $i$,
$\operatorname{FS}(\\{x_{t}:t\leq i,t\notin\bigcup_{q\leq
i}F_{q}\\})=\operatorname{FS}(\\{x_{t}:t<i,t\notin\bigcup_{q<i}F_{q}\\})\cup(x_{i}+\operatorname{FS}(\\{x_{t}:\
t<i,t\notin\bigcup_{q<i}F_{q}\\}))\subseteq\operatorname{FS}(X)$;

  * (g2)

if $a>0$, $t_{0}<\ldots<t_{a}\leq i$ and $t_{b}\notin\bigcup_{q\leq i}F_{q}$,
then either $t_{a}<i$ and $\sum_{b\leq a}x_{t_{b}}\in
x_{t_{0}}+\operatorname{FS}(D_{t_{0}})$ (by item (g2) applied to $i-1$) or
$t_{a}=i$ and $\sum_{b\leq a}x_{t_{b}}\in
x_{t_{0}}+\operatorname{FS}(D_{t_{0}})$ as $\sum_{b<a}x_{t_{b}}\in
x_{t_{0}}+\operatorname{FS}(D_{t_{0}})$ and
$x_{i}\in\operatorname{FS}(D_{i-1})\subseteq\operatorname{FS}(D_{t_{0}})\setminus\\{y\in\operatorname{FS}(D_{t_{0}}):\
\exists_{j<i}\
\alpha_{D_{t_{0}}}(y)\cap\alpha_{D_{t_{0}}}(x_{j})\neq\emptyset\\}$ by item
(c3) for $i$ and items (b1), (f) and (g2) for $i-1$.

Hence, all the requirements are met. This finishes the case of $j_{i}=0$.

Note that $x_{i}\neq x_{j}$ for $i\neq j$ (it follows from items (a2), (c3)
and (d3a)). Once the whole recursive construction is completed, define
$A=\\{x_{i}:i\notin\bigcup_{q\in\omega}F_{q}\\}$. We need to show two facts:

  * (i)

$A$ is infinite;

  * (ii)

$f[\operatorname{FS}(A)]\in\mathcal{R}$.

Note that this will finish the proof as item (i) together with
$\operatorname{FS}(A)\subseteq\operatorname{FS}(X)$ (by item (g1)) guarantee
that $\operatorname{FS}(A)\notin\mathcal{H}\restriction\operatorname{FS}(X)$.

(i): Since $x_{i}\neq x_{j}$ for $i\neq j$, we only need to show that there
are infinitely many $t\in\omega$ such that $x_{t}\in A$. Assume to the
contrary that there is $p\in\omega$ such that $x_{t}\notin A$ for all $t\geq
p$. Without loss of generality we may assume that $p$ is minimal with that
property. Since $x_{p}\notin A$, we have that $p\in\bigcup_{q\in\omega}F_{q}$,
hence there is $q$ such that $p\in F_{q}$. By items (c2) and (d2), we know
that $j_{q}=1$, $q>p$ and, by minimality of $p$, $F_{q}=\\{p,p+1,\ldots
q-1\\}$. Again, as $x_{q}\notin A$ (because $q>p$), there should be $r$ such
that $q\in F_{r}=\\{k_{r},k_{r}+1,\ldots,r-1\\}$ (so $k_{r}\leq q<r$) and
$k_{r}\geq p$ (by minimality of $p$). However, this is impossible as item (d1)
gives us:

$k_{r}\in\\{u<r:u\notin\bigcup_{w<r}F_{w},j_{u}=0\\}\cap\\{0,1,\ldots,q\\}\subseteq$
$\subseteq\\{u\leq q:u\notin F_{q}\\}\cap\\{u\leq q:j_{u}=0\\}=$
$=(\\{0,1,\ldots,p-1\\}\cup\\{q\\})\cap\\{u\leq
q:j_{u}=0\\}\subseteq\\{0,1,\ldots,p-1\\}$

(here, if $p=0$ then $\\{0,1,\ldots,p-1\\}=\emptyset$).

(ii): We have:

$\begin{split}\operatorname{FS}(A)=&\bigcup_{i\in
B}(\\{x_{i}\\}\cup(x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})))\\\
&\cup\bigcup_{i\in
C}(\\{x_{i}\\}\cup(x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\}))),\end{split}$

where $B=\\{i\in\omega:i\notin\bigcup_{q\in\omega}F_{q},j_{i}=0\\}$ and
$C=\\{i\in\omega:i\notin\bigcup_{q\in\omega}F_{q},j_{i}=1\\}$.

At first we will show that $f[\bigcup_{i\in
C}(\\{x_{i}\\}\cup(x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})))]\in\mathcal{R}$.
Note that $f(x_{i})=(n_{i},n_{k_{i}})$ (by item (d3a)) and
$f[x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})]\subseteq
f[x_{i}+\operatorname{FS}(D_{i})]=\\{(n_{i},n_{k_{i}})\\}$ for each $i\in C$
(by items (d4) and (g2)). Moreover, the sequence $(n_{i})_{i\in\omega}$ is
injective (by item (a1)). Hence, $f[\bigcup_{i\in
C}(\\{x_{i}\\}\cup(x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})))]\in\mathcal{R}$,
as any set intersecting each $(\\{n\\}\times\omega)\cap\Gamma$ on at most one
point belongs to $\mathcal{R}$.

Now we will show that $f[\bigcup_{i\in
B}(\\{x_{i}\\}\cup(x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})))]\in\mathcal{R}$.
By items (c3), (c4) and (g2),
$f[\\{x_{i}\\}\cup(x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\}))]\subseteq
f[\\{x_{i}\\}\cup(x_{i}+\operatorname{FS}(D_{i}))]\subseteq(\omega\times\\{n_{i}\\})\cap\Gamma$
for all $i\in B$. Note that $\bigcup_{i\in B}f[\\{x_{i}\\}]\in\mathcal{R}$
(from (a1), as each set intersecting each $(\omega\times\\{n\\})\cap\Gamma$ on
at most one point belongs to $\mathcal{R}$). Suppose that
$Z^{2}\cap\Gamma\subseteq\bigcup_{i\in
B}f[x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})]$ for some
$Z\in[\omega]^{\omega}$.

Firstly, we will show that $|Z\setminus\\{n_{i}:i\in B\\}|\leq 1$. Suppose
that there are $z,w\in Z\setminus\\{n_{i}:i\in B\\}$ such that $z>w$. Then
there is $i\in B$ such that $(z,w)\in
f[x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})]$, hence
$x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})\subseteq
f^{-1}[\\{(z,w)\\}]$. But by (c4) and (g2) we have
$x_{i}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{i}\\})\subseteq
x_{i}+\operatorname{FS}(D_{i})\subseteq
f^{-1}[(\omega\times\\{n_{i}\\})\cap\Gamma]$. So
$(z,w)\in(\omega\times\\{n_{i}\\})\cap\Gamma$, i.e., $w=n_{i}$. A
contradiction.

By the previous paragraph, since $Z$ is infinite, there are $i,j\in B$ such
that $j<i$ and $n_{i},n_{j}\in Z$. We will show that
$(n_{i},n_{j})\notin\bigcup_{k\in
B}f[x_{k}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{k}\\})]$. This will
contradict $Z^{2}\cap\Gamma\subseteq\bigcup_{k\in
B}f[x_{k}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{k}\\})]$ and finish the
proof.

Suppose that $(n_{i},n_{j})\in\bigcup_{k\in
B}f[x_{k}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{k}\\})]$. From (c4) and
(g2), for every $k\neq j$, $k\in B$ we have
$f[x_{k}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{k}\\})]\subseteq
f[x_{k}+\operatorname{FS}(D_{k})]\subseteq(\omega\times\\{n_{k}\\})\cap\Gamma$,
so $(n_{i},n_{j})\notin
f[x_{k}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{k}\\})]$. Hence,
$(n_{i},n_{j})\in
f[x_{j}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{j}\\})]$. Let $y\in
x_{j}+\operatorname{FS}(A\setminus\\{0,1,\ldots,x_{j}\\})$ be such that
$f(y)=(n_{i},n_{j})$. Then $y=x_{j}+x_{s_{0}}+\ldots+x_{s_{p}}$ for some
$j<s_{0}<\ldots<s_{p}$. We have five cases:

  * •

If $s_{p}<i$, then from item (a2) we have
$f(y)\in[\\{0,\ldots,n_{i}-1\\}]^{2}$. A contradiction.

  * •

If $s_{p}=i$, then $y=(x_{j}+\ldots+x_{s_{p-1}})+x_{i}$ and from item (e3)
(applied to $x=(x_{j}+\ldots+x_{s_{p-1}})$) we get $f(y)\neq(n_{i},n_{j})$, a
contradiction.

  * •

If there exists $k<p$ such that $s_{k}=i$, then
$y=(x_{j}+\ldots+x_{s_{k-1}})+x_{i}+(x_{s_{k+1}}+\ldots+x_{s_{p}})\in(x_{j}+\ldots+x_{s_{k-1}})+x_{i}+\operatorname{FS}(D_{i})$
by item (g2) and from item (e1) (applied to $x=(x_{j}+\ldots+x_{s_{k-1}})$) we
get a contradiction.

  * •

If there exists $k\leq p$ such that $s_{k-1}<i<s_{k}$, then
$y=(x_{j}+\ldots+x_{s_{k-1}})+(x_{s_{k}}+\ldots+x_{s_{p}})\in(x_{j}+\ldots+x_{s_{k-1}})+\operatorname{FS}(D_{i})$
by item (g2) and from item (e2) we get a contradiction.

  * •

If $i<s_{0}$, then $y=x_{j}+(x_{s_{0}}+\ldots+x_{s_{p}})\in
x_{j}+\operatorname{FS}(D_{i})$ by item (g2) and from item (e2) we get a
contradiction.

Thus, $f[\operatorname{FS}(A)]\in\mathcal{R}$ and the proof is finished. ∎

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