Some implications of Ramsey Choice for families of $\varvec{n}$-element sets

Abstract

For $n\in \omega$, the weak choice principle $\textrm{RC}_n$ is defined as follows:

For every infinite set X there is an infinite subset $Y\subseteq X$ with a choice function on $[Y]^n:=\{z\subseteq Y:|z|=n\}$.

The choice principle $\textrm{C}_n^-$ states the following:

For every infinite family of n-element sets, there is an infinite subfamily ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a choice function.

The choice principles $\textrm{LOC}_n^-$ and $\textrm{WOC}_n^-$ are the same as $\textrm{C}_n^-$, but we assume that the family ${\mathcal {F}}$ is linearly orderable (for $\textrm{LOC}_n^-$) or well-orderable (for $\textrm{WOC}_n^-$). In the first part of this paper, for $m,n\in \omega$ we will give a full characterization of when the implication $\textrm{RC}_m\Rightarrow \textrm{WOC}_n^-$ holds in ${\textsf {ZF}}$. We will prove the independence results by using suitable Fraenkel-Mostowski permutation models. In the second part, we will show some generalizations. In particular, we will show that $\textrm{RC}_5\Rightarrow \textrm{LOC}_5^-$ and that $\textrm{RC}_6\Rightarrow \textrm{C}_3^-$, answering two open questions from Halbeisen and Tachtsis (Arch Math Logik 59(5):583–606, 2020). Furthermore, we will show that $\textrm{RC}_6\Rightarrow \textrm{C}_9^-$ and that $\textrm{RC}_7\Rightarrow \textrm{LOC}_7^-$.

1 Definitions and terminology

The notation we use is standard and follows that of [5]. Now we list some definitions that shall be used in the sequel:

Definition 1.1

Let n be an arbitrary positive natural number.

• 1. $\textrm{C}_n^-$ states that every infinite family ${\mathcal {F}}$ of sets of size n has an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a choice function.

• 2. $\textrm{LOC}_n^-$ states that every infinite, linearly orderable family ${\mathcal {F}}$ of sets of size n has an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a choice function.

• 3. $\textrm{WOC}_n^-$ states that every infinite, well-orderable family ${\mathcal {F}}$ of sets of size n has an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a choice function.

• 4. $\textrm{RC}_n$ states that every infinite set X has an infinite subset $Y\subseteq X$ such that the set

  $$\begin{aligned} {[}Y]^n=\{ z\subseteq Y:|z|=n\} \end{aligned}$$

  has a choice function.

• 5. Let ${\mathcal {F}}$ be an infinite family of n-element sets. A Kinna-Wagner selection function of ${\mathcal {F}}$ is a function f with $\textrm{dom}(f)={\mathcal {F}}$ such that for all $p\in {\mathcal {F}}$, $\emptyset \not =f(p)\subsetneq p$.

• 6. $\textrm{KW}_n^-$ states that every infinite family ${\mathcal {F}}$ of sets of size n has an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a Kinna-Wagner selection function.

• 7. $\textrm{LOKW}_n^-$ states that every infinite, linearly orderable family ${\mathcal {F}}$ of sets of size n has an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a Kinna-Wagner selection function.

In 1995, Montenegro proved in [6] that $\textrm{RC}_n\Rightarrow \textrm{C}_n^-$ for all $n\in \{2,3,4\}$. It is still unknown whether this implication holds for any $n\ge 5$. In 2017, Halbeisen and Tachtsis found interesting results concerning the implications $\textrm{RC}_m\Rightarrow \textrm{C}_n^-$ and $\textrm{RC}_m\Rightarrow \textrm{RC}_n$ for $m,n\in \omega \setminus \{0,1\}$ (see [4]). Among other results they proved that the following statements are consistent with ${\textsf {ZF}}$ or provable in ${\textsf {ZF}}$, respectively:

($\alpha$):

If $m,n\in \omega \setminus \{0,1\}$ are such that there is a prime p with $p\not \mid m$ and $p\mid n$, then

$$\begin{aligned} \textrm{RC}_m\nRightarrow \textrm{RC}_n\text { and } \textrm{RC}_m\nRightarrow \textrm{C}_n^-. \end{aligned}$$

($\beta$):

$\textrm{RC}_5\nRightarrow \textrm{LOC}_2^-$ and $\textrm{RC}_5\nRightarrow \textrm{LOC}_3^-$.

($\gamma$):

For every $n\in \omega \setminus \{0,1\}$ we have that $\textrm{C}_n^-\Rightarrow \textrm{LOC}_n^- \Rightarrow \textrm{WOC}_n^-$ but none of these implications is reversible.

($\delta$):

For every $n\in \omega \setminus \{0,1\}$ the implication $\textrm{RC}_{2n}\Rightarrow \textrm{LOKW}_n^-$ holds. In particular we have that $\textrm{RC}_6\Rightarrow \textrm{LOC}_3^-$ (notice that $\textrm{LOKW}_3^-\Leftrightarrow \textrm{LOC}_3^-$).

In Sect. 2 of this paper, we will give a full characterization of when the implication $\text {RC}_m\Rightarrow \text {WOC}_n^-$ (for $m,n\in \omega \setminus \{0,1\}$) is provable in ${\textsf {ZF}}$. To be more precise, it will be shown (see Theorem 2.10) that for every $m,n\in \omega \setminus \{0,1\}$, $\textrm{RC}_m\Rightarrow \textrm{WOC}_n^-$ is provable in ${\textsf {ZF}}$ if an only if the following condition holds: Whenever we can write n in the form

$$\begin{aligned} n=\sum _{i< k}a_ip_i, \end{aligned}$$

where $p_0,\ldots ,p_{k-1}$ are prime numbers and $a_0,\ldots ,a_{k-1}\in \omega \setminus \{0\}$, then we find $b_0,\dots , b_{k-1}\in \omega$ with

$$\begin{aligned} m=\sum _{i< k} b_ip_i. \end{aligned}$$

In order to prove the independence of this implication with ${\textsf {ZF}}$, we shall use permutation models (see [5] for an introduction to permutation models and to models of ${\textsf {ZFA}}$). With Pincus’ transfer theorems (see [7]), we are able to transfer the results obtained in ${\textsf {ZFA}}$ to ${\textsf {ZF}}$. Furthermore, Theorem 2.10 gives us the following three special cases:

• 1. For all $n\in \omega$ we have that $\textrm{RC}_n \Rightarrow \textrm{WOC}_n^-$ (see Corollary 2.3).

• 2. Let p be a prime number, $m\in \omega \setminus \{0\}$ and $n\in \omega \setminus \{0,1\}.$ Then

  $$\begin{aligned} \textrm{RC}_{p^m}\Rightarrow \textrm{WOC}_n^- \end{aligned}$$

  if and only if $n\mid p^m$ or $p=2$, $m=1$ and $n=4$ (see Corollary 2.12).

• 3. If $\textrm{RC}_m\not \Rightarrow \textrm{WOC}_n^-$, we also have that $\textrm{RC}_m\not \Rightarrow \textrm{RC}_n^-$ and $\textrm{RC}_m\not \Rightarrow \textrm{C}_n^-$ (see Corollary 2.11). This generalizes Halbeisens and Tachtsis’ result ($\alpha$).

In Sect. 3, we will give some insights into the question what happens when we weaken the assumption that our family of n-element sets is well-ordered. We will prove that $\textrm{RC}_6\Rightarrow \textrm{C}_n^-$ for $n\in \{3,9\}$ and that $\textrm{RC}_n\Rightarrow \textrm{LOC}_n^-$ for $n\in \{5,7\}$.

2 On the implication $\textbf{RC}_{{\varvec{m}}}\Rightarrow \textbf{WOC}_{{\varvec{n}}}^-$

2.1 When is $\textbf{RC}_{{\varvec{m}}}\Rightarrow \textbf{WOC}_{{\varvec{n}}}^-$ provable in ${\textsf {ZF}}$?

In this section, we will characterise the values m and n for which the implication $\textrm{RC}_{m}\Rightarrow \textrm{WOC}_{n}^-$ is provable in ${\textsf {ZF}}$. However, before we state and prove the main result of this section, we introduce some notation and prove an auxiliary result.

Two finite partitions $\{x_i:0\le i\le l\}$ and $\{y_j:0\le j\le k\}$ of sets of the same cardinality are of the same type, if $l=k$ and for each $0\le i\le l$ we have $|x_i|=|y_i|$.

Let k be a positive integer and let $n=\sum _{i< k}a_ip_i$, where $p_0,\ldots ,p_{k-1}$ are prime numbers and $a_0,\ldots ,a_{k-1}\in \omega \setminus \{0\}$. Furthermore, for an infinite, well-ordered set $\lambda$, let ${\mathcal {F}}=\{F_\alpha :\alpha \in \lambda \}$ be an infinite family of pairwise disjoint n-element sets, where for each $\alpha \in \lambda$, $F_\alpha$ is partitioned into sets $F_{\alpha ,i}$ ($i< k$), where $|F_{\alpha ,i}|=a_ip_i$, i.e.,

$$\begin{aligned} F_\alpha =\bigcup _{i< k}F_{\alpha ,i}\quad \text {and}\quad F_{\alpha ,i}\cap F_{\alpha ,i'}=\emptyset \ \text {whenever}\ i\ne i'. \end{aligned}$$

In particular, for any $\alpha ,\alpha '\in \lambda$, the partitions $\{F_{\alpha ,i}:i<k\}$ and $\{F_{\alpha ',i}:i<k\}$ are of the same type.

For $\alpha \in \lambda$ we say that $d\subseteq F_\alpha$ diagonalises $F_\alpha$ if for all $i<k$, $|F_{\alpha ,i}\cap d|=1$. Let

$$\begin{aligned} {\mathcal {D}}_\alpha :=\{d\subseteq F_\alpha : d\ \text {diagonalises}\ F_\alpha \} \end{aligned}$$

and for each $\alpha \in \lambda$ let $D_\alpha$ be a non-empty subset of ${\mathcal {D}}_\alpha$ such that for any $\alpha ,\alpha ' \in \lambda$ we have $|D_{\alpha }|=|D_{\alpha '}|$.

Finally, for some positive integer $t\ge 1$ and some prime number p, for each $\alpha \in \lambda$ let $\{D_{\alpha ,j}^p: j<t\}$ be a partition of $[D_\alpha ]^p$ such that for any $\alpha ,\alpha '\in \lambda$, the partitions $\{D_{\alpha ,j}^p:j<t\}$ and $\{D_{\alpha ',j}^{p}:j<t\}$ are of the same type.

Lemma 2.1

Let $n=\sum _{i< k}a_ip_i$, ${\mathcal {F}}=\{F_{\alpha }: \alpha \in \lambda \}$, $F_{\alpha }=\bigcup \{F_{\alpha ,i}:i<k\}$, $D_{\alpha }$, and $\{D_{\alpha ,j}^p:j<t\}$ be as above. Furthermore, let $p:=p_{i_0}$ for some $p_{i_0}\in \{p_0,\ldots ,p_{k-1}\}$, and assume that for some integer $l\ge 0$ there is a choice function

$$\begin{aligned} h:\left[ \bigcup _{\,\alpha \in \lambda }D_\alpha \right] ^{l+p} \rightarrow \bigcup _{\alpha \in \lambda }D_\alpha \,. \end{aligned}$$

Then there is an infinite subset $\lambda '\subseteq \lambda$ such that we are in at least one of the following cases:

• (a) There is a choice function

  $$\begin{aligned} h':\left[ \bigcup _{\,\alpha \in \lambda '}D_\alpha \right] ^{l} \rightarrow \bigcup _{\alpha \in \lambda '}D_\alpha \,. \end{aligned}$$

• (b) We can simultaneously refine the partitions on $\{F_{\alpha }:\alpha \in \lambda '\}$ to partitions of the same type (and extend accordingly the corresponding sets $D_\alpha$).

• (c) We can simultaneously refine the partitions on $\{[D_{\alpha }]^p:\alpha \in \lambda '\}$ to partitions of the same type.

• (d) For each $\alpha \in \lambda '$ we can choose a non-empty proper subset $D_\alpha '$ of $D_\alpha$, i.e.,

  $$\begin{aligned} \emptyset \ne D_\alpha ' \subsetneq D_\alpha , \end{aligned}$$

  such that for all $\alpha ,\beta \in \lambda '$ we have $|D_\alpha '|=|D_{\beta }'|$.

Proof

Recall that for all $\alpha ,\alpha '\in \lambda$ we have $|D_{\alpha }|=|D_{\alpha '}|$. Now, assume that there is a $j_0<k$ such that for $n_{j_0}:=a_{j_0}p_{j_0}$ and all $\alpha \in \lambda$ we have

$$\begin{aligned} n_{j_0}\not \mid |D_\alpha |\,. \end{aligned}$$

For all $\alpha \in \lambda$ and all $z\in F_\alpha$ define

$$\begin{aligned} \#z:=\big |\big \{ X\in D_\alpha : z\in X\big \}\big |. \end{aligned}$$

Since $\sum _{z\in F_{\alpha ,j_0}}\#z=|D_\alpha |$, $\vert F_{\alpha ,j_0}\vert =n_{j_0}$ and $n_{j_0}\not \mid |D_\alpha |$, it follows that

$$\begin{aligned} \emptyset \not =\big \{ z\in F_{\alpha ,j_0}: \forall z^\prime \in F_{\alpha ,j_0} (\#z\le \#z^\prime )\big \}\subsetneq F_{\alpha ,j_0}. \end{aligned}$$

Therefore, we can simultaneously refine the partition on each $F_\alpha$ for $\alpha \in \lambda$. Moreover, notice that since $n_{j_0}$ is finite, we find an infinite set $\lambda '\subseteq \lambda$ such that for each $\alpha \in \lambda '$, the block $F_{\alpha ,j_0}$ is partitioned into two non-empty blocks $F_{\alpha ,j_1}$ and $F_{\alpha ,j_2}$ where for all $\alpha ,\beta \in \lambda '$, $|F_{\alpha ,j_1}|=|F_{\beta ,j_1}|$ and $|F_{\alpha ,j_2}|=|F_{\beta ,j_2}|$. This shows that all the refined partitions are of the same type and we are in Case (b).

So, we can assume that for all $i<k$ and all $\alpha \in \lambda$ we have

$$\begin{aligned} n_i\mid |D_\alpha |\end{aligned}$$

where $n_i:=a_ip_i$.

We consider now the following four cases:

Case 1: There is a $Z_0\in \left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^{l}$ and an infinite subset $\lambda '\subseteq \lambda$ such that

$$\begin{aligned} \forall \alpha \in \lambda '~\forall X\in [D_\alpha ]^p~\bigl (h(Z_0\cup X) \in X\bigr ). \end{aligned}$$

By shrinking $\lambda '$ if necessary, we may assume that $Z_0\cap \bigcup _{\alpha \in \lambda '}D_\alpha =\emptyset$. For every $\alpha \in \lambda '$ and all $d\in D_\alpha$ define

$$\begin{aligned} \textrm{deg}_\alpha (d):=\big |\big \{ X\in [D_\alpha ]^p:h(Z_0\cup X)=d\big \}\big |. \end{aligned}$$

Note that $\sum _{d\in D_\alpha }\textrm{deg}_\alpha (d)=\big |[D_\alpha ]^p\big |=\begin{pmatrix} |D_\alpha |\\ p \end{pmatrix}$. Since $p=p_{i_0}$ and since $n_{i_0}\mid |D_\alpha |$, we have $p\mid |D_\alpha |$. Hence, it follows that $|D_\alpha |\not \mid \begin{pmatrix}|D_\alpha |\\ p\end{pmatrix}$. To see this, let $D:=|D_\alpha |$ and notice that if $D=ap^s$ for some positive integers a, s where $p\not \mid a$, then

$$\begin{aligned} \left( {\begin{array}{c}D\\ p\end{array}}\right) =\frac{ap^s\cdot (ap^s-1) \cdot \,\ldots \,\cdot (ap^s-p+1)}{1\cdot 2\cdot \,\ldots \,\cdot p} =\frac{ap^{s-1}\cdot (ap^s-1)\cdot \,\ldots \,\cdot (ap^s-p+1)}{1\cdot 2 \cdot \,\ldots \,\cdot (p-1)}\,. \end{aligned}$$

Hence, $p^s\not \mid \left( {\begin{array}{c}D\\ p\end{array}}\right)$ and in particular we have $D\not \mid \left( {\begin{array}{c}D\\ p\end{array}}\right)$.

Thus, for each $\alpha \in \lambda '$ we can choose

$$\begin{aligned} \emptyset \not =D_\alpha ':=\big \{ d\in D_\alpha : \forall d^\prime \in D_\alpha (\textrm{deg}_\alpha (d)\le \textrm{deg}_\alpha (d^\prime ))\big \}\subsetneq D_\alpha . \end{aligned}$$

Moreover, notice that since $D_\alpha$ is finite, by shrinking $\lambda '$ if necessary, we can assume that for all $\alpha ,\beta \in \lambda '$ we have $|D_\alpha '|=|D_\beta '|$, and we are in Case (d).

Case 2: There is a set $Z_0\in \left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^{l}$, a non-negative integer $j_0<t$, and an infinite subset $\lambda '\subseteq \lambda$ such that $Z_0\cap \bigcup _{\alpha \in \lambda '}D_\alpha =\emptyset$ and

$$\begin{aligned} \forall \alpha \in \lambda '~\exists X,X^\prime \in D_{\alpha ,j_0}^p~\bigl (h(Z_0\cup X) \in Z_0\wedge h(Z_0\cup X^\prime )\in X^\prime \bigr ). \end{aligned}$$

In this case, we can simultaneously refine the partition on $[D_\alpha ]^p$ for each $\alpha \in \lambda '$. Moreover, since $[D_\alpha ]^p$ is finite (for all $\alpha \in \lambda '$), by shrinking $\lambda '$ if necessary, we can assume that for all $\alpha ,\beta \in \lambda '$, the partition on $[D_\alpha ]^p$ has the same type as the partition on $[D_\beta ]^p$, and we are in Case (c).

Case 3: There is a set $Z_0\in \left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^{l}$, a non-negative integer $j_0<t$, and an infinite subset $\lambda '\subseteq \lambda$ such that $Z_0\cap \bigcup _{\alpha \in \lambda '}D_\alpha =\emptyset$ and

$$\begin{aligned} \forall \alpha \in \lambda '\Bigl (\bigl ( \forall X\in D_{\alpha ,j_0}^p~h(Z_0\cup X)\in Z_0\bigr )\wedge \exists X,X^\prime \in D_{\alpha ,j}^p\bigl (h(Z_0\cup X) \ne h(Z_0\cup X^\prime )\bigr )\Bigr ). \end{aligned}$$

In this case, we can simultaneously refine the partition on $[D_\alpha ]^p$ for each $\alpha \in \lambda '$. Moreover, by shrinking $\lambda '$ if necessary, we can assume that all partitions are of the same type and we are again in Case (c).

Case 4: For all $Z\in \left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^l$ and for all but finitely many $\alpha \in \lambda$ we have

Then, for each $Z\in \left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^l$ let $\alpha _Z\in \lambda$ be the least element with respect to the well-ordering on $\lambda$ such that () holds for $\alpha =\alpha _Z$. Furthermore, for every $Z\in \left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^l$ let $j_Z< t$ be the least integer such that () holds for $\alpha =\alpha _Z$ and $j=j_Z$. So, for every $Z\in \left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^l$ we have

Finally, we define a function $h':\left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^l\rightarrow \bigcup _{\alpha \in \lambda } D_\alpha$ by stipulating

$$\begin{aligned} h':\quad \left[ \bigcup _{\alpha \in \lambda } D_\alpha \right] ^l&\longrightarrow \bigcup _{\alpha \in \lambda } D_\alpha \\ Z&\longmapsto h(Z\cup X) \end{aligned}$$

where X is an arbitrary element of  $D_{\alpha _Z,j_Z}^p$. Note that by (), $h'$ is a well-defined choice function and we are in Case (a). $\square$

Now, we are ready to prove the main result of this section.

Proposition 2.2

Let $m,n\in \omega \setminus \{0,1\}$ and assume that whenever we can write n in the form

$$\begin{aligned} n=\sum _{i< k}a_ip_i, \end{aligned}$$

where $p_0,\ldots ,p_{k-1}$ are prime numbers and $a_0,\ldots ,a_{k-1}$ are positive integers, then we find $b_0,\dots , b_{k-1}\in \omega$ with

$$\begin{aligned} m=\sum _{i< k} b_ip_i. \end{aligned}$$

Then, in ${\textsf {ZF}}$ we have

$$\begin{aligned} \textrm{RC}_m\Rightarrow \textrm{WOC}_n^-. \end{aligned}$$

Proof

Let ${\mathcal {F}}=\{F_\alpha :\alpha \in \lambda \}$ be an infinite, well-ordered family of pairwise disjoint n-element sets. The goal is to construct an infinite subfamily of ${\mathcal {F}}$ with a choice function.

Applying $\textrm{RC}_m$ to the set $X_0:=\bigcup _{\alpha \in \lambda } F_\alpha$, we obtain an infinite set $Y_0\subseteq X_0$ such that the set $[Y_0]^m$ has a choice function. For $1\le j\le n$, let

$$\begin{aligned} \lambda _{j}:=\big \{\alpha \in \lambda : |F_\alpha \cap Y_0|=j\big \}. \end{aligned}$$

Since n is finite and $\lambda$ is infinite, there exists a $j_0$ with $1\le j_0\le n$ such that $\lambda _{j_0}\subseteq \lambda$ is infinite. If $j_0=1$ we are done since $\{F_\alpha :\alpha \in \lambda _{1}\}\subseteq {\mathcal {F}}$ has a choice function. If $1<j_0<n$, we apply $\textrm{RC}_m$ to the set

$$\begin{aligned} X_1:=\bigcup \big \{ F_\alpha \setminus Y_0:\alpha \in \lambda _{j_0}\big \} \end{aligned}$$

and obtain an infinite set $Y_1\subseteq X_1$ such the set $[Y_1]^m$ has a choice function. As above, for $1\le j\le n-j_0$, let

$$\begin{aligned} \lambda _{j_0,j}:=\big \{\alpha \in \lambda _{j_0}: |F_\alpha \cap Y_1|=j\big \}. \end{aligned}$$

Then there exists a $j_1$ with $1\le j_1\le n-j_0$ such that $\lambda _{j_0,j_1}\subseteq \lambda$ is infinite. If $j_1=1$, then the infinite family $\{F_\alpha :\alpha \in \lambda _{j_0,1}\}\subseteq {\mathcal {F}}$ has a choice function. Proceeding this way, we either find an infinite subfamily of ${\mathcal {F}}$ with a choice function, or for an infinite subset $\lambda _0\subseteq \lambda$, for all $\alpha \in \lambda _0$ we can simultaneously partition the sets $F_\alpha$ into sets $F_{\alpha ,i}$ with $i<k$ for some $k\ge 1$. Since for each $i<k$, $|F_{\alpha ,i}|\ge 2$, we have $|F_{\alpha ,i}|=a_ip_i$, where $p_i$ is prime and $a_i>0$. Finally, for each $\alpha \in \lambda _0$, let let $D_\alpha :=\{d\subseteq F_\alpha : d\ \text {diagonalises}\ F_\alpha \}$.

Now, since $n=\sum _{i<k}a_ip_i$, by our assumption we find $b_0,\ldots , b_{k-1}\in \omega$ with $m=\sum _{i<k} b_ip_i$, and since $m\ge 2$, there is an $i_0<k$ with $b_{i_0}\ne 0$. In particular, we have $m\ge p_{i_0}$. Let $p:=p_{i_0}$ and $l:=m-p$, where $l\ge 0$. Furthermore, for $t=1$, $\{D_{\alpha ,j}: j<t\}=[D_\alpha ]^p$ is the trivial partition of $[D_\alpha ]^p$. Thus, by $\textrm{RC}_m$, there is an infinite set $\lambda \subseteq \lambda _0$ and a choice function

$$\begin{aligned} h:\left[ \bigcup _{\,\alpha \in \lambda } D_\alpha \right] ^{l+p}\rightarrow \bigcup _{\alpha \in \lambda } D_\alpha \,. \end{aligned}$$

So, we have all the requirements to apply Lemma 2.1 iteratively until — after finitely many steps — the partitions of the $F_\alpha$’s or of the $[D_\alpha ]^p$’s contain a block with just one element, or the sets $D_\alpha$ are singletons: To see this, notice first that if we are in one of the cases (b), (c), or (d), or if $l=0$, then we can either refine the partition of the $F_\alpha$’s or of the $[D_\alpha ]^p$’s. Now, if we are in case (a) for $l>0$, then, by the properties of

$$\begin{aligned} m=\sum _{i<k} b_ip_i \end{aligned}$$

and since we start with $l=m-p$, $l\ge p_i$ (for some $i<k$) and we can proceed with $l':=l-p_i$.

So, after finitely many steps — in particular after finitely many choices of sets $Z_0$ — we are in the situation where the partitions of the $F_\alpha$’s or of the $[D_\alpha ]^p$’s contain a block with just one element, or the $D_\alpha$’s are reduced to singletons, which gives us an algorithm to select an element from each of the remaining $F_\alpha$’s — where in the case when $|D_\alpha |=1$, we choose the element in $D_\alpha \cap F_{\alpha ,0}$. $\square$

Corollary 2.3

For every $n\in \omega$ we have that

$$\begin{aligned} \textrm{RC}_n\Rightarrow \textrm{WOC}_n^-. \end{aligned}$$

2.2 When is $\textbf{RC}_{\varvec{m}}\nRightarrow \textbf{WOC}_{\varvec{n}}^-$ consistent with ${\textsf {ZF}}$?

In this section we will show that for all $n,m\in \omega \setminus \{0,1\}$ which do not satisfy the conditions of Proposition 2.2 we get that

$$\begin{aligned} \textrm{RC}_m\not \Rightarrow \textrm{WOC}_n^- \end{aligned}$$

is consistent with ${\textsf {ZF}}$. In a first step we will construct suitable Fraenkel-Mostowski permutation models — similar to those constructed in [2, Sec. 6] — in which we have $\textrm{RC}_m\not \Rightarrow \textrm{WOC}_n^-$. We will then see that both statements, $\textrm{RC}_m$ and $\textrm{WOC}_n^-$, are injectively boundable. So, by [7, Theorem 3A3] the result is transferable to ${\textsf {ZF}}$.

Let $p_0$ and $p_1$ be two prime numbers. We start with a ground model ${\mathcal {M}}_{p_0,p_1}$ of ${\textsf {ZFA}}+{\textsf {AC}}$ with a set of atoms

$$\begin{aligned} {\mathcal {A}}:=\bigcup \{A_i: i\in \omega \}\cup \bigcup \{B_j: j\in \omega \}, \end{aligned}$$

where for all $i,j\in \omega$ the sets $A_i$ and $B_j$ are called blocks. These blocks have the following properties:

• For all $i\in \omega$, $A_i=\{ a_{i,k}: k< p_0\}$ and $B_i=\{b_{i,l}: l< p_1\}$ with $|A_i|=p_0$ and $|B_i|=p_1$.

• The blocks are pairwise disjoint.

For all $i,j\in \omega$ we define a permutation on ${\mathcal {A}}$ as follows:

• For all $i\in \omega$ and all $k< p_0$ let

  $$\begin{aligned} \alpha _i(a_{i,k}):={\left\{ \begin{array}{ll} a_{i,0} &{}\text {if}\ k=p_0-1,\\ a_{i,k+1} &{}\text {if}\ k< p_0-1, \end{array}\right. } \end{aligned}$$

  and $\alpha _i(a)=a$ for all $a\in {\mathcal {A}}\setminus A_i$. Analogously for all $j\in \omega$ and all $l< p_1$ let

  $$\begin{aligned} \beta _j(b_{j,l}):= {\left\{ \begin{array}{ll} b_{i,0} &{}\text {if}\ l=p_1-1,\\ b_{j,l+1} &{}\text {if}\ l< p_1-1, \end{array}\right. } \end{aligned}$$

  and $\beta _j(b)=b$ for all $b\in {\mathcal {A}}\setminus B_j$.

Now we define an abelian group G of permutations of ${\mathcal {A}}$ by requiring

$$\begin{aligned} \phi \in G\iff \phi =\alpha \circ \beta , \end{aligned}$$

where

$$\begin{aligned} \alpha =\prod _{i\in \omega }\alpha _i^{k_i}\text { with } k_i< p_0 \text { for each }i\in \omega \end{aligned}$$

and

$$\begin{aligned} \beta =\prod _{j\in \omega }\beta _j^{l_j}\text { with }l_j< p_1 \text { for each }j\in \omega . \end{aligned}$$

Let ${\mathcal {F}}$ be the normal filter on G generated by the subgroups

$$\begin{aligned} \textrm{fix}_G(E)=\{ \phi \in G: \forall a\in E(\phi (a)=a)\} \end{aligned}$$

with $E\in \textrm{fin}({\mathcal {A}}):=\{A\subseteq {\mathcal {A}}:|A|\in \omega \}$. Let ${\mathcal {V}}_{p_0,p_1}$ be the class of all hereditarily symmetric sets.

Remark 2.4

We can also work with k blocks of size $p_0,\dots , p_{k-1}$, where $p_i$ is a prime number for every $i< k$. The corresponding permutation model is denoted by ${\mathcal {V}}_{p_0,\dots , p_{k-1}}$.

Definition 2.5

A set $E\in \textrm{fin}({\mathcal {A}})$ is closed if and only if for all $i,j\in \omega$ we have that

$$\begin{aligned} A_i\cap E\not =\emptyset \Rightarrow A_i\subseteq E\qquad \text {and} \qquad B_j\cap E\not =\emptyset \Rightarrow B_j\subseteq E. \end{aligned}$$

We now define a well-ordering on the set of closed sets.

Definition 2.6

Let $C_1$ and $C_2$ be two blocks in $\{A_i: i\in \omega \}\cup \{ B_j: j\in \omega \}$. We define

$$\begin{aligned} C_1<C_2:\iff {\left\{ \begin{array}{ll} C_1=A_i\;\wedge \; C_2=B_j, \text { or}\\ C_1=A_i\;\wedge \; C_2=A_j\;\wedge \; i<j, \text { or}\\ C_1=B_i\;\wedge \; C_2=B_j\;\wedge \; i<j. \end{array}\right. } \end{aligned}$$

Moreover, for distinct closed sets $E=\bigcup \{F_0,\dots F_n\}\in \textrm{fin}({\mathcal {A}})$ and $E^\prime =\bigcup \{ F_0^\prime ,\dots , F_m^\prime \}\in \textrm{fin}({\mathcal {A}})$ with blocks $F_0,\dots , F_n,$ $F_0^\prime ,\dots , F_m^\prime$ let

$$\begin{aligned}{} & {} E\prec E^\prime \ :\iff \text { The}\ <\text {-least block in the symmetric difference }\\{} & {} \qquad \{F_0,\dots , F_n\}\Delta \{F^\prime _0,\dots , F_m^\prime \} \text { belongs to }E. \end{aligned}$$

Note that this defines a well-ordering on the set of closed sets and therefore on the set of all closed supports.

Lemma 2.7

Let $n\in \omega \setminus \{0,1\}$ and let $p_0$ and $p_1$ be two prime numbers such that

$$\begin{aligned} n=cp_0+dp_1\not =0 \end{aligned}$$

for $c,d\in \omega$. Then we have that

$$\begin{aligned} {\mathcal {V}}_{p_0,p_1}\models \lnot \textrm{WOC}_n^-. \end{aligned}$$

Proof

Define

$$\begin{aligned} {\mathcal {F}}:=\Big \{ A_l\cup A_{l+1}\cup \dots \cup A_{l+c-1}\cup B_{l+c}\cup \dots \cup B_{l+c+d-1}: l=k(c+d)\text { for a }k\in \omega \Big \}. \end{aligned}$$

Then ${\mathcal {F}}$ is an infinite family of pairwise disjoint n-element sets. Since the empty set is a support of ${\mathcal {F}}$, we have that ${\mathcal {F}} \in {\mathcal {V}}_{p_0,p_1}$. Moreover, ${\mathcal {F}}$ is well-orderable in ${\mathcal {V}}_{p_0,p_1}$. Assume towards contradiction that there is an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a choice function

$$\begin{aligned} g:{\mathcal {G}}\rightarrow \bigcup {\mathcal {G}} \end{aligned}$$

in ${\mathcal {V}}_{p_0,p_1}$. Let $E_g\in \textrm{fin}({\mathcal {A}})$ be a closed support of g. Since $E_g$ is finite, there is a $G_0\in {\mathcal {G}}$ such that $G_0\cap E_g=\emptyset$. Then there are $i,j\in \omega$ with

$$\begin{aligned} g(G_0)\in A_{l+i}\cup B_{l+c+j}. \end{aligned}$$

Define $\gamma _0:=\alpha _{l+i}\circ \beta _{l+c+j}$. We have that

$$\begin{aligned} g(\gamma _0(G_0))=g(G_0)\not = \gamma _0(g(G_0)). \end{aligned}$$

So $E_g$ is not a support of g which is a contradiction. $\square$

Lemma 2.8

Let $m\in \omega \setminus \{0,1\}$ and let $p_0,p_1$ be prime numbers such that

$$\begin{aligned} m\not =cp_0+dp_1 \end{aligned}$$

for all $c,d\in \omega$. Then we have:

$$\begin{aligned} {\mathcal {V}}_{p_0,p_1}\models \textrm{RC}_m \end{aligned}$$

Proof

Let $x\in {\mathcal {V}}_{p_0,p_1}$ be an infinite set with closed support $E_x\in \textrm{fin}({\mathcal {A}})$. If there is an $E\in \textrm{fin}({\mathcal {A}})$ such that

$$\begin{aligned} y:=\big \{ z\in x: E\text { is a support of }z\big \} \end{aligned}$$

is an infinite set, then y can be well-ordered in ${\mathcal {V}}_{p_0,p_1}$ and we can define a choice function on $[y]^m$ by choosing the least element with respect to that well-ordering.

So, assume that for all $E\in \textrm{fin}({\mathcal {A}})$ there are only finitely many $z\in x$ with support E. For every closed set $E\in \textrm{fin}({\mathcal {A}})$ with $E_x\subsetneq E$ define

$$\begin{aligned} M_E:=\big \{ z\in x: E \text { is the minimal closed support of}\ z\ \text {with } E_x\subseteq E\big \}. \end{aligned}$$

Since E is a support of $M_E$, the sets $M_E$ belong to ${\mathcal {V}}_{p_0,p_1}$, and by our assumption, the sets $M_E$ are finite. Now, for each $z\in M_E$ define

$$\begin{aligned} {[}z]:=\{ \phi (z): \phi \in \textrm{fix}_G(E_x)\}\subseteq M_E. \end{aligned}$$

To see that $[z]\subseteq M_E$, notice that since $E\in \textrm{fin}({\mathcal {A}})$ is closed, for all $\phi \in G$ we have $\phi (E)=E$.

We consider the following two cases:

Case 1: For infinitely many $M_E$ there is a $z\in M_E$ with

$$\begin{aligned} {[}z]=M_E. \end{aligned}$$

Let $y:=\bigcup \{M_E: E_x\subsetneq E\wedge \exists z\in M_E(M_E=[z])\}.$ The set y is in ${\mathcal {V}}_{p_0,p_1}$ because $E_x$ is a support of y. Let $t\subseteq y$ with $|t|=m$ and let E be a smallest closed set such that $M_E\subseteq y$ and $|t\cap M_E|$ is not of the form $k_0p_0+k_1p_1$ with $k_0,k_1\in \omega$. To see that such a set t exists, notice that for $[z]=M_E$ and $[z']=M_{E'}$, if $[z]\cap [z']\ne \emptyset$, then $M_E=M_{E'}$.

Define $t_{-1}:=t\cap M_E$. Since $E\setminus E_x\not =\emptyset$ there are blocks $A_{i_0},\dots , A_{i_{u-1}},B_{j_u},\dots , B_{j_{u+v-1}}$ with

$$\begin{aligned} E\setminus E_x=\bigcup \{ A_{i_0},A_{i_1}\dots , A_{i_{u-1}},B_{j_u},B_{j_{u+1}},\dots , B_{j_{u+v-1}}\}. \end{aligned}$$

Define

$$\begin{aligned} \tilde{G}:=\left\{ \prod _{k\in u}\alpha _{i_k}^{\kappa _{i_k}}\circ \prod _{l\in v} \beta _{j_{u+l}}^{\lambda _{j_{u+l}}}: \forall k< u~\forall l< v~\Bigl (\kappa _{i_k}< p_0\wedge \lambda _{j_{u+l}}< p_1\Bigr )\right\} . \end{aligned}$$

Let $\phi =\alpha _{i_0}^{\kappa _{i_0}}\circ \dots \circ \alpha _{i_{u-1}}^{\kappa _{i_{u-1}}}\circ \beta _{j_u}^{\lambda _{j_u}}\circ \dots \circ \beta _{j_{u+v-1}}^{\lambda _{j_{u+v-1}}}\in \tilde{G}$. Define

$$\begin{aligned} \phi \vert _r:=\kappa _{i_r}\text { if } r< u\text { and } \phi \vert _r :=\lambda _{j_r} \text { if } u\le r< u+v. \end{aligned}$$

The elements in $\tilde{G}$ can be ordered lexicographically. We call this well-ordering $\le _{\tilde{G}}$. For all $s,s^\prime < t_{-1}$ and all $r< u+v$ define

$$\begin{aligned} \textrm{dist}_r(\langle s,s^\prime \rangle ):=\phi \vert _r, \end{aligned}$$

where $\phi$ is the $\le _{\tilde{G}}$-smallest element in $\tilde{G}$ with $\phi (s)=s^\prime$.

The rest of the proof can be done as in [2, Proposition 6.6]. For the sake of completeness, we will redo it here:

Claim 1: For all $s,s^\prime , s^{\prime \prime }< t_{-1}$ and all $r< u+v$ we have that

$$\begin{aligned} \textrm{dist}_r(\langle s,s^\prime \rangle )+_p \textrm{dist}_r(\langle s^\prime , s^{\prime \prime }\rangle ) =\textrm{dist}_r(\langle s,s^{\prime \prime }\rangle ), \end{aligned}$$

where $p=p_0$ if $r< u$ and $p=p_1$ if $u\le r<u+v$. Moreover, $+_p$ denotes addition modulo p.

Proof of Claim 1

Let $\phi _0,\phi _1,\phi \in \tilde{G}$ be $\le _{\tilde{G}}$-minimal with

$$\begin{aligned} \phi _0(s)=s^\prime , \phi _1(s^\prime )=s^{\prime \prime }\text { and } \phi (s)=s^{\prime \prime }. \end{aligned}$$

Assume that $\phi \not =\phi _1\circ \phi _0$. So we have that $\phi ^{-1}\circ \phi _1\circ \phi _0\not = \textrm{id}$ and

$$\begin{aligned} \phi ^{-1}\circ \phi _1\circ \phi _0(s)=s. \end{aligned}$$

Let $l< u+v$ be the largest number such that

$$\begin{aligned} \phi ^{-1}\circ \phi _1\circ \phi _0\vert _{l}\not =0. \end{aligned}$$

Without loss of generality we assume that $l< u$. Then let $m\in \omega$ with

$$\begin{aligned} (\phi ^{-1}\circ \phi _1\circ \phi _0)^m\vert _l=1. \end{aligned}$$

Note that $(\phi ^{-1}\circ \phi _1\circ \phi _0)^m\not =\alpha _{i_l}$ because otherwise we would have that $\alpha _{i_l}(s)=s$ which is a contradiction to the fact that E is the minimal support of s with $E_x\subseteq E$. So there is a $\varphi \in \tilde{G}\setminus \{ \textrm{id}\}$ with

$$\begin{aligned} (\phi ^{-1}\circ \phi _1\circ \phi _0)^m=\varphi \circ \alpha _{i_l}\text { and } \varphi <_{\tilde{G}}\alpha _{i_l}. \end{aligned}$$

Then $\varphi \circ \alpha _{i_l}(s)=s\Rightarrow \alpha _{i_l}(s)=\varphi ^{-1}(s)$. Note that $\varphi ^{-1}<_{\tilde{G}}\alpha _{i_l}$. We have that $\phi _0\vert _{l}\not =0$ or $\phi _1\vert _{l}\not =0$ or $\phi \vert _{l}\not = 0$. Without loss of generality we assume that $\phi _0\vert _{l}\not =0$. Then

$$\begin{aligned} \phi _0\circ \alpha _{i_l}^{-1}\circ \varphi ^{-1}<_{\tilde{G}} \phi _0 \end{aligned}$$

and

$$\begin{aligned} \phi _0\circ \alpha _{i_l}^{-1}\circ \varphi ^{-1}(s)=\phi _0 \circ \alpha _{i_l}^{-1}\circ \alpha _{i_l}(s)=\phi _0(s)=s^\prime . \end{aligned}$$

This contradicts the minimality of $\phi _0$. ${\dashv }_{\textrm{Claim}\ 1}$

For all $\tilde{t}\subseteq t_{-1}$, all $s< \tilde{t}$ and all $r< u+v$ define

$$\begin{aligned} \chi _r(s,\tilde{t}):=\{ \textrm{dist}_r(\langle s,s^\prime \rangle ) : s^\prime \in \tilde{t}\}. \end{aligned}$$

These sets have the following properties:

Claim 2: For all $\tilde{t}\subseteq t_{-1}$ and all $s,s^\prime <\tilde{t}$ we have that

• 1. $1\le |\chi _r(s,\tilde{t})|\le p_0$ for all $r< u$ and $1\le |\chi _r(s,\tilde{t})|\le p_1$ for all $u\le r< u+v$.

• 2. for all $r< u+v$ there is a $k_r\in \omega$ such that $\chi _r(s,\tilde{t})=\chi _r(s^\prime , \tilde{t})+_p k_r$, where $p=p_0$ if $r< u$ and $p=p_1$ if $u\le r< u+v$.

• 3. $|\chi _r(s,\tilde{t})|= |\chi _r(s^\prime ,\tilde{t})|$.

• 4. if $s\not = s^\prime$ there is an $r< u+v$ such that $\chi _r(s,\tilde{t})\not =\chi _r(s^\prime , \tilde{t})$.

Proof of Claim 2

• 1. Note that $0<\chi _r(s,\tilde{t})$ since $\textrm{dist}_r(\langle s, s\rangle )=0$.

• 2. Set $k_r:=\phi \vert _r$, where $\phi$ is $\le _{\tilde{G}}$-minimal with $\phi (s)=s^\prime$ and use Claim 1.

• 3. This follows from 2.

• 4. Let $s,s^\prime <\tilde{t}$ and let $\phi$ be $\le _{\tilde{G}}$-minimal with $\phi (s)=s^\prime$. If $\chi _r(s,\tilde{t})=\chi _r(s^\prime ,\tilde{t})$ for all $r< u+v$ it follows that $\phi \vert _r=k_r=0$ for all $r< u+v$. So $\phi =\textrm{id}$ and therefore $s=s^\prime$.

${\dashv }_{\textrm{Claim}\ 2}$

We define an ordering $\preceq$ on the sets $\chi _r(s,\tilde{t})$ as follows: $\chi _r(s,\tilde{t})\preceq \chi _r(s^\prime ,\tilde{t})$ if and only if $\chi _r(s,\tilde{t})=\chi _r(s^\prime ,\tilde{t})$ or the smallest integer in the symmetric difference $\chi (s,\tilde{t})\Delta \chi _r(s^\prime ,\tilde{t})$ belongs to $\chi _r(s,\tilde{t})$.

For all non-empty sets $\tilde{t}\subseteq t_{-1},$ all $r< u+v$ and all natural numbers n define $\lambda _{r,n}(\tilde{t})$ as follows: Let $\lambda _{r,0}(\tilde{t}):=\emptyset$ and for every $n\in \omega \setminus \{ 0\}$ let

$$\begin{aligned} \lambda _{r,n}(\tilde{t}):=\left\{ s\in \tilde{t}\setminus \bigcup _{i=0}^{n-1} \lambda _{r,i}(\tilde{t}): \forall s^\prime \in \tilde{t} \setminus \bigcup _{i=0}^{n-1}\lambda _{r,i}(\tilde{t}) \left( \chi _r(s,\tilde{t})\preceq \chi _r(s^\prime ,\tilde{t})\right) \right\} . \end{aligned}$$

Note that $\bigcup _{n\in \omega }\lambda _{r,n}(\tilde{t})=\tilde{t}$ and only finitely many $\lambda _{r,n}(\tilde{t})$ are non-empty. Assume that $t_{r-1}$ is defined for an $r< u+v$. Then let

$$\begin{aligned} t_r:=\lambda _{r,n_0}(t_{r-1}), \end{aligned}$$

where $n_0\in \omega$ is the smallest natural number such that $\lambda _{r,n_0}(t_{r-1})$ is not of the form

$$\begin{aligned} cp_0+dp_1 \end{aligned}$$

with $c,d\in \omega$. By Claim 2, $t_{u+v-1}$ is a one-element set, i.e., there is an $s< t$ with

$$\begin{aligned} t_{u+v-1}=\{ s\}. \end{aligned}$$

So we choose s from t. This shows that $\textrm{RC}_m$ holds in ${\mathcal {V}}_{p_0,p_1}$.

Case 2: There are infinitely many $M_E$ such that there are $z,z^\prime \in M_E$ with

$$\begin{aligned} {[}z]\cap [z^\prime ]=\emptyset . \end{aligned}$$

Our goal is to reduce this case to Case 1. For every $E\in \textrm{fin}({\mathcal {A}})$ with $E_x\subsetneq E$ define

$$\begin{aligned} {[}M_E] :=\{[z]: z\in M_E\}. \end{aligned}$$

Furthermore, choose a $w_0$ in the ground model ${\mathcal {M}}_{p_0,p_1}\models {\textsf {ZFA}}+{\textsf {AC}}$ such that

$$\begin{aligned} w_0\setminus \bigcup _{\begin{array}{c} E\in \textrm{fin}({\mathcal {A}})\\ E_x\subsetneq E \end{array}} [M_E] = \emptyset \end{aligned}$$

and

$$\begin{aligned} \textit{for all closed sets}\ E\in \textrm{fin}({\mathcal {A}}) \ \textit{with}\ E_x\subsetneq E\ \textit{and}\ M_E\not =\emptyset \ \textit{we have}\ |w_0 \cap [M_E]|=1. \end{aligned}$$

In other words, $w_0$ picks exactly one element from each non-empty $[M_E]$. Note that $E_x$ is a support of $w_0$. So $w_0\in {\mathcal {V}}_{p_0,p_1}$. Choose

$$\begin{aligned} M_E^\prime :=M_E\cap w_0. \end{aligned}$$

This reduces Case 2 to Case 1. $\square$

Proposition 2.9

Let $m,n\in \omega \setminus \{0,1\}$, $k\in \omega$, and let $p_0,\dots , p_{k-1}$ be prime numbers such that

$$\begin{aligned} m\not = \sum _{i< k}c_ip_i \end{aligned}$$

for all $c_i\in \omega$, $i< k$, and

$$\begin{aligned} n=\sum _{i< k}d_ip_i \end{aligned}$$

for some $d_i\in \omega ,~i\in k$. Then

$$\begin{aligned} \textrm{RC}_m\not \Rightarrow \textrm{WOC}_n^- \end{aligned}$$

is consistent with ${\textsf {ZF}}$.

Proof

Similar as in Lemmas 2.7 and 2.8 we can prove that

$$\begin{aligned} {\mathcal {V}}_{p_0,\dots , p_{k-1}}\models \textrm{RC}_m\wedge \lnot \textrm{WOC}_n^-. \end{aligned}$$

(1)

In order to transfer this statement to ${\textsf {ZF}}$, we have to show that $\textrm{RC}_n$ and $\textrm{WOC}_n^-$ are injectively boundable for all $n\in \omega$. Then we can use Pincus’ transfer theorem [7, Theorem 3A3]. The terms “boundable” and “injectively boundable” are defined in [7].

For a set x we define the injective cardinality

$$\begin{aligned} |x|_-:=\{\alpha \in \Omega : \text { there is an injection from}\ \alpha \ \text {into}\ x\}, \end{aligned}$$

where $\Omega$ is the class of all ordinal numbers. Moreover let $\varphi (x)$ denote the following property:

$$\begin{aligned} \textit{if}\ x\ \textit{is an infinite set, there is an infinite} \ y\subseteq x\ \textit{with a choice function on}\ [y]^n. \end{aligned}$$

Note that $\varphi (x)$ is boundable. Since $\varphi (x)$ holds when $|x|_->\omega$, it follows that

$$\begin{aligned} \textrm{RC}_n\iff \forall x(|x|_- \le \omega \Rightarrow \varphi (x)). \end{aligned}$$

So, $\textrm{RC}_n$ is injectively boundable. Furthermore, we have that $\lnot \textrm{WOC}_n^-$ is boundable. So, (1) is transferable into ${\textsf {ZF}}$. $\square$

Propostion 2.2 together with Propostion 2.9 gives us the following result:

Theorem 2.10

Let $m,n\in \omega \setminus \{0,1\}$. Then $\textrm{RC}_m$ implies $\textrm{WOC}_n^-$ if an only if the following condition holds: For all prime numbers $p_0,\dots , p_{k-1}$ such that there are positive integers $a_0,\dots ,a_{k-1}$ with

$$\begin{aligned} n=\sum _{i< k}a_ip_i, \end{aligned}$$

we can find $b_0,\dots , b_{k-1}\in \omega$ with

$$\begin{aligned} m=\sum _{i< k} b_ip_i. \end{aligned}$$

We conclude this section by giving a few consequences. Since $\lnot \textrm{WOC}_n^-\Rightarrow \lnot \textrm{RC}_n$, Proposition 2.9 gives us:

Corollary 2.11

Let $m,n\in \omega \setminus \{0,1\}$ and let $p_0,\dots , p_{k-1}$ be $k\in \omega$ prime numbers such that

$$\begin{aligned} m\not = \sum _{i< k}c_ip_i \end{aligned}$$

for all $c_i\in \omega$, $i< k$, and

$$\begin{aligned} n=\sum _{i< k}d_ip_i \end{aligned}$$

for some $d_i\in \omega ,~i< k$. Then

$$\begin{aligned} \textrm{RC}_m\not \Rightarrow \textrm{RC}_n \end{aligned}$$

in ${\textsf {ZF}}$.

Proof

This follows from $\textrm{RC}_n\Rightarrow \textrm{WOC}_n^-$ (Corollary 2.3) and $\textrm{RC}_m\nRightarrow \textrm{WOC}_n^-$ (Proposition 2.9). $\square$

Corollary 2.12

Let p be a prime number, let $m\in \omega \setminus \{0\}$ and $n\in \omega \setminus \{0,1\}$. Then we have that

$$\begin{aligned} \textrm{RC}_{p^m}\Rightarrow \textrm{WOC}_n^- \end{aligned}$$

if and only if $n\mid p^m$ or $p=2$, $m=1$ and $n=4$.

Proof

If n is divisible by a prime $q\not =p$ we have that

$$\begin{aligned} {\mathcal {V}}_q\models \textrm{RC}_{p^m}\wedge \lnot \textrm{WOC}_n^-. \end{aligned}$$

Therefore, $\textrm{RC}_{p^m}\not \Rightarrow \textrm{WOC}_n^-$ in ${\textsf {ZF}}$. So we can assume that $n=p^k$ for a $k\in \omega \setminus \{0\}.$

Case 1: $m\ge k$

Let $r\in \omega$ and let $p_0,p_1\dots , p_{r-1}$ be prime numbers such that there are $a_0,a_1,\dots , a_{r-1}\in \omega$ with

$$\begin{aligned} n=p^k=\sum _{i< r}a_ip_i. \end{aligned}$$

Then

$$\begin{aligned} p^m=p^{m-k}p^k=\sum _{i< r}p^{m-k}a_ip_i. \end{aligned}$$

So by Proposition 2.2 we have that

$$\begin{aligned} \textrm{RC}_{p^m}\Rightarrow \textrm{WOC}_n^-. \end{aligned}$$

Case 2: $m<k$

First, assume that $p\not =2$. By Bertrand’s postulate there is a prime number $q_0$ with

$$\begin{aligned} p^m<q_0<2p^m. \end{aligned}$$

Note that $p^k-q_0>p^k-2p^{m}\ge p$ and $q_0\not =p$. So there is a prime number $q_1\not = p$ with

$$\begin{aligned} q_1\mid (p^k-q_0). \end{aligned}$$

By construction, $p^k$ can be written as a sum of multiples of $q_0$ and $q_1$. Since $q_1\not \mid p^m$ and $p^m<q_0$, we have that

$$\begin{aligned} p^m\not = aq_0+bq_1. \end{aligned}$$

for all $a,b\in \omega$. So by Proposition 2.9 we have that

$$\begin{aligned} \textrm{RC}_{p^m}\not \Rightarrow \textrm{WOC}_{p^k}^-. \end{aligned}$$

Now, let $p=2$ and $k\ge 3$. Then there is a prime number $q_0$ with

$$\begin{aligned} 2^{k-1}-1<q_0<2^k-2. \end{aligned}$$

It follows that

$$\begin{aligned} 2<2^{k-1}<q_0<2^k-2. \end{aligned}$$

So, $2^k-q_0>2$ and with the same argumentation as above we see that

$$\begin{aligned} \textrm{RC}_{2^n}\not \Rightarrow \textrm{WOC}_{2^k}. \end{aligned}$$

Now we assume that $p=2$ and $k=2$ (i.e., $m=1$). This is the only remaining case. By Proposition 2.2 we have that

$$\begin{aligned} \textrm{RC}_2\Rightarrow \textrm{WOC}_4^-. \end{aligned}$$

$\square$

3 Results provable in ${\textsf {ZF}}$

In this section we shall prove four results which are provable in ${\textsf {ZF}}$. The first two results are about the implications $\textrm{RC}_6\Rightarrow \textrm{C}_n^-$ for $n\in \{3,9\}$, and the second two results are about the implications $\textrm{RC}_n\Rightarrow \textrm{LOC}_m^-$ for $m\in \{5,7\}$.

3.1 $\textbf{RC}_{\varvec{6}}$ implies $\textbf{C}_{\varvec{3}}^-$

In the proof of the next result, we will closely follow the proof of $\textrm{RC}_4\Rightarrow \textrm{C}_4^-$ given in [6].

Proposition 3.1

${\textsf {ZF}} \vdash \textrm{RC}_6\Rightarrow \textrm{C}_3^-$, i.e., it is provable in ${\textsf {ZF}}$ that $\textrm{RC}_6$ implies $\textrm{C}_3^-$.

Proof

Let ${\mathcal {F}}$ be an infinite family of pairwise disjoint sets of size 3. We apply $\textrm{RC}_6$ to the set $\bigcup {\mathcal {F}}$. This gives us an infinite subset $Y\subseteq \bigcup {\mathcal {F}}$ with a choice function on $[Y]^6$. For every $i\in \{1,2,3\}$ we define

$$\begin{aligned} {\mathcal {G}}_i:=\{ u\in {\mathcal {F}}:|u\cap Y|=i\}. \end{aligned}$$

Without loss of generality we can assume that ${\mathcal {G}}:={\mathcal {G}}_3$ is infinite, since otherwise, we can easily define a choice function on an infinite subset of ${\mathcal {F}}$. So, there is a choice function

$$\begin{aligned} f:\left[ \bigcup {\mathcal {G}}\right] ^6\rightarrow \bigcup {\mathcal {G}}. \end{aligned}$$

We define a directed graph on ${\mathcal {G}}$ by putting a directed edge from v to u (i.e., $v\rightarrow u$), if and only if $f(u\cup v)\in u$. If there is direct edge from v to u we will say that the edge points from v to u. With this graph we carry out the same construction as in [6]. So, there is an infinite subset ${\mathcal {H}}\subseteq {\mathcal {G}}$ which is partitioned into finite sets $(A_n)_{n\in \omega }$ such that for every $n\in \omega$, all elements in $A_n$ have outdegree n. Moreover, for all $n\in \omega$ we have that $|A_n|$ is odd, and for all $n<m$, the edges between $A_n$ and $A_m$ all point from $A_m$ to $A_n$. We can assume that we are in one of the following two cases:

Case 1: There are infinitely many $n\in \omega$ with $3\not \mid |A_n|$.

In this case we follow the proof of the Claim in [6, p. 60]: Without loss of generality we can assume that $3\not \mid |A_n|$ for every $n\in \omega$. Let $n_0\in \omega$ and $p_0=\{ x_0, x_1, x_2\}\in A_{n_0}$. For each $i\le 2$ we define

$$\begin{aligned} \textrm{deg}(x_i):=|\{q\in A_{n_0+1}:f(q\cup p_0)=x_i\}|. \end{aligned}$$

Since $3\not \mid |A_{n_0+1}|$ we have that $3\not \mid (\textrm{deg}(x_0)+\textrm{deg}(x_1)+\textrm{deg}(x_2))$. Therefore, we can choose one element from $p_0$.

Case 2: For all $n\in \omega$ we have that $3\mid |A_n|$.

Let $p_0\in {\mathcal {H}}$ and let $n\in \omega$ be the unique natural number with $p_0\in A_n$. There is an $s\in \omega$ with $|A_n|=2s+1$. We want to find the number of elements in $A_n$ with edges pointing to $p_0$. There are $\left( {\begin{array}{c}|A_n|\\ 2\end{array}}\right)$ edges in $A_n$. Since the number of edges in $A_n$ that point to an element in $A_n$ is the same for every element of $A_n$, we have that the indegree of $p_0$ in $A_n$ is given by

$$\begin{aligned} \textrm{indegree}_{A_n}(p_0)=\frac{1}{|A_n|} \left( {\begin{array}{c}|A_n|\\ 2\end{array}}\right) =\frac{1}{2}(|A_n|-1)=s. \end{aligned}$$

By assumption we have that $3\mid |A_n|=2s+1.$ Therefore, $3\not \mid s$. Assume that $p_0=\{ x_0,x_1, x_2\}$. For every $i\le 2$ we define

$$\begin{aligned} A_n^{x_i}:=\{ v\in A_n:f(v\cup p_0)=x_i\}. \end{aligned}$$

Since $3\not \mid (|A_n^{x_0}|+|A_n^{x_1}|+|A_n^{x_2}|)=s$, we can choose an element from $p_0$. $\square$

3.2 $\textbf{RC}_{\varvec{6}}$ implies $\textbf{C}_{\varvec{9}}^-$

Lemma 3.2

Let ${\mathcal {F}}$ be an infinite family of pairwise disjoint 4-element sets. If there is a choice function

$$\begin{aligned} f:\left[ \bigcup {\mathcal {F}}\right] ^6\rightarrow \bigcup {\mathcal {F}}, \end{aligned}$$

then there is a function h with $h(p\cup q)\in p\cup q$ for all $p\not = q$ in ${\mathcal {F}}$.

Proof

Let $p\not = q$ be elements of ${\mathcal {F}}$. We will show that we can choose exactly one element from $p\cup q$. There are

$$\begin{aligned} \begin{pmatrix} 8\\ 6 \end{pmatrix}=28 \end{aligned}$$

6-element subsets of $p\cup q$. From each of these subsets we can choose one point with the choice function f. Let A be the set of all elements in $p\cup q$ which are chosen the most times. Note that $1\le |A|\le 7$, because 8 does not divide 28.

• If $|A|=1$ we are done.

• If $|A|=2$, choose $f((p\cup q)\setminus A)$.

• If $|A|=3$ and $A\subseteq p$ or $A\subseteq q$ we are done because we can choose the point in $p\setminus A$ or in $q\setminus A$. Otherwise, $|p\cap A|=1$ or $|q\cap A|=1$ and we are also done.

• If $|A|\in \{5,6,7\}$, replace A by $(p\cup q)\setminus A$. So we are in one of the cases above.

• If $|A|=4$, the set $[ (p\cup q)\setminus A]^2$ contains $\begin{pmatrix} 4 \\ 2 \end{pmatrix}=6$ elements. For each $B\in [ (p\cup q){\setminus } A]^2$ choose $f(A\cup B)$. Let $C_0$ and $C_1$ be the sets of all elements in $p\cup q$ which are chosen the most and the least often. Note that either $C_0$ or $C_1$ does not contain 4 elements. By the cases above we are done.

So there is a choice function

$$\begin{aligned} h:\{ p\cup q:p,q\in {\mathcal {F}}\}\rightarrow \bigcup {\mathcal {F}}. \end{aligned}$$

$\square$

Lemma 3.3

Let $\{ A_n:n\in \omega \}$ be a countable family of pairwise disjoint non-empty finite sets of pairwise disjoint sets of size 2, and let ${\mathcal {F}}:=\bigcup _{n\in \omega } A_n$ be the corresponding infinite family of 2-element sets. If

$$\begin{aligned} f:\left[ \bigcup {\mathcal {F}}\right] ^6\rightarrow \bigcup {\mathcal {F}}. \end{aligned}$$

is a choice function, then there is an infinite subfamily ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a choice function.

Proof

By using a bijection between $\omega$ and an infinite subset of $\omega$, without loss of generality we are in one of the following four cases:

Case 1: For all $n\in \omega$ we have that $2\not \mid |A_n|$.

Let $k\in \omega$. Then there are natural numbers $l_0,l_1$ and $l_2$ such that

$$\begin{aligned} |A_{3k}|=2l_0+1,~|A_{3k+1}|=2l_1+1~\text { and }~|A_{3k+2}|=2l_2+1. \end{aligned}$$

For every $a\in A_{3k}\cup A_{3k+1}\cup A_{3k+2}$ define

$$\begin{aligned} \# a:=|\{ (a_0,a_1,a_2)\in A_{3k}\times A_{3k+1}\times A_{3k+2}:f(a_0\cup a_1\cup a_2)\in a\}|. \end{aligned}$$

If $\# a$ is odd, we can choose an element from a, for example the element in a we choose more often than the other. Since

$$\begin{aligned} 2\not \mid \prod _{i\le 2} (2l_i+1)~\text { and } ~\sum _{a\in A_{3k}\cup A_{3k+1}\cup A_{3k+2}} \#a~~~~=\prod _{i\le 2} (2l_i+1), \end{aligned}$$

we have that for every $k\in \omega$ there is at least one $a\in A_{3k}\cup A_{3k+1}\cup A_{3k+2}$ such that $\# a$ is odd. So, we can find a choice function on the infinite set

$$\begin{aligned} {\mathcal {G}}:=\{ a\in {\mathcal {F}}:\# a \text { is odd}\}. \end{aligned}$$

Case 2: For all $n\in \omega$ we have that $|A_n|=2$.

For every $k\in \omega$ let $A_{2k}=\{a_{2k},b_{2k}\}$ and $B_0:=\{a_{2k}\}\cup A_{2k+1}$ and $B_1:=\{b_{2k}\}\cup A_{2k+1}$. For every $a\in A_{2k}\cup A_{2k+1}$ we define

$$\begin{aligned} \# a:=\left|\left\{ i\in \{0,1\}:f\left( \bigcup B_i\right) \in a\right\} \right|. \end{aligned}$$

Note that if $\# a=1$, we can choose an element from a and we are done. So, if there are infinitely many $a\in {\mathcal {F}}$ such that $\# a$ is odd, we are done. Otherwise, there is an infinite subset $I\subseteq \omega$ such that for all $k\in I$ there is a unique $a_k\in A_{2k}\cup A_{2k+1}$ with $\# a_k=2$. Then we are in the first case for the family $\{ \{ a_k\}:k\in I\}.$

Case 3: For all $n\in \omega$ we have that $|A_n|\ge 3$, $4\not \mid |A_n|$ and $2\mid |A_n|$.

Let $n\in \omega$. Then, by the properties of $|A_n|$ we have $|A_n|=2t$ for some odd t, and therefore we have that $\begin{pmatrix}|A_n|\\ 2\end{pmatrix}$ is odd. For every $k\in \omega$ we look at the 4-element subsets of $A_{2k}\cup A_{2k+1}$ with two elements in $A_{2k}$ and two elements in $A_{2k+1}$. Note that the number of such subsets, as the product of two odd numbers, is odd. Let h be the choice function we found in Lemma 3.2. Then for every $k\in \omega$ there is at least one $a\in A_{2k}\cup A_{2k+1}$ such that

$$\begin{aligned} \# a:=|\{ (\{ a_0,a_1\}, \{ b_0, b_1\})\in [A_{2k}]^2 \times [A_{2k+1}]^2: h(a_0\cup a_1\cup b_0\cup b_1)\in a\}|\end{aligned}$$

is odd. So again we found a choice function on the infinite set

$$\begin{aligned} {\mathcal {G}}:=\{ a\in {\mathcal {F}}: \# a\text { is odd}\}. \end{aligned}$$

Case 4: For all $n\in \omega$ we have that $|A_n|\ge 3$ and $4\mid |A_n|$.

Let $n\in \omega$. Then there is a $k\in \omega$ with $|A_n|= 4k$. We have that

$$\begin{aligned} 2|A_n|\not \mid \left( {\begin{array}{c}|A_n|\\ 3\end{array}}\right) , \end{aligned}$$

(2)

since otherwise we would have that

$$\begin{aligned} \frac{|A_n|(|A_n|-1) (|A_n|-2)}{2\cdot |A_n|\cdot 2\cdot 3}= \frac{ 2(4k^2-3k)+1}{2\cdot 3}\in \omega , \end{aligned}$$

but this is not the case since the numerator is odd. We define

$$\begin{aligned} \# a:=|\{ \{ a_0,a_1,a_2\}\in [A_n]^3: f(a_0\cup a_1\cup a_2)\in a\}|\end{aligned}$$

and for all $y \in \bigcup A_n$ let

$$\begin{aligned} \#(y):=|\{ \{ a_0,a_1,a_2\} \in [A_n]^3: f(a_0\cup a_1\cup a_2)=y\}|. \end{aligned}$$

Note that by (2)

$$\begin{aligned} \left|\left\{ y\in \bigcup A_n: \#(y) =\max \left\{ \#(z): z\in \bigcup A_n\right\} \right\} \right|<2|A_n|. \end{aligned}$$

If there is an $a=\{ a_0,a_1\}\in A_n$ with

$$\begin{aligned} \#(a_0)\not = \#(a_1) \end{aligned}$$

choose the element $a_i$ with lower $\#(a_i)$. Otherwise we have that

$$\begin{aligned} B_n:=\{ a\in A_n: \# a=\max \{ \# b: b\in A_n\}\}\subsetneq A_n. \end{aligned}$$

Repeat the procedure with $A_n:= B_n$ until either $4\not \mid |A_n|$ or there is an $a=\{ a_0,a_1\}\in A_n$ with

$$\begin{aligned} \#(a_0)\not = \#(a_1). \end{aligned}$$

Note that we have to repeat the procedure at most $|A_n|$ times. In the end we either found a choice function on an infinite subset of ${\mathcal {F}}$ or we reduced Case 4 to one of the other cases. $\square$

Corollary 3.4

Let ${\mathcal {F}}$ be an infinite family of pairwise disjoint 4-element sets. If

$$\begin{aligned} f:\left[ \bigcup {\mathcal {F}}\right] ^6\rightarrow \bigcup {\mathcal {F}} \end{aligned}$$

is a choice function, then there is an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a choice function on ${\mathcal {G}}$.

Proof

Let h be the choice function we found in Lemma 3.2. We can define a complete, directed graph on ${\mathcal {F}}$ by putting an edge from p to q if and only if $h(p\cup q)\in q$. With this graph we can do the same construction as in [6]. So, we can find an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ such that we can choose exactly 1 or 2 elements from each $G\in {\mathcal {G}}$. So either we found a choice function on an infinite subset of ${\mathcal {G}}$ or we can find an infinite family of 2-element sets ${\mathcal {H}}$. Then we apply Lemma 3.3 to ${\mathcal {H}}$ and we are done. $\square$

Lemma 3.5

Let ${\mathcal {F}}:=\{F_{\lambda }:\lambda \in \Lambda \}$ be an infinite family of 10-element sets. Assume that each $F_\lambda \in {\mathcal {F}}$ is a disjoint union of five 2-element sets $F_{\lambda ,i}$, $0\le i\le 4$. Moreover, assume that

$$\begin{aligned} f:\left[ \bigcup {\mathcal {F}} \right] ^6\rightarrow \bigcup {\mathcal {F}} \end{aligned}$$

is a choice function. Then there is an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a Kinna-Wagner selection function.

Proof

For all 4-element sets $A\subseteq \bigcup {\mathcal {F}}$, we define the degree of A by

$$\begin{aligned} \textrm{deg}(A):=|\{ F_{\lambda ,i}: F_{\lambda }\in {\mathcal {F}} \wedge i\le 4\wedge F_{\lambda ,i}\cap A=\emptyset \wedge f(A\cup F_{\lambda ,i})\in F_{\lambda ,i}\}|. \end{aligned}$$

If there is an $A_0\in \left[ \bigcup {\mathcal {F}}\right] ^4$ with infinite degree we are done, because then the set

$$\begin{aligned} {\mathcal {G}}:=\{ F_{\lambda }\in {\mathcal {F}}: \exists i\le 4~(f(A_0\cup F_{\lambda ,i})\in F_{\lambda ,i})\} \end{aligned}$$

is infinite and from every $G\in {\mathcal {G}}$ we can choose the set

$$\begin{aligned} \{ f(A_0\cup G_i): i\le 4\}\cap G\subsetneq G. \end{aligned}$$

Thus, we can assume that each $A\in \left[ \bigcup {\mathcal {F}}\right] ^4$ has finite degree. Define ${\mathcal {F}}^2=\{ F_{\lambda ,i}: F_{\lambda }\in {\mathcal {F}}\wedge i\le 4\}$ and for all $F_{\lambda }\in {\mathcal {F}}$ let $F_{\lambda }^2:=\{ F_{\lambda ,i}: i\le 4\}$.

Case 1: There is an $n\in \omega$ such that for infinitely many $\lambda \in \Lambda$ there are distinct $A,B\in F_{\lambda }^2$ with $\textrm{deg}(A\cup B)=n$.

Let ${\mathcal {G}}:=\{ F_{\lambda }\in {\mathcal {F}}: \exists A,B\in F_{\lambda }^2(\textrm{deg}(A\cup B)=n)\}$. By assumption this is an infinite set. Choose an $(n+3)$-element set $\{X_i: i\le n+2\}\subseteq {\mathcal {F}}^2$. For all $G\in {\mathcal {G}}$ and all $A,B\in G^2$ with $\textrm{deg}(A\cup B)=n$ put an edge pointing from A to B if and only if

$$\begin{aligned} f(A\cup B\cup X_{i_0})\in B, \end{aligned}$$

where

$$\begin{aligned} i_0:=\min \{ i\le n+2: f(A\cup B\cup X_i)\notin X_i\}. \end{aligned}$$

Notice that this gives us a directed graph with at least one edge in each $G^2$ with $G\in {\mathcal {G}}$. If for infinitely many $G\in {\mathcal {G}}$ not all elements of $G^2$ have the same outdegree, we are done. So, we either have a cycle on infinitely many $G^2$ or we have a complete graph in which every node has outdegree 2. In the former case we can choose a point in each $A\cup B$, where $A,B\in G^2$ are neighbours. Thus, we can choose 5 elements in each $G\in {\mathcal {G}}$. In the latter case, we can choose 5 edges as follows: For the node $A\in G^2$, let $B,C\in G^2$ be the two successors of A in the graph. Consider the edge which connects B and C (see Fig. 1). If this edge points to C, then we go to B and consider the two successors of B. Proceeding this way, we obtain a cycle on infinitely many $G^2$’s and can again choose 5 elements from G.

Fig. 1

How to choose the edges

Case 2: For all $n\in \omega$ there are only finitely many $\lambda \in \Lambda$ such that there are $A,B\in F_{\lambda }^2$ with $\textrm{deg}(A\cup B)=n$.

Let $A_{-1}:=\emptyset$ and for every $n\in \omega$ define

$$\begin{aligned} A_n:=\{A\in {\mathcal {F}}^2: \exists B\in {\mathcal {F}}^2(\textrm{deg} (A\cup B)=n)\}\setminus A_{n-1}. \end{aligned}$$

Note that these sets are pairwise disjoint families of 2-element sets. So we can apply Lemma 3.3 and we are done. $\square$

Now, we are ready to prove the following:

Proposition 3.6

${\textsf {ZF}} \vdash \textrm{RC}_6\Rightarrow \textrm{C}_9^-$.

Proof

Let ${\mathcal {F}}$ be an infinite family of pairwise disjoint sets of size 9. Since $\textrm{RC}_6$ holds, there is an infinite set $Y\subseteq \bigcup {\mathcal {F}}$ with a choice function

$$\begin{aligned} f:[Y]^6\rightarrow Y. \end{aligned}$$

For all $0\le i\le 9$ let

$$\begin{aligned} {\mathcal {G}}_i:=\{F\cap Y: F\in {\mathcal {F}} \wedge |F\cap Y|=i\}. \end{aligned}$$

There is a $1\le i\le 9$ such that ${\mathcal {G}}_i$ is an infinite set.

Case 1: ${\mathcal {G}}_1$ or ${\mathcal {G}}_8$ is infinite.

In the case ${\mathcal {G}}_8$ is infinite, we look at the complements.

Case 2: ${\mathcal {G}}_3$ or ${\mathcal {G}}_6$ is infinite.

Use Proposition 3.1.

Case 3: ${\mathcal {G}}_4$ is infinite.

Use Corollary 3.4.

Case 4: ${\mathcal {G}}_5$ is infinite.

Apply $\textrm{RC}_6$ to the complements. Then we are either in one of the preceding cases or the complements are partitioned into two sets of size two. We look at the 10 edges between the first 5 elements and the second two elements and use Lemma 3.5.

Case 5: ${\mathcal {G}}_7$ is infinite.

For all $G\in {\mathcal {G}}_7$ let $\overline{G}$ be the complement of G in the sense that for the $F\in {\mathcal {F}}$ with $G\subseteq F$ we have that

$$\begin{aligned} \overline{G}:=F\setminus G. \end{aligned}$$

Note that $|\overline{G}|=2$. Let

$$\begin{aligned} {\mathcal {E}}:=\{ \{x,y\}: \exists G\in {\mathcal {G}}_7 (x\in G\text { and }y\in \overline{G})\}. \end{aligned}$$

Apply $\textrm{RC}_6$ to ${\mathcal {E}}$. Without loss of generality we can assume that we find a choice function

$$\begin{aligned} g:[{\mathcal {E}}]^6\rightarrow {\mathcal {E}}, \end{aligned}$$

because otherwise we are in one of the preceding cases. So, for every $G\in {\mathcal {G}}_7$ there are 14 edges between G and $\overline{G}$. Hence, there are

$$\begin{aligned} \begin{pmatrix} 14\\ 6 \end{pmatrix} =3\cdot 7\cdot 11\cdot 13 \end{aligned}$$

6-element subsets. From each of them g chooses one element. Since $\begin{pmatrix}14\\ 6\end{pmatrix}$ is not divisible by 14, we can choose less than 14 edges and we are in one of the preceding cases.

Case 6: ${\mathcal {G}}_9$ is infinite.

With the choice function f we can choose an element from each 6-element subset of a $G\in {\mathcal {G}}_9$. There are $\begin{pmatrix}9\\ 6\end{pmatrix}$ subsets of size 6. Since $9\not \mid \begin{pmatrix}9\\ 6\end{pmatrix}$ we can reduce this case to one of the cases above.

Case 7: ${\mathcal {G}}_2$ is infinite.

We iteratively apply $\textrm{RC}_6$ to the complements. So, we can reduce this case to one of the cases above. $\square$

3.3 $\textbf{RC}_{\varvec{5}}$ implies $\textbf{LOC}_{\varvec{5}}^-$

We will now show that $\textrm{RC}_5$ implies $\textrm{LOC}_5^-$. The beginning of the proof will be as usual: Let ${\mathcal {F}}$ be an infinite, linearly orderable family of 5-element sets. We apply $\textrm{RC}_5$ to $\bigcup {\mathcal {F}}$. This will give us an infinite subfamily ${\mathcal {G}}\subseteq {\mathcal {F}}$ such that each $p\in {\mathcal {G}}$ is partitioned into two parts. If one of these parts is of size one, we have a choice function and we are done. Otherwise, the two parts are of size 2 and 3. So if we could show that $\textrm{RC}_5$ implies $\textrm{LOC}_2^-$ or $\textrm{LOC}_3^-$, the proof would be finished. However, Halbeisen’s and Tachtsis’ result ($\beta$) shows that this idea will not lead to success — which is the reason why we will work with the set of edges between the two parts.

Theorem 3.7

${\textsf {ZF}} \vdash \textrm{RC}_5\Rightarrow \textrm{LOC}_5^-$.

Proof

Let ${\mathcal {F}}$ be an infinite, linearly orderable collection of pairwise disjoint sets of size 5. We fix a linear order on ${\mathcal {F}}$ and apply $\textrm{RC}_5$ on the set $X:=\bigcup {\mathcal {F}}$ to find an infinite subset $Y\subseteq X$ with a choice function $f:[Y]^5\rightarrow Y$. For every $i\le 5$ we define

$$\begin{aligned} {\mathcal {F}}_i:=\{ p\in {\mathcal {F}}: |p\cap Y|=i\}. \end{aligned}$$

The only non-trivial case is when the elements p of an infinite subfamily ${\mathcal {G}}\subseteq {\mathcal {F}}$ are partitioned into a set with two elements and a set with three elements, namely $p=\{ a_p,b_p,c_p\}\cup \{ x_p,y_p\}$.

Now we look at the set Z of all non-directed edges between a point in $\{a_p,b_p,c_p\}$ and one in $\{x_p,y_p\}$. For every $p\in {\mathcal {G}}$ let $p^*$ be the set of all edges in Z belonging to p and for each subset ${\mathcal {H}} \subseteq {\mathcal {F}}$ we define ${\mathcal {H}}^*:=\{ p^*: p\in {\mathcal {H}}\}$.

Claim 1: Assume that there is an infinite subset ${\mathcal {H}}\subseteq {\mathcal {G}}$ such that we can choose between 1 and 5 elements from each $p^*\in {\mathcal {H}}^*$. Then there is a choice function

$$\begin{aligned} h:{\mathcal {H}}\rightarrow \bigcup {\mathcal {H}}. \end{aligned}$$

Fig. 2

The partitions of a $p^*$ into $\gamma _0^p$, $\gamma _1^p$ on the left and into $\beta _0^p$, $\beta _1^p$ and $\beta _2^p$ on the right

Proof of Claim 1

Let $p\in {\mathcal {H}}$ and assume that we can choose $k\in \{ 1,2,3,4,5\}$ elements from $p^*$. We look at p as a graph with k edges. If $2\not \mid k$, $x_p$ and $y_p$ do not have the same degree and we can choose the element with lower degree. Otherwise we have that $3\not \mid k$ and we can choose an element from $\{ a_p,b_p,c_p\}$. ${\dashv }_{\textrm{Claim}\ 1}$

Now we apply $\textrm{RC}_5$ on the set Z. Then there is an infinite subset $Q\subseteq Z$ with a choice function $g:[Q]^5\rightarrow Q$. By Claim 1 we can without loss of generality assume that $p^*\subseteq Q$ for every p in some infinite ${\mathcal {H}}\subseteq {\mathcal {G}}$.

We can partition each $p^*\in {\mathcal {H}}^*$ as follows into two sets $\gamma _0^p$ and $\gamma _1^p$ of size three (Fig. 2):

$$\begin{aligned} \gamma _0^p:=\{\{ a_p, x_p\},\{b_p,x_p\},\{ c_p,x_p\}\}~\text { and } ~\gamma _1^p:=\{\{ a_p, y_p\},\{b_p,y_p\},\{ c_p,y_p\}\}. \end{aligned}$$

Analogously we can partition $p^*$ into three sets $\beta _0^p, \beta _1^p, \beta _2^p$ of size two as follows (Fig. 2):

$$\begin{aligned}{} & {} \beta _0^p:=\{\{a_p,x_p\},\{a_p,y_p\}\}, ~\beta _1^p: =\{\{b_p,x_p\},\{b_p,y_p\}\}~\text { and }\\{} & {} \beta _2^p :=\{\{c_p,x_p\},\{c_p,y_p\}\}. \end{aligned}$$

Let

$$\begin{aligned} {\mathcal {H}}_3^*:=\{\gamma _i^p: i\le 1\wedge p\in {\mathcal {H}}\} \end{aligned}$$

be the sets of size three appearing in the partition of a $p^*\in {\mathcal {H}}^*$ and let

$$\begin{aligned} {\mathcal {H}}_2^*:=\{\beta _i^p: i\le 2\wedge p\in {\mathcal {H}}\} \end{aligned}$$

be the family of sets of size two which appear in the partition of a $p^*\in {\mathcal {H}}^*$. If there is a $\gamma \in {\mathcal {H}}_3^*$ such that for infinitely many $\beta \in {\mathcal {H}}_2^*$

$$\begin{aligned} g(\gamma \cup \beta )\in \beta , \end{aligned}$$

(3)

we are done by Claim 1. Otherwise, for every $\gamma \in {\mathcal {H}}_3^*$ there are only finitely many $\beta \in {\mathcal {H}}_2^*$ with (3) and we define

$$\begin{aligned} \textrm{deg}(\gamma ):=|\{\beta \in {\mathcal {H}}_2^*: g(\gamma \cup \beta )\in \beta \}|\in \omega . \end{aligned}$$

We are in one of the following two cases:

Case 1: There is an $n\in \omega$ such that $\textrm{deg}(\gamma )=n$ for infinitely many $\gamma \in {\mathcal {H}}_3^*$.

Let ${\mathcal {I}}_3^*:=\{\gamma \in {\mathcal {H}}_3^*: \textrm{deg}(\gamma )=n\}$. Choose an $(n+4)$-element set $\{\beta _i: i\le n+3\}\subseteq {\mathcal {H}}_2^*$. For every $\gamma \in {\mathcal {I}}_3^*$ we define

$$\begin{aligned} j(\gamma ):=\min \{i\le n+3: g(\gamma \cup \beta _i)\in \gamma \}. \end{aligned}$$

So from every $\gamma \in {\mathcal {I}}_3^*$ we choose the element

$$\begin{aligned} g(\gamma \cup \beta _{j(\gamma )})\in \gamma \end{aligned}$$

and we are done by Claim 1.

Case 2: For each $n\in \omega$ there are only finitely many $\gamma \in {\mathcal {H}}_3$ with $\textrm{deg}(\gamma )=n$.

For every $n\in \omega$ we define

$$\begin{aligned}{} & {} A_n:=\{\gamma \in {\mathcal {H}}_3^*:\textrm{deg}(\gamma )=n\} \text { and } \\{} & {} B_n:=\{\beta \in {\mathcal {H}}_2^*:\exists \gamma \in A_n\exists p^*\in {\mathcal {H}}^*(\gamma \subseteq p^* \wedge \beta \subseteq p^*)\}. \end{aligned}$$

If there are infinitely many $p\in {\mathcal {H}}$ such that $\gamma ^p_0\in A_n$ and $\gamma _1^p\in A_m$ with $n\not =m$ we are done by Claim 1 since we can choose three edges from each of these infinitely many p’s. So we can assume that for every $p\in {\mathcal {H}}$ both, $\gamma _0^p$ and $\gamma _1^p$, have the same degree and we define

$$\begin{aligned} C_n:=\{ p\in {\mathcal {H}}: \{\gamma _0^p,\gamma _1^p\}\subseteq A_n\} \end{aligned}$$

for every $n\in \omega$. Moreover, let

$$\begin{aligned} \textrm{out}(\beta ):=\left\{ \gamma \in \bigcup _{m>n} A_m: g(\beta \cup \gamma )\in \gamma \right\} . \end{aligned}$$

for every $n\in \omega$ and every $\beta \in B_n$. If there is a $\beta \in \bigcup _{n\in \omega } B_n$ with $|\textrm{out}(\beta )|=\infty$ we are done by Claim 1. So assume that $|\textrm{out}(\beta )|\in \omega$ for all $\beta \in \bigcup _{n\in \omega }B_n$.

Claim 2: We can find an infinite subset ${\mathcal {K}}\subseteq {\mathcal {H}}$ with a partition ${\mathcal {K}}=\bigcup _{n\in \omega } K_n$ where each $K_n$ is finite and non-empty. Moreover, we can assume that for all natural numbers $n>m$, all $p\in K_n$, all $q\in K_m$ and all $j\le 2$

$$\begin{aligned} g(\gamma _0^p\cup \beta _j^q)=g(\gamma _1^p\cup \beta _j^q)\in \beta _j^q. \end{aligned}$$

Proof of Claim 2

For every $n\in \omega$ we define $R_n$ to be the set of all $p\in \bigcup _{k>n} C_k$ such that there are a $q\in C_n$, an $i\in \{0,1\}$ and a $j\in \{0,1,2\}$ with

$$\begin{aligned} g(\gamma _i^p\cup \beta _j^q)\in \gamma _i^p. \end{aligned}$$

Since $|\textrm{out}(\beta )|$ is finite for all $\beta \in \bigcup _{n\in \omega } B_n$, the set $R_n$ is finite. Let $J_n:=C_n\setminus R_n$. Define $S_n$ to be the set of all $p\in \bigcup _{k>n}J_k$ such that there are a $q\in J_n$, and a $j\in \{0,1,2\}$ with

$$\begin{aligned} g(\gamma _0^p\cup \beta _j^q)\not = g(\gamma _1^p\cup \beta _j^q). \end{aligned}$$

First of all assume that there is an $n_0\in \omega$ such that $S_{n_0}$ is infinite. Since $J_{n_0}$ is finite, we can then find a $q_0\in J_{n_0}$ and a $j_0\in \{0,1,2\}$ such that for infinitely many $p\in S_{n_0}$

$$\begin{aligned} \beta _{j_0}^{q_0}\ni g(\gamma _0^p\cup \beta _{j_0}^{q_0}) \not = g(\gamma _1^p\cup \beta _{j_0}^{q_0})\in \beta _{j_0}^{q_0} \end{aligned}$$

and we can choose the set of edges $\gamma _0^p$ or $\gamma _1^p$ depending on the choice in $\beta _{j_0}^{p_0}$. With Claim 1 we are done. Therefore, we can assume that each $S_n$ is finite. In this case we define $K_n:=J_n\setminus S_n$ for all $n\in \omega$. Infinitely many sets $K_n$ are non-empty. By renumbering the sets $K_n$ we can assume that each $K_n$ is non-empty. ${\dashv }_{\textrm{Claim}\ 2}$

With the same construction we did in the proof of Claim 2 we can find an infinite subset ${\mathcal {K}}\subseteq {\mathcal {J}}$ with a partition ${\mathcal {K}}=\bigcup _{n\in \omega } K_n,$ where each $K_n$ is finite and non-empty. Moreover, we can assume that for all natural numbers $n>m$, all $p\in I_n$, all $q\in I_m$ and all $j\le 2$

$$\begin{aligned} g(\gamma _0^p\cup \beta _j^q)=g(\gamma _1^p\cup \beta _j^q)\in \beta _j^q. \end{aligned}$$

Note: Up to now we nowhere used the assumption that our infinite family ${\mathcal {F}}$ of sets of size five is linearly ordered. In the last step we will need this assumption.

For each $n\in \omega$, let $p_n\in K_n$ be the smallest element in $K_n$ with respect to the linear order on ${\mathcal {F}}$. Note that such a smallest element exists since each $K_n$ is finite and non-empty. We define

$$\begin{aligned} h^*:\{ p_n^*: n\in \omega \}&\rightarrow \left[ \bigcup _{n\in \omega } p_n^*\right] ^3\\ p_n^*&\mapsto \left\{ g\left( \gamma _0^{p_{n+1}}\cup \beta _j^{p_n}\right) : j \le 2\right\} . \end{aligned}$$

By Claim 1 we are done. $\square$

3.4 $\textbf{RC}_{\varvec{7}}$ implies $\textbf{LOC}_{\varvec{7}}^-$

Before we prove our last result, we shall prove three lemmata.

Lemma 3.8

Let ${\mathcal {F}}$ be a linearly orderable family of pairwise disjoint 6-element sets. Assume that we can partition each $p\in {\mathcal {F}}$ in a unique way into three 2-element sets $\beta _0^p,\beta _1^p$ and $\beta _2^p$ and in a unique way into two 3-element sets $\gamma _0^p,\gamma _1^p$. Further assume that there is a choice function

$$\begin{aligned} f:\left[ \bigcup {\mathcal {F}}\right] ^7\rightarrow \bigcup {\mathcal {F}}. \end{aligned}$$

Then there is an infinite subfamily ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a Kinna-Wagner selection function.

Proof

We define

$$\begin{aligned} {\mathcal {F}}_3:=\big \{\gamma _i^p: i\in \{ 0,1\}\wedge p\in {\mathcal {F}}\big \} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {F}}_4:=\big \{\beta _i^p\cup \beta _j^p: \{i,j\}\in [3]^2 \wedge p\in {\mathcal {F}}\big \}. \end{aligned}$$

For every $\gamma \in {\mathcal {F}}_3$ let

$$\begin{aligned} \textrm{deg}(\gamma ):=\big |\big \{ \delta \in {\mathcal {F}}_4:\delta \cap \gamma =\emptyset \wedge f(\delta \cup \gamma )\in \delta \big \}\big |. \end{aligned}$$

If there is a $\gamma \in {\mathcal {F}}_3$ with $\textrm{deg}(\gamma )=\infty$, then we are done because we can choose between one and three elements from infinitely many $p\in {\mathcal {F}}$. The rest of the proof is similar to the proof of Theorem 3.7. $\square$

Lemma 3.9

Let ${\mathcal {F}}$ be a linearly orderable family of pairwise disjoint 12-element sets. Assume that we can partition each $p\in {\mathcal {F}}$ in a unique way into three 4-element sets $\delta _0,\delta _1$ and $\delta _2$ and in a unique way into four 3-element sets $\gamma _0,\gamma _1,\gamma _2$ and $\gamma _3$. Further assume that there is a choice function

$$\begin{aligned} f:\left[ \bigcup {\mathcal {F}}\right] ^7\rightarrow \bigcup {\mathcal {F}}. \end{aligned}$$

Then there is an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a Kinna-Wagner selection function.

Proof

The proof is similar to the proof of Theorem 3.7. $\square$

Lemma 3.10

Let ${\mathcal {F}}$ be a linearly orderable family of pairwise disjoint 10-element sets. Assume that we can partition each $p\in {\mathcal {F}}$ in a unique way into two 5-element sets $\epsilon _0$ and $\epsilon _1$ and in a unique way into five 2-element sets $\beta _i,~i\le 4$. Further assume that there is a choice function

$$\begin{aligned} f:\left[ \bigcup {\mathcal {F}}\right] ^7\rightarrow \bigcup {\mathcal {F}}. \end{aligned}$$

Then there is an infinite subset ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a Kinna-Wagner selection function.

Proof

The proof is similar to the proof of Theorem 3.7. $\square$

Proposition 3.11

${\textsf {ZF}} \vdash \textrm{RC}_7\Rightarrow \textrm{LOC}_7^-$.

Proof

Let ${\mathcal {F}}$ be a linearly orderable, infinite family of sets of size 7. We apply $\textrm{RC}_7$ on the set $X:=\bigcup {\mathcal {F}}$ to find an infinite subset $Y\subseteq X$ with a choice function $f:[Y]^7\rightarrow Y$. For every $i\le 7$ we define

$$\begin{aligned} {\mathcal {F}}_i:=\big \{ p\in {\mathcal {F}}: |p\cap Y|=i\big \}. \end{aligned}$$

Note that we can without loss of generality assume that ${\mathcal {F}}_2$ or ${\mathcal {F}}_3$ has infinite cardinality.

Case 1: ${\mathcal {F}}_3$ has infinite cardinality.

For every $p\in {\mathcal {F}}_3$ let

$$\begin{aligned} p^*:=\big \{\{ a,x\}\in [p]^2: a\in p\cap Y\wedge x\in p\setminus Y\big \} \end{aligned}$$

and apply $\textrm{RC}_7$ on the set $X^*:=\bigcup \{p^*: p\in {\mathcal {F}}_3\}$. We get an infinite subset $Y^*\subseteq X^*$ with a choice function $g:[Y^*]^7\rightarrow Y^*$. For every $1\le i\le 12$ define

$$\begin{aligned} {\mathcal {F}}_i^*:=\{ p^*: p\in {\mathcal {F}}_3\wedge |p^*\cap Y^*|=i\}. \end{aligned}$$

There is an i with $1\le i\le 12$ such that $|{\mathcal {F}}_i^*|=\infty$. If $i\notin \{ 6,12\}$ we can choose an element from each p with $p^*\in {\mathcal {F}}_i^*$ and therefore we are done. If $i=6$, the only case in which we cannot choose an element from all p with $p^*\in {\mathcal {F}}_6^*$ is the one illustrated in Fig. 3:

But in this case we are done by Lemma 3.8. And if $i=12$ we are done by Lemma 3.9.

Case 2: ${\mathcal {F}}_2$ has infinite cardinality.

For every $1\le i\le 10$ we define ${\mathcal {F}}_i^*$ as in Case 1. The only i for which we cannot choose one element from each p with $p^*\in {\mathcal {F}}_i^*$ or for which we cannot choose three elements from each p with $p^*\in {\mathcal {F}}_i^*$ in order to reduce it to Case 1, is $i=10$. But in this case we are done by Lemma 3.10. $\square$

Fig. 3

Case $i=6$

4 Open questions

• 1. By [6] we have that $\textrm{RC}_n\Rightarrow \textrm{C}_n^-$ in ZF for every $n\in \{2,3,4\}$. Does this implication hold for any other $n\in \omega \setminus \{0,1\}$?

• 2. By [6], Proposition 3.11 and Theorem 3.7 we have that $\textrm{RC}_n\Rightarrow \textrm{LOC}_n^-$ in ZF for any $n\in \{2,3,4,5,7\}$. Does this implication hold for any other $n\in \omega \setminus \{0,1\}$?

• 3. For every $n\in \omega \setminus \{0,1\}$ the following weak choice principle was introduced in [8]:

  $\textrm{nC}_{<\aleph _0}^-$: For every infinite family ${\mathcal {F}}$ of finite sets with cardinality at least n there is an infinite subfamily ${\mathcal {G}}\subseteq {\mathcal {F}}$ with a selection function $f:{\mathcal {G}}\rightarrow \left[ \bigcup {\mathcal {G}}\right] ^n$ such that $f(G)\in [G]^n$ for all $G\in {\mathcal {G}}$.

  Moreover, as in [1] we can define a restricted version of $\textrm{nC}_{<\aleph _0}^-$ as follows:

  $\mathrm{nRC_{fin}}$: Given any infinite set x, there is an infinite subset $y\subseteq x$ and a selection function f that chooses an n-element subset from every $z\subseteq y$ containing at least n elements.

  The relationship of $\textrm{RC}_n$ and $\mathrm{nRC_{fin}}$ to $\textrm{kC}_{<\aleph _0}^-$ and $\textrm{C}_j^-$ has already been studied in [3]. However, the following question is still open: For every $n\in \{2,3,4,6\}$ we have that $\mathrm{nRC_{fin}}\Rightarrow \textrm{nC}_{\aleph _0}^-$ in ZF. Does this implication hold for any other $n\in \omega \setminus \{0,1\}$?