Katětov order between Hindman, Ramsey and summable ideals

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 $f{: }Y\rightarrow X$ 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.

1 Introduction

The Katětov order is an efficient tool for studying ideals over countable sets [17,18,19,20, 31, 33]. Originally, the Katětov order (introduced by Katětov [21] in 1968) was used to study convergence in topological spaces, and our interest in Katětov order between the Hindman, Ramsey 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 and Hindman’s finite sums theorem [3, 4, 22,23,24,25]. It is known [10, 27] 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 incomparable 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 [18]. For instance, an ultrafilter $\mathcal {U}$ is a P-point if and only if the dual ideal $\mathcal {U}^*$ is not Katětov above $\textrm{Fin}^2$ (equivalently $\mathcal {U}$ is a $\textrm{Fin}^2$-ultrafilter as defined by Baumgartner [2]). It is known [9] that an existence of an ultrafilter which distinguishes between some classes of ultrafilters is reducible to a question whether particular ideals are incomparable in the Katětov order.

Below we describe the results obtained in this paper and introduce the 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 $\textrm{Fin}(X)$ we denote the family of all finite subsets of X and we write $\textrm{Fin}$ instead of $\textrm{Fin}(\omega )$.

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. ${{\,\textrm{FS}\,}}(B)=\{\sum _{n\in F}n: F\in [B]^{<\omega }{\setminus }\{\emptyset \}\}$.

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

• the Ramsey ideal

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

• the Hindman ideal

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

• the summable ideal

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

The Ramsey ideal was introduced by Meza-Alcántara [30, p. 28] where he used the notation $\mathcal {G}_c$ for this ideal. Moreover, the author 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. The Hindman ideal was introduced by Kojman and Shelah [24, p. 1620] where they used the notation $\mathcal {I}_{IP}$. The summable ideal is a particular instance of the so-called summable ideals which seem to be “ancient” compared to previously mentioned ideals as they were introduced in 1972 by Mathias [28, 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 [21] if there is a function $f:Y\rightarrow X$ 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, 14,15,16,17,18,19,20, 30,31,33, 36].

The aim of this paper is to prove the following

Theorem 1.1

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

One more ideal is closely related to the above ones – the van der Waerden ideal

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

Similarly to the Hindman ideal, van der Waerden ideal was introduced by Kojman and Shelah [24, p. 1620] where they used the notation $\mathcal {I}_{AP}$ for this ideal.

It is known that the ideals $\mathcal {R}$, $\mathcal {H}$ and $\mathcal {I}_{1/n}$ are not below the ideal $\mathcal {W}$ in the Katětov order (see [10, Theorem 7.7], [24, Lemma 1] and [12, Lemma 3.1], respectively). However, 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 [29, 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 $\iff$ $\mathcal {I}\le _K\mathcal {J}$ for every ideal $\mathcal {J}$ $\iff$ $\mathcal {I}\le _K\textrm{Fin}$ $\iff$ $\mathcal {I}\restriction A=\textrm{Fin}(A)$ 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 $\phi :X\rightarrow Y$ such that $A\in \mathcal {I}\iff \phi [A] \in \mathcal {J}$ for each $A\subseteq X$. An ideal $\mathcal {I}$ is homogeneous [26, Definition 1.3] if the ideals $\mathcal {I}$ and $\mathcal {I}\restriction A$ are isomorphic for every $A\in \mathcal {I}^+$.

Proposition 2.2

[26, 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 $A-n=\{a-n:a\in A,a\ge n\}$.

A set $D \subseteq \omega$ is sparse [22, p. 1598] if for each $x\in {{\,\textrm{FS}\,}}(D)$ 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 text, we write $\alpha (x)$ instead of $\alpha _E(x)$.

A set $D \subseteq \omega$ is very sparse [11, p. 894] if it is sparse and

$$\begin{aligned} \forall x,y\in {{\,\textrm{FS}\,}}(D)\, \left( \alpha _D(x)\cap \alpha _D(y) \ne \emptyset \implies x+y\notin {{\,\textrm{FS}\,}}(D)\right) . \end{aligned}$$

In the text, we will use the following

Lemma 2.3

[11, Lemma 2.2] For every infinite set $D \subseteq \omega$ there is an infinite set $D'\subseteq D$ which is very sparse.

Moreover, the following versions of Ramsey and Hindman’s theorems will be used in some proofs.

Theorem 2.4

Canonical Ramsey’s Theorem [8, Theorem II]; see also [13, Theorem 2 at p. 129] For every function $\phi :\omega \rightarrow \omega$ there exists an infinite set $T\subseteq \omega$ such that one of the following four cases holds:

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

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

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

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

Theorem 2.5

Canonical Hindman’s Theorem [34, Theorem 2.1]; see also [13, Theorem 5 at p. 133] For every function $\phi :\omega \rightarrow \omega$ there exists 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) $\forall x,y\in {{\,\textrm{FS}\,}}(C) (\phi (x)=\phi (y))$,

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

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

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

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

The ordinary Ramsey’s Theorem (Hindman’s Theorem, resp.) says that if the range of $\phi$ is finite in the Canonical Ramsey’s Theorem (Canonical Hindman’s Theorem, resp.) then we can always require that item (1) holds. There is also the Canonical van der Waerden Theorem, however we will not use it here, so let us just recall the ordinary version.

Theorem 2.6

van der Waerden’s Theorem [35]; see also [13, Theorem 1 at p. 29] For every function $\phi :\omega \rightarrow \omega$ with finite range there exists an infinite set $A\subseteq \omega$ which contains arbitrarily long finite arithmetic progressions and $\phi (x)=\phi (y)$ for every $x,y\in A$.

3 Summable and Hindman ideals are incomparable

Theorem 3.1

$\mathcal {H}\not \le _K\mathcal {I}_{1/n}$ and $\mathcal {I}_{1/n} \not \le _K \mathcal {H}$.

Proof

In [10, Theorem 7.7], the authors proved $\mathcal {H}\not \le _K\mathcal {I}_{1/n}$, while $\mathcal {I}_{1/n} \not \le _K \mathcal {H}$ is proved in [11, Theorem 3.2]. Below we provide a simpler proof of the latter inequality which is based on the Canonical Hindman’s Theorem.

Let $\phi :\omega \rightarrow \omega$ be an arbitrary function. Using the Canonical Hindman’s Theorem (Theorem 2.5) we can find 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) $\forall x,y\in {{\,\textrm{FS}\,}}(C) (\phi (x)=\phi (y))$,

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

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

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

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

We will show that $\phi$ is not a witness for $\mathcal {I}_{1/n}\le _K \mathcal {H}$ i.e. we will find an infinite set $D\subseteq \omega$ such that $\phi [{{\,\textrm{FS}\,}}(D)]\in \mathcal {I}_{1/n}$. To be precise, we will construct (separately in each case) a strictly increasing sequence $\{k_n:n\in \omega \}$ such that the set $D = \{c_{k_n}:n\in \omega \}$ will work.

Case (1) In this case we put $k_n=n$. Then $D=C$ and we see that the set $\phi [{{\,\textrm{FS}\,}}(D)]$ has only one element, so it belongs to $\mathcal {I}_{1/n}$.

Case (2) (Case (3), resp.) We construct $k_n$ 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 $k\in \omega$, $\min \alpha (c_k)\ne \min \alpha (c_l)$ ($\max \alpha (c_k)\ne \max \alpha (c_l)$, resp.) 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 recursive construction of $k_n$.

We put $A_n = \phi [c_{k_n}+{{\,\textrm{FS}\,}}(\{c_{k_i}:i>n\})]$ ($A_n = \phi [c_{k_n}+{{\,\textrm{FS}\,}}(\{c_{k_i}:i<n\})]$, resp.) for every $n\in \omega$. Using the properties of $c_{k_n}$’s we can see that $A_n = \{\phi (c_{k_n})\}$ for every $n\in \omega$, so

$$\begin{aligned} \begin{aligned} \sum _{y\in \phi [{{\,\text {FS}\,}}(D)]} \frac{1}{y+1}&= \sum _{n\in \omega }\!\left( \sum _{y\in \{\phi (c_{k_n})\}\cup A_n}\frac{1}{y\!+\!1}\right) \!=\! \sum _{n \in \omega }\frac{1}{\phi (c_{k_n})\!+\!1}\! \le \!\sum _{n \in \omega } \frac{1}{2^{n}+1}\! <\! \infty . \end{aligned} \end{aligned}$$

Case (4) We construct $k_n$ such that

$$\begin{aligned} \forall n\in \omega \,\forall i<n\, \left( \phi (c_{k_n})>n2^{n} \wedge \phi (c_{k_n}+c_{k_i})>n2^{n}\right) . \end{aligned}$$

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 $\min \alpha (c_k+c_{k_i})\ne \min \alpha (c_k+c_{k_j})$ and $\min \alpha (c_k)\ne \min \alpha (c_k+c_{k_i})$ for every $k>k_{n-1}$ and $i<j\le n-1$. 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 recursive construction of $k_n$.

We put $A_{m,n} =\phi [c_{k_m}+{{\,\textrm{FS}\,}}(\{c_{k_i}:m<i<n\})+c_{k_n}]$ for every $m,n\in \omega$, $m<n$. Using the properties of $c_{k_n}$’s we can see that $A_{m,n} = \{\phi (c_{k_m}+c_{k_n})\}$ for every $m<n$, $m,n\in \omega$, so

$$\begin{aligned} \begin{aligned} \sum _{y\in \phi [{{\,\textrm{FS}\,}}(D)]} \frac{1}{y+1}&= \sum _{n\in \omega }\frac{1}{\phi (c_{k_n})+1} + \sum _{n\in \omega }\sum _{m<n}\left( \sum _{y\in \{\phi (c_{k_m}+c_{k_n})\}\cup A_{m,n}}\frac{1}{y+1}\right) \\ {}&= \sum _{n\in \omega } \frac{1}{\phi (c_{k_n})+1} + \sum _{n\in \omega }\sum _{m<n}\frac{1}{\phi (c_{k_m}+c_{k_n})+1} \\ {}&\le \sum _{n\in \omega } \frac{1}{n2^{n}+1} + \sum _{n\in \omega }\sum _{m<n} \frac{1}{n2^{n}+1} <\infty . \end{aligned} \end{aligned}$$

Case (5) We construct $k_n$ such that

$$\begin{aligned} \forall n\in \omega \, \forall x\in {{\,\textrm{FS}\,}}(\{c_{k_i}:i<n\})\, \left( \phi (c_{k_n})> 2^{2n} \wedge \phi (c_{k_n}+x)> 2^{2n}\right) . \end{aligned}$$

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 $x\in {{\,\textrm{FS}\,}}(\{c_{k_i}:i<n\})$. Since $\phi \restriction {{\,\textrm{FS}\,}}(C)$ 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 $x\in {{\,\textrm{FS}\,}}(\{c_{k_i}:i<n\})$ we have $c_{k_n}+x>c_{k_n}>\max F$, so $\phi (c_{k_n}+x)> m>2^{2n}$. That finishes the recursive construction of $k_n$.

We put $A_{n} =\phi [c_{k_n}+{{\,\textrm{FS}\,}}(\{c_{k_i}:i<n\})]$ for every $n\in \omega$. Using the properties of $c_{k_n}$’s we can see that:

$$\begin{aligned} \begin{aligned} \sum _{y\in \phi [{{\,\textrm{FS}\,}}(D)]} \frac{1}{y+1}&= \sum _{n\in \omega }\left( \frac{1}{\phi (c_{k_n})+1} + \sum _{y\in A_n}\frac{1}{y+1}\right) \\ {}&\le \sum _{n\in \omega }\left( \frac{1}{2^{2n}+1}+\sum _{y\in A_n}\frac{1}{2^{2n} +1}\right) \\ {}&\le \sum _{n\in \omega }\left( \frac{1}{2^{2n}+1}+(2^n-1)\cdot \frac{1}{2^{2n} +1}\right) <\infty . \end{aligned} \end{aligned}$$

$\square$

4 Summable and Ramsey ideals are incomparable

Theorem 4.1

$\mathcal {R}\not \le _K\mathcal {I}_{1/n}$ and $\mathcal {I}_{1/n} \not \le _K \mathcal {R}$.

Proof

In [10, Theorem 7.7], the authors proved $\mathcal {R}\not \le _K\mathcal {I}_{1/n}$. Below we provide a proof of $\mathcal {I}_{1/n} \not \le _K \mathcal {R}$ which is based on the Canonical Ramsey’s Theorem.

Let $\phi :[\omega ]^2\rightarrow \omega$ be an arbitrary function. Using the Canonical Ramsey’s Theorem (Theorem 2.4) we can find an infinite set $T\subseteq \omega$ such that one of the following four cases holds:

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

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

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

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

We will show that $\phi$ is not a witness for $\mathcal {I}_{1/n}\le _K \mathcal {R}$ i.e. we will find an infinite set $H\subseteq \omega$ such that $\phi [[H]^2]\in \mathcal {I}_{1/n}$.

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) (Case (3), resp.) In this case, for every $t\in T$ the restriction $\phi \restriction \{\{t,s\}: s\in T, s>t\}$ ($\phi \restriction \{\{t,s\}: s\in T, s<t\}$, resp.) is constant with distinct values for distinct t. Thus, for every $t\in T$ there is $k_t$ such that $\{k_t\}=\phi [\{\{t,s\}:s\in T, s>t\}]$ ($\{k_t\}=\phi [\{\{t,s\}:s\in T, s<t\}]$, resp.).

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 $n\in \omega$.

Let $H=\{t_n:n\in \omega \}$ and $A_n = \phi [\{\{t_n,t_i\}:i>n\}]$ ($A_n = \phi [\{\{t_i,t_n\}:i<n\}]$, resp.) for every $n\in \omega$. Then

$$\begin{aligned} \sum _{k\in \phi [[H]^2]}\frac{1}{k+1} = \sum _{n=0}^\infty \left( \sum _{k\in A_n}\frac{1}{k+1}\right) = \sum _{n=0}^\infty \frac{1}{k_{t_n}+1} \le \sum _{n=0}^\infty \frac{1}{ 2^n} <\infty . \end{aligned}$$

Case (4) We construct recursively 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 recursive construction of $t_n$.

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

$$\begin{aligned} \sum _{k\in \phi [[H]^2]}\frac{1}{k+1} = \sum _{n=0}^\infty \sum _{i<n}\frac{1}{\phi (\{t_i,t_n\})+1} \le \sum _{n=0}^\infty \sum _{i<n}\frac{1}{n\cdot 2^n+1} \le \sum _{n=0}^\infty \frac{1}{2^n}<\infty . \end{aligned}$$

$\square$

5 Hindman ideal is not below Ramsey ideal

Lemma 5.1

If D is very sparse, then $\{x\in {{\,\textrm{FS}\,}}(D): \alpha _D(x)\cap \alpha _D(y)\ne \emptyset \}\in \mathcal {H}$ for every $y\in {{\,\textrm{FS}\,}}(D)$.

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

$$\begin{aligned} \Bigg \{x\in {{\,\textrm{FS}\,}}(D): \alpha _D(x)\cap \alpha _D(y)\ne \emptyset \Big \} = \bigcup _{i\le n}\Big \{x\in {{\,\textrm{FS}\,}}(D): k_i\in \alpha _D(x)\Bigg \}, \end{aligned}$$

we only need to show that $\{x\in {{\,\textrm{FS}\,}}(D): k_i\in \alpha _D(x)\}\in \mathcal {H}$ for every $i\le n$.

If $y,z\in \{x\in {{\,\textrm{FS}\,}}(D): k_i\in \alpha _D(x)\}$, then $k_i\in \alpha _D(y)\cap \alpha _D(z)\ne \emptyset$, so $y+z\notin {{\,\textrm{FS}\,}}(D)$ (since D is very sparse). Thus, there is no infinite (even two-element) set C such that ${{\,\textrm{FS}\,}}(C)\subseteq \{x\in {{\,\textrm{FS}\,}}(D): k_i\in \alpha _D(x)\}$. $\square$

Theorem 5.2

$\mathcal {H}\not \le _K \mathcal {R}$.

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 $\mathcal {H}\restriction {{\,\textrm{FS}\,}}(D)\not \le _K\mathcal {R}$.

Assume to the contrary that there exists $f:[\omega ]^2\rightarrow {{\,\textrm{FS}\,}}(D)$ which witnesses $\mathcal {H}\restriction {{\,\textrm{FS}\,}}(D) \le _K\mathcal {R}$.

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:

• (a) $b_n\in B_n$, $b_{n+1}>b_n$,

• (b) $B_{n+1}\subseteq B_{n}$, $B_{0}=\omega$,

• (c) for each $y \in f\left[ [\{b_i: i<n\}]^2\right]$ we have

  $$\begin{aligned} f[[B_n]^2]\cap \{x\in {{\,\textrm{FS}\,}}(D):\ \alpha _D(x)\cap \alpha _D(y)\ne \emptyset \}=\emptyset , \end{aligned}$$

• (d) for each $y \in f\left[ [\{b_i: i< n\}]^2\right]$ and $i< n$ we have

  $$\begin{aligned} f\left[ \{\{b_i,b\}:b\in B_n\}\right] -y\in \mathcal {H}. \end{aligned}$$

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 $\{x\in {{\,\textrm{FS}\,}}(D): \alpha _D(x)\cap \alpha _D(y)\ne \emptyset \}\in \mathcal {H}$ for every $y\in f\left[ [\{b_i: i<n\}]^2\right] \subseteq {{\,\textrm{FS}\,}}(D)$ (by Lemma 5.1), $[B_{n-1}]^2\in \mathcal {R}^+$ and we assumed that f witnesses $\mathcal {H}\restriction {{\,\textrm{FS}\,}}(D)\le _K\mathcal {R}$, there exists an infinite set $B \subseteq \omega$ such that

$$\begin{aligned}{}[B]^2 \subseteq [B_{n-1}]^2\setminus \bigcup _{y\in f\left[ [\{b_i: i< n\}]^2\right] } f^{-1}\left[ \{x\in {{\,\textrm{FS}\,}}(D):\ \alpha _D(x)\cap \alpha _D(y)\ne \emptyset \}\right] . \end{aligned}$$

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 $g:E{\setminus } \{b\}\rightarrow \omega$ be given by $g(x) = f(\{b,x\}) - y$. Since $\mathcal {H}$ is a tall ideal (Proposition 2.1), $\mathcal {H}\not \le _K \textrm{Fin}(E\setminus \{b\})$. Thus, there is $C\notin \textrm{Fin}(E{\setminus } \{b\})$ 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 $\mathcal {H}\restriction {{\,\textrm{FS}\,}}(D)\le _K\mathcal {R}$, $f[[B]^2]\in \mathcal {H}^+\restriction {{\,\textrm{FS}\,}}(D)$, and consequently there exists an infinite set $C\subseteq \omega$ such that ${{\,\textrm{FS}\,}}(C)\subseteq f[[B]^2]$.

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

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

Let $Y = \{\{b_i,b_k\}: i\le n < k\}\}$. 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 ${{\,\textrm{FS}\,}}(C{\setminus }\{c\}) \cap \left( f\left[ Z \right] - c\right) = \emptyset$. Suppose to the contrary that there exists $a\in {{\,\textrm{FS}\,}}(C{\setminus }\{c\})\cap \left( f\left[ Z \right] - c\right)$. Then $a+c \in {{\,\textrm{FS}\,}}(C)\cap f[Z]\subseteq {{\,\textrm{FS}\,}}(D)\cap f[[B_{n+1}]^2]$, so by item (c) applied to $y=c$, $\alpha _D(c)\cap \alpha _D(a+c)=\emptyset$. On the other hand, $a,c\in {{\,\textrm{FS}\,}}(D)$, D is very sparse and $a+c\in {{\,\textrm{FS}\,}}(D)$, so $\alpha _D(a)\cap \alpha _D(c)=\emptyset$. Consequently, $\alpha _D(a+c) = \alpha _D(a)\cup \alpha _D(c)$, so $\alpha _D(c)\cap \alpha _D(a+c) = \alpha _D(c) \ne \emptyset$, a contradiction.

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

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

a contradiction. $\square$

6 Ramsey ideal is not below Hindman ideal

Theorem 6.1

$\mathcal {R}\not \le _K \mathcal {H}$.

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 $\mathcal {R}\not \le _K\mathcal {H}\restriction {{\,\textrm{FS}\,}}(X)$. Fix any $f:{{\,\textrm{FS}\,}}(X)\rightarrow \Gamma$ and assume to the contrary that it witnesses $\mathcal {R}\le _K\mathcal {H}$. There are two possible cases.

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

In this case we recursively pick $\{x_n: n\in \omega \}\subseteq {{\,\textrm{FS}\,}}(D)$ 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) $x_n\in {{\,\textrm{FS}\,}}(D_{n-1})\setminus \big (\{x_i: i<n\}\cup \bigcup _{i<n}\bigcup _{j<n} \{y\in {{\,\textrm{FS}\,}}(D_j): \alpha _{D_j}(y)\cap \alpha _{D_j}(x_i)\ne \emptyset \}\big )$ (here we put $\alpha _{D_j}(x_i)=\emptyset$ whenever $x_i\notin D_j$);

• (b) $D_n$ is very sparse;

• (c) ${{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_n\})\subseteq {{\,\textrm{FS}\,}}(D)$;

• (d) ${{\,\textrm{FS}\,}}(D_n)\subseteq {{\,\textrm{FS}\,}}(D_{n-1})\subseteq {{\,\textrm{FS}\,}}(D)$;

• (e) $\left( f^{-1}[(\omega \times \{k+i\})\cap \Gamma ]-x\right) \cap {{\,\textrm{FS}\,}}(D_n)=\emptyset$ for every $x\in {{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_n\})$ and $1\le i\le n+1$;

• (f) $f^{-1}[(\omega \times \{k+i\})\cap \Gamma ]\cap {{\,\textrm{FS}\,}}(D_n)=\emptyset$ for all $1\le i\le n+1$.

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 $x_n\in {{\,\textrm{FS}\,}}(D_{n-1})$ such that ${{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_n\})\subseteq {{\,\textrm{FS}\,}}(D)$ and $x_n\ne x_i$ for all $i<n$. This is possible since it suffices to pick any point from the set

$$\begin{aligned} {{\,\textrm{FS}\,}}(D_{n-1})\setminus \bigcup _{i<n}\bigcup _{j<n} \{y\in {{\,\textrm{FS}\,}}(D_j): \alpha _{D_j}(y)\cap \alpha _{D_j}(x_i)\ne \emptyset \}, \end{aligned}$$

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

$$\begin{aligned} \bigcup _{i<n}\bigcup _{j<n} \{y\in {{\,\textrm{FS}\,}}(D_j): \alpha _{D_j}(y)\cap \alpha _{D_j}(x_i)\ne \emptyset \}\in \mathcal {H}\restriction {{\,\textrm{FS}\,}}(X) \end{aligned}$$

by Lemma 5.1 and item (b) for all $j<n$ (here we put $\alpha _{D_j}(x_i)=\emptyset$ whenever $x_i\notin D_j$).

Enumerate ${{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_n\})=\{c_0,c_1,\ldots ,c_{2^{n+1}-2}\}$. We will define sets $E_t\in [\omega ]^\omega$ for $-1\le t\le n$ such that $E_{-1}=D_{n-1}$ and for all $0\le t\le n$:

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

• $\left( \bigcup _{1\le i\le n+1}f^{-1}[(\omega \times \{k+i\})\cap \Gamma ]-c_l\right) \cap {{\,\textrm{FS}\,}}(E_t)=\emptyset$ for every $0\le l\le 2^{n+1}-2$.

Such construction is possible. Indeed, since we are in Case 1 and each $c_l\in {{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_n\})\subseteq {{\,\textrm{FS}\,}}(D)$, we know that:

$$\begin{aligned} \begin{aligned}&\left( \bigcup _{1\le i\le n+1}f^{-1}[(\omega \times \{k+i\})\cap \Gamma ]-c_l\right) \cap \left( \{y\in {{\,\textrm{FS}\,}}(D): \alpha _D(c_l)\subseteq \alpha _D(y)\}-c_l\right) \\ {}&\quad = \left( \bigcup _{1\le i\le n+1}f^{-1}[(\omega \times \{k+i\})\cap \Gamma ] \cap \{y\in {{\,\textrm{FS}\,}}(D): \alpha _D(c_l)\subseteq \alpha _D(y)\}\right) \\&\quad -c_l \in \mathcal {H}\restriction {{\,\textrm{FS}\,}}(D). \end{aligned} \end{aligned}$$

On the other hand, we get:

$$\begin{aligned} \begin{aligned}&{{\,\textrm{FS}\,}}(E_{t-1})\cap \left( \{y\in {{\,\textrm{FS}\,}}(D):\alpha _D(c_l)\subseteq \alpha _D(y)\}-c_l\right) \\&\quad \supseteq {{\,\textrm{FS}\,}}(E_{t-1})\cap \left( {{\,\textrm{FS}\,}}(D)\setminus \{y\in {{\,\textrm{FS}\,}}(D): \alpha _D(c_l)\cap \alpha _D(y)\ne \emptyset \}\right) \\&\quad \supseteq {{\,\textrm{FS}\,}}(E_{t-1})\setminus \{y\in {{\,\textrm{FS}\,}}(D): \alpha _D(c_l)\cap \alpha _D(y)\ne \emptyset \}\notin \mathcal {H}\restriction {{\,\textrm{FS}\,}}(D), \end{aligned} \end{aligned}$$

as $\{y\in {{\,\textrm{FS}\,}}(D):\alpha _D(c_l)\cap \alpha _D(y)\ne \emptyset \}\in \mathcal {H}\restriction {{\,\textrm{FS}\,}}(D)$ (by Lemma 5.1). Then ${{\,\textrm{FS}\,}}(E_{t-1}){\setminus } \left( \bigcup _{1\le i\le n+1}f^{-1}[ (\omega \times \{k+i\})\cap \Gamma ]-c_l\right)$ contains the set

$$\begin{aligned}{} & {} \left( {{\,\textrm{FS}\,}}(E_{t-1})\cap \left( \{y\in {{\,\textrm{FS}\,}}(D): \alpha _D(c_l)\subseteq \alpha _D(y)\}-c_l\right) \right) \\{} & {} \quad \setminus \left( \left( \bigcup _{1\le i\le n+1}f^{-1}[(\omega \times \{k+i\})\cap \Gamma ]-c_l\right) \cap \left( \{y\in {{\,\textrm{FS}\,}}(D):\alpha _D(c_l)\subseteq \alpha _D(y)\}-c_l\right) \right) \end{aligned}$$

which does not belong to $\mathcal {H}\restriction {{\,\textrm{FS}\,}}(D)$. Thus, there is $E_t\in [\omega ]^\omega$ as needed.

Once all $E_t$ are defined, observe that

$$\begin{aligned} \bigcup _{1\le i\le n+1}(\omega \times \{k+i\})\cap \Gamma \in \mathcal {R}. \end{aligned}$$

Since we assumed that f witnesses $\mathcal {R}\le _K\mathcal {H}$,

$$\begin{aligned} {{\,\textrm{FS}\,}}(E_n)\setminus \bigcup _{1\le i\le n+1}f^{-1}[(\omega \times \{k+i\})\cap \Gamma ]\notin \mathcal {H}. \end{aligned}$$

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

$$\begin{aligned} {{\,\textrm{FS}\,}}(D_n)\subseteq {{\,\textrm{FS}\,}}(E_n)\setminus \bigcup _{1\le i\le n+1}f^{-1}[(\omega \times \{k+i\})\cap \Gamma ] \end{aligned}$$

(by Lemma 2.3). Note that ${{\,\textrm{FS}\,}}(D_n)\subseteq {{\,\textrm{FS}\,}}(E_n)\subseteq {{\,\textrm{FS}\,}}(D_{n-1})\subseteq {{\,\textrm{FS}\,}}(D)$ and

$$\begin{aligned} \bigcup _{1\le i\le n+1}\left( f^{-1}[(\omega \times \{k+i\})\cap \Gamma ]-x\right) \cap {{\,\textrm{FS}\,}}(D_n)=\emptyset \end{aligned}$$

for all $x\in {{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_n\})$.

This finishes the construction of $\{x_n: n\in \omega \}\subseteq {{\,\textrm{FS}\,}}(D)$ and $\{D_n:n\in \omega \cup \{-1\}\}\subseteq [\omega ]^\omega$.

Define $B={{\,\textrm{FS}\,}}(\{x_n:\ n\in \omega \})$. Obviously, $B\notin \mathcal {H}\restriction {{\,\textrm{FS}\,}}(X)$ as ${{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_n\})\subseteq {{\,\textrm{FS}\,}}(D)\subseteq {{\,\textrm{FS}\,}}(X)$ 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 $\bigcup _{n\le k}(\omega \times \{n\})\cap \Gamma \in \mathcal {R}$ 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 $x=x_{n_0}+\cdots +x_{n_t}\in B$, where $n_0<\ldots <n_t$. If $n_0>m$, then $x\in {{\,\textrm{FS}\,}}(\{x_n:\ n>m\})\subseteq {{\,\textrm{FS}\,}}(D_m)$ which contradicts $f(x)\in (\omega \times \{k+m+1\})\cap \Gamma$ (by item (f)). If $n_0\le m$ but $J=\{j\le t: n_j>m\}\ne \emptyset$, then let $j=\min J$ and note that $x\in \sum _{i<j}x_{n_i}+{{\,\textrm{FS}\,}}(D_m)$. As $\sum _{i<j}x_{n_i}\in {{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_m\})$, item (e) gives us a contradiction with $f(x)\in (\omega \times \{k+m+1\})\cap \Gamma$. Hence, the only possibility is that $n_j\le m$ for all $j\le t$. Thus, $f[B]\cap (\omega \times \{k+m+1\})\cap \Gamma \subseteq f[{{\,\textrm{FS}\,}}(\{x_0,\ldots ,x_m\})]$, which is a finite set.

Case (2) For every $k\in \omega$ and very sparse $D\in [\omega ]^\omega$, ${{\,\textrm{FS}\,}}(D)\subseteq {{\,\textrm{FS}\,}}(X)$, there are $n>k$ and $x\in {{\,\textrm{FS}\,}}(D)$ such that:

$$\begin{aligned} (f^{-1}[(\omega \times \{n\})\cap \Gamma ]\cap \{y\in {{\,\textrm{FS}\,}}(D): \alpha _D(x)\subseteq \alpha _D(y)\})-x\notin \mathcal {H}\restriction {{\,\textrm{FS}\,}}(X). \end{aligned}$$

In this case we will pick $\{n_i: i\in \omega \}\subseteq \omega$, $\{j_i: i\in \omega \}\subseteq \{0,1\}$, $\{x_i: i\in \omega \}\subseteq {{\,\textrm{FS}\,}}(D)$, $\{D_i: i\in \omega \cup \{-1\}\}\subseteq [\omega ]^\omega$, $\{k_i: i\in \omega \}\subseteq \omega \cup \{-1\}$ and $\{F_i:i\in \omega \}\subseteq \textrm{Fin}$ 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) $n_{i}>\min \{a\in \omega : f[{{\,\textrm{FS}\,}}(\{x_j: j<i\})]\subseteq \{0,1,\ldots ,a\}^2\cap \Gamma \}$;

• (b)  

  • (b1) ${{\,\textrm{FS}\,}}(D_i)\subseteq {{\,\textrm{FS}\,}}(D_{i-1})\subseteq {{\,\textrm{FS}\,}}(X)$;

  • (b2) $D_i$ is very sparse;

• (c) if $j_i=0$, then:

  • (c1) $k_i=-1$;

  • (c2) $F_i=\emptyset$;

  • (c3) $x_i\in {{\,\textrm{FS}\,}}(D_{i-1})\cap f^{-1}[(\omega \times \{n_i\})\cap \Gamma ]$;

  • (c4) $x_i+{{\,\textrm{FS}\,}}(D_i)\subseteq f^{-1}[(\omega \times \{n_i\})\cap \Gamma ]$;

• (d) if $j_i=1$, then:

  • (d1) $k_i\in \{0\le u<i: u\notin \bigcup _{q<i}F_q,j_u=0\}$;

  • (d2) $F_i=\{k_i,k_i+1,\ldots ,i-1\}$;

  • (d3)  

    • (d3a) $x_i\in f^{-1}[\{(n_i,n_{k_i})\}]$;

    • (d3b) $x_i\in x_{k_i}+(\{0\}\cup {{\,\textrm{FS}\,}}(\{x_r: k_i<r<i, r\notin \bigcup _{q<i}F_q\}))+{{\,\textrm{FS}\,}}(D_{i-1})$;

  • (d4) $x_i+{{\,\textrm{FS}\,}}(D_i)\subseteq f^{-1}[\{(n_i,n_{k_i})\}]$;

• (e) if $x=\sum _{b\le a}x_{t_b}$ for some $0\le t_0<\ldots<t_a<i$, $t_b\notin \bigcup _{q\le i}F_q$ (so $x\in {{\,\textrm{FS}\,}}(\{x_t: t<i,\ t\notin \bigcup _{q\le i}F_q\})$), then:

  • (e1) $(x+x_i+{{\,\textrm{FS}\,}}(D_i))\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$;

  • (e2) $(x+{{\,\textrm{FS}\,}}(D_i))\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$;

  • (e3) $(x+x_i)\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$;

• (f) ${{\,\textrm{FS}\,}}(D_i)\cap \{y\in {{\,\textrm{FS}\,}}(D_{t}): \alpha _{D_{t}}(y)\cap \alpha _{D_{t}}(x_u)\ne \emptyset \}=\emptyset$ for all $-1\le t<i$ and $0\le u\le i$ such that $x_u\in {{\,\textrm{FS}\,}}(D_t)$;

• (g)  

  • (g1) ${{\,\textrm{FS}\,}}(\{x_t: t\le i, t\notin \bigcup _{q\le i}F_q\})\subseteq {{\,\textrm{FS}\,}}(X)$;

  • (g2) $\sum _{b\le a}x_{t_b}\in x_{t_{0}}+{{\,\textrm{FS}\,}}(D_{t_0})$ for every $a>0$, $0\le t_0<\ldots <t_a\le i$, $t_b\notin \bigcup _{q\le i}F_q$;

In the 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 $x'_0\in {{\,\textrm{FS}\,}}(X)$ such that: $(f^{-1}[(\omega \times \{n_0\})\cap \Gamma ]\cap \{y\in {{\,\textrm{FS}\,}}(X): \alpha _X(x'_0)\subseteq \alpha _X(y)\})-x'_0\notin \mathcal {H}\restriction {{\,\textrm{FS}\,}}(X).$ Hence, there is $D'_0\in [\omega ]^\omega$ such that: $x'_0+{{\,\textrm{FS}\,}}(D'_0)\subseteq f^{-1}[(\omega \times \{n_0\})\cap \Gamma ]\cap \{y\in {{\,\textrm{FS}\,}}(X): \alpha _X(x'_0)\subseteq \alpha _X(y)\}\subseteq {{\,\textrm{FS}\,}}(X).$ Put $j_0=0$, $k_0=-1$ and $F_0=\emptyset$ (note that (c1) and (c2) are satisfied). Moreover, define $x_0=x'_0+\min (D'_0)$ (note that (c3) and (g1) are satisfied, because $x_0\in x'_0+{{\,\textrm{FS}\,}}(D'_0)\subseteq {{\,\textrm{FS}\,}}(X)$ and $x'_0+{{\,\textrm{FS}\,}}(D'_0)\subseteq f^{-1}[(\omega \times \{n_0\})\cap \Gamma ]$) and using Lemma 2.3 find a very sparse $D_0\in [\omega ]^\omega$ such that

$$\begin{aligned} {{\,\textrm{FS}\,}}(D_0)\subseteq {{\,\textrm{FS}\,}}(D'_0\setminus \{\min (D'_0)\})\setminus \{y\in {{\,\textrm{FS}\,}}(X): \alpha _X(y)\cap \alpha _X(x_0)\ne \emptyset \}, \end{aligned}$$

which is possible as $\{y\in {{\,\textrm{FS}\,}}(X):\alpha _X(y)\cap \alpha _X(x_0)\ne \emptyset \}\in \mathcal {H}\restriction {{\,\textrm{FS}\,}}(X)$ by Lemma 5.1 (note that (c4), (f) and (b) are satisfied, because $x_0+{{\,\textrm{FS}\,}}(D_0)\subseteq x'_0+{{\,\textrm{FS}\,}}(D'_0)\subseteq f^{-1}[(\omega \times \{n_0\})\cap \Gamma ]$ and ${{\,\textrm{FS}\,}}(D_0)\subseteq {{\,\textrm{FS}\,}}(D'_0)\subseteq \{y\in {{\,\textrm{FS}\,}}(X):\alpha _X(x'_0)\subseteq \alpha _X(y)\}-x'_0\subseteq {{\,\textrm{FS}\,}}(X)$). In conditions (e) and (g2) there is nothing to check. Thus, all the requirements are met.

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

$$\begin{aligned} \left( f^{-1}[(\omega \times \{n_i\})\!\cap \!\Gamma ]\!\cap \!\{y\!\in \! {{\,\textrm{FS}\,}}(D_{i\!-\!1}):\alpha _{D_{i\!-\!1}}(x_i')\!\subseteq \! \alpha _{D_{i\!-\!1}}(y)\}\right) \!-\!x'_i\notin \mathcal {H}\restriction {{\,\textrm{FS}\,}}(X). \end{aligned}$$

Hence, there is $D'_i\in [\omega ]^\omega$ such that: $x'_i+{{\,\textrm{FS}\,}}(D'_i)\subseteq f^{-1}[(\omega \times \{n_i\})\cap \Gamma ]\cap \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(x_i')\subseteq \alpha _{D_{i-1}}(y)\}.$ In particular, ${{\,\textrm{FS}\,}}(D'_i)\subseteq \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(x_i')\subseteq \alpha _{D_{i-1}}(y)\}-x'_i= \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(x_i')\cap \alpha _{D_{i-1}}(y)=\emptyset \}\subseteq {{\,\textrm{FS}\,}}(D_{i-1})$. There are two possibilities.

Assume first that there is $x=\sum _{b\le a}x_{t_b}$ for some $t_0<\ldots<t_a<i$, $t_b\notin \bigcup _{q<i}F_q$ such that either $x+x'_i+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq f^{-1}[\{(n_i,n_{t_0})\}]$ or $x+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq f^{-1}[\{(n_i,n_{t_0})\}]$ for some $\bar{D}_i\in [\omega ]^\omega$ such that ${{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq {{\,\textrm{FS}\,}}(D'_i)$. 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 $k_i\in \{0\le u<i:\ u\notin \bigcup _{q<i}F_q\}$. We will show that $j_{k_i}=0$ (i.e., (d1) is satisfied). Suppose that $j_{k_i}=j_{t_0}=1$. Observe that $x'_i+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq {{\,\textrm{FS}\,}}(D_{i-1})$ (by ${{\,\textrm{FS}\,}}(\bar{D}_i)\cap \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(y)\cap \alpha _{D_{i-1}}(x_i')\ne \emptyset \}=\emptyset$) and consequently $x'_i+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq {{\,\textrm{FS}\,}}(D_{t_0})$ (by item (b1)). Then items (b1), (f) and (g2) give us:

• $x+x'_i+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq x_{t_0}+{{\,\textrm{FS}\,}}(D_{t_0})$,

• $x+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq x_{t_0}+{{\,\textrm{FS}\,}}(D_{t_0})$.

Then from (d4) we have:

• $f[x+x'_i+{{\,\textrm{FS}\,}}(\bar{D}_i)]\subseteq \{(n_{t_0},n_{k_{t_0}})\}$,

• $f[x+{{\,\textrm{FS}\,}}(\bar{D}_i)]\subseteq \{(n_{t_0},n_{k_{t_0}})\}$.

This contradicts $x+x'_i+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq f^{-1}[\{(n_i,n_{t_0})\}]$ or $x+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq f^{-1}[\{(n_i,n_{t_0})\}]$, 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 $\bar{x}_i=x+x'_i$ (or $\bar{x}_i=x$ if $x+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq f^{-1}[\{(n_i,n_{t_0})\}]$).

To define $D_i$ and $x_i$, note that by the choice of $k_i$, for each $y=\sum _{b\le a}x_{t_b}$, $t_0<\ldots<t_a<i$, $t_b\notin \bigcup _{q\le i}F_q$ (so in fact $t_a<k_i$) we know that $(y+\bar{x}_i+{{\,\textrm{FS}\,}}(E))\not \subseteq f^{-1}[\{(n_i,n_{t_0})\}]$ and $(y+{{\,\textrm{FS}\,}}(E))\not \subseteq f^{-1}[\{(n_i,n_{t_0})\}]$ for every $E\in [\omega ]^\omega$ such that ${{\,\textrm{FS}\,}}(E)\subseteq {{\,\textrm{FS}\,}}(D'_i)$. In other words, $f^{-1}[\{(n_i,n_{t_0})\}]-(y+\bar{x}_i)\in \mathcal {H}\restriction {{\,\textrm{FS}\,}}(D'_i)$ and $f^{-1}[\{(n_i,n_{t_0})\}]-y\in \mathcal {H}\restriction {{\,\textrm{FS}\,}}(D'_i)$, for every such y. Thus, we can find $\tilde{D}_i\in [\omega ]^\omega$ such that:

• ${{\,\textrm{FS}\,}}(\tilde{D}_i)\subseteq {{\,\textrm{FS}\,}}(\bar{D}_i)$;

• $(y+\bar{x}_i+{{\,\textrm{FS}\,}}(\tilde{D}_i))\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$ and $(y+{{\,\textrm{FS}\,}}(\tilde{D}_i))\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$ for every $y=\sum _{b\le a}x_{t_b}$, where $t_0<\ldots<t_a<i$, $t_b\notin \bigcup _{q\le i}F_q$;

• ${{\,\textrm{FS}\,}}(\tilde{D}_i)\cap \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(y)\cap \alpha _{D_{i-1}}(x_i')\ne \emptyset \}=\emptyset$;

(the last item is trivial, as ${{\,\textrm{FS}\,}}(\tilde{D}_i)\subseteq {{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq {{\,\textrm{FS}\,}}(D'_i)$ and ${{\,\textrm{FS}\,}}(D'_i)\subseteq \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(x_i')\cap \alpha _{D_{i-1}}(y)=\emptyset \}$).

Define $x_i=\bar{x}_i+\min (\tilde{D}_i)$ and let $D_i\in [\omega ]^\omega$ be very sparse such that ${{\,\textrm{FS}\,}}(D_i)\subseteq {{\,\textrm{FS}\,}}(\tilde{D}_i{\setminus }\{\min (\tilde{D}_i)\})$ and $D_i$ satisfies item (f). It is possible using Lemma 2.3, as $\{y\in {{\,\textrm{FS}\,}}(D_{t}):\alpha _{D_{t}}(y)\cap \alpha _{D_{t}}(x_u)\ne \emptyset \}\in \mathcal {H}\restriction {{\,\textrm{FS}\,}}(X)$ by Lemma 5.1 and item (b2) for all $-1\le t<i$. Then (b2) is satisfied. Observe that other conditions are met:

• (b1) ${{\,\textrm{FS}\,}}(D_i)\subseteq {{\,\textrm{FS}\,}}(\tilde{D}_i)\subseteq {{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq {{\,\textrm{FS}\,}}(D'_i)\subseteq {{\,\textrm{FS}\,}}(D_{i-1})\subseteq {{\,\textrm{FS}\,}}(X)$;

• (d3b) if $\bar{x}_i=x+x'_i$, then $x_i=\bar{x}_i+\min (\tilde{D}_i)=x_{k_i}+(x-x_{k_i})+x'_i+\min (\tilde{D}_i)\in x_{k_i}+(\{0\}\cup {{\,\textrm{FS}\,}}(\{x_r:\ k_i<r<i, r\notin \bigcup _{q<i}F_q\}))+{{\,\textrm{FS}\,}}(D_{i-1})$ by the fact that ${{\,\textrm{FS}\,}}(\tilde{D}_i)\cap \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(y)\cap \alpha _{D_{i-1}}(x_i')\ne \emptyset \}=\emptyset$ (if $\bar{x}_i=x$ this is even easier to show);

• (d3a) if $\bar{x}_i=x+x'_i$, then $x_i\in x+x'_i+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq f^{-1}[\{(n_i,n_{k_i})\}]$ (if $\bar{x}_i=x$ this is also true);

• (d4) if $\bar{x}_i=x+x'_i$, then $x_i+{{\,\textrm{FS}\,}}(D_i)\subseteq x+x'_i+{{\,\textrm{FS}\,}}(\bar{D}_i)\subseteq f^{-1}[\{(n_i,n_{k_i})\}]$ (if $\bar{x}_i=x$ this is also true);

• (e) for (e3), if $y=\sum _{b\le a}x_{t_b}$, $t_0<\ldots<t_a<i$, $t_b\notin \bigcup _{q\le i}F_q$, then note that $y+x_i=y+\bar{x}_i+\min (\tilde{D}_i)\in y+\bar{x}_i+{{\,\textrm{FS}\,}}(\tilde{D}_i)$ and recall that $(y+\bar{x}_i+{{\,\textrm{FS}\,}}(\tilde{D}_i))\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$, thus $y+x_i\notin f^{-1}[\{(n_i,n_{t_0})\}]$ ((e1) and (e2) are similar);

• (g1) ${{\,\textrm{FS}\,}}(\{x_t: t\le i, t\notin \bigcup _{q\le i}F_q\})\subseteq {{\,\textrm{FS}\,}}(\{x_t: t<i, t\notin \bigcup _{q<i}F_q\})\cup (x_i+{{\,\textrm{FS}\,}}(\{x_t: t<i, t\notin \bigcup _{q<i}F_q\})\subseteq {{\,\textrm{FS}\,}}(X)$ by items (f) and (g1) applied to $i-1$ and item (d3b) applied to i;

• (g2) if $a>0$, $t_{0}<\ldots <t_{a}\le i$, $t_b\notin \bigcup _{q\le i}F_q$ then either $t_a<i$ and $\sum _{b\le a}x_{t_b}\in x_{t_0}+{{\,\textrm{FS}\,}}(D_{t_0})$ (by (g2) applied to $i-1$) or $t_a=i$ and $\sum _{b\le a}x_{t_b}=\sum _{b<a}x_{t_b}+x_i\in x_{t_0}+(\{0\}\cup {{\,\textrm{FS}\,}}(\{x_r: t_0<r<k_i, r\notin \bigcup _{q<i}F_q\}))+x_{k_i}+(\{0\}\cup {{\,\textrm{FS}\,}}(\{x_r: k_i<r<i, r\notin \bigcup _{q<i}F_q\}))+{{\,\textrm{FS}\,}}(D_{i-1})\subseteq x_{t_0}+{{\,\textrm{FS}\,}}(D_{t_0})$ 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 $x=\sum _{b\le a}x_{t_b}$, $t_0<\ldots<t_a<i$, $t_b\notin \bigcup _{q<i}F_q$ we have $x+x'_i+{{\,\textrm{FS}\,}}(E)\not \subseteq f^{-1}[\{(n_i,n_{t_0})\}]$ and $x+{{\,\textrm{FS}\,}}(E)\not \subseteq f^{-1}[\{(n_i,n_{t_0})\}]$ for all $E\in [\omega ]^\omega$ such that ${{\,\textrm{FS}\,}}(E)\subseteq {{\,\textrm{FS}\,}}(D'_i)$. 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:

• ${{\,\textrm{FS}\,}}(\tilde{D}_i)\subseteq {{\,\textrm{FS}\,}}(D'_i)$;

• $(x+x'_i+{{\,\textrm{FS}\,}}(\tilde{D}_i))\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$ and $(x+{{\,\textrm{FS}\,}}(\tilde{D}_i))\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$ for every $x=\sum _{b\le a}x_{t_b}$, $t_0<\ldots<t_a<i$, $t_b\notin \bigcup _{q\le i}F_q$;

• ${{\,\textrm{FS}\,}}(\tilde{D}_i)\cap \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(y)\cap \alpha _{D_{i-1}}(x_i')\ne \emptyset \}=\emptyset$.

Define $x_i=x'_i+\min (\tilde{D}_i)$ and let $D_i\in [\omega ]^\omega$ be very sparse such that ${{\,\textrm{FS}\,}}(D_i)\subseteq {{\,\textrm{FS}\,}}(\tilde{D}_i{\setminus }\{\min (\tilde{D}_i)\})$ and $D_i$ satisfies item (f) (which is possible by Lemmas 2.3 and 5.1 and (b2) applied to all $-1\le t<i$). Note that (b2) is satisfied. Observe that other conditions are met:

• (b1) ${{\,\textrm{FS}\,}}(D_i)\subseteq {{\,\textrm{FS}\,}}(\tilde{D}_i)\subseteq {{\,\textrm{FS}\,}}(D'_i)\subseteq {{\,\textrm{FS}\,}}(D_{i-1})\subseteq {{\,\textrm{FS}\,}}(X)$;

• (c3) $x_i\in {{\,\textrm{FS}\,}}(D_{i-1})$ as ${{\,\textrm{FS}\,}}(\tilde{D}_i)\cap \{y\in {{\,\textrm{FS}\,}}(D_{i-1}):\alpha _{D_{i-1}}(y)\cap \alpha _{D_{i-1}}(x_i')\ne \emptyset \}=\emptyset$, $x_i\in x'_i+{{\,\textrm{FS}\,}}(D'_i)\subseteq f^{-1}[(\omega \times \{n_i\})\cap \Gamma ]$;

• (c4) $x_i+{{\,\textrm{FS}\,}}(D_i)\subseteq x'_i+{{\,\textrm{FS}\,}}(D'_i)\subseteq f^{-1}[(\omega \times \{n_i\})\cap \Gamma ]$;

• (e) for (e3), if $x=\sum _{b\le a}x_{t_b}$, $t_0<\ldots<t_a<i$, $t_b\notin \bigcup _{q\le i}F_q$, then note that $x+x_i=x+x'_i+\min (\tilde{D}_i)\in x+x'_i+{{\,\textrm{FS}\,}}(\tilde{D}_i)$ and recall that $(x+x'_i+{{\,\textrm{FS}\,}}(\tilde{D}_i))\cap f^{-1}[\{(n_i,n_{t_0})\}]=\emptyset$, 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, ${{\,\textrm{FS}\,}}(\{x_t: t\le i,t\notin \bigcup _{q\le i}F_q\})={{\,\textrm{FS}\,}}(\{x_t: t<i,t\notin \bigcup _{q<i}F_q\})\cup (x_i+{{\,\textrm{FS}\,}}(\{x_t:\ t<i,t\notin \bigcup _{q< i}F_q\}))\subseteq {{\,\textrm{FS}\,}}(X)$;

• (g2) if $a>0$, $t_{0}<\ldots <t_{a}\le i$ and $t_b\notin \bigcup _{q\le i}F_q$, then either $t_a<i$ and $\sum _{b\le a}x_{t_b}\in x_{t_0}+{{\,\textrm{FS}\,}}(D_{t_0})$ (by item (g2) applied to $i-1$) or $t_a=i$ and $\sum _{b\le a}x_{t_b}\in x_{t_0}+{{\,\textrm{FS}\,}}(D_{t_0})$ as $\sum _{b<a}x_{t_b}\in x_{t_0}+{{\,\textrm{FS}\,}}(D_{t_0})$ and $x_i\in {{\,\textrm{FS}\,}}(D_{i-1})\subseteq {{\,\textrm{FS}\,}}(D_{t_0}){\setminus } \{y\in {{\,\textrm{FS}\,}}(D_{t_0}):\ \exists _{j<i}\ \alpha _{D_{t_0}}(y)\cap \alpha _{D_{t_0}}(x_j)\ne \emptyset \}$ 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 $x_i\ne x_j$ for $i\ne j$ (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) $f[{{\,\textrm{FS}\,}}(A)]\in \mathcal {R}$.

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

(i) Since $x_i\ne x_j$ for $i\ne j$, 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 $t\ge p$. 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 $k_r\le q<r$) and $k_r\ge p$ (by minimality of p). However, this is impossible as item (d1) gives us:

$$\begin{aligned}{} & {} k_r\in \left\{ u<r: u\notin \bigcup _{w<r} F_w, j_u=0\right\} \cap \{0,1,\ldots ,q\} \\{} & {} \quad \subseteq \{u\le q: u\notin F_q\}\cap \{u\le q: j_u=0\} \\{} & {} \quad =(\{0,1,\ldots ,p-1\}\cup \{q\})\cap \{u\le q: j_u=0\}\subseteq \{0,1,\ldots ,p-1\} \end{aligned}$$

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

(ii) We have:

$$\begin{aligned} \begin{aligned} {{\,\textrm{FS}\,}}(A) =&\bigcup _{i\in B}(\{x_i\}\cup (x_i+{{\,\textrm{FS}\,}}(A\setminus \{0,1,\ldots ,x_i\}))) \\ {}&\cup \bigcup _{i\in C}(\{x_i\}\cup (x_i+{{\,\textrm{FS}\,}}(A\setminus \{0,1,\ldots ,x_i\}))), \end{aligned} \end{aligned}$$

where $B=\{i\in \omega : i\notin \bigcup _{q\in \omega }F_q, j_i=0\}$ and $C=\{i\in \omega : i\notin \bigcup _{q\in \omega }F_q, j_i=1\}$.

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

Now we will show that $f[\bigcup _{i\in B}(\{x_i\}\cup (x_i+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_i\})))]\in \mathcal {R}$. By items (c3), (c4) and (g2), $f[\{x_i\}\cup (x_i+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_i\}))]\subseteq f[\{x_i\}\cup (x_i+{{\,\textrm{FS}\,}}(D_i))]\subseteq (\omega \times \{n_i\})\cap \Gamma$ 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 $Z^2\cap \Gamma \subseteq \bigcup _{i\in B} f[x_i+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_i\})]$ for some $Z\in [\omega ]^\omega$.

First we will show that $|Z\setminus \{n_i:i\in B\}|\le 1$. 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 $(z,w)\in f[x_i+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_i\})]$, hence $x_i+{{\,\textrm{FS}\,}}(A\setminus \{0,1,\ldots ,x_i\})\subseteq f^{-1}[\{(z,w)\}]$. But by (c4) and (g2) we have $x_i+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_i\})\subseteq x_i+{{\,\textrm{FS}\,}}(D_i)\subseteq f^{-1}[(\omega \times \{n_i\})\cap \Gamma ]$. 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 $(n_i,n_j)\notin \bigcup _{k\in B} f[x_k+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_k\})]$. This will contradict $Z^2\cap \Gamma \subseteq \bigcup _{k\in B} f[x_k+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_k\})]$ and finish the proof.

Suppose that $(n_i,n_j)\in \bigcup _{k\in B} f[x_k+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_k\})]$. From (c4) and (g2), for every $k\ne j$, $k\in B$ we have $f[x_k+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_k\})]\subseteq f[x_k+{{\,\textrm{FS}\,}}(D_k)] \subseteq (\omega \times \{n_k\})\cap \Gamma$, so $(n_i,n_j)\notin f[x_k+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_k\})]$. Hence, $(n_i,n_j)\in f[x_j+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_j\})]$. Let $y\in x_j+{{\,\textrm{FS}\,}}(A{\setminus }\{0,1,\ldots ,x_j\})$ be such that $f(y)=(n_i,n_j)$. Then $y=x_j+x_{s_0}+\cdots +x_{s_p}$ 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 $y=(x_j+\cdots +x_{s_{p-1}})+x_{i}$ and from item (e3) (applied to $x=(x_j+\cdots +x_{s_{p-1}})$) we get $f(y)\ne (n_i,n_j)$, a contradiction.

• If there exists $k<p$ such that $s_k=i$, then $y=(x_j+\cdots +x_{s_{k-1}})+x_{i}+(x_{s_{k+1}}+\cdots +x_{s_p})\in (x_j+\cdots +x_{s_{k-1}})+x_{i}+{{\,\textrm{FS}\,}}(D_i)$ by item (g2) and from item (e1) (applied to $x=(x_j+\cdots +x_{s_{k-1}})$) we get a contradiction.

• If there exists $k\le p$ such that $s_{k-1}<i<s_k$, then $y=(x_j+\cdots +x_{s_{k-1}})+(x_{s_k}+\cdots +x_{s_p})\in (x_j+\cdots +x_{s_{k-1}})+{{\,\textrm{FS}\,}}(D_i)$ by item (g2) and from item (e2) we get a contradiction.

• If $i<s_0$, then $y=x_j+(x_{s_0}+\cdots +x_{s_p})\in x_j+{{\,\textrm{FS}\,}}(D_i)$ by item (g2) and from item (e2) we get a contradiction.

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