Infinities within Finitely Supported Structures

TL;DR

This paper explores the concept of infinity within finitely supported algebraic structures, extending classical set theory notions to atomic objects with finite support, and characterizes finitely supported countable sets.

Contribution

It introduces and analyzes different types of infinities and cardinalities in finitely supported structures, extending classical properties and providing new characterizations.

Findings

Extended properties of infinite cardinalities to finitely supported structures

Compared alternative definitions of infinite finitely supported sets

Characterized finitely supported countable sets

Abstract

The theory of finitely supported algebraic structures is related to Pitts theory of nominal sets (by equipping finitely supported sets with finitely supported internal algebraic laws). It represents a reformulation of Zermelo Fraenkel set theory obtained by requiring every set theoretical construction to be finitely supported according to a certain action of a group of permutations of some basic elements named atoms. Its main purpose is to let us characterize infinite algebraic structures, defined involving atoms, only by analyzing their finite supports. The first goal of this paper is to define and study different kinds of infinities and the notion of `cardinality' in the framework of finitely supported structures. We present several properties of infinite cardinalities. Some of these properties are extended from the non-atomic Zermelo Fraenkel set theory into the world of atomic objects with finite support, while other properties are specific to finitely supported structures. We also compare alternative definitions of `infinite finitely supported set', and we finally provide a characterization of finitely supported countable sets.

Equations

$$h(x)=\left\{\begin{array}[]{ll}f(x),&\text{for}\>x\in T;\\ g^{-1}(x),&\text{for}\>x\in B-T.\end{array}\right.$$

$$h(x)=\left\{\begin{array}[]{ll}f(x),&\text{for}\>x\in T;\\ g^{-1}(x),&\text{for}\>x\in B_{1}-T.\end{array}\right.$$

$$f(y)=\left\{\begin{array}[]{ll}y,&\text{if}\>\text{$y$ is an injective non-empty tuple};\\ (a),&\text{if}\>\text{$y=\bar{\emptyset}$}\>.\end{array}\right.$$

$$g(y)=\left\{\begin{array}[]{ll}\bar{\emptyset},&\text{if}\>\text{$y$ is a tuple with exactly one element};\\ y^{\prime},&\text{otherwise}\>;\end{array}\right.$$

$$s(U)=\left\{\begin{array}[]{ll}u,&\text{if}\>\text{$U$ is an one-element set \{u\} };\\ y,&\text{if}\>\text{$U$ has more than one element}\>.\end{array}\right.$$

$$f^{\prime}(y)=\left\{\begin{array}[]{ll}f^{-1}(y),&\text{if}\>\text{$y\in Im(f)$ };\\ x_{0},&\text{if}\>\text{$y\notin Im(f)$}\>.\end{array}\right.$$

$$g(u)=\left\{\begin{array}[]{ll}(0,f(x)),&\text{if}\>u=(0,x)\>\text{with}\>x\in X;\\ (1,z),&\text{if}\>u=(1,z)\>\text{with}\>z\in Z.\end{array}\right.$$

$$g(u)=\left\{\begin{array}[]{ll}(x,y_{0}),&\text{if}\>u=(0,x)\>\text{with}\>x\in X,x\neq x_{0};\\ (x_{0},y),&\text{if}\>u=(1,y)\>\text{with}\>y\in Y;\\ (x_{1},y_{1}),&\text{if}\>u=(0,x_{0})\end{array}\right.$$

$$s(X)=\left\{\begin{array}[]{ll}x,&\text{if}\>\text{$X$ is an one-element set $\{x\}$ };\\ a,&\text{if}\>\text{$X$ is not an one-element set,}\end{array}\right.$$

$$i(x)=\left\{\begin{array}[]{ll}\{x,a_{1},a_{2},\ldots,a_{n-1}\},&\text{if}\>\text{$x\neq a_{1},\ldots,a_{n-1}$ };\\ \{a^{1}_{1},\ldots,a^{n}_{1}\},&\text{if}\>\text{$x=a_{1}$ }\\ \vdots\\ \{a^{1}_{n-1},\ldots,a^{n}_{n-1}\},&\text{if}\>\text{$x=a_{n-1}$ }\end{array}\right.$$

$$s(X)=\left\{\begin{array}[]{ll}i^{-1}(X),&\text{if}\>\text{$X\in Im(i)$ };\\ a,&\text{if}\>\text{$X\notin Im(i)$}\end{array}\right.$$

$$g(X)=\left\{\begin{array}[]{ll}X,&\text{if}\>\text{$X\in\wp_{n}(A)$ };\\ \{x_{1},\ldots x_{n}\},&\text{if}\>\text{$X\notin\wp_{n}(A)$}\end{array}\right.$$

$$s(X)=\left\{\begin{array}[]{ll}x,&\text{if}\>\text{$X$ is an one-element set $\{x\}$ };\\ a,&\text{if}\>\text{$X$ is not an one-element set}\end{array}\right.$$

$$g(X)=\left\{\begin{array}[]{ll}X,&\text{if}\>\text{$X\in\wp_{fin}(A)$ };\\ \{b\},&\text{if}\>\text{$X\notin\wp_{fin}(A)$}\end{array}\right.$$

$$s(X)=\left\{\begin{array}[]{ll}a,&\text{if}\>\text{$X$ is an one-element set \{a\} };\\ x,&\text{if}\>\text{$X$ has more than one element}\>.\end{array}\right.$$

$$g(X)=\left\{\begin{array}[]{ll}f^{-1}(X),&\text{if}\>\text{$X\in Im(f)$ };\\ (U,V),&\text{if}\>\text{$X\notin Im(f)$}\>.\end{array}\right.$$

$$f(u)=\left\{\begin{array}[]{ll}\{x\},&\text{if}\>u=(0,x)\>\text{with}\>x\in A;\\ \{a,y\},&\text{if}\>u=(1,y)\>\text{with}\>y\in A,y\neq a;\\ \{b,c\},&\text{if}\>u=(1,a)\end{array}\right.$$

$$g(U)=\left\{\begin{array}[]{ll}U,&\text{if}\>\text{$U\in\wp_{fin}(A)$ };\\ \{a\},&\text{if}\>\text{$U\notin\wp_{fin}(A)$}\>.\end{array}\right.$$

$$g(x)=\left\{\begin{array}[]{ll}f(n+1),&\text{if}\>\text{there exists $n\in\mathbb{N}$ with $x=f(n)$};\\ x,&\text{if}\>\text{$x\notin Im(f)$}\>.\end{array}\right.$$

$$Y_{n}=\left\{\begin{array}[]{ll}X_{n^{\prime}}\setminus\underset{m<n}{\cup}Y_{m},&\text{if}\>\{X_{k}\setminus(X_{n^{\prime}}\cup\underset{m<n}{\cup}Y_{m})\,|\,k>n^{\prime}\}\>\text{is infinite};\\ (X\setminus X_{n^{\prime}})\setminus\underset{m<n}{\cup}Y_{m},&\text{otherwise}.\end{array}\right.$$

$$j(x)=\left\{\begin{array}[]{ll}i^{-1}(x),&\text{if}\>\text{$x\in Im(i)$ };\\ n_{0},&\text{if}\>\text{$x\notin Im(i)$}\>.\end{array}\right.$$

$$h(u)=\left\{\begin{array}[]{ll}f(n)&\text{if}\>u=(n,0)\\ \\ g(n)&\text{if}\>u=(n,1)\end{array}\right.,$$

$$f(n)=\left\{\begin{array}[]{ll}g^{-1}(n)&\text{if}\>n\in g(Y)\\ \\ t&\text{if}\>n\in\mathbb{N}\setminus g(Y)\end{array}\right.,$$

Full text

1 Introduction

The theory of finitely supported algebraic structures which is known under the name of ‘nominal sets’ (when dealing with computer science applications) or ‘Finitely Supported Mathematics’ (in some pure set theoretical papers related to the foundations of mathematics) represents an alternative framework for working with infinite structures hierarchically constructed by involving some basic elements (called atoms) by dealing only with a finite number of entities that form their supports. The theory of nominal sets is presented in a categorical manner as a Zermelo-Fraenkel (ZF) alternative to Fraenkel and Mostowski 1930s permutation models of set theory with atoms [7]. A nominal set is defined as a usual ZF set endowed with a group action of the group of (finitary) permutations over a certain fixed countable ZF set $A$ (also called the set of atoms by analogy with the Fraenkel and Mostowski framework) formed by elements whose internal structure is not taken into consideration (i.e. by elements that can be checked only for equality), satisfying a finite support requirement. This requirement states that for any element in a nominal set there should exist a finite set of atoms such that any permutation fixing pointwise this set of atoms also leaves the element invariant under the related group action. Nominal sets represents a categorical mathematical theory of names studying scope, binding, freshness and renaming in formal languages based upon symmetry. Inductively defined finitely supported sets (that are finitely supported elements in the powerset of a nominal set) involving the name-abstraction together with Cartesian product and disjoint union can encode syntax modulo renaming of bound variables. In this way, the standard theory of algebraic data types can be extended to include signatures involving binding operators. In particular, there is an associated notion of structural recursion for defining syntax-manipulating functions and a notion of proof by structural induction. Various generalizations of nominal were used in order to study automata, languages or Turing machines that operate over infinite alphabets; for this a relaxed notion of finiteness, called ‘orbit finiteness’, was defined and means ‘having a finite number of orbits under a certain group action’ [2].

Finitely Supported Mathematics (FSM) is an alternative name for nominal algebraic structures, used in theoretical papers focused on the foundations of set theory (rather than on applications in computer science). In order to describe FSM as a theory of finitely supported algebraic structures (that is finitely supported sets together with finitely supported internal algebraic laws), we use nominal sets (without the requirement that the set $A$ of atoms is countable) which by now on will be called invariant sets motivated by Tarski’s approach regarding logicality (i.e. a logical notion is defined by Tarski as one that is invariant under the permutations of the universe of discourse). The cardinality of the set of atoms cannot be internally compared with any other ZF cardinality, and so we just say that atoms form an infinite set without any specifications regarding its cardinality. In FSM we actually study the finitely supported subsets of invariant sets together with finitely supported relations (order relations, functions, algebraic laws etc), and so FSM becomes a theory of atomic algebraic structures constructed/defined according to the finite support requirement. The requirement of being finitely supported under a canonical action of the group of permutation of atoms (constructed under the rules in Proposition 2.5) is actually an axiom adjoined to ZF, and so non-finitely supported structures are not allowed (they do not exist) in FSM.

FSM contains the family of ‘non-atomic’ (ordinary) ZF sets (which are proved to be trivial FSM sets) and the family of ‘atomic’ sets with finite supports (hierarchically constructed from the empty set and the fixed ZF set $A$). The main question now is whether a classical ZF result (obtained in ZF framework for non-atomic sets) can be adequately reformulated by replacing ‘non-atomic element/set’ with ‘atomic finitely supported element/set’ (according to the canonical actions of the group of one-to-one transformations of $A$ onto itself) in order to be valid also for atomic sets with finite supports. The (non-atomic) ZF results cannot be directly translated into the framework of atomic finitely supported sets, unless we are able to reprove their new formulations internally in FSM, i.e. by involving only finitely supported structures even in the intermediate steps of the proof. This is because the family of finitely supported sets is not closed under subset constructions, and we cannot use something outside FSM in order to prove something in FSM.

The meta-theoretical techniques for the translation of a result from non-atomic structures to atomic structures are fully described in [1] (or in [7], with the mention that, working on foundations of mathematics, and so we use a slightly different terminology for the same concept). They are based on a refinement of the finite support principle form [7] called “$S$-finite supports principle" claiming that for any finite set $S$ of atoms, anything that is definable in higher order logic from $S$-supported structures using $S$-supported constructions is also $S$-supported. The formal involvement of the $S$-finite support principles implies a constructive method for defining the support of a structure by employing the supports of the sub-structures of a related structure.

In this paper we introduce the notion of ‘cardinality’ of a finitely supported set, and we prove several properties of this concept. Some properties are naturally extended from the non-atomic ZF into the world of atomic structures. In this sense we prove that Cantor-Schröder-Bernstein theorem for cardinalities is still valid in FSM. Several other cardinality properties are preserved from ZF. However, although Cantor-Schröder-Bernstein theorem can be successfully translated into FSM, its ZF dual is no longer valid in FSM. Other specific FSM properties of cardinalities (that do not have related ZF correspondents) are also emphasized. We introduce various definition for infinity and we compare them, providing relevant examples of atomic sets verifying the conditions of each such definition. Finally, we introduce and study the concept of countability in FSM.

2 Finitely Supported Sets

A ZF finite set is referred to a set for which there is a bijection with a finite ordinal; a ZF infinite set is a set that is not finite. Adjoin to ZF a special infinite set $A$ (called ‘the set of atoms’; despite classical set theory with atoms we do not need to modify the axiom of extensionality). Actually, atoms are entities whose internal structure is considered to be irrelevant which are considered as basic for a higher-order construction, i.e. their internal structure is not taken into consideration.

A transposition is a function $(a\,b):A\to A$ given by $(a\,b)(a)=b$, $(a\,b)(b)=a$ and $(a\,b)(n)=n$ for $n\neq a,b$. A (finitary) permutation of $A$ in FSM is a one-to-one transformation of $A$ onto itself (a bijection of $A$) generated by composing finitely many transpositions. We denote by $S_{A}$ the set of all finitary permutations of $A$. According to Proposition 2.6 from [1], a function $f:A\to A$ is a bijection on $A$ in FSM if and only if it leaves unchanged all but finitely many elements of $A$. Thus, in FSM a function is a one-to-one transformation of $A$ onto itself if and only if it is a (finitary) permutation of $A$. Thus, the notions ‘permutation (bijection) of $A$’ and ‘finitary permutation of $A$’ coincide in FSM.

Definition 2.1

Let $X$ be a ZF set.

An $S_{A}$-action on $X$ is a function $\cdot:S_{A}\times X\rightarrow X$ having the properties that $Id\cdot x=x$ and $\pi\cdot(\pi^{\prime}\cdot x)=(\pi\circ\pi^{\prime})\cdot x$ for all $\pi,\pi^{\prime}\in S_{A}$ and $x\in X$, where $Id$ is the identity mapping on $A$. An $S_{A}$-set is a pair $(X,\cdot)$ where $X$ is a ZF set, and $\cdot:S_{A}\times X\to X$ is an $S_{A}$-action on $X$. 2. 2.

Let $(X,\cdot)$ be an $S_{A}$-set. We say that $S\subset A$ supports $x$ whenever for each $\pi\in Fix(S)$ we have $\pi\cdot x=x$, where $Fix(S)=\{\pi\,|\,\pi(a)=a,\forall a\in S\}$. The least finite set supporting $x$ (which exists according to Proposition 2.2) is called the support of $x$ and is denoted by $supp(x)$. An empty supported element is called equivariant; this means that $x\in X$ is equivariant if and only if $\pi\cdot x=x$, $\forall\pi\in S_{A}$. 3. 3.

*Let $(X,\cdot)$ be an $S_{A}$-set. We say that $X$ is an invariant set if for each $x\in X$ there exists a finite set **$S_{x}\subset A$ *which supports $x$.

Proposition 2.2

[1]** Let $X$ be an $S_{A}$-set and let $x\in X$. If there exists a finite set supporting $x$ (particularly, if $X$ is an invariant set), then there exists a least finite set supporting $x$ which is constructed as the intersection of all finite sets supporting $x$.

Proposition 2.3

[1]** Let $(X,\cdot)$ be an $S_{A}$-set, and $\pi\in S_{A}$. If $x\in X$ is finitely supported, then $\pi\cdot x$ is finitely supported and $supp(\pi\cdot x)=\pi(supp(x))$.

Example 2.4

The set $A$ of atoms is an $S_{A}$-set with the $S_{A}$-action $\cdot:S_{A}\times A\rightarrow A$ defined by $\pi\cdot a:=\pi(a)$ for all $\pi\in S_{A}$ and $a\in A$. $(A,\cdot)$ is an invariant set because for each $a\in A$ we have that $\{a\}$ supports $a$. Furthermore, $supp(a)=\{a\}$ for each $a\in A$. 2. 2.

The set $S_{A}$ is an $S_{A}$-set with the $S_{A}$-action $\cdot:S_{A}\times S_{A}\rightarrow S_{A}$ defined by $\pi\cdot\sigma:=\pi\circ\sigma\circ\pi^{-1}$ for all $\pi,\sigma\in S_{A}$. $(S_{A},\cdot)$ is an invariant set because for each $\sigma\in S_{A}$ we have that the finite set $\{a\in A\,|\,\sigma(a)\neq a\}$ supports $\sigma$. Furthermore, $supp(\sigma)=\{a\in A\,|\,\sigma(a)\neq a\}$ for each $\sigma\in S_{A}$. 3. 3.

Any ordinary (non-atomic) ZF-set $X$ (such as $\mathbb{N},\mathbb{Z},\mathbb{Q}$ or $\mathbb{R}$ for example) is an invariant set with the single possible $S_{A}$-action $\cdot:S_{A}\times X\rightarrow X$ defined by $\pi\cdot x:=x$ for all $\pi\in S_{A}$ and $x\in X$.

Proposition 2.5

Let $(X,\cdot)$ and $(Y,\diamond)$ be $S_{A}$-sets.

The Cartesian product $X\times Y$ is also an $S_{A}$-set with the $S_{A}$-action $\otimes:S_{A}\times(X\times Y)\rightarrow(X\times Y)$ defined by $\pi\otimes(x,y)=(\pi\cdot x,\pi\diamond y)$ for all $\pi\in S_{A}$ and all $x\in X$, $y\in Y$. If $(X,\cdot)$ and $(Y,\diamond)$ are invariant sets, then $(X\times Y,\otimes)$ is also an invariant set. 2. 2.

The powerset $\wp(X)=\{Z\,|\,Z\subseteq X\}$ is also an $S_{A}$-set with the $S_{A}$-action $\star:S_{A}\times\wp(X)\rightarrow\wp(X)$ defined by $\pi\star Z:=\{\pi\cdot z\,|\,z\in Z\}$ for all $\pi\in S_{A}$, and all $Z\subseteq X$. For each invariant set $(X,\cdot)$, we denote by $\wp_{fs}(X)$ the set formed from those subsets of $X$ which are finitely supported according to the action $\star$ . $(\wp_{fs}(X),\star|_{\wp_{fs}(X)})$ is an invariant set, where $\star|_{\wp_{fs}(X)}$ represents the action $\star$ restricted to $\wp_{fs}(X)$. 3. 3.

The finite powerset of $X$ $\wp_{fin}(X)=\{Y\subseteq X\,|\,Y\text{finite}\}$ and the cofinite powerset of $X$ $\wp_{cofin}(X)=\{Y\subseteq X\,|\,X\setminus Y\text{finite}\}$ are $S_{A}$-sets with the $S_{A}$-action $\star$ defined as in item 2. If $X$ is an invariant set, then both $\wp_{fin}(X)$ and $\wp_{cofin}(X)$ are invariant sets. 4. 4.

Let $(X,\cdot)$ and $(Y,\diamond)$ be $S_{A}$-sets. We define the disjoint union of $X$ and $Y$ by $X+Y=\{(0,x)\,|\,x\in X\}\cup\{(1,y)\,|\,y\in Y\}$. $X+Y$ is an $S_{A}$-set with the $S_{A}$-action $\star:S_{A}\times(X+Y)\rightarrow(X+Y)$ defined by $\pi\star z=(0,\pi\cdot x)$ if $z=(0,x)$ and $\pi\star z=(1,\pi\diamond y)$ if $z=(1,y)$. If $(X,\cdot)$ and $(Y,\diamond)$ are invariant sets, then $(X+Y,\star)$ is also an invariant set: each $z\in X+Y$ is either of the form $(0,x)$ and supported by the finite set supporting $x$ in $X$, or of the form $(1,y)$ and supported by the finite set supporting $y$ in $Y$.

Definition 2.6

Let $(X,\cdot)$ be an $S_{A}$-set. A subset $Z$ of $X$ is called finitely supported if and only if $Z\in\wp_{fs}(X)$ with the notations from Proposition 2.5. A subset $Z$ of $X$ is uniformly supported if all the elements of $Z$ are supported by the same set $S$ (and so $Z$ is itself supported by $S$ as an element of $\wp_{fs}(X)$). Generally, an FSM set is a finitely supported subset (possibly equivariant) of an invariant set. 2. 2.

Let $(X,\cdot)$ be a finitely supported subset of an $S_{A}$- set $(Y,\cdot)$. A subset $Z$ of $Y$ is called finitely supported subset of $X$ (and we denote this by $Z\in\wp_{fs}(X)$) if and only if $Z\in\wp_{fs}(Y)$ and $Z\subseteq X$. Similarly, we say that a uniformly supported subset of $Y$ contained in $X$ is a uniformly supported subset of $X$.

From Definition 2.1, a subset $Z$ of an invariant set $(X,\cdot)$ is finitely supported by a set $S\subseteq A$ if and only if $\pi\star Z\subseteq Z$ for all $\pi\in Fix(S)$. This is because any permutation of atoms should have finite order.

Proposition 2.7

Let $X$ be a finite subset of an invariant set $(U,\cdot)$. Then $X$ is finitely supported and $supp(X)=\cup\{supp(x)\,|\,x\in X\}$. 2. 2.

Let $X$ be a uniformly supported subset of an invariant set $(U,\cdot)$. Then $X$ is finitely supported and $supp(X)=\cup\{supp(x)\,|\,x\in X\}$.

Proof. 1. Let $X=\left\{x_{1},\ldots,x_{k}\right\}$, and $S=supp(x_{1})\cup\ldots\cup supp(x_{k})$. Obviously, $S$ supports $X$. Indeed, let us consider $\pi\in Fix(S)$. We have that $\pi\in Fix(supp(x_{i}))$ for each $i\in\{1,\ldots,k\}$. Therefore, $\pi\cdot x_{i}=x_{i}$ for each $i\in\{1,\ldots,k\}$ because $supp(x_{i})$ supports $x_{i}$ for each $i\in\{1,\ldots,k\}$, and so $supp(X)\subseteq S$. It remains to prove that $S\subseteq supp(X)$. Consider $a\in S$. This means there exists $j\in\{1,\ldots,k\}$ such that $a\in supp(x_{j})$. Let $b$ be an atom such that $b\notin supp(X)$ and $b\notin supp(x_{i})$, $\forall i\in\{1,\ldots,k\}$. Such an atom exists because $A$ is infinite, while $supp(X)$ and $supp(x_{i})$, $i\in\{1,\ldots,k\}$, are all finite. We prove by contradiction that $(b\;a)\cdot x_{j}\notin X$. Indeed, suppose that $(b\;a)\cdot x_{j}\in X$. In this case there is $y\in X$ with $(b\;a)\cdot x_{j}=y$. Since $a\in supp(x_{j})$, we have $b\in(b\;a)(supp(x_{j}))$. However, according to Proposition 2.3, we have $supp(y)=(b\;a)(supp(x_{j}))$. We obtain that $b\in supp(y)$ for some $y\in X$, which is a contradiction with the choice of $b$. Therefore, $(b\;a)\star X\neq X$, where $\star$ is the standard $S_{A}$-action on $\wp(U)$ is defined in Proposition 2.5(2). Since $b\notin supp(X)$, we prove by contradiction that $a\in supp(X)$. Indeed, suppose that $a\notin supp(X)$. It follows that the transposition $(b\;a)$ fixes each element from $supp(X)$, i.e. $(b\;a)\in Fix(supp(X))$. Since $supp(X)$ supports $X$, by Definition 2.1, it follows that $(b\;a)\star X=X$, which is a contradiction. Thus, $a\in supp(X)$, and so $S\subseteq supp(X)$.

2. Since $X$ is uniformly supported, there exists a finite subset of atoms $T$ such that $T$ supports every $x\in X$, i.e. $supp(x)\subseteq T$ for all $x\in X$. Thus, $\cup\{supp(x)\,|\,x\in X\}\subseteq T$. Clearly, $supp(X)\subseteq\cup\{supp(x)\,|\,x\in X\}$. Conversely, let $a\in\cup\{supp(x)\,|\,x\in X\}$. Thus, there exists $x_{0}\in X$ such that $a\in supp(x_{0})$. Let $b$ be an atom such that $b\notin supp(X)$ and $b\notin T$. Such an atom exists because $A$ is infinite, while $supp(X)$ and $T$ are both finite. We prove by contradiction that $(b\;a)\cdot x_{0}\notin X$. Indeed, suppose that $(b\;a)\cdot x_{0}=y\in X$. Since $a\in supp(x_{0})$, we have $b=(b\;a)(a)\in(b\;a)(supp(x_{0}))=supp((b\;a)\cdot x_{0})=supp(y)$. Since $supp(y)\subseteq T$, we get $b\in T$, a contradiction. Therefore, $(b\;a)\star X\neq X$. Since $b\notin supp(X)$, we have that $a\in supp(X)$ as in the above item.

Corollary 2.8

Let $X$ be a uniformly supported subset of an invariant set. Then $X$ is uniformly supported by $supp(X)$.

Proof. Since $supp(X)=\cup\{supp(x)\,|\,x\in X\}$, we have $supp(x)\subseteq supp(X)$ for all $x\in X$ which means $supp(X)$ supports every $x\in X$.

Proposition 2.9

We have $\wp_{fs}(A)=\wp_{fin}(A)\cup\wp_{cofin}(A)$.

Proof. We know that $B$ is finitely supported with $supp(B)=B$ whenever $B\subset A$ and $B$ is finite. If $C\subseteq A$ and $C$ is cofinite, then $C$ is finitely supported by $A\setminus C$ with $supp(C)=A\setminus C$. However, if $D\subsetneq A$ is neither finite nor cofinite, then $D$ is not finitely supported. Indeed, assume by contradiction that there exists a finite set of atoms $S$ supporting $D$. Since $S$ is finite and both $D$ and its complementary $C_{D}$ are infinite, we can take $a\in D\setminus S$ and $b\in C_{D}\setminus S$. Then the transposition $(a\,b)$ fixes $S$ pointwise, but $(a\,b)\star D\neq D$ because $(a\,b)(a)=b\notin D$; this contradicts the assertion that $S$ supports $D$. Therefore, $\wp_{fs}(A)=\wp_{fin}(A)\cup\wp_{cofin}(A)$.

Definition 2.10

Let $X$ and $Y$ be invariant sets.

A function $f:X\rightarrow Y$ is finitely supported if $f\in\wp_{fs}(X\times Y)$. The set of all finitely supported functions from $X$ to $Y$ is denoted by $Y^{X}_{fs}$. 2. 2.

Let $Z$ be a finitely supported subset of $X$ and $T$ a finitely supported subset of $Y$. A function $f:Z\rightarrow T$ is finitely supported if $f\in\wp_{fs}(X\times Y)$. The set of all finitely supported functions from $Z$ to $T$ is denoted by $T^{Z}_{fs}$.

Proposition 2.11

[1]** Let $(X,\cdot)$ and $(Y,\diamond)$ be two invariant sets.

$Y^{X}$* (i.e. the set of all functions from $X$ to $Y$) is an $S_{A}$-set with the $S_{A}$-action $\widetilde{\star}:S_{A}\times Y^{X}\rightarrow Y^{X}$ defined by $(\pi\widetilde{\star}f)(x)=\pi\diamond(f(\pi^{-1}\cdot x))$ for all $\pi\in S_{A}$, $f\in Y^{X}$ and $x\in X$. A function $f:X\rightarrow Y$ is finitely supported in the sense of Definition 2.10 if and only if it is finitely supported with respect the permutation action $\widetilde{\star}$.* 2. 2.

Let $Z$ be a finitely supported subset of $X$ and $T$ a finitely supported subset of $Y$. A function $f:Z\rightarrow T$ is supported by a finite set $S\subseteq A$ if and only if for all $x\in Z$ and all $\pi\in Fix(S)$ we have $\pi\cdot x\in Z$, $\pi\diamond f(x)\in T$ and $f(\pi\cdot x)=\pi\diamond f(x)$. Particularly, a function $f:X\rightarrow Y$ is supported by a finite set $S\subseteq A$ if and only if for all $x\in X$ and all $\pi\in Fix(S)$ we have $f(\pi\cdot x)=\pi\diamond f(x)$.

3 Cardinalities and Order Properties

Definition 3.1

• •

An invariant partially ordered set (invariant poset) is an invariant set $(P,\cdot)$ together with an equivariant partial order relation $\sqsubseteq$ on $P$. An invariant poset is denoted by $(P,\sqsubseteq,\cdot)$ or simply $P$.

• •

A finitely supported partially ordered set (finitely supported poset) is a finitely supported subset $X$ of an invariant set $(P,\cdot)$ together with a partial order relation $\sqsubseteq$ on $X$ that is finitely supported as a subset of $P\times P$.

Two FSM sets $X$ and $Y$ are called equipollent if there exists a finitely supported bijection $f:X\to Y$. The FSM cardinality of $X$ is defined as the equivalence class of all FSM sets equipollent to $X$, and is denoted by $|X|$. This means that for two FSM sets $X$ and $Y$ we have $|X|=|Y|$ if and only if there exists a finitely supported bijection $f:X\to Y$. On the family of cardinalities we can define the relations:

• •

$\leq$ by:  $|X|\leq|Y|$ if and only if there is a finitely supported injective mapping $f:X\to Y$;

• •

$\leq^{*}$ by: $|X|\leq^{*}|Y|$ if and only if there is a finitely supported surjective mapping $f:Y\to X$.

Theorem 3.2

The relation $\leq$ is equivariant, reflexive, anti-symmetric and transitive, but it is not total. 2. 2.

The relation $\leq^{*}$ is equivariant, reflexive and transitive, but it is not anti-symmetric, nor total.

Proof.

• •

$\leq$ and $\leq^{*}$ are equivariant because for any FSM sets $X$ and $Y$, whenever there is a finitely supported injection/ surjection $f:X\to Y$, according to Proposition 2.3, we have that $\pi\star f:\pi\star X\to\pi\star Y$, defined by $(\pi\star f)(\pi\cdot x)=\pi\cdot f(x)$ for all $x\in X$, is a finitely supported injective/surjective mapping, and so $\pi\star X$ is comparable with $\pi\star Y$ (under $\leq$ or $\leq^{*}$, after case).

• •

$\leq$ and $\leq^{*}$ are obviously reflexive because for each FSM set $X$, the identity of $X$ is an equivariant bijection from $X$ to $X$.

• •

$\leq$ and $\leq^{*}$ are transitive because for any FSM sets $X$, $Y$ and $Z$, whenever there are two finitely supported injections/surjections $f:X\to Y$ and $g:Y\to Z$, there exists an injection/surjection $g\circ f:X\to Z$ which is finitely supported by $supp(f)\cup supp(g)$.

• •

The anti-symmetry of $\leq$.

Lemma 3.3

Let $(B,\cdot)$ and   $(C,\diamond)$ be two invariant sets. If there exist a finitely supported injective mapping $f:B\to C$ and a finitely supported injective mapping $g:C\to B$, then there exists a finitely supported bijective mapping $h:B\to C$. Furthermore, $supp(h)\subseteq supp(f)\cup supp(g)$.

Proof of Lemma 3.3. Let us define $F:\wp_{fs}(B)\to\wp_{fs}(B)$ by $F(X)=B-g(C-f(X))$ for all finitely supported subsets $X$ of $B$.

Claim 1: $F$ is correctly defined, i.e. $Im(F)\subseteq\wp_{fs}(B)$.

For every finitely supported subset $X$ of $B$, we have that $f(X)$ is supported by $supp(f)\cup supp(X)$. Indeed, let $\pi\in Fix(supp(f)\cup supp(X))$. Let $y$ be an arbitrary element from $f(X)$; then $y=f(x)$ for some $x\in X$. However, because $\pi\in Fix(supp(X))$, it follows that $\pi\cdot x\in X$ and so, because $supp(f)$ supports $f$ and $\pi$ fixes $supp(f)$ pointwise, from Proposition 2.11 we get $\pi\diamond y=\pi\diamond f(x)=f(\pi\cdot x)\in f(X)$. Thus $\pi\widetilde{\star}f(X)=f(X)$, where $\widetilde{\star}$ is the $S_{A}$-action on $\wp_{fs}(C)$ defined as in Proposition 2.5. Analogously, $g(Y)$ is finitely supported by $supp(g)\cup supp(Y)$ for all $Y\in\wp_{fs}(C)$. It is easy to remark that for every finitely supported subset $X$ of $B$ we have that $C-f(X)$ is also supported by $supp(f)\cup supp(X)$, $g(C-f(X))$ is supported by $supp(g)\cup supp(f)\cup supp(X)$, and $B-g(C-f(X))$ is supported by $supp(g)\cup supp(f)\cup supp(X)$. Thus, $F$ is well-defined.

Claim 2: $F$ is a finitely supported function.

We prove that $F$ is finitely supported by $supp(f)\cup supp(g)$. Let us consider $\pi\in Fix(supp(f)\cup supp(g))$. Since $\pi\in Fix(supp(f))$ and $supp(f)$ supports $f$, according to Proposition 2.11 we have that $f(\pi\cdot x)=\pi\diamond f(x)$ for all $x\in B$. Thus, for every finitely supported subset $X$ of $B$ we have $f(\pi\star X)=\{f(\pi\cdot x)\;|\;x\in X\}=\{\pi\diamond f(x)\;|\;x\in X\}=\pi\widetilde{\star}f(X)$, where $\star$ is the $S_{A}$-action on $\wp_{fs}(B)$ and $\widetilde{\star}$ is the $S_{A}$-action on $\wp_{fs}(C)$. Similarly, $g(\pi\widetilde{\star}Y)=\pi\star g(Y)$ for any finitely supported subset $Y$ of $C$. Therefore, $F(\pi\star X)=B-g(C-f(\pi\star X))=B-g(C-\pi\widetilde{\star}f(X))\overset{\pi\widetilde{\star}C=C}{=}B-g(\pi\widetilde{\star}(C-f(X)))=B-(\pi\star g(C-f(X)))\overset{\pi\star B=B}{=}\pi\star(B-g(C-f(X)))=\pi\star F(X)$. From Proposition 2.11 it follows that $F$ is finitely supported. Moreover, because $supp(F)$ is the least set of atoms supporting $F$, we have $supp(F)\subseteq supp(f)\cup supp(g)$.

Claim 3: For any $X,Y\in\wp_{fs}(B)$ with $X\subseteq Y$, we have $F(X)\subseteq F(Y)$. This remark follows by direct calculation.

Claim 4: The set $S:=\{X\;|\;X\in\wp_{fs}(B),X\subseteq F(X)\}$ is a non-empty finitely supported subset of $\wp_{fs}(B)$. Obviously, $\emptyset\in S$. We claim that $S$ is supported by $supp(F)$. Let $\pi\in Fix(supp(F))$, and $X\in S$. Then $X\subseteq F(X)$. From the definition of $\star$ (see Proposition 2.5) we have $\pi\star X\subseteq\pi\star F(X)$. According to Proposition 2.11, because $supp(F)$ supports $F$, we have $\pi\star X\subseteq\pi\star F(X)=F(\pi\star X)$, and so $\pi\star X\in S$. It follows that $S$ is finitely supported, and $supp(S)\subseteq supp(F)$.

Claim 5: $T:=\underset{X\in S}{\cup}X$ is finitely supported by $supp(S)$.

Let $\pi\in Fix(supp(S))$, and $t\in T$. Since $T=\underset{X\in S}{\cup}X$, we have that there exists $Z\in S$ such that $t\in Z$. Therefore, $\pi\cdot t\in\pi\star Z$. However, since $\pi$ fixes $supp(S)$ pointwise and $supp(S)$ supports $S$, we have that $\pi\star Z\in S$. Thus, there exists $Y\in S$ such that $\pi\star Z=Y$. Therefore $\pi\cdot t\in Y$, and so $\pi\cdot t\in\underset{X\in S}{\cup}X$. It follows that $\underset{X\in S}{\cup}X$ is finitely supported, and so $T=\underset{X\in S}{\cup}X\in\wp_{fs}(B)$. Furthermore, $supp(T)\subseteq supp(S)$.

Claim 6: We prove that $F(T)=T$.

Let $X\in S$ arbitrary. We have $X\subseteq F(X)\subseteq F(T)$. By taking the supremum on $S$, this leads to $T\subseteq F(T)$. However, because $T\subseteq F(T)$, from Claim 3 we also have $F(T)\subseteq F(F(T))$. Furthermore, $F(T)$ is supported by $supp(F)\cup supp(T)$ (i.e. by $supp(f)\cup supp(g)$), and so $F(T)\in S$. According to the definition of $T$, we get $F(T)\subseteq T$.

We get $T=B-g(C-f(T))$, or equivalently, $B-T=g(C-f(T))$. Since $g$ is injective, we obtain that for each $x\in B-T$, $g^{-1}(x)$ is a set containing exactly one element. Let us define $h:B\to C$ by

[TABLE]

Claim 7: We claim that $h$ is supported by the set $supp(f)\cup supp(g)\cup supp(T)$ (more exactly, by $supp(f)\cup supp(g)$, according to the previous claims). Let $\pi\in Fix(supp(f)\cup supp(g)\cup supp(T))$, and $x$ an arbitrary element of $B$.

If $x\in T$, because $\pi\in Fix(supp(T))$ and $supp(T)$ supports $T$, we have $\pi\cdot x\in T$. Thus, from Proposition 2.11 we get $h(\pi\cdot x)=f(\pi\cdot x)=\pi\diamond f(x)=\pi\diamond h(x)$.

If $x\in B-T$, we have $\pi\cdot x\in B-T$. Otherwise, we would obtain the contradiction $x=\pi^{-1}\cdot(\pi\cdot x)\in T$ because $\pi^{-1}$ also fixes $supp(T)$ pointwise. Thus, because $g$ is finitely supported, according to Proposition 2.11 we have $h(\pi\cdot x)=g^{-1}(\pi\cdot x)=\{y\in C\;|\;g(y)$=$\pi\cdot x\}=\{y\in C\;|\;\pi^{-1}\cdot g(y)$=$x\}=\{y\in C\;|\;g(\pi^{-1}\diamond y)$= $x\}\overset{\pi^{-1}\diamond y:=z}{=}\{\pi\diamond z\in C\;|\;g(z)$=$x\}=\pi\diamond\{z\in C\;|\;g(z)$=$x\}=\pi\diamond g^{-1}(x)=\pi\diamond h(x)$. We obtained $h(\pi\cdot x)=\pi\diamond h(x)$ for all $\pi\in Fix(supp(f)\cup supp(g)\cup supp(T))$ and all $x\in B$. According to Proposition 2.11, we get that $h$ is finitely supported. Furthermore, we also have that supp(h)\subseteq supp(f)\cup supp(g)\cup supp(T)\overset{Claim\;5}{\subseteq}supp(f)\cup supp(g)\cup supp(S)$$\overset{Claim\;4}{\subseteq}supp(f)\cup supp(g)\cup supp(F)\overset{Claim\;2}{\subseteq}supp(f) $\cup supp(g)$.

Claim 8: $h$ is a bijective function.

First we prove that $h$ is injective. Let us suppose that $h(x)=h(y)$. We claim that either $x,y\in T$ or $x,y\in B-T$. Indeed, let us suppose that $x\in T$ and $y\notin T$ (the case $x\notin T$, $y\in T$ is similar). We have $h(x)=f(x)$ and $h(y)=g^{-1}(y)$. If we denote $g^{-1}(y)=z$, we have $g(z)=y$. However, we supposed that $y\in B-T$, and so there exists $u\in C-f(T)$ such that $y=g(u)$. Since $y=g(z)$, from the injectivity of $g$ we get $u=z$. This is a contradiction because $u\notin f(T)$, while $z=f(x)\in f(T)$. Since we proved that both $x,y$ are contained either in $T$ or in $B-T$, the injectivity of $h$ follows from the injectivity of $f$ or $g$, respectively.

Now we prove that $h$ is surjective. Let $y\in C$ be arbitrarily chosen. If $y\in f(T)$, then there exists $z\in T$ such that $y=f(z)$, and so $y=h(z)$. If $y\in C-f(T)$, and because $g(C-f(T))=B-T$, there exists $x\in B-T$ such that $g(y)=x$. Thus, $y\in g^{-1}(x)$. Since $g$ is injective, and so $g^{-1}(x)$ is a one-element set, we can say that $g^{-1}(x)=y$ with $x\in B-T$. Thus we have $y=h(x)$.

Lemma 3.4

Let $(B,\cdot)$ and   $(C,\diamond)$ be two invariant sets (in particular, $B$ and $C$ could coincide), $B_{1}$ a finitely supported subset of $B$ and $C_{1}$ a finitely supported subset of $C$. If there exist a finitely supported injective mapping $f:B_{1}\to C_{1}$ and a finitely supported injective mapping $g:C_{1}\to B_{1}$, then there exists a finitely supported bijective mapping $h:B_{1}\to C_{1}$. Furthermore, $supp(h)\subseteq supp(f)\cup supp(g)\cup supp(B_{1})\cup supp(C_{1})$.

Proof of Lemma 3.4. We follow the proof of Lemma 3.3. We define $F:\wp_{fs^{\prime}}(B_{1})\to\wp_{fs^{\prime}}(B_{1})$ by $F(X)=B_{1}-g(C_{1}-f(X))$ for all $X\in\wp_{fs^{\prime}}(B_{1})$, where $\wp_{fs^{\prime}}(B_{1})$ is a finitely supported subset of the invariant set $\wp_{fs}(B)$ (supported by $supp(B_{1})$) defined by $\wp_{fs^{\prime}}(B_{1})=\{X\in\wp_{fs}(B)\;|\;X\subseteq B_{1}\}$. As in the previous lemma, but using Proposition 2.11, we get that $F$ is well-defined, i.e. for every $X\in\wp_{fs^{\prime}}(B_{1})$ we have that $F(X)$ is supported by $supp(f)\cup supp(g)\cup supp(B_{1})\cup supp(C_{1})\cup supp(X)$ which means $F(X)\in\wp_{fs^{\prime}}(B_{1})$. Moreover, $F$ is itself finitely supported (in the sense of Definition 2.10) by $supp(f)\cup supp(g)\cup supp(B_{1})\cup supp(C_{1})$. The set $S:=\{X\;|\;X\in\wp_{fs^{\prime}}(B_{1}),\ X\subseteq F(X)\}$ is contained in $\wp_{fs^{\prime}}(B_{1})$ and it is supported by $supp(F)$ as a subset of $\wp_{fs}(B)$. The set $T:=\underset{X\in S}{\cup}X\in\wp_{fs^{\prime}}(B_{1})$ is finitely supported by $supp(S)$, and it is a fixed point of $F$.

As in the proof of Lemma 3.3, we define the bijection $h:B_{1}\to C_{1}$ by

[TABLE]

According to Proposition 2.11, we obtain that $h$ is finitely supported by $supp(f)\cup supp(g)\cup supp(B_{1})\cup supp(C_{1})\cup supp(T)$, and $supp(h)\subseteq supp(f)\cup supp(g)\cup supp(B_{1})\cup supp(C_{1})$. Thus, $h$ is the required finitely supported bijection between $B_{1}$ and $C_{1}$.

The anti-symmetry of $\leq$ follows from Lemma 3.3 and Lemma 3.4 because FSM sets are actually finitely supported subsets of invariant sets. It is worth noting that $\leq^{*}$ is not anti-symmetric.

Lemma 3.5

There are two invariant sets $B$ and $C$ such that there exist both a finitely supported surjective mapping $f:C\to B$ and a finitely supported surjective mapping $g:B\to C$, but it does not exist a finitely supported bijective mapping $h:B\to C$.

Proof of Lemma 3.5. Let us consider the invariant set $(A,\cdot)$ of atoms. The family $T_{fin}(A)=\{(x_{1},\ldots,x_{m})\subseteq(A\times\ldots\times A)\,|\,m\geq 0\}$ of all finite injective tuples from $A$ (including the empty tuple denoted by $\bar{\emptyset}$) is an $S_{A}$-set with the $S_{A}$-action $\star:S_{A}\times T_{fin}(A)\rightarrow T_{fin}(A)$ defined by $\pi\star(x_{1},\ldots,x_{m})=(\pi\cdot x_{1},\ldots,\pi\cdot x_{m})$ for all $(x_{1},\ldots,x_{m})\in T_{fin}(A)$ and all $\pi\in S_{A}$. Since $A$ is an invariant set, we have that $T_{fin}(A)$ is an invariant set. Whenever $X$ is an invariant set, we have that each injective tuple $(x_{1},\ldots,x_{m})$ of elements belonging to $X$ is finitely supported, and, furthermore, $supp(x_{1},\ldots,x_{m})=supp(x_{1})\cup\ldots\cup supp(x_{m})$. Particularly, we obtain that $supp(a_{1},\ldots,a_{m})=\{a_{1},\ldots,a_{m}\}$, for any injective tuple of atoms $(a_{1},\ldots,a_{m})$ (similarly as in Proposition 2.2 from [1]).

Since $supp(\bar{\emptyset})=\emptyset$, it follows that $T^{*}_{fin}(A)=T_{fin}(A)\setminus\bar{\emptyset}$ is an equivariant subset of $T_{fin}(A)$, and is itself an invariant set. Let us fix an atom $a\in A$. We define $f:T_{fin}(A)\to T_{fin}(A)\setminus\bar{\emptyset}$ by

[TABLE]

Clearly, $f$ is surjective. We claim that $f$ is supported by $supp(a)$. Let $\pi\in Fix(supp(a))$, i.e. $a=\pi(a)=\pi\star(a)$. If $y$ is a non-empty tuple of atoms, we obviously have $f(\pi\star y)=\pi\star y=\pi\star f(y)$. If $y=\bar{\emptyset}$, we have $\pi\star y=\bar{\emptyset}$, and so $f(\pi\star y)=(a)=\pi(a)=\pi\star f(y)$. Thus, $f(\pi\star y)=\pi\star f(y)$ for all $y\in T_{fin}(A)$. According to Proposition 2.11, we have that $f$ is finitely supported.

We define an equivariant surjective function $g:T_{fin}(A)\setminus\bar{\emptyset}\to T_{fin}(A)$ by

[TABLE]

where $y^{\prime}$ is a new tuple formed by deleting the first element in tuple $y$ (the first position in a finite injective tuple exists without requiring any form of choice). Clearly, $g$ is surjective. Indeed, $\bar{\emptyset}=g((a))$ for some one-element tuple $(a)$ ($A$ is non-empty, and so it has at least one atom). For a fixed finite injective non-empty $m$-tuple $y$, we have that $y$ can be seen as being “contained" in an injective $(m+1)$-tuple $z$ of form $(b,y)$ (whose first element is a certain atom $b$, and the following elements are precisely the elements of $y$). The related atom $b$ exists because $y$ is finite, while $A$ is infinite (generally, we can always find an atom $b\notin supp(y)=\{y\}$ according to the finite support requirement in FSM - more details in Section 2.9 of [1]). We get $y=g(z)$. For proving the surjectivity of $g$ we do not need to ‘choose’ a precise such an element $b$ (we do not need to define an inverse function for $g$); it is sufficient to ascertain that $g(b,y)=y$ for every $b\in A\setminus\{y\}$ and $A\setminus\{y\}$ is non-empty (the axiom of choice is not required because for proving only the surjectivity of $g$ we do not involve the construction of a system of representatives for the family $(g^{-1}(y))_{y\in T_{fin}(A)}$).

We claim now that $g$ is equivariant. Let $(x)$ be a one-element tuple from $A$ and $\pi$ an arbitrary permutation from $S_{A}$. We have that $\pi\star(x)=(\pi(x))$ is a one-element tuple from $A$, and so $g(\pi\star(x))=\bar{\emptyset}=\pi\star\bar{\emptyset}=\pi\star g((x))$. Now, let us consider $(x_{1},\ldots,x_{m})\in T_{fin}(A),m\geq 2$ and $\pi\in S_{A}$. We have $g(\pi\star(x_{1},\ldots,x_{m}))=g((\pi\cdot x_{1},\ldots,\pi\cdot x_{m}))=g((\pi(x_{1}),\ldots,\pi(x_{m})))=(\pi(x_{2}),\ldots,\pi(x_{m}))=\pi\star(x_{2},\ldots,x_{m})=\pi\star g(x_{1},\ldots,x_{m})$. According to Proposition 2.11, we have that $g$ is empty-supported (equivariant).

We prove by contradiction that there could not exist a finitely supported injective $h:T_{fin}(A)\to T_{fin}(A)\setminus\bar{\emptyset}$. Let us suppose there is a finitely supported injection $h:T_{fin}(A)\rightarrow T_{fin}(A)\setminus\bar{\emptyset}$. We have $\bar{\emptyset}\notin Im(h)$ because $Im(h)\subseteq T_{fin}(A)\setminus\bar{\emptyset}$. We can form an infinite sequence $\mathcal{F}$ which has the first term $y_{0}=\bar{\emptyset}$, and the general term $y_{n+1}=h(y_{n})$ for all $n\in\mathbb{N}$. Since $\bar{\emptyset}\notin Im(h)$, it follows that $\bar{\emptyset}\neq h(\bar{\emptyset})$. Since $h$ is injective and $\bar{\emptyset}\notin Im(h)$, we obtain by induction that $h^{n}(\bar{\emptyset})\neq h^{m}(\bar{\emptyset})$ for all $n,m\in\mathbb{N}$ with $n\neq m$.

We prove now that for each $n\in\mathbb{N}$ we have that $y_{n+1}$ is supported by $supp(h)\cup supp(y_{n})$. Let $\pi\in Fix(supp(h)\cup supp(y_{n}))$. According to Proposition 2.11, because $\pi\in Fix(supp(h))$ we have $h(\pi\star y_{n})=\pi\star h(y_{n})$. Since $\pi\in Fix(supp(y_{n}))$ we have $\pi\star y_{n}=y_{n}$, and so $h(y_{n})=\pi\star h(y_{n})$. Thus, $\pi\star y_{n+1}=\pi\star h(y_{n})=h(y_{n})=y_{n+1}$. Furthermore, because $supp(y_{n+1})$ is the least set supporting $y_{n+1}$, we have $supp(y_{n+1})\subseteq supp(h)\cup supp(y_{n})$ for all $n\in\mathbb{N}$. Since each $y_{n}$ is a finite injective tuple of atoms, it follows that $supp(y_{n})=\{y_{n}\}$ for all $n\in\mathbb{N}$ (where by $\{y_{n}\}$ we denoted the set of atoms forming $y_{n}$). We get $\{y_{n+1}\}=supp(y_{n+1})\subseteq supp(h)\cup supp(y_{n})=supp(h)\cup\{y_{n}\}$. By repeatedly applying this result, we get $\{y_{n}\}\subseteq supp(h)\cup\{y_{0}\}=supp(h)\cup\emptyset=supp(h)$ for all $n\in\mathbb{N}$. Since $supp(h)$ has only a finite number of subsets, we contradict the statement that the infinite sequence $(y_{n})_{n}$ never repeats. Thus, there does not exist a finitely supported bijection between $T_{fin}(A)\setminus\bar{\emptyset}$ and $T_{fin}(A)$.

• •

$\leq$ and $\leq^{\star}$ are not total.

We prove that whenever $X$ is an infinite ordinary (non-atomic) ZF-set, for any finitely supported function $f:A\to X$ and any finitely supported function $g:X\to A$, $Im(f)$ and $Im(g)$ are finite. As a direct consequence there are no finitely supported injective mappings and no finitely supported surjective mappings between $A$ and $X$.

Let us consider a finitely supported mapping $f:A\to X$. Let let us fix an element $b\in A$ with $b\notin supp(f)$. Let $c$ be an arbitrary element from $A\setminus supp(f)$. Since $b\notin supp(f)$, we have that $(b\,c)$ fixes every element from $supp(f)$, i.e. $(b\,c)\in Fix(supp(f))$. However, $supp(f)$ supports $f$, and so, by Proposition 2.11, we have $f((b\,c)(a))=(b\,c)\diamond f(a)=f(a)$ for all $a\in A$. In particular, $f(c)=f((b\,c)(b))=f(b)$. Since $c$ has been chosen arbitrarily from $A\setminus supp(f)$, it follows that $f(c)=f(b)$, for all $c\in A\setminus supp(f)$. If $supp(f)=\{a_{1},\ldots,a_{n}\}$, then $Im(f)=\{f(a_{1})\}\cup\ldots\cup\{f(a_{n})\}\cup\{f(b)\}$. Thus, $Im(f)$ is finite (because it is a finite union of singletons).

Let $g:X\to A$ be a finitely supported function. Assume by contradiction that $Im(g)$ is infinite. Pick any atom $a\in Im(g)\setminus supp(g)$ (such an atom exists because $supp(g)$ is finite). There exists an $x\in X$ such that $g(x)=a$. Now pick any atom $b\in Im(g)\setminus(supp(g)\cup\{a\})$, The transposition $(a\,b)$ fixes $supp(g)$ pointwise, and so $g(x)=g((a\,b)\diamond x)=(a\,b)\cdot g(x)=(a\,b)(a)=b$, contradicting the fact that $g$ is a function. Thus, $Im(g)$ is finite.

Corollary 3.6

There exist two invariant sets $B$ and $C$ such that there is a finitely supported bijection between $\wp_{fs}(B)$ and $\wp_{fs}(C)$, but there is no finitely supported bijection between $B$ and $C$.

Proof. Firstly we prove the following lemma.

Lemma 3.7

Let $X$ and $Y$ be two FSM sets and $f:X\to Y$ a finitely supported surjective function. Then the mapping $g:\wp_{fs}(Y)\to\wp_{fs}(X)$ defined by $g(V)=f^{-1}(V)$ for all $V\in\wp_{fs}(Y)$ is well defined, injective and finitely supported by $supp(f)\cup supp(X)\cup supp(Y)$.

Proof of Lemma 3.7. Let $V$ be an arbitrary element from $\wp_{fs}(Y)$. We claim that $f^{-1}(V)\in\wp_{fs}(X)$. Indeed we prove that the set $f^{-1}(V)$ is supported by $supp(f)\cup supp(V)\cup supp(X)\cup supp(Y)$. Let $\pi\in Fix(supp(f)\cup supp(V)\cup supp(X)\cup supp(Y))$, and $x\in f^{-1}(V)$. This means $f(x)\in V$. According to Proposition 2.11, and because $\pi$ fixes $supp(f)$ pointwise and $supp(f)$ supports $f$, we have $f(\pi\cdot x)=\pi\cdot f(x)\in\pi\star V=V$, and so $\pi\cdot x\in f^{-1}(V)$ (we denoted the actions on $X$ and $Y$ generically by $\cdot$, and the actions on their powersets by $\star$). Therefore, $f^{-1}(V)$ is finitely supported, and so the function $g$ is well defined. We claim that $g$ is supported by $supp(f)\cup supp(X)\cup supp(Y)$. Let $\pi\in Fix(supp(f)\cup supp(X)\cup supp(Y))$. For any arbitrary $V\in\wp_{fs}(Y)$ we get $\pi\star V\in\wp_{fs}(Y)$ and $\pi\star g(V)\in\wp_{fs}(X)$, and by Proposition 2.11 we have that $\pi^{-1}\in Fix(supp(f))$, and so $f(\pi^{-1}\cdot x)=\pi^{-1}\cdot f(x)$ for all $x\in X$. For any arbitrary $V\in\wp_{fs}(Y)$, we have that $z\in g(\pi\star V)=f^{-1}(\pi\star V)\Leftrightarrow f(z)\in\pi\star V\Leftrightarrow\pi^{-1}\cdot f(z)\in V\Leftrightarrow f(\pi^{-1}\cdot z)\in V\Leftrightarrow\pi^{-1}\cdot z\in f^{-1}(V)\Leftrightarrow z\in\pi\star f^{-1}(V)=\pi\star g(V)$. If follows that $g(\pi\star V)=\pi\star g(V)$ for all $V\in\wp_{fs}(Y)$, and so $g$ is finitely supported. Moreover, because $f$ is surjective, a simple calculation shows us that $g$ is injective. Indeed, let us suppose that $g(U)=g(V)$ for some $U,V\in\wp_{fs}(Y)$. We have $f^{-1}(U)=f^{-1}(V)$, and so $f(f^{-1}(U))=f(f^{-1}(V))$. Since $f$ is surjective, we get $U=f(f^{-1}(U))=f(f^{-1}(V))=V$.

We start the proof of Corollary 3.6. As in Lemma 3.5, we consider the sets $B=T_{fin}(A)\setminus\bar{\emptyset}$ and $C=T_{fin}(A)$. According to Lemma 3.5 there exists a finitely supported surjective function $f:C\to B$ and a finitely supported (equivariant) surjection $g:B\to C$. Thus, according to Lemma 3.7, there exist a finitely supported injective function $f^{\prime}:\wp_{fs}(B)\to\wp_{fs}(C)$ and a finitely supported injective function $g^{\prime}:\wp_{fs}(C)\to\wp_{fs}(B)$. According to Lemma 3.3, there is a finitely supported bijection between $\wp_{fs}(B)$ and $\wp_{fs}(C)$. However, we proved in Lemma 3.5 that there is no finitely supported bijection between $B=T_{fin}(A)\setminus\bar{\emptyset}$ and $C=T_{fin}(A)$.

The following result communicated by Levy in 1965 for non-atomic ZF sets can be reformulated in the world of finitely supported atomic structures.

Corollary 3.8

Let $X$ and $Y$ be two invariant sets with the property that whenever $|2^{X}_{fs}|=|2^{Y}_{fs}|$ we have $|X|=|Y|$. If $|X|\leq^{\star}|Y|$ and $|Y|\leq^{\star}|X|$, then $|X|=|Y|$.

Proof. According to the hypothesis and to Lemma 3.7 there exist two finitely supported injective functions $f:\wp_{fs}(Y)\to\wp_{fs}(X)$ and $g:\wp_{fs}(X)\to\wp_{fs}(Y)$. According to Lemma 3.3, there is a bijective mapping $h:\wp_{fs}(X)\to\wp_{fs}(Y)$ . According to Theorem 3.12, we get $|2^{X}_{fs}|=|2^{Y}_{fs}|$, and so we get $|X|=|Y|$.

Proposition 3.9** (Cantor)**

Let $X$ be a finitely supported subset of an invariant set $(Y,\cdot)$. Then $|X|\lneq|\wp_{fs}(X)|$ and $|X|\lneq^{*}|\wp_{fs}(X)|$

Proof. First we prove that there is no finitely supported bijection between $X$ and $\wp_{fs}(X)$, and so their cardinalities cannot be equal. Assume, by contradiction, that there is a finitely supported surjective mapping $f:X\to\wp_{fs}(X)$. Let us consider $Z=\{x\in X\,|\,x\notin f(x)\}$. We claim that $supp(X)\cup supp(f)$ supports $Z$. Let $\pi\in Fix(supp(X)\cup supp(f))$. Let $x\in Z$. Then $\pi\cdot x\in X$ and $\pi\cdot x\notin\pi\star f(x)=f(\pi\cdot x)$. Thus, $\pi\cdot x\in Z$, and so $Z\in\wp_{fs}(X)$. Therefore, since $f$ is surjective there is $x_{0}\in X$ such that $f(x_{0})=Z$. However, from the definition of $Z$ we have $x_{0}\in Z$ if and only if $x_{0}\notin f(x_{0})=Z$, which is a contradiction.

Now, it is clear that the mapping $i:X\to\wp_{fs}(X)$ defined by $i(x)=\{x\}$ is injective and supported by $supp(X)$. Thus, $|X|\lneq|\wp_{fs}(X)|$. Let us fix an atom $y\in X$. We define $s:\wp_{fs}(X)\to X$ by

[TABLE]

Clearly, $s$ is surjective. We claim that $s$ is supported by $supp(y)\cup supp(X)$. Let $\pi\in Fix(supp(y)\cup supp(X))$. Thus, $y=\pi\cdot y$. If $U$ is of form $U=\{u\}$, we obviously have $s(\pi\star U)=s(\{\pi\cdot u\})=\pi\cdot u=\pi\cdot s(U)$. If $U$ has more than one element, then $\pi\star U$ has more than one element, and we have $s(\pi\star U)=y=\pi\cdot y=\pi\cdot s(U)$. Thus, $\pi\star U\in\wp_{fs}(X)$, $\pi\cdot s(U)\in X$, and $s(\pi\star U)=\pi\cdot s(U)$ for all $U\in\wp_{fs}(X)$ . According to Proposition 2.11, we have that $s$ is finitely supported. Therefore, $|X|\lneq^{*}|\wp_{fs}(X)|$.

In Proposition 3.9 we used a technique for constructing a surjection starting from an injection defined in the opposite way, that can be generalized as follows.

Proposition 3.10

Let $X$ and $Y$ be finitely supported subsets of an invariant set $U$. If If $|X|\leq|Y|$, then $|X|\leq^{\star}|Y|$. The converse is not valid. However, if $|X|\leq^{\star}|Y|$, then $|X|\leq|\wp_{fs}(Y)|$.

Proof. Suppose there exists a finitely supported injective mapping $f:X\to Y$. We consider the case $Y\neq\emptyset$ (otherwise, the result follows trivially). Fix $x_{0}\in X$. Define the mapping $f^{\prime}:Y\to X$ by

[TABLE]

Since $f$ is injective, it follows that $f^{-1}(y)$ is an one-element set for each $y\in Im(f)$, and so $f^{\prime}$ is a function. Clearly, $f^{\prime}$ is surjective. We claim that $f^{\prime}$ is supported by the set $supp(f)\cup supp(x_{0})\cup supp(X)\cup supp(Y)$. Indeed, let us consider $\pi\in Fix(supp(f)\cup supp(x_{0})\cup supp(X)\cup supp(Y))$. Whenever $y\in Im(f)$ we have $y=f(z)$ for some $z\in X$ and $\pi\cdot y=\pi\cdot f(z)=f(\pi\cdot z)\in Im(f)$, which means $Im(f)$ is finitely supported by $supp(f)$. Consider an arbitrary $y_{0}\in Im(f)$, and thus $\pi\cdot y_{0}\in Im(f)$. Then $f^{\prime}(y_{0})=f^{-1}(y_{0})=z_{0}$ with $f(z_{0})=y_{0}$, and so $f(\pi\cdot z_{0})=\pi\cdot f(z_{0})=\pi\cdot y_{0}$, which means $f^{\prime}(\pi\cdot y_{0})=f^{-1}(\pi\cdot y_{0})=\pi\cdot z_{0}=\pi\cdot f^{-1}(y_{0})=\pi\cdot f^{\prime}(y_{0})$. Now, for $y\notin Im(f)$ we have $\pi\cdot y\notin Im(f)$, which means $f^{\prime}(\pi\cdot y)=x_{0}=\pi\cdot x_{0}=\pi\cdot f(y)$ since $\pi$ fixes $x_{0}$ pointwise. Thus, $|X|\leq^{\star}|Y|$. Conversely, from the proof of Lemma 3.5, we know that there is a finitely supported surjection $g:T_{fin}(A)\setminus\bar{\emptyset}\to T_{fin}(A)$, but there does not exist a finitely supported injection $h:T_{fin}(A)\to T_{fin}(A)\setminus\bar{\emptyset}$.

Assume now there is a finitely supported surjective mapping $f:Y\to X$. We proceed similarly as in the proof of Lemma 3.7. Fix $x\in X$. Then $f^{-1}(\{x\})$ is supported by $supp(f)\cup supp(x)\cup supp(X)$. Indeed, let $\pi\in Fix(supp(f)\cup supp(x)\cup supp(X))$, and $y\in f^{-1}(\{x\})$. This means $f(y)=x$. According to Proposition 2.11, we have $f(\pi\cdot y)=\pi\cdot f(y)=\pi\cdot x=x$, and so $\pi\cdot y\in f^{-1}(\{x\})$. Define $g:X\to\wp_{fs}(Y)$ by $g(x)=f^{-1}(\{x\})$. We claim that $g$ is supported by $supp(f)\cup supp(X)$. Let $\pi\in Fix(supp(f)\cup supp(X))$. For any arbitrary $x\in X$, we have that $z\in g(\pi\cdot x)=f^{-1}(\{\pi\cdot x\})\Leftrightarrow f(z)=\pi\cdot x\Leftrightarrow\pi^{-1}\cdot f(z)=x\Leftrightarrow f(\pi^{-1}\cdot z)=x\Leftrightarrow\pi^{-1}\cdot z\in f^{-1}(\{x\})\Leftrightarrow z\in\pi\star f^{-1}(\{x\})=\pi\star g(x)$. From Proposition 2.11 it follows that $g$ is finitely supported. Since $g$ is also injective, we get $|X|\leq|\wp_{fs}(Y)|$.

Proposition 3.11

Let $X,Y,Z$ be finitely supported subsets of an invariant set $U$. The following properties hold.

If $|X|\leq|Y|$, then $|X|+|Z|\leq|Y|+|Z|$; 2. 2.

If $|X|\leq|Y|$, then $|X|\cdot|Z|\leq|Y|\cdot|Z|$; 3. 3.

If $|X|\leq|Y|$, then $|X^{Z}_{fs}|\leq|Y^{Z}_{fs}|$; 4. 4.

If $|X|\leq|Y|$ and $Z\neq\emptyset$, then $|Z^{X}_{fs}|\leq|Z^{Y}_{fs}|$; 5. 5.

$|X|+|Y|\leq|X|\cdot|Y|$ whenever both $X$ and $Y$ have more than two elements.

Proof. 1. Suppose there is a finitely supported injective $f:X\to Y$, and define the injection $g:X+Z\to Y+Z$ by

[TABLE]

Since $f$ is finitely supported we have that $f(\pi\cdot x)=\pi\cdot f(x)$ for all $x\in X$ and $\pi\in Fix(supp(f))$. By using Proposition 2.11, i.e verifying that $g(\pi\star u)=\pi\star g(u)$ for all $u\in X+Z$ and all $\pi\in Fix(supp(f)\cup supp(X)\cup supp(Y)\cup supp(Z))$, we have that $g$ is also finitely supported.

2. Suppose there exists a finitely supported injective mapping $f:X\to Y$. Define the injection $g:X\times Z\to Y\times Z$ by $g((x,z))=(f(x),z)$ for all $(x,z)\in X\times Z$. Clearly $g$ is injective. Since $f$ is finitely supported we have that $f(\pi\cdot x)=\pi\cdot f(x)$ for all $x\in X$ and $\pi\in Fix(supp(f))$, and so $g(\pi\otimes(x,z))=g((\pi\cdot x,\pi\cdot z))=(f(\pi\cdot x),\pi\cdot z)=(\pi\cdot f(x),\pi\cdot z)=\pi\otimes g((x,z))$ for all $(x,z)\in X\times Z$ and $\pi\in Fix(supp(f)\cup supp(X)\cup supp(Y)\cup supp(Z))$, which means $g$ is supported by $supp(f)\cup supp(X)\cup supp(Y)\cup supp(Z)$.

3. Suppose there exists a finitely supported injective mapping $f:X\to Y$. Define $g:X^{Z}_{fs}\to Y^{Z}_{fs}$ by $g(h)=f\circ h$. We have that $g$ is injective and for any $\pi\in Fix(supp(f))$ we have $\pi\widetilde{\star}f=f$, and so $g(\pi\widetilde{\star}h)=f\circ(\pi\widetilde{\star}h)=(\pi\widetilde{\star}f)\circ(\pi\widetilde{\star}h)=\pi\widetilde{\star}(f\circ h)=\pi\widetilde{\star}g(h)$ for all $h\in X^{Z}_{fs}$. We used the relation $(\pi\widetilde{\star}f)\circ(\pi\widetilde{\star}h)=\pi\widetilde{\star}(f\circ h)$ for all $\pi\in S_{A}$. This can be proved as follows. Fix $x\in Z$, we have $(\pi\widetilde{\star}(f\circ h))(x)=\pi\cdot(f(h(\pi^{-1}\cdot x)))$. Also, if we denote $(\pi\widetilde{\star}h)(x)=y$ we have $y=\pi\cdot(h(\pi^{-1}\cdot x))$ and $((\pi\widetilde{\star}f)\circ(\pi\widetilde{\star}h))(x)=(\pi\widetilde{\star}f)(y)=\pi\cdot(f(\pi^{-1}\cdot y))=\pi\cdot(f((\pi^{-1}\circ\pi)\cdot h(\pi^{-1}\cdot x)))=\pi\cdot(f(h(\pi^{-1}\cdot x)))$. We finally obtain that $g$ is supported by $supp(f)\cup supp(X)\cup supp(Y)\cup supp(Z)$.

4. Suppose there exists a finitely supported injective mapping $f:X\to Y$. According to Proposition 3.10, there is a finitely supported surjective mapping $f^{\prime}:Y\to X$. Define the injective mapping $g:Z^{X}_{fs}\to Z^{Y}_{fs}$ by $g(h)=h\circ f^{\prime}$. As in item 3 one can prove that $g$ is finitely supported by $supp(f^{\prime})\cup supp(X)\cup supp(Y)\cup supp(Z)$.

5. Fix $x_{0},x_{1}\in X$ with $x_{0}\neq x_{1}$ and $y_{0},y_{1}\in Y$ with $y_{0}\neq y_{1}$. Define the injection $g:X+Y\to X\times Y$ by

[TABLE]

It follows that $g$ is supported by $supp(x_{0})\cup supp(y_{0})\cup supp(x_{1})\cup supp(y_{1})\cup supp(X)\cup supp(Y)$, and $g$ is injective.

Theorem 3.12

Let $(X,\cdot)$ be a finitely supported subset of an invariant set $(Z,\cdot)$. There exists a one-to-one mapping from $\wp_{fs}(X)$ onto $\{0,1\}^{X}_{fs}$ which is finitely supported by $supp(X)$, where $\wp_{fs}(X)$ is considered the family of those finitely supported subsets of $Z$ contained in $X$.

Proof. Let $Y$ be a finitely supported subset of $Z$ contained in $X$, and $\varphi_{Y}$ be the characteristic function on $Y$, i.e. $\varphi_{Y}:X\to\{0,1\}$ is defined by

\varphi_{Y}(x)\overset{def}{=}\left\{\begin{array}[]{ll}1&\text{for}\>x\in Y\\ 0&\text{for}\>x\in X\setminus Y\end{array}\right..

We prove that $\varphi_{Y}$ is a finitely supported function from $X$ to $\{0,1\}$ (according to Proposition 2.5, $\{0,1\}$ is a trivial invariant set), and the mapping $Y\mapsto\varphi_{Y}$ defined on $\wp_{fs}(X)$ is also finitely supported in the sense of Definition 2.10.

First we prove that $\varphi_{Y}$ is supported by $supp(Y)\cup supp(X)$. Let $\pi\in Fix(supp(Y)\cup supp(X))$. Thus $\pi\star Y=Y$ (where $\star$ represents the canonical permutation action on $\wp(Z)$), and so $\pi\cdot x\in Y$ if and only if $x\in Y$. Since we additionally have $\pi\star X=X$, we obtain $\pi\cdot x\in X\setminus Y$ if and only if $x\in X\setminus Y$. Thus, $\varphi_{Y}(\pi\cdot x)=\varphi_{Y}(x)$ for all $x\in X$. Furthermore, because $\pi$ fixes $supp(X)$ pointwise we have $\pi\cdot x\in X$ for all $x\in X$, and from Proposition 2.11 we get that $\varphi_{Y}$ is supported by $supp(Y)\cup supp(X)$.

We remark that $\{0,1\}^{X}_{fs}$ is a finitely supported subset of the set $(\wp_{fs}(Z\times\{0,1\}),\widetilde{\star})$. Let $\pi\in Fix(supp(X))$ and $f:X\to\{0,1\}$ finitely supported. We have $\pi\widetilde{\star}f=\{(\pi\cdot x,$ $\pi\diamond y)\,|\,(x,y)\in f\}=\{(\pi\cdot x,$ $y)\,|\,(x,y)\in f\}$ because $\diamond$ is the trivial action on $\{0,1\}$. Thus, $\pi\widetilde{\star}f$ is a function with the domain $\pi\star X=X$ which is finitely supported as an element of $(\wp(Z\times\{0,1\}),\widetilde{\star})$ according to Proposition 2.3. Moreover, $(\pi\widetilde{\star}f)(\pi\cdot x)=f(x)$ for all $x\in X$ (1).

According to Proposition 2.11, to prove that the function $g:=Y\mapsto\varphi_{Y}$ defined on $\wp_{fs}(X)$ (with the codomain contained in $\{0,1\}^{X}_{fs}$) is supported by $supp(X)$, we have to prove that $\pi\widetilde{\star}g(Y)=g(\pi\star Y)$ for all $\pi\in Fix(supp(X))$ and all $Y\in\wp_{fs}(X)$ (where $\widetilde{\star}$ symbolizes the induced $S_{A}$-action on $\{0,1\}^{X}_{fs}$). This means that we need to verify the relation $\pi\widetilde{\star}\varphi_{Y}=\varphi_{\pi\star Y}$ for all $\pi\in Fix(supp(X))$ and all $Y\in\wp_{fs}(X)$. Let us consider $\pi\in Fix(supp(X))$ (which means $\pi\cdot x\in X$ for all $x\in X$) and $Y\in\wp_{fs}(X)$. For any $x\in X$, we know that $x\in\pi\star Y$ if and only if $\pi^{-1}\cdot x\in Y$. Thus, $\varphi_{Y}(\pi^{-1}\cdot x)=\varphi_{\pi\star Y}(x)$ for all $x\in X$, and so $(\pi\widetilde{\star}\varphi_{Y})(x)\overset{(1)}{=}\varphi_{Y}(\pi^{-1}\cdot x)=\varphi_{\pi\star Y}(x)$ for all $x\in X$. Moreover, from Proposition 2.3, $\pi\star Y$ is a finitely supported subset of $Z$ contained in $\pi\star X=X$, and $\{0,1\}^{X}_{fs}$ can be represented as a finitely supported subset of $\wp_{fs}(Z\times\{0,1\})$ (supported by $supp(X)$). According to Proposition 2.11 we have that $g$ is a finitely supported function from $\wp_{fs}(X)$ to $\{0,1\}^{X}_{fs}$.

Obviously, $g$ is one-to-one. Now we prove that $g$ is onto. Let us consider an arbitrary finitely supported function $f:X\to\{0,1\}$. Let $Y_{f}\overset{def}{=}\{x\in X\;|\;f(x)=1\}$. We claim that $Y_{f}\in\wp_{fs}(X)$. Let $\pi\in Fix(supp(f))$. According to Proposition 2.11 we have $\pi\cdot x\in X$ and $f(\pi\cdot x)=f(x)$ for all $x\in X$. Thus, for each $x\in Y_{f}$, we have $\pi\cdot x\in Y_{f}$. Therefore $\pi\star Y_{f}=Y_{f}$, and so $Y_{f}$ is finitely supported by $supp(f)$ as a subset of $Z$, and it is contained in $X$. A simple calculation show us that $g(Y_{f})=f$, and so $g$ is onto.

One can easy verify that the properties of $\leq$ presented in Proposition 3.11 (1), (2) and (4) also hold for $\leq^{\star}$. We left the details to the reader.

Theorem 3.13

There exists an invariant set $X$ (particularly the set $A$ of atoms) having the following properties.

$|X\times X|\nleq^{*}|\wp_{fs}(X)|$; 2. 2.

$|X\times X|\nleq|\wp_{fs}(X)|$; 3. 3.

$|X\times X|\nleq^{*}|X|$; 4. 4.

$|X\times X|\nleq|X|$; 5. 5.

For each $n\in\mathbb{N},n\geq 2$ we have $|X|\lneq|\wp_{n}(X)|\lneq|\wp_{fs}(X)|$, where $\wp_{n}(X)$ is the family of all $n$-sized subsets of $X$; 6. 6.

For each $n\in\mathbb{N}$ we have $|X|\lneq^{*}|\wp_{n}(X)|\lneq^{*}|\wp_{fs}(X)|$; 7. 7.

$|X|\lneq|\wp_{fin}(X)|\lneq|\wp_{fs}(X)|$; 8. 8.

$|X|\lneq^{*}|\wp_{fin}(X)|\lneq^{*}|\wp_{fs}(X)|$; 9. 9.

$|\wp_{fs}(X)\times\wp_{fs}(X)|\nleq^{*}|\wp_{fs}(X)|$; 10. 10.

$|\wp_{fs}(X)\times\wp_{fs}(X)|\nleq|\wp_{fs}(X)|$; 11. 11.

$|X+X|\lneq^{*}|X\times X|$; 12. 12.

$|X+X|\lneq|X\times X|$.

Proof.

1. We prove that that there does not exist a finitely supported surjective mapping $f:\wp_{fs}(A)\to A\times A$. Suppose, by contradiction, that there is a finitely supported surjective mapping $f:\wp_{fs}(A)\to A\times A$. Let us consider two atoms $a,b\notin supp(f)$ with $a\neq b$. These atoms exist because $A$ is infinite, while $supp(f)\subseteq A$ is finite. It follows that the transposition $(a\,b)$ fixes each element from $supp(f)$, i.e. $(a\,b)\in Fix(supp(f))$. Since $f$ is surjective, it follows that there exists an element $X\in\wp_{fs}(A)$ such that $f(X)=(a,b)$. Since $supp(f)$ supports $f$ and $(a\,b)\in Fix(supp(f))$, from Proposition 2.11 we have $f((a\,b)\star X)=(a\,b)\otimes f(X)=(a\,b)\otimes(a,b)=((a\,b)(a),(a\,b)(b))=(b,a)$. Due to the functionality of $f$ we should have $(a\,b)\star X\neq X$. Otherwise, we would obtain $(a,b)=(b,a)$.

We claim that if both $a,b\in supp(X)$, then $(a\,b)\star X=X$. Indeed, suppose $a,b\in supp(X)$. Since $X$ is a finitely supported subset of $A$, then $X$ is either finite or cofinite. If $X$ is finite, then $supp(X)=X$, and so $a,b\in X$. Moreover, $(a\,b)(a)=b$, $(a\,b)(b)=a$, and $(a\,b)(c)=c$ for all $c\in X$ with $c\neq a,b$. Therefore, $(a\,b)\star X=\{(a\,b)(x)\,|\,x\in X\}=\{(a\,b)(a)\}\cup\{(a\,b)(b)\}\cup\{(a\,b)(c)\,|\,c\in X\setminus\{a,b\}\}=\{b\}\cup\{a\}\cup(X\setminus\{a,b\})=X$. Now, if $X$ is cofinite, then $supp(X)=A\setminus X$, and so $a,b\in A\setminus X$. Since $a,b\notin X$, we have $a,b\neq x$ for all $x\in X$, and so $(a\,b)(x)=x$ for all $x\in X$. Thus, in this case we also have $(a\,b)\star X=X$.

Since when both $a,b\in supp(X)$ we have $(a\,b)\star X=X$, it follows that one of $a$ or $b$ does not belong to $supp(X)$. Suppose $b\notin supp(X)$ (the other case is analogue). Let us consider $c\neq a,b$, $c\notin supp(f)$, $c\notin supp(X)$. Then $(b\,c)\in Fix(supp(X))$, and, because $supp(X)$ supports $X$, we have $(b\,c)\star X=X$. Furthermore, $(b\,c)\in Fix(supp(f))$, and by Proposition 2.11 we have $(a,b)=f(X)=f((b\,c)\star X)=(b\,c)\otimes f(X)=(b\,c)\otimes(a,b)=((b\,c)(a),(b\,c)(b))=(a,c)$ which is a contradiction because $b\neq c$. Thus, $|A\times A|\nleq^{*}|\wp_{fs}(A)|$.

2. We prove that there does not exist a finitely supported injective mapping $f:A\times A\to\wp_{fs}(A)$. Suppose, by contradiction, that there is a finitely supported injective mapping $f:A\times A\to\wp_{fs}(A)$. According to Proposition 3.10, one can define a finitely supported surjection $g:\wp_{fs}(A)\to A\times A$. This contradicts the above item. Thus, $|A\times A|\nleq|\wp_{fs}(A)|$.

3. We prove that there does not exist a finitely supported surjection $f:A\to A\times A$. Since there exists a surjection $s$ from $\wp_{fs}(A)$ onto $A$ defined by

[TABLE]

where $a$ is a fixed atom, and $s$ is finitely supported (by$\{a\}$), the result follows from item 1. Thus, $|A\times A|\nleq^{*}|A|$.

4. We prove that there does not exist a finitely supported injection $f:A\times A\to A$. Since there exists an equivariant injection from $A$ into $\wp_{fs}(A)$ defined as $x\mapsto\{x\}$, the result follows from item 2. Thus, $|A\times A|\nleq|A|$;

Alternatively, one can prove that there does not exist a one-to-one mapping from $A\times A$ to $A$ (and so neither a finitely supported one). Suppose, by contradiction, that there is a an injective mapping $i:A\times A\to A$. Let us fix two atoms $x$ and $y$ with $x\neq y$. The sets $\{i(a,x)\,|\,a\in A\}$ and $\{i(a,y)\,|\,a\in A\}$ are disjoint and infinite. Thus, $\{i(a,x)\,|\,a\in A\}$ is a infinite and coinfinite subset of $A$, which contradicts the fact that any subset of $A$ is either finite or cofinite.

5. We prove that $|A|\lneq|\wp_{n}(A)|\lneq|\wp_{fs}(A)|$ for all $n\in\mathbb{N},n\geq 2$. Consider $a_{1},a_{2},\ldots,a_{n-1}$ $,a^{1}_{1},\ldots,a^{n}_{1},\ldots,a^{1}_{n-1},\ldots,a^{n}_{n-1}\in A$ a family of pairwise different elements. Then $i:A\to\wp_{n}(A)$ defined by

[TABLE]

is obviously an injective mapping from $(A,\cdot)$ to $(\wp_{n}(A),\star)$. Furthermore, we can easy check that $i$ is supported by the finite set {$a_{1},a_{2},\ldots,a_{n-1}$ $,a^{1}_{1},\ldots,a^{n}_{1},\ldots,a^{1}_{n-1},\ldots,a^{n}_{n-1}\}$, and so $|A|\leq|\wp_{n}(A)|$ in FSM.

We claim that there does not exist a finitely supported injection from $\wp_{n}(A)$ into $A$. Assume on the contrary that there exists an finitely supported injection $f:\wp_{n}(A)\to A$.

First, we claim that, for any $Y\in\wp_{n}(A)$ which is disjoint from $supp(f)$, we have $f(Y)\notin Y$. Assume by contradiction that $f(Y)\in Y$ for a fixed $Y$ with $Y\cap supp(f)=\emptyset$. Let $\pi$ be a permutation of atoms which fixes $supp(f)$ pointwise, and interchanges all the elements of $Y$ (e.g. $\pi$ is a cyclic permutation of $Y$). Since $\pi$ permutes all the elements of $Y$, we have $\pi\cdot f(Y)=\pi(f(Y))\neq f(Y)$. However, $\pi\star Y=\{\pi(a_{1}),\ldots,\pi(a_{n})\}=\{a_{1},\ldots,a_{n}\}=Y$. Since $\pi$ fixes $supp(f)$ pointwise and $supp(f)$ supports $f$, we have $\pi(f(Y))=\pi\cdot f(Y)=f(\pi\star Y)=f(Y)$, a contradiction.

Since $supp(f)$ is finite, there are infinitely many such $Y$ with the property that $Y\cap supp(f)=\emptyset$. Thus, because it is injective, $f$ takes infinitely many values on those $Y$. Since $supp(f)$ is finite, there should exist at least one element in $\wp_{n}(A)$, denoted by $Z$ such that $Z\cap supp(f)=\emptyset$ and $f(Z)\notin supp(f)$. Thus, $f(Z)=a$ for some $a\in A\setminus(Z\cup supp(f))$. Let $b\in A\setminus(supp(f)\cup Z\cup\{a\})$ and also let $\pi=(a\,b)$. Then $\pi\in Fix(supp(f)\cup Z)$, and hence $f(Z)=f((a\,b)\star Z)=(a\,b)(f(Z))=b$, a contradiction. We obtained that $|A|\neq|\wp_{n}(A)|$ in FSM, and so $|A|<|\wp_{n}(A)|$.

We obviously have $|\wp_{n}(A)|\leq|\wp_{fs}(A)|$. We prove below that there does not exist a finitely supported injective mapping from $\wp_{fs}(A)$ onto one of its finitely supported proper subsets, i.e. any finitely supported injection $f:\wp_{fs}(A)\rightarrow\wp_{fs}(A)$ is also surjective. Let us consider a finitely supported injection $f:\wp_{fs}(A)\rightarrow\wp_{fs}(A)$. Suppose, by contradiction, $Im(f)\subsetneq\wp_{fs}(A)$. This means that there exists $X_{0}\in\wp_{fs}(A)$ such that $X_{0}\notin Im(f)$. Since $f$ is injective, we can define an infinite sequence $\mathcal{F}=(X_{n})_{n}$ starting from $X_{0}$, with distinct terms of form $X_{n+1}=f(X_{n})$ for all $n\in\mathbb{N}$. Furthermore, according to Proposition 2.11, for a fixed $k\in\mathbb{N}$ and $\pi\in Fix(supp(f)\cup supp(X_{k}))$, we have $\pi\star X_{k+1}=\pi\star f(X_{k})=f(\pi\star X_{k})=f(X_{k})=X_{k+1}$. Then, $supp(X_{n+1})\subseteq supp(f)\cup supp(X_{n})$ for all $n\in\mathbb{N}$, and by induction on $n$ we have that $supp(X_{n})\subseteq supp(f)\cup supp(X_{0})$ for all $n\in\mathbb{N}$. We obtained that each element $X_{n}\in\mathcal{F}$ is supported by the same finite set $S:=supp(f)\cup supp(X_{0})$. However, there could exist only finitely many subsets of $A$ (i.e. only finitely many elements in $\wp_{fs}(A)$) supported by $S$, namely the subsets of $S$ and the supersets of $A\setminus S$ (where a superset of $A\setminus S$ is of form $A\setminus X$ with $X\subseteq S$). We contradict the statement that the infinite sequence $(X_{n})_{n}$ never repeats. Thus, $f$ is surjective, and so there could not exist a bijection between $\wp_{fin}(A)$ and $\wp_{fs}(A)$, which means $|\wp_{n}(A)|\neq|\wp_{fs}(A)|$.

6. Fix $n\in\mathbb{N}$. As in the above item there does not exist neither a finitely supported bijection between $\wp_{n}(A)$ and $\wp_{fs}(A)$, nor a finitely supported bijection between $A$ and $\wp_{n}(A)$. However, there exists a finitely supported injection $i:A\to\wp_{n}(A)$. Fix an atom $a\in A$. The mapping $s:\wp_{n}(A)\to A$ defined by

[TABLE]

is supported by $supp(i)\cup\{a\}$ and is surjective.

Now, fix $n$ atoms $x_{1},\ldots,x_{n}$. The mapping $g:\wp_{fs}(A)\to\wp_{n}(A)$ defined by

[TABLE]

is supported by $\{x_{1},\ldots x_{n}\}$ and is surjective.

7. We prove that $|A|\lneq|\wp_{fin}(A)|\lneq|\wp_{fs}(A)|$. We obviously have that $|A|\leq|\wp_{fin}(A)|$ by taking the equivariant injective mapping $f:A\to\wp_{fin}(A)$ defined by $f(a)=\{a\}$ for all $a\in A$. We prove, by contradiction, that there is no finitely supported surjection from $A$ onto $\wp_{fin}(A)$. Assume that $g:A\to\wp_{fin}(A)$ is a finitely supported surjection. Let us fix two atoms $x$ and $y$. We define the function $h:\wp_{fin}(A)\to\wp_{2}(A)$ by h(X)=\left\{\begin{array}[]{ll}X,&\text{if}\>\text{|X|=2 };\\ \{x,y\},&\text{if}\>\text{|X|\neq 2}\>.\end{array}\right.. Since for every $\pi\in S_{A}$ and $X\in\wp_{fin}(A)$ we have $|\pi\star X|=|X|$, we conclude that $h$ is finitely supported by $\{x,y\}$. Thus, $h\circ g$ is a surjection from $A$ onto $\wp_{2}(A)$ supported by $supp(g)\cup\{x,y\}$, which contradicts the previous item. Therefore, $|A|<|\wp_{fin}(A)|$. Since every element in $\wp_{fin}(A)$ belongs to $\wp_{fs}(A)$, but there does not exist a finitely supported injective mapping from $\wp_{fs}(A)$ onto one of its finitely supported proper subsets, we also have $|\wp_{fin}(A)|<|\wp_{fs}(A)|$.

8. As in the above item there does not exist neither a finitely supported bijection between $\wp_{fin}(A)$ and $\wp_{fs}(A)$, nor a finitely supported bijection between $A$ and $\wp_{fin}(A)$. Fix an atom $a\in A$. The mapping $s:\wp_{fin}(A)\to A$ defined by

[TABLE]

is supported by $\{a\}$ and is surjective.

Now, fix an atom $b$. The mapping $g:\wp_{fs}(A)\to\wp_{fin}(A)$ defined by

[TABLE]

is supported by $\{b\}$ and is surjective.

9. According to Theorem 3.13(1) there is no finitely supported surjection from $\wp_{fs}(A)$ onto $A\times A$. Suppose there is a finitely supported surjective mapping $f:\wp_{fs}(A)\to\wp_{fs}(A)\times\wp_{fs}(A)$. Obviously, there exists a supported surjection $s:\wp_{fs}(A)\to A$ defined by

[TABLE]

where $x$ is a fixed atoms of $A$. The surjection $s$ is supported by $supp(x)=x$. Thus, we can define a surjection $g:\wp_{fs}(A)\times\wp_{fs}(A)\to A\times A$ by $g(X,Y)=(s(X),s(Y))$ for all $X,Y\in\wp_{fs}(A)$. Let $\pi\in Fix(supp(s))$. Since $supp(s)$ supports $s$, by Proposition 2.11 we have $g(\pi\otimes_{\star}(X,Y))=g(\pi\star X,\pi\star Y)=(s(\pi\star X),s(\pi\star Y))=(\pi\cdot s(X),\pi\cdot s(Y))=\pi\otimes(s(X),s(Y))$ for all $X,Y\in\wp_{fs}(A)$, where $\otimes_{\star}$ and $\otimes$ represent the $S_{A}$-actions on $\wp_{fs}(A)\times\wp_{fs}(A)$ and $A\times A$, respectively. Thus, $supp(s)$ supports $g$, and so $supp(g)\subseteq supp(s)$. Furthermore, the function $h=g\circ f:\wp_{fs}(A)\to A\times A$ is surjective and finitely supported by $supp(s)\cup supp(f)$. This is a contradiction, and so $|\wp_{fs}(A)\times\wp_{fs}(A)|\nleq^{*}|\wp_{fs}(A)|$.

10. Suppose, by contradiction, that there is a finitely supported injective mapping $f:\wp_{fs}(A)\times\wp_{fs}(A)\to\wp_{fs}(A)$. In the view of Proposition 3.10, let us fix two finitely supported subsets of $A$, namely $U$ and $V$. We define the function $g:\wp_{fs}(A)\to\wp_{fs}(A)\times\wp_{fs}(A)$ by

[TABLE]

Clearly, $g$ is surjective. Furthermore, $g$ is supported by $supp(f)\cup supp(U)\cup supp(V)$ (the proof uses the fact that $Im(f)$ is a subset of $\wp_{fs}(A)$ supported by $supp(f)$). This contradicts the above item, and so $|\wp_{fs}(A)\times\wp_{fs}(A)|\nleq|\wp_{fs}(A)|$.

11. In the view of Proposition 3.11(5) there is a finitely supported injection from $A+A$ into $A\times A$, and a finitely supported surjection from $A\times A$ onto $A+A$ according to Proposition 3.10. Thus $|A+A|\leq|A\times A|$ and $|A+A|\leq^{*}|A\times A|$ Fix three different atoms $a,b,c\in A$. Define the mapping $f:A+A\to\wp_{fs}(A)$ by

[TABLE]

One can directly prove that $f$ is injective and supported by $\{a,b,c\}$. According to Proposition 3.10, we have $|A+A|\leq^{*}|\wp_{fs}(A)|$. If we had $|A\times A|=|A+A|$, we would obtain $|A\times A|\leq^{*}|\wp_{fs}(A)|$ which contradicts item 1.

12. According to the above item $|A+A|\leq|\wp_{fs}(A)|$. If we had $|A\times A|=|A+A|$, we would obtain $|A\times A|\leq|\wp_{fs}(A)|$ which contradicts item 2.

Proposition 3.14

There exists an invariant set $X$ having the following properties:

$|X|\lneq|X|+|X|$; 2. 2.

$|X|\lneq^{*}|X|+|X|$.

Proof.

1. First we prove that in FSM we have $|\wp_{fs}(A)|=2|\wp_{fin}(A)|$. Let us consider the function $f:\wp_{fin}(A)\to\wp_{cofin}(A)$ defined by $f(U)=A\setminus U$ for all $U\in\wp_{fin}(A)$. Clearly, $f$ is bijective. We claim that $f$ is equivariant. Indeed, let $\pi\in S_{A}$. To prove that $f(\pi\star U)$=$\pi\star f(U)$ for all $U\in\wp_{fin}(A)$, we have to prove that $A\setminus(\pi\star U)=\pi\star(A\setminus U)$ for all $U\in\wp_{fin}(A)$. Let $y\in A\setminus(\pi\star U)$. We can express $y$ as $y=\pi\cdot(\pi^{-1}\cdot y)$. If $\pi^{-1}\cdot y\in U$, then $y\in\pi\star U$, which is a contradiction. Thus, $\pi^{-1}\cdot y\in(A\setminus U)$, and so $y\in\pi\star(A\setminus U)$. Conversely, if $y\in\pi\star(A\setminus U)$, then $y=\pi\cdot x$ with $x\in A\setminus U$. Suppose $y\in\pi\star U$. Then $y=\pi\cdot z$ with $z\in U$. Thus, $x=z$ which is a contradiction, and so $y\in A\setminus(\pi\star U)$. Since $f$ is equivariant and bijective, it follows that $|\wp_{fin}(A)|=|\wp_{cofin}(A)|$. However, every finitely supported subset of $A$ is either finite or cofinite, and so $\wp_{fs}(A)$ is the union of the disjoint subsets $\wp_{fin}(A)$ and $\wp_{cofin}(A)$. Thus, $|\wp_{fs}(A)|=2|\wp_{fin}(A)|$. Moreover, there exists an equivariant injection $i:\wp_{fin}(A)\to\wp_{fs}(A)$ defined by $i(U)=U$ for all $U\in\wp_{fin}(A)$. However, there does not exist a finitely supported one-to-one mapping from $\wp_{fs}(A)$ onto one of its finitely supported proper subsets. Thus, there could not exist a bijection $f:\wp_{fs}(A)\to\wp_{fin}(A)$. Therefore, $|\wp_{fin}(A)|\neq|\wp_{fs}(A)|=2|\wp_{fin}(A)|$. We can consider $X=\wp_{fin}(A)$ or $X=\wp_{cofin}(A)$.

2. It remains to prove that there is a finitely supported surjection from $\wp_{fs}(A)$ onto $\wp_{fin}(A)$. We either use Proposition 3.10 or effectively construct the surjection as below. Fix $a\in A$. We define $g:\wp_{fs}(A)\to\wp_{fin}(A)$ by

[TABLE]

Clearly, $g$ is supported by $\{a\}$ and surjective. We can consider $X=\wp_{fin}(A)$ or $X=\wp_{cofin}(A)$.

4 Forms of Infinite in Finitely Supported Structures

The equivalence of various definitions for infinity is provable in ZF under the consideration of the axiom of choice. Since in FSM the axiom of choice fails, our goal is to study various FSM forms of infinite and to provide several relations between them.

Definition 4.1

Let $X$ be a finitely supported subset of an invariant set.

$X$* is called FSM usual infinite if $X$ does not correspond one-to-one and onto to a finite ordinal. We simply call infinite an FSM usual infinite set.* 2. 2.

$X$* is FSM covering infinite if there is a finitely supported directed family $\mathcal{F}$ of finitely supported sets with the property that $X$ is contained in the union of the members of $\mathcal{F}$, but there does not exist $Z\in\mathcal{F}$ such that $X\subseteq Z$.* 3. 3.

$X$* is called FSM Tarski I infinite if there exists a finitely supported one-to-one mapping of $X$ onto $X\times X$.* 4. 4.

$X$* is called FSM Tarski II infinite if there exists a finitely supported family of finitely supported subsets of $X$, totally ordered by inclusion, having no maximal element.* 5. 5.

$X$* is called FSM Tarski III infinite if $|X|=2|X|$.* 6. 6.

$X$* is called FSM Mostowski infinite if there exists an infinite finitely supported totally ordered subset of $X$.* 7. 7.

$X$* is called FSM Dedekind infinite if there exist a finitely supported one-to-one mapping of $X$ onto a finitely supported proper subset of $X$.* 8. 8.

$X$* is FSM ascending infinite if there is a finitely supported increasing countable chain of finitely supported sets $X_{0}\subseteq X_{1}\subseteq\ldots\subseteq X_{n}\subseteq\ldots$ with $X\subseteq\cup X_{n}$, but there does not exist $n\in\mathbb{N}$ such that $X\subseteq X_{n}$;*

Note that in the definition of FSM Tarski II infinity for a certain $X$, the existence of a finitely supported family of finitely supported subsets of $X$ is required, while in the definition of FSM ascending infinity for $X$, the related family of finitely supported subsets of $X$ has to be FSM countable (i.e. the mapping $n\mapsto X_{n}$ should be finitely supported). It is immediate that if $X$ is FSM ascending infinite, then it is also FSM Tarski II infinite.

Theorem 4.2

Let $X$ be a finitely supported subset of an invariant set. Then $X$ is FSM usual infinite if and only if $X$ is FSM covering infinite.

Proof. Let us suppose that $X$ is FSM usual infinite. Let $\mathcal{F}$ be the family of all FSM usual non-infinite (FSM usual finite) subsets of $X$ ordered by inclusion. Since $X$ is finitely supported, it follows that $\mathcal{F}$ is supported by $supp(X)$. Moreover, since all the elements of $\mathcal{F}$ are finite sets, it follows that all the elements of $\mathcal{F}$ are finitely supported. Clearly, $\mathcal{F}$ is directed and $X$ is the union of the members of $\mathcal{F}$. Suppose by contradiction, that $X$ is not FSM covering infinite. Then there exists $Z\in\mathcal{F}$ such that $X\subseteq Z$. Therefore, $X$ should by FSM usual finite which is a contradiction with our original assumption.

Conversely, assume that $X$ is FSM covering infinite. Suppose, by contradiction that $X$ is FSM usual finite, i.e. $X=\{x_{1},\ldots x_{n}\}$. Let $\mathcal{F}$ be a directed family such that $X$ is contained in the union of the members of $\mathcal{F}$ (at least one such a family exists, for example $\wp_{fs}(X)$). Then for each $i\in\{1,\ldots,n\}$ there exists $F_{i}\in\mathcal{F}$ such that $x_{i}\in F_{i}$. Since $\mathcal{F}$ is directed, there is $Z\in\mathcal{F}$ such that $F_{i}\subseteq Z$ for all $i\in\{1,\ldots,n\}$, and so $X\subseteq Z$ with $Z\in\mathcal{F}$, which is a contradiction.

Theorem 4.3

The following properties of FSM Dedekind infinite sets hold.

Let $X$ be a finitely supported subset of an invariant set $Y$. Then $X$ is FSM Dedekind infinite if and only if there exists a finitely supported one-to-one mapping $f:\mathbb{N}\to X$. As a consequence, an FSM superset of an FSM Dedekind infinite set is FSM Dedekind infinite, and an FSM subset of an FSM set that is not Dedekind infinite is also not FSM Dedekind infinite. 2. 2.

Let $X$ be an infinite finitely supported subset of an invariant set $Y$. Then the sets $\wp_{fs}(\wp_{fin}(X))$ and $\wp_{fs}(T_{fin}(X))$ are FSM Dedekind infinite. 3. 3.

Let $X$ be an infinite finitely supported subset of an invariant set $Y$. Then the set $\wp_{fs}(\wp_{fs}(X))$ is FSM Dedekind infinite. 4. 4.

Let $X$ be a finitely supported subset of an invariant set $Y$ such that $X$ does not contain an infinite subset $Z$ with the property that all the elements of $Z$ are supported by the same set of atoms. Then $X$ is not FSM Dedekind infinite. 5. 5.

Let $X$ be a finitely supported subset of an invariant set $Y$ such that $X$ does not contain an infinite subset $Z$ with the property that all the elements of $Z$ are supported by the same set of atoms. Then $\wp_{fin}(X)$ is not FSM Dedekind infinite. 6. 6.

Let $X$ and $Y$ be two finitely supported subsets of an invariant set $Z$. If neither $X$ nor $Y$ is FSM Dedekind infinite, then $X\times Y$ is not FSM Dedekind infinite. 7. 7.

Let $X$ and $Y$ be two finitely supported subsets of an invariant set $Z$. If neither $X$ nor $Y$ is FSM Dedekind infinite, then $X+Y$ is not FSM Dedekind infinite. 8. 8.

Let $X$ be a finitely supported subset of an invariant set $Y$. Then $\wp_{fs}(X)$ is FSM Dedekind infinite if and only if $X$ is FSM ascending infinite. 9. 9.

Let $X$ be a finitely supported subset of an invariant set $Y$. If $X$ is FSM Dedekind infinite, then $X$ is FSM ascending infinite. The reverse implication is not valid.

Proof.

Let us suppose that $(X,\cdot)$ is FSM Dedekind infinite, and $g:X\rightarrow X$ is an injection supported by the finite set $S\subsetneq A$ with the property that $Im(g)\subsetneq X$. This means that there exists $supp(g)\subseteq S$ and there exists $x_{0}\in X$ such that $x_{0}\notin Im(g)$. We can form a sequence of elements from $X$ which has the first term $x_{0}$ and the general term $x_{n+1}=g(x_{n})$ for all $n\in\mathbb{N}$. Since $x_{0}\notin Im(g)$ it follows that $x_{0}\neq g(x_{0})$. Since $g$ is injective and $x_{0}\notin Im(g)$, by induction we obtain that $g^{n}(x_{0})\neq g^{m}(x_{0})$ for all $n,m\in\mathbb{N}$ with $n\neq m$. Furthermore, $x_{n+1}$ is supported by $supp(g)\cup supp(x_{n})$ for all $n\in\mathbb{N}$. Indeed, let $\pi\in Fix(supp(g)\cup supp(x_{n}))$. According to Proposition 2.11, $\pi\cdot x_{n+1}=\pi\cdot g(x_{n})=g(\pi\cdot x_{n})=g(x_{n})=x_{n+1}$. Since $supp(x_{n+1})$ is the least set supporting $x_{n+1}$, we obtain $supp(x_{n+1})\subseteq supp(g)\cup supp(x_{n})$ for all $n\in\mathbb{N}$. By finite recursion, we have $supp(x_{n})\subseteq supp(g)\cup supp(x_{0})$ for all $n\in\mathbb{N}$. Since all $x_{n}$ are supported by the same set of atoms $supp(g)\cup supp(x_{0})$, we have that the function $f:\mathbb{N}\to X$, defined by $f(n)=x_{n}$, is also finitely supported (by the set $supp(g)\cup supp(x_{0})\cup supp(X)$ not depending on $n$). Indeed, for any $\pi\in Fix(supp(g)\cup supp(x_{0})\cup supp(X))$ we have $f(\pi\diamond n)=f(n)=x_{n}=\pi\cdot x_{n}=\pi\cdot f(n)$, $\forall n\in\mathbb{N}$, where by $\diamond$ we denoted the trivial $S_{A}$-action on $\mathbb{N}$. Furthermore, because $\pi$ fixes $supp(X)$ pointwise we have $\pi\cdot f(n)\in X$ for all $n\in\mathbb{N}$. From Proposition 2.11 we have that $f$ is finitely supported. Obviously, $f$ is also injective.

Conversely, suppose there exists a finitely supported injective mapping $f:\mathbb{N}\to X$. According to Proposition 2.11, it follows that for any $\pi\in Fix(supp(f))$ we have $\pi\cdot f(n)=f(\pi\diamond n)=f(n)$ and $\pi\cdot f(n)\in X$ for all $n\in\mathbb{N}$. Let us define $g:X\to X$ by

[TABLE]

We claim that $g$ is supported by $supp(f)\cup supp(X)$. Indeed, let us consider $\pi\in Fix(supp(f)\cup supp(X))$ and $x\in X$. If there is some $n$ such that $x=f(n)$, we have that $\pi\cdot x=\pi\cdot f(n)=f(n)$, and so $g(\pi\cdot x)=g(f(n))=f(n+1)=\pi\cdot f(n+1)=\pi\cdot g(x)$. If $x\notin Im(f)$, we prove by contradiction that $\pi\cdot x\notin Im(f)$. Indeed, suppose that $\pi\cdot x\in Im(f)$. Then there is $y\in\mathbb{N}$ such that $\pi\cdot x=f(y)$ or, equivalently, $x=\pi^{-1}\cdot f(y)$. However, since $\pi\in Fix(supp(f))$, from Proposition 2.11 we have $\pi^{-1}\cdot f(y)=f(\pi^{-1}\diamond y)$, and so we get $x=f(\pi^{-1}\diamond y)=f(y)\in Im(f)$ which contradicts the assumption that $x\notin Im(f)$. Thus, $\pi\cdot x\notin Im(f)$, and so $g(\pi\cdot x)=\pi\cdot x=\pi\cdot g(x)$. We obtained that $g(\pi\cdot x)=\pi\cdot x=\pi\cdot g(x)$ for all $x\in X$ and all $\pi\in Fix(supp(f)\cup supp(X))$. Furthermore, $\pi\cdot g(x)\in\pi\star X=X$ (where by $\star$ we denoted the $S_{A}$-action on $\wp_{fs}(Y)$), and so $g$ is finitely supported. Since $f$ is injective, it follows immediately that $g$ is injective. Furthermore, $Im(g)=X\setminus\{f(0)\}$ which is a proper subset of $X$, finitely supported by $supp(f(0))\cup supp(X)=supp(f)\cup supp(X)$. 2. 2.

The family $\wp_{fin}(X)$ represents the family of those finite subsets of $X$ (these subsets of $X$ are finitely supported as subsets of the invariant set $Y$ in the sense of Definition 2.6). Obviously, $\wp_{fin}(X)$ is a finitely supported subset of the invariant set $\wp_{fs}(Y)$, supported by $supp(X)$. This is because whenever $Z$ is an element of $\wp_{fin}(X)$ (i.e. whenever $Z$ is a finite subset of $X$) and $\pi$ fixes $supp(X)$ pointwise, we have that $\pi\star Z$ is also a finite subset of $X$. The family $\wp_{fs}(\wp_{fin}(X))$ represents the family of those subsets of $\wp_{fin}(X)$ which are finitely supported as subsets of the invariant set $\wp_{fs}(Y)$ in the sense of Definition 2.6. As above, according to Proposition 2.3, we have that $\wp_{fs}(\wp_{fin}(X))$ is a finitely supported subset of the invariant set $\wp_{fs}(\wp_{fs}(Y))$, supported by $supp(\wp_{fin}(X))\subseteq supp(X)$.

Let $X_{i}$ be the set of all $i$-sized subsets from $X$, i.e. $X_{i}=\{Z\subseteq X\,|\,|Z|=i\}$. Since $X$ is infinite, it follows that each $X_{i},i\geq 1$ is non-empty. Obviously, we have that any $i$-sized subset $\{x_{1},\ldots,x_{i}\}$ of $X$ is finitely supported (as a subset of $Y$) by $supp(x_{1})\cup\ldots\cup supp(x_{i})$. Therefore, $X_{i}\subseteq\wp_{fin}(X)$ and $X_{i}\subseteq\wp_{fs}(Y)$ for all $i\in\mathbb{N}$. Since $\cdot$ is a group action, the image of an $i$-sized subset of $X$ under an arbitrary permutation is an $i$-sized subset of $Y$. However, any permutation of atoms that fixes $supp(X)$ pointwise also leaves $X$ invariant, and so for any permutation $\pi\in Fix(supp(X))$ we have that $\pi\star Z$ is an $i$-sized subset of $X$ whenever $Z$ is an $i$-sized subset of $X$. Thus, each $X_{i}$ is a subset of $\wp_{fin}(X)$ finitely supported by $supp(X)$, and so $X_{i}\in\wp_{fs}(\wp_{fin}(X))$.

We define $f:\mathbb{N}\to\wp_{fs}(\wp_{fin}(X))$ by $f(n)=X_{n}$. We claim that $supp(X)$ supports $f$. Indeed, let $\pi\in Fix(supp(X))$. Since $supp(X)$ supports $X_{n}$ for all $n\in\mathbb{N}$, we have $\pi\star f(n)=\pi\star X_{n}=X_{n}=f(n)=f(\pi\diamond n)$ (where $\diamond$ is the trivial $S_{A}$-action on $\mathbb{N}$) and $\pi\star f(n)=\pi\star X_{n}=X_{n}\in\wp_{fs}(\wp_{fin}(X))$ for all $n\in\mathbb{N}$. According to Proposition 2.11, we have that $f$ is finitely supported. Furthermore, $f$ is injective and, by item 1, we have that $\wp_{fs}(\wp_{fin}(X))$ is FSM Dedekind infinite.

If we consider $Y_{i}$ the set of all $i$-sized injective tuples formed by elements of $X$, we have that each $Y_{i}$ is a subset of $T_{fin}(X)$ supported by $supp(X)$, and the family $(Y_{i})_{i\in\mathbb{N}}$ is a countably infinite, uniformly supported, subset of $\wp_{fs}(T_{fin}(X))$. From item 1 we get that $\wp_{fs}(T_{fin}(X))$ is FSM Dedekind infinite. 3. 3.

The proof is actually the same as in the above item because every $X_{i}\in\wp_{fs}(\wp_{fs}(A))$. 4. 4.

If there does not exist a uniformly supported subset of $X$, then there does not exist a finitely supported injective mapping $f:\mathbb{N}\to X$, and so $f$ cannot be FSM Dedekind infinite. 5. 5.

We prove the following lemma:

Lemma 4.4

Let $X$ be a finitely supported subset of an invariant set $Y$ such that $X$ does not contain an infinite uniformly supported subset. Then the set $\wp_{fin}(X)=\{Z\!\subseteq\!X\,|\,Z\,\text{finite}\}$ does not contain an infinite uniformly supported subset.

Proof of Lemma 4.4. Suppose, by contradiction, that the set $\wp_{fin}(X)$ contains an infinite subset $\mathcal{F}$ such that all the elements of $\mathcal{F}$ are different and supported by the same finite set $S$. Therefore, we can express $\mathcal{F}$ as $\mathcal{F}=(X_{i})_{i\in I}\subseteq\wp_{fin}(X)$ with the properties that $X_{i}\neq X_{j}$ whenever $i\neq j$ and $supp(X_{i})\subseteq S$ for all $i\in I$. Fix an arbitrary $j\in I$. However, from Proposition 2.7, because $supp(X_{j})=\underset{x\in X_{j}}{\cup}supp(x)$, we have that $X_{j}$ has the property that $supp(x)$ $\subseteq S$ for all $x\in X_{j}$. Since $j$ has been arbitrarily chosen from $I$, it follows that every element from every set of form $X_{i}$ is supported by $S$, and so $\underset{i}{\cup}X_{i}$ is an uniformly supported subset of $X$ (all its elements being supported by $S$). Furthermore, $\underset{i\in I}{\cup}X_{i}$ is infinite because the family $(X_{i})_{i\in I}$ is infinite and $X_{i}\neq X_{j}$ whenever $i\neq j$. Otherwise, if $\underset{i}{\cup}X_{i}$ was finite, the family $(X_{i})_{i\in I}$ would be contained in the finite set $\wp(\underset{i}{\cup}X_{i})$, and so it couldn’t be infinite with the property that $X_{i}\neq X_{j}$ whenever $i\neq j$. We were able to construct an infinite uniformly supported subset of $X$, namely $\underset{i}{\cup}X_{i}$, and this contradicts the hypothesis that $X$ does not contain an infinite uniformly supported subset.

Proof of this item According to the above lemma, if $X$ does not contain an infinite uniformly supported subset, then $\wp_{fin}(X)$ does not contain an infinite uniformly supported subset. Suppose, by contradiction, that $\wp_{fin}(X)$ is FSM Dedekind infinite. According to item 1, there exists a finitely supported injective mapping $f:\mathbb{N}\to\wp_{fin}(X)$. Thus, because $\mathbb{N}$ is a trivial invariant set, according to Proposition 2.11, there exists an infinite injective (countable) sequence $f(\mathbb{N})=(X_{i})_{i\in\mathbb{N}}\subseteq\wp_{fin}(X)$ having the property $supp(X_{i})\subseteq supp(f)$ for all $i\in\mathbb{N}$. We obtained that $\wp_{fin}(X)$ contains an infinite uniformly supported subset $(X_{i})_{i\in\mathbb{N}}$, which is a contradiction. 6. 6.

Suppose, by contradiction, that $X\times Y$ is FSM Dedekind infinite. According to item 1, there exists a finitely supported injective mapping $f:\mathbb{N}\to X\times Y$ Thus, according to Proposition 2.11, there exists an infinite injective sequence $f(\mathbb{N})=((x_{i},y_{i}))_{i\in\mathbb{N}}\subseteq X\times Y$ with the property that $supp((x_{i},y_{i}))\subseteq supp(f)$ for all $i\in\mathbb{N}$ (1). Fix some $j\in\mathbb{N}$. We claim that $supp((x_{j},y_{j}))=supp(x_{j})\cup supp(y_{j})$. Let $U=(x_{j},y_{j})$, and $S=supp(x_{j})\cup supp(y_{j})$. Obviously, $S$ supports $U$. Indeed, let us consider $\pi\in Fix(S)$. We have that $\pi\in Fix(supp(x_{j}))$ and also $\pi\in Fix(supp(y_{j}))$ Therefore, $\pi\cdot x_{j}=x_{j}$ and $\pi\cdot y_{j}=y_{j}$, and so $\pi\otimes(x_{j},y_{j})=(\pi\cdot x_{j},\pi\cdot y_{j})=(x_{j},y_{j})$, where $\otimes$ represent the $S_{A}$ action on $X\times Y$ described in Proposition 2.5. Thus, $supp(U)\subseteq S$. It remains to prove that $S\subseteq supp(U)$. Fix $\pi\in Fix(supp(U))$. Since $supp(U)$ supports $U$, we have $\pi\otimes(x_{j},y_{j})=(x_{j},y_{j})$, and so $(\pi\cdot x_{j},\pi\cdot y_{j})=(x_{j},y_{j})$, from which we get $\pi\cdot x_{j}=x_{j}$ and $\pi\cdot y_{j}=y_{j}$. Thus, $supp(x_{j})\subseteq supp(U)$ and $supp(y_{j})\subseteq supp(U)$. Hence $S=supp(x_{j})\cup supp(y_{j})\subseteq supp(U)$.

According to relation (1) we obtain, $supp(x_{i})\cup supp(y_{i})\subseteq supp(f)$ for all $i\in\mathbb{N}$. Thus, $supp(x_{i})\subseteq supp(f)$ for all $i\in\mathbb{N}$ and $supp(y_{i})\subseteq supp(f)$ for all $i\in\mathbb{N}$ (2). Since the sequence $((x_{i},y_{i}))_{i\in\mathbb{N}}$ is infinite and injective, then at least one of the sequences $(x_{i})_{i\in\mathbb{N}}$ and $(y_{i})_{i\in\mathbb{N}}$ is infinite. Assume that $(x_{i})_{i\in\mathbb{N}}$ is infinite. Then there exists an infinite subset $B$ of $\mathbb{N}$ such that $(x_{i})_{i\in B}$ is injective, and so there exists an injection $u:B\to X$ defined by $u(i)=x_{i}$ for all $i\in B$ which is supported by $supp(f)$ (according to relation (2) and Proposition 2.11). However, since $B$ is an infinite subset of $\mathbb{N}$, there exists a ZF bijection $h:\mathbb{N}\to B$. The construction of $h$ requires only the fact that $\mathbb{N}$ is well-ordered which is obtained from the Peano construction of $\mathbb{N}$ and does not involve a form of the axiom of choice. Since both $B$ and $\mathbb{N}$ are trivial invariant sets, it follows that $h$ is equivariant. Thus, $u\circ h$ is an injection from $\mathbb{N}$ to $X$ which is finitely supported by $supp(u)\subseteq supp(f)$. This contradicts the assumption that $X$ is not FSM Dedekind infinite.

Remark 4.5

Analogously, using the relation $supp(x)\cup supp(y)=supp((x,y))$ for all $x\in X$ and $y\in Y$ derived from Proposition 2.7, it can be proved that $X\times Y$ does not contain an infinite uniformly supported subset if neither $X$ nor $Y$ contain an infinite uniformly supported subset. 7. 7.

Suppose, by contradiction, that $X+Y$ is FSM Dedekind infinite. According to item 1, there exists a finitely supported injective mapping $f:\mathbb{N}\to X+Y$. Thus, there exists an infinite injective sequence $(z_{i})_{i\in\mathbb{N}}\subseteq X+Y$ such that $supp(z_{i})\subseteq supp(f)$ for all $i\in\mathbb{N}$. According to the construction of the disjoint union of two $S_{A}$-sets (see Proposition 2.5), as in the proof of item 6, there should exist an infinite subsequence of $(z_{i})_{i}$ of form $((0,x_{j}))_{x_{j}\in X}$ which is uniformly supported by $supp(f)$, or an infinite sequence of form $((1,y_{k}))_{y_{k}\in Y}$ which is uniformly supported by $supp(f)$. Since [math] and $1$ are constants, this means there should exist at least an infinite uniformly supported sequence of elements from $X$, or an infinite uniformly supported sequence of elements from $Y$. This contradicts the hypothesis neither $X$ nor $Y$ is FSM Dedekind infinite.

Remark 4.6

Analogously, it can be proved that $X+Y$ does not contain an infinite uniformly supported subset if neither $X$ nor $Y$ contain an infinite uniformly supported subset. 8. 8.

Assume, by contradiction, that $(X_{n})_{n\in\mathbb{N}}$ is an infinite countable family of different subsets of $X$ such that the mapping $n\mapsto X_{n}$ is finitely supported. Thus, each $X_{n}$ is supported by the same set $S=supp(n\mapsto X_{n})$. We define a countable family $(Y_{n})_{n\in\mathbb{N}}$ of subsets of $X$ that are non-empty and pairwise disjoint. A ZF construction of such a family belongs to Kuratowski and can also be found in Lemma 4.11 from [4]. This approach works also in FSM in the view of the $S$-finite support principle because every $Y_{k}$ is defined only involving elements in the family $(X_{n})_{n\in\mathbb{N}}$, and so whenever $(X_{n})_{n\in\mathbb{N}}$ is uniformly supported (meaning that all $X_{n}$ are supported by the same set of atoms), we get that $(Y_{n})_{n\in\mathbb{N}}$ is uniformly supported. Formally the sequence $(Y_{n})_{n\in\mathbb{N}}$ is recursively constructed as below. For $n\in\mathbb{N}$, assume that $Y_{m}$ is defined for any $m<n$ such that the set $\{X_{k}\setminus\underset{m<n}{\cup}Y_{m}\,|\,k\geq n\}$ is infinite. Define $n^{\prime}=min\{k\,|\,k\geq n\>\text{and}\>X_{k}\setminus\underset{m<n}{\cup}Y_{m}\neq\emptyset\>\text{and}\>(X\setminus X_{k})\setminus\underset{m<n}{\cup}Y_{m}\neq\emptyset\}$. We define

[TABLE]

Obviously, $Y_{1}$ is supported by $S\cup supp(X)$. By induction, assume that $Y_{m}$ is supported by $S\cup supp(X)$ for each $m<n$. Since $Y_{n}$ is defined as a set combination of $X_{i}$’s (which are all $S$-supported) and $Y_{m}$’s with $m<n$, we get that $Y_{n}$ is supported by $S\cup supp(X)$ according to the $S$-finite support principle. Therefore the family $(Y_{i})_{i\in\mathbb{N}}$ is uniformly supported by $S\cup supp(X)$. Let $U_{i}=Y_{0}\cup\ldots\cup Y_{i}$ for all $i\in\mathbb{N}$. Clearly all $U_{i}$ are supported by $S\cup supp(X)$, and $U_{0}\subsetneq U_{1}\subsetneq U_{2}\subsetneq\ldots\subsetneq X$. Let $V_{n}=(X\setminus\underset{i\in\mathbb{N}}{\cup}U_{i})\cup U_{n}$. Clearly, $X=\underset{n\in\mathbb{N}}{\cup}V_{n}$. Moreover, $V_{n}$ is supported by $S\cup supp(X)$ for all $n\in\mathbb{N}$. Therefore, the mapping $n\mapsto V_{n}$ is finitely supported. Obviously, $V_{0}\subsetneq V_{1}\subsetneq V_{2}\subsetneq\ldots\subsetneq X$. However, there does not exist $n\in\mathbb{N}$ such that $X=V_{n}$, and so $X$ is FSM ascending infinite.

The converse holds since if $X$ is FSM ascending infinite, there is a finitely supported increasing countable chain of finitely supported sets $X_{0}\subseteq X_{1}\subseteq\ldots\subseteq X_{n}\subseteq\ldots$ with $X\subseteq\cup X_{n}$, but there does not exist $n\in\mathbb{N}$ such that $X\subseteq X_{n}$. In this sequence there should exist infinitely many different elements of form $X_{i}$ (otherwise their union will be a term of the sequence), and the result follows from Proposition 5.4. 9. 9.

Suppose $X$ is FSM Dedekind infinite. Therefore, $\wp_{fs}(X)$ is FSM Dedekind infinite. According to item 8, we have that $X$ is FSM ascending infinite. The reverse implication is not valid because, as it is proved in Proposition 4.18, $\wp_{fin}(A)$ is FSM ascending infinite, but not FSM Dedekind infinite.

Corollary 4.7

The following sets and all of their FSM usual infinite subsets are FSM usual infinite, but they are not FSM Dedekind infinite.

The invariant set $A$ of atoms. 2. 2.

The powerset $\wp_{fs}(A)$ of the set of atoms. 3. 3.

The set $T_{fin}(A)$ of all finite injective tuples of atoms. 4. 4.

The invariant set $A^{A}_{fs}$ of all finitely supported functions from $A$ to $A$. 5. 5.

The invariant set of all finitely supported functions $f:A\to A^{n}$, where $n\in\mathbb{N}$. 6. 6.

The invariant set of all finitely supported functions $f:A\to T_{fin}(A)$. 7. 7.

The invariant set of all finitely supported functions $f:A\to\wp_{fs}(A)$. 8. 8.

The sets $\wp_{fin}(A)$, $\wp_{cofin}(A)$, $\wp_{fin}(\wp_{fs}(A))$, $\wp_{fin}(\wp_{cofin}(A))$, $\wp_{fin}(\wp_{fin}(A))$, $\wp_{fin}(A^{A}_{fs})$. 9. 9.

Any construction of finite powersets of form $\wp_{fin}(\ldots\wp_{fin}(A))$, $\wp_{fin}(\ldots\wp_{fin}(P(A)))$, or $\wp_{fin}(\ldots\wp_{fin}(\wp_{fs}(A)))$. 10. 10.

Every finite Cartesian combination between the set $A$, $\wp_{fin}(A)$, $\wp_{cofin}(A)$, $\wp_{fs}(A)$ and $A^{A}_{fs}$ . 11. 11.

The disjoint unions $A+A^{A}_{fs}$, $A+\wp_{fs}(A)$, $\wp_{fs}(A)+A^{A}_{fs}$ and $A+\wp_{fs}(A)+A^{A}_{fs}$ and all finite disjoint unions between $A$, $A^{A}_{fs}$ and $\wp_{fs}(A)$.

Proof.

$A$ does not contain an infinite uniformly supported subset, and so it is not FSM Dedekind infinite (according to Theorem 4.3(4)). 2. 2.

$\wp_{fs}(A)$ does not contain an infinite uniformly supported subset because for any finite set $S$ of atoms there exist only finitely many elements of $\wp_{fs}(A)$ supported by $S$, namely the subsets of $S$ and the supersets of $A\setminus S$. Thus, $\wp_{fs}(A)$ it is not FSM Dedekind infinite (Theorem 4.3(4)). 3. 3.

$T_{fin}(A)$ does not contain an infinite uniformly supported subset because the finite injective tuples of atoms supported by a finite set $S$ are only those injective tuples formed by elements of $S$, being at most $1+A_{|S|}^{1}+A_{|S|}^{2}+\ldots+A_{|S|}^{|S|}$ such tuples, where $A_{n}^{k}=n(n-1)\ldots(n-k+1)$. 4. 4.

We prove the following lemmas.

Lemma 4.8

Let $S=\{s_{1},\ldots,s_{n}\}$ be a finite subset of an invariant set $(U,\cdot)$ and $X$ a finitely supported subset of an invariant set $(V,\diamond)$. Then if $X$ is does not contain an infinite uniformly supported subset, we have that $X^{S}_{fs}$ does not contain an infinite uniformly supported subset.

Proof of Lemma 4.8. First we prove that there is an FSM injection $g$ from $X^{S}_{fs}$ into $X^{|S|}$. For $f\in X^{S}_{fs}$ define $g(f)=(f(s_{1}),\ldots,f(s_{n}))$. Clearly $g$ is injective (and it is also surjective). Let $\pi\in Fix(supp(s_{1})\cup\ldots\cup supp(s_{n})\cup supp(X))$. Thus, $g(\pi\widetilde{\star}f)=(\pi\diamond f(\pi^{-1}\cdot s_{1}),\ldots,\pi\diamond f(\pi^{-1}\cdot s_{n}))=(\pi\diamond f(s_{1}),\ldots,\pi\diamond f(s_{n}))$ $=\pi\otimes g(f)$ for all $f\in X^{S}_{fs}$, where $\otimes$ is the $S_{A}$-action on $X^{|S|}$ defined as in Proposition 2.5. Hence $g$ is finitely supported, and the conclusion follows from Theorem 4.3(1) and by repeatedly applying similar arguments as in Theorem 4.3(6) (if we slightly modify the proof of the theorem, using the fact that $supp(x)\cup supp(y)=supp((x,y))$ for all $x,y\in X$, we show that the $|S|$-time Cartesian product of $X$, i.e. $X^{|S|}$ does not contain an infinite uniformly supported subset; otherwise $X$ should contain itself an infinite uniformly supported subset, which contradicts the hypothesis).

Lemma 4.9

Let $S=\{s_{1},\ldots,s_{n}\}$ be a finite subset of an invariant set $(U,\cdot)$ and $X$ a finitely supported subset of an invariant set $(V,\diamond)$. Then if $X$ is not FSM Dedekind infinite, we have that $X^{S}_{fs}$ is not FSM Dedekind-infinite.

Proof of Lemma 4.9 First we proved that there is an FSM injection $g$ from $X^{S}_{fs}$ into $X^{|S|}$. The conclusion follows from Theorem 4.3(1) and by repeatedly applying Theorem 4.3(6) (from which we know that the $|S|$-time Cartesian product of $X$, i.e. $X^{|S|}$, is not FSM Dedekind infinite).

Lemma 4.10

Let $f:A\to A$ be a function that is finitely supported by a certain finite set of atoms $S$. Then either $f|_{A\setminus S}=Id$ or $f|_{A\setminus S}$ is an one-element subset of $S$.

Proof of Lemma 4.10 Let $f:A\to A$ be a function that is finitely supported by the finite set of atoms $S$. We distinguish two cases:

I. There is $a\notin S$ with $f(a)=a$. Then for each $b\notin S$ we have that $(a\,b)\in Fix(S)$, and so $f(b)=f((a\,b)(a))=(a\,b)(f(a))=(a\,b)(a)=b$. Thus, $f|_{A\setminus S}=Id$.

II. For all $a\notin S$ we have $f(a)\neq a$. We claim that $f(a)\in S$ for all $a\notin S$. Suppose, by contradiction, that $f(a)=b\in A\setminus S$ for a certain $a\notin S$. Thus, $(a\,b)\in Fix(S)$, and so $f(b)=f((a\,b)(a))=(a\,b)(f(a))=(a\,b)(b)=a$. Let us consider $c\in A\setminus S$, $c\neq a,b$. Thus, $(a\,c)\in Fix(S)$, and so $f(c)=f((a\,c)(a))=(a\,c)(f(a))=(a\,c)(b)=b$. Furthermore, $(b\,c)\in Fix(S)$, and so $f(b)=f((b\,c)(c))=(b\,c)(f(c))=(b\,c)(b)=c$. However, $f(b)=a$ which contradicts the functionality of $f$. Thus $f(a)\in S$ for any $a\notin S$. If $x,y\notin S$, then we should have $f(x),f(y)\in S$, and so, because $(x\,y)\in Fix(S)$, we get $f(x)=f((x\,y)(y))=(x\,y)(f(y))=f(y)$ since both $x$ and $y$ belong to $A\setminus S$ which means they are different from $f(y)$ belonging to $S$. Therefore there is $x_{0}\in S$ such that $f|_{A\setminus S}$ $=\{x_{0}\}$.

Proof of this item. Assume, by contradiction, that $A^{A}_{fs}$ contains an infinite, uniformly supported subset, meaning that there are infinitely many functions from $A$ to $A$ supported by the same finite set $S$. According to Lemma 4.10, any $S$-supported function $f:A\to A$ should have the property that either $f|_{A\setminus S}=Id$ or $f|_{A\setminus S}$ is an one-element subset of $S$. A function from $A$ to $A$ is precisely characterized by the set of values it takes on the elements of $S$ and on the elements of $A\setminus S$, respectively. For each possible definition of such an $f$ on $S$ we have at most $|S|+1$ possible ways to define $f$ on $A\setminus S$. Since we assumed that there exist infinitely many finitely supported functions from $A$ to $A$ supported by the same set $S$, there should exist infinitely many finitely supported functions from $S$ to $A$ supported by the set $S$. But this is a contradiction according to Lemma 4.8 which states that $A^{S}_{fs}$ is does not contain an infinite uniformly supported subset (because $A$ does not contain an infinite uniformly supported subset). 5. 5.

There is an equivariant bijective mapping between $(A^{n})^{A}_{fs}$ and $(A^{A}_{fs})^{n}$ defined as follows. If $f:A\to A^{n}$ is a finitely supported function with $f(a)=(a_{1},\ldots,a_{n})$, we associate to $f$ the Cartesian pair $(f_{1},\ldots,f_{n})$ where for each $i\in\mathbb{N}$, $f_{i}:A\to A$ is defined by $f_{i}(a)=a_{i}$ for all $a\in A$. We omit technical details since they are based only on the application of Proposition 2.11. We proved above that $A^{A}_{fs}$ does not contain an infinite uniformly supported subset, and so neither $(A^{A}_{fs})^{n}$ contains an infinite uniformly supported subset by involving a similar proof as of Theorem 4.3(6) (see the proof of Lemma 4.8). 6. 6.

Assume by contradiction that $T_{fin}(A)^{A}$ contains an infinite $S$-uniformly supported subset. If $f:A\to T_{fin}(A)$ is a function supported by $S$, then consider $f(a)=x$ for some $a\notin S$. For $b\notin S$ we have $(a\,b)\in Fix(S)$, and so $f(b)=f((a\,b)(a))=(a\,b)\otimes f(a)=(a\,b)\otimes x$ which means $|f(a)|=|f(b)|$ for all $a,b\notin S$. Each $S$-supported function $f:A\to T_{fin}(A)$ is fully described the values it takes on the elements of $S$ and on the elements of $A\setminus S$, respectively, i.e., by the elements of $f(S)$ and of $f(A\setminus S)$. More precisely, each $S$-supported function $f:A\to T_{fin}(A)$ can be uniquely decomposed into two $S$-supported functions $f|_{S}$ and $f|_{A\setminus S}$ (this follows from Proposition 2.11 and because both $S$ and $A\setminus S$ are supported by $S$). However, $f(A\setminus S)\subseteq A^{\prime n}$ for some $n\in\mathbb{N}$, where $A^{\prime n}$ is the set of all injective $n$-tuples of $A$. According to Lemma 4.8 we have at most finitely many $S$-supported functions from $S$ to $T_{fin}(A)$. According to item 5, we have at most finitely many $S$-supported functions from $A\setminus S$ to $A^{\prime n}$ for each fixed $n\in\mathbb{N}$. This is because $A^{\prime n}$ is a subset of $A^{n}$ and $A\setminus S$ is a subset of $A$, and so by involving Proposition 3.11(3) and (4) we find a finitely supported injection $\varphi$ from $(A^{\prime n})^{A\setminus S}$ and $(A^{n})^{A}$; if $\mathcal{K}$ was an infinite subset in $(A^{\prime n})^{A\setminus S}$ uniformly supported by $T$, then $\varphi(\mathcal{K})$ would be an infinite subset of $(A^{n})^{A}$ uniformly supported by $T\cup supp(\varphi)$. Therefore, there should exist an infinite subset $M\subseteq\mathbb{N}$ such that we have at least one $S$-supported function $g:A\setminus S\to A^{\prime k}$ for any $k\in M$. We do not need to find a set of representatives for such $g$’s; we consider all of them. Fix $a\in A\setminus S$. For each of the above $g$’s (that form an $S$-supported family $\mathcal{F}$) we have that $g(a)$’s form an uniformly supported family (by $S\cup\{a\}$) of $T_{fin}(A)$, which is also infinite because tuples having different cardinalities are different and $M$ is infinite. However, we contradict the proof of item 3 stating that $T_{fin}(A)$ does not contain an infinite uniformly supported subset. Alternatively, one can remark that if $|S\cup\{a\}|=l$ with $l$ fixed, then there is $m\in M$ fixed with $m>l$. Moreover, $g(a)$ for some $g:A\setminus S\to A^{\prime m}$ in $\mathcal{F}$ (we need to select only a function from those functions $g:A\setminus S\to A^{\prime m}$ with $m$ fixed depending only on the fixed $l$, and not a set of representatives for the entire family $(\{g:A\setminus S\to A^{\prime k}\})_{k\in M}$), which is an injective $m$-tuple of atoms, cannot be supported by $S\cup\{a\}$; thus, the set of all $g(a)$’s cannot be infinite and uniformly supported. 7. 7.

We can use a similar approach as in item 6, to prove that there exist at most finitely many $S$-supported functions from $A$ to $\wp_{fin}(A)$. For this we just replace $A^{\prime n}$ with the set of all $n$-sized subsets of $A$, $\wp_{n}(A)$. All it remains is to prove that, for each $n\in\mathbb{N}$, there cannot exist infinitely many functions $g:A\to\wp_{n}(A)$ supported by the same set $S^{\prime}$. Fix $n\in\mathbb{N}$. Assume, by contradiction that there exist infinitely many functions $g:A\to\wp_{n}(A)$ supported by the same set $S^{\prime}$. According to Lemma 4.8 there are only finitely many functions from $S^{\prime}$ to $\wp_{n}(A)$ supported by the same set of atoms, and so there should exist infinitely many functions $g:(A\setminus S^{\prime})\to\wp_{n}(A)$ supported by $S^{\prime}$. For such a $g$, let us fix an element $a\in A$ with $a\notin S^{\prime}$. There exist $x_{1},\ldots,x_{n}\in A$ fixed (depending only on the fixed $a$) and different such that $g(a)=\{x_{1},\ldots,x_{n}\}$. Let $b$ be an arbitrary element from $A\setminus S^{\prime}$, and so $(a\,b)\in Fix(S^{\prime})$ which means $g(b)=g((a\,b)(a))=(a\,b)\star g(a)=(a\,b)\star\{x_{1},\ldots,x_{n}\}=\{(a\,b)(x_{1}),\ldots,(a\,b)(x_{n})\}$.

We analyze the two possibilities:

Case 1: One of $x_{1},\ldots,x_{n}$ coincides to $a$. Suppose $x_{1}=a$. We claim that $x_{2},\ldots,x_{n}\in S^{\prime}$. Assume the contrary, that is, there exists $i\in\{2,\ldots,n\}$ such that $x_{i}\notin S^{\prime}$. Without losing the generality suppose $x_{2}\notin S^{\prime}$, which means $(a\,x_{2})\in Fix(S^{\prime})$, and so $g(x_{2})=g((a\,x_{2})(a))=(a\,x_{2})\star g(a)=(a\,x_{2})\star\{a,x_{2},\ldots,x_{n}\}=\{a,x_{2},\ldots,x_{n}\}$. Let $c\in A\setminus S^{\prime}$ with $c$ different from $a,x_{2},\ldots,x_{n}$. We have $g(c)=g((a\,c)(a))=(a\,c)\star g(a)=(a\,c)\star\{a,x_{2},\ldots,x_{n}\}=\{c,x_{2},\ldots,x_{n}\}$, and hence $g(x_{2})=g((c\,x_{2})(c))=(c\,x_{2})\star g(c)=(c\,x_{2})\star\{c,x_{2},\ldots,x_{n}\}=\{c,x_{2},\ldots,x_{n}\}$ which contradicts the functionality of $g$. Therefore, $g(b)=(b,x_{2},\ldots,x_{n})$ for all $b\in A\setminus S^{\prime}$, and so only the selection of $x_{2},\ldots x_{n}$ provides the distinction between $g$’s. Since $S^{\prime}$ is finite, $\{x_{2},\ldots,x_{n}\}$ can be selected in $C_{|S^{\prime}|}^{n-1}$ ways if $|S^{\prime}|\geq n-1$, or in [math] ways otherwise.

Case 2: Consider now that all $x_{1},\ldots,x_{n}$ are different from $a$.

Then g(b)=\left\{\begin{array}[]{ll}\{x_{1},\ldots,x_{n}\},&\text{if}\>\text{b\neq x_{1},\ldots,x_{n} };\\ \{a,x_{2},\ldots,x_{n}\},&\text{if}\>\text{x_{1}\notin S^{\prime}$and$b=x_{1} };\\ \ldots\\ \{x_{1},\ldots,x_{n-1},a\},&\text{if}\>\text{x_{n}\notin S^{\prime}$and$b=x_{n}}\>.\end{array}\right.

Since $x_{1},\ldots,x_{n},a$ are fixed atoms, then $g(A\setminus S^{\prime})$ is finite. However, $Im(g)$ should be supported by $S^{\prime}$. According to Proposition 2.7, since $Im(g)$ is finite, it should be uniformly supported by $S^{\prime}$. We obtain that $x_{1},\ldots,x_{n}\in S^{\prime}$, and so $g(A\setminus S^{\prime})=\{x_{1},\ldots,x_{n}\}$. Otherwise, if some $x_{i}\notin S^{\prime}$, we would get $\{x_{1},\ldots,a,\ldots,x_{n}\}\in Im(g)$ (where $a$ replaces $x_{i}$) and so $\{x_{1},\ldots,a,\ldots,x_{n}\}$ is supported by $S^{\prime}$. Again by Proposition 2.7 we would have that $a$ is supported by $S^{\prime}$ which means $\{a\}=supp(a)\subseteq S^{\prime}$ contradicting the choice of $a$. Alternatively, for proving that all $x_{1},\ldots,x_{n}\in S^{\prime}$, assume by contradiction that one of them (say $x_{1}$) does not belong to $S^{\prime}$. Let $c$ be an atom from $A\setminus S^{\prime}$ with $c$ different from $a,x_{1},x_{2},\ldots,x_{n}$. We have $g(c)=g((a\,c)(a))=(a\,c)\star g(a)=(a\,c)\star\{x_{1},\ldots,x_{n}\}=\{x_{1},\ldots,x_{n}\}$, and hence $g(x_{1})=g((c\,x_{1})(c))=(c\,x_{1})\star g(c)=(c\,x_{1})\star\{x_{1},x_{2},\ldots,x_{n}\}=\{c,x_{2},\ldots,x_{n}\}$. However $g(x_{1})=\{a,x_{2},\ldots,x_{n}\}$ which contradicts the functionality of $g$. Since $S^{\prime}$ is finite, $\{x_{1},\ldots,x_{n}\}$ can be selected in $C_{|S^{\prime}|}^{n}$ ways $|S^{\prime}|\geq n$ or in [math] ways otherwise.

In either case, there couldn’t exist infinitely many $g$’s supported by $S^{\prime}$, and so for each $n\in\mathbb{N}$, there exist at most finitely many functions from $A$ to $\wp_{n}(A)$ supported by the same set of atoms.

Assume by contradiction that $\wp_{fin}(A)^{A}$ contains an infinite $S$-uniformly supported subset. If $f:A\to\wp_{fin}(A)$ is a function supported by $S$, then we have $|f(a)|=|(a\,b)\star f(a)|=|f((a\,b)(a))|=|f(b)|$ for all $a,b\notin S$. According to Proposition 2.11 (since both $S$ and $A\setminus S$ are supported by $S$), each $S$-supported function $f:A\to\wp_{fin}(A)$ is uniquely decomposed into two $S$-supported functions $f|_{S}$ and $f|_{A\setminus S}$. However $f(A\setminus S)\subseteq\wp_{n}(A)$ for some $n\in\mathbb{N}$. According to Lemma 4.8 there are at most finitely many $S$-supported functions from $S$ to $\wp_{fin}(A)$. Furthermore, there exist at most finitely many $S$-supported functions from $A\setminus S$ to $\wp_{n}(A)$ for each fixed $n\in\mathbb{N}$. Therefore, there should exist an infinite subset $M\subseteq\mathbb{N}$ such that we have at least one $S$-supported function $g:A\setminus S\to\wp_{k}(A)$ for any $k\in M$. Fix $a\in A\setminus S$. For each of the above $g$’s (that form an $S$-supported family $\mathcal{F}$) we have that $g(a)$’s form an uniformly supported family (by $S\cup\{a\}$) of $\wp_{fin}(A)$. If $|S\cup\{a\}|=l$ with $l$ fixed, then there is $m\in M$ fixed with $m>l$. Moreover, $g(a)$ for $g:A\setminus S\to\wp_{m}(A)\in\mathcal{F}$, which is an $m$-sized subset of atoms, cannot be supported by $S\cup\{a\}$ (according to Proposition 2.7); thus, the set of all $g(a)$’s cannot be infinite and uniformly supported.

Analogously, there there exist at most finitely many $S$-supported functions from $A$ to $\wp_{cofin}(A)$ (using eventually the fact that there is an equivariant bijection $X\mapsto A\setminus X$ between $\wp_{fin}(A)$ and $\wp_{cofin}(A)$). Assume by contradiction that $\wp_{fs}(A)^{A}$ contains an infinite $S$-uniformly supported subset. If $f:A\to\wp_{fs}(A)$ is a function supported by $S$, then consider $f(a)=X$ for some $a\notin S$. For $b\notin S$ we have $f(b)=(a\,b)\star X$ which means $f(A\setminus S)$ is formed only by finite subsets of atoms if $X$ is finite, and $f(A\setminus S)$ is formed only by cofinite subsets of atoms if $X$ is cofinite. Thus, whenever $f:A\to\wp_{fs}(A)$ is a function supported by $S$, we have either $f(A\setminus S)\subseteq\wp_{fin}(A)$ or $f(A\setminus S)\subseteq\wp_{cofin}(A)$. Each $S$-supported function $f:A\to\wp_{fs}(A)$ is fully described by $f(S)$ and $f(A\setminus S)$. According to Lemma 4.8 we have at most finitely many $S$-supported functions from $S$ to $\wp_{fs}(A)$. Furthermore, we have at most finitely many $S$-supported functions from $A\setminus S$ to $\wp_{fin}(A)$, and at most finitely many $S$-supported functions from $A\setminus S$ to $\wp_{cofin}(A)$. Thus, $\wp_{fs}(A)^{A}$ does not contain an infinite uniformly supported subset. 8. 8.

The sets $\wp_{fin}(A)$, $\wp_{cofin}(A)$, $\wp_{fin}(\wp_{fs}(A))$, $\wp_{fin}(\wp_{cofin}(A))$, $\wp_{fin}(\wp_{fin}(A))$, $\wp_{fin}(A^{A}_{fs})$ do not contain infinite uniformly supported subsets, and so they are not FSM Dedekind infinite (Theorem 4.3(5)). 9. 9.

Directly from Theorem 4.3(5). 10. 10.

According to Theorem 4.3(6). 11. 11.

According to Theorem 4.3(7).

Corollary 4.11

There exist two FSM sets that whose cardinalities incomparable via the relation $\leq$ on cardinalities, and none of them is FSM Dedekind infinite.

Proof. According to Corollary 4.7, none of the sets $A\times A$ and $\wp_{fs}(A)$ is FSM Dedekind infinite. According to Theorem 3.13, there does not exist a finitely supported injective mapping $f:A\times A\to\wp_{fs}(A)$. According to Lemma 11.10 from [5] that is preserved in FSM (proof omitted) there does not exist a finitely supported injective mapping $f:\wp_{fs}(A)\to A\times A$.

Corollary 4.12

The following sets and all of their supersets, their powersets and the families of their finite subsets, are both FSM usual infinite and FSM Dedekind infinite.

The invariant sets $\wp_{fs}(\wp_{fs}(A))$, $\wp_{fs}(\wp_{fin}(A))$ and $\mathbb{N}$. 2. 2.

The set of all finitely supported mappings from $X$ to $Y$, and the set of all finitely supported mappings from $Y$ to $X$, where $X$ is a finitely supported subset of an invariant set with at least two elements, and $Y$ is an FSM Dedekind infinite set. 3. 3.

The set of all finitely supported functions $f:\wp_{fin}(Y)\to X$ and the set of all finitely supported functions $f:\wp_{fs}(Y)\to X$, where $Y$ is an infinite finitely supported subset of an invariant set, and $X$ is a finitely supported subset of an invariant set with at least two elements. 4. 4.

The set $T^{\delta}_{fin}(A)=\underset{n\in\mathbb{N}}{\cup}A^{n}$ of all finite tuples of atoms (not necessarily injective).

Proof.

This follows from Theorem 4.3(3) and Theorem 4.3 (2). 2. 2.

Let $(y_{n})_{n\in\mathbb{N}}$ be an injective, uniformly supported, countable sequence in $Y$ (that exists from Theorem 4.3(1)). Thus, each $y_{n}$ is supported by the same set $S$ of atoms. In $Y^{X}$ we consider the injective family $(f_{n})_{n\in\mathbb{N}}$ of functions from $X$ to $Y$ where for each $i\in\mathbb{N}$ we define $f_{i}(x)=y_{i}$ for all $x\in X$. According to Proposition 2.11, each $f_{i}$ is supported by $S$, and so is the infinite family $(f_{n})_{n\in\mathbb{N}}$, meaning that there is an $S$-supported injective mapping from $\mathbb{N}$ to $Y^{X}$. In this case it is necessary to require only that $X$ is non-empty.

Fix two different elements $x_{1},x_{2}\in X$. Take $\mathcal{F}=(y_{n})_{n\in\mathbb{N}}$ an injective, uniformly supported, countable sequence in $Y$. In $X^{Y}$ we consider the injective family $(g_{n})_{n\in\mathbb{N}}$ of functions from $Y$ to $X$ where for each $i\in\mathbb{N}$ we define g_{i}(y)=\left\{\begin{array}[]{ll}x_{1}&\text{if}\>y=y_{i}\\ x_{2}&\text{if}\>y=y_{j}\;\text{with}\;j\neq i,\;\text{or}\;y\notin\mathcal{F}\end{array}\right.. According to Proposition 2.11, each $g_{i}$ is supported by the finite set $supp(x_{1})\cup supp(x_{2})\cup supp(\mathcal{F})$, and so the infinite family $(g_{n})_{n\in\mathbb{N}}$ is uniformly supported meaning that there is an injective mapping from $\mathbb{N}$ to $X^{Y}$ supported by $supp(x_{1})\cup supp(x_{2})\cup supp(\mathcal{F})$. 3. 3.

From Theorem 3.12, there exists a one-to-one mapping from $\wp_{fs}(U)$ onto $\{0,1\}^{U}_{fs}$ for an arbitrary finitely supported subset of an invariant set $U$. Fix two distinct elements $x_{1},x_{2}\in X$. There exists a finitely supported (by $supp(x_{1})\cup supp(x_{2})$) bijective mapping from $\{0,1\}^{U}_{fs}$ to $\{x_{1},x_{2}\}^{U}_{fs}$ which associates to each $f\in\{0,1\}^{U}_{fs}$ an element $g\in\{x_{1},x_{2}\}^{U}_{fs}$ defined by g(x)=\left\{\begin{array}[]{ll}x_{1}&\text{for}\>f(x)=0\\ x_{2}&\text{for}\>f(x)=1\end{array}\right. for all $x\in U$ and supported by $supp(x_{1})\cup supp(x_{2})\cup supp(f)$. Obviously, there is a finitely supported injection between $\{x_{1},x_{2}\}^{U}_{fs}$ and $X^{U}_{fs}$. Thus, there is a finitely supported injection from $\wp_{fs}(U)$ into $X^{U}_{fs}$. If we take $U=\wp_{fin}(Y)$ or $U=\wp_{fs}(Y)$, the result follows from Theorem 4.3(1), Theorem 4.3(2) and Theorem 4.3(3). 4. 4.

Fix $a\in A$ and $i\in\mathbb{N}$. We consider the tuple $x_{i}=(a,\ldots,a)\in A^{i}$. Clearly $x_{i}$ is supported by $\{a\}$ for each $i\in\mathbb{N}$, and so $(x_{n})_{n\in\mathbb{N}}$ is a uniformly supported subset of $T^{\delta}_{fin}(A)$.

Proposition 4.13

Let $X$ be a finitely supported subset of an invariant set such that $\wp_{fs}(X)$ is not FSM Dedekind infinite. Then each finitely supported surjective mapping $f:X\to X$ should be injective.

Proof. Let $f:X\to X$ be a finitely supported surjection. Since $f$ is surjective, we can define the function $g:\wp_{fs}(X)\rightarrow\wp_{fs}(X)$ by $g(Y)=f^{-1}(Y)$ for all $Y\in\wp_{fs}(X)$ which is finitely supported and injective according to Lemma 3.7. Since $\wp_{fs}(X)$ is not FSM Dedekind infinite, it follow that $g$ is surjective.

Now let us consider two elements $a,b\in X$ such that $f(a)=f(b)$. We prove by contradiction that $a=b$. Suppose that $a\neq b$. Let us consider $Y=\{a\}$ and $Z=\{b\}$. Obviously, $Y,Z\in\wp_{fs}(X)$. Since $g$ is surjective, for $Y$ and $Z$ there exist $Y_{1},Z_{1}\in\wp_{fs}(X)$ such that $f^{-1}(Y_{1})=g(Y_{1})=Y$ and $f^{-1}(Z_{1})=g(Z_{1})=Z$. We know that $f(Y)\cap f(Z)=\{f(a)\}$. Thus, $f(a)\in f(Y)=f(f^{-1}(Y_{1}))\subseteq Y_{1}$. Similarly, $f(a)=f(b)\in f(Z)=f(f^{-1}(Z_{1}))\subseteq Z_{1}$, and so $f(a)\in Y_{1}\cap Z_{1}$. Thus, $a\in f^{-1}(Y_{1}\cap Z_{1})=f^{-1}(Y_{1})\cap f^{-1}(Z_{1})=Y\cap Z$. However, since we assumed that $a\neq b$, we have that $Y\cap Z=\emptyset$, which represents a contradiction. It follows that $a=b$, and so $f$ is injective.

Proposition 4.14

Let $X$ be a finitely supported subset of an invariant set. If $\wp_{fin}(X)$ is FSM Dedekind infinite, then $X$ should be FSM non-uniformly amorphous, meaning that $X$ should contain two disjoint, infinite, uniformly supported subsets. 2. 2.

Let $X$ be a finitely supported subset of an invariant set. If $\wp_{fs}(X)$ is FSM Dedekind infinite, then $X$ should be FSM non-amorphous, meaning that $X$ should contain two disjoint, infinite, finitely supported supported subsets. The reverse implication is not valid.

Proof. 1. Assume that $(X_{n})_{n\in\mathbb{N}}$ is a countable family of different finite subsets of $X$ such that the mapping $n\mapsto X_{n}$ is finitely supported. Thus, each $X_{n}$ is supported by the same set $S=supp(n\mapsto X_{n})$. Since each $X_{n}$ is finite (and the support of a finite set coincides with the union of the supports of its elements), as in the proof of Lemma 4.4, we have that $\underset{n\in\mathbb{N}}{\cup}X_{n}$ is uniformly supported by $S$. Furthermore, $\underset{n\in\mathbb{N}}{\cup}X_{n}$ is infinite since all $X_{i}$ are pairwise different. Moreover, the countable sequence $(Y_{n})_{n\in\mathbb{N}}$ defined by $Y_{n}=X_{n}\setminus\underset{m<n}{\cup}X_{m}$ is a uniformly supported (by $S$) sequence of pairwise disjoint uniformly supported sets with $\underset{n\in\mathbb{N}}{\cup}X_{n}=\underset{n\in\mathbb{N}}{\cup}Y_{n}$. Again since each $Y_{n}$ is finite (and the support of a finite set coincides with the union of the supports of its elements), any element belonging to a set from the sequence $(Y_{n})_{n\in\mathbb{N}}$ is $S$-supported. Since the union of all $Y_{n}$ is infinite, and each $Y_{n}$ is finite, there should exist infinitely many terms from the sequence $(Y_{n})_{n\in\mathbb{N}}$ that are non-empty. Assume that $(Y_{n})_{n\in M\subseteq\mathbb{N}}$ with $M$ infinite is a subset of $(Y_{n})_{n\in\mathbb{N}}$ formed by non-empty terms. Let $U_{1}=\{\cup Y_{k}\,|\,k\in M,k\;\text{is odd}\}$ and $U_{2}=\{\cup Y_{k}\,|\,k\in M,k\;\text{is even}\}$. Then $U_{1}$ and $U_{2}$ are disjoint, uniformly $S$-supported and infinite subsets of $X$.

2. Assume that $\wp_{fs}(X)$ is FSM Dedekind infinite. As in the proof of Theorem 4.3(8), we can define a uniformly supported, countable family $(Y_{n})_{n\in\mathbb{N}}$ of subsets of $X$ that are non-empty and pairwise disjoint. Let $V_{1}=\{\cup Y_{k}\,|\,k\;\text{is odd}\}$ and $V_{2}=\{\cup Y_{k}\,|\,k\;\text{is even}\}$. Then $V_{1}$ and $V_{2}$ are disjoint, infinite subsets of $X$. Since each $Y_{i}$ is supported by $S^{\prime}=supp(n\mapsto Y_{n})$ we have $\pi\star Y_{i}=Y_{i}$ for all $i\in\mathbb{N}$ and $\pi\in Fix(S^{\prime})$. Fix $\pi\in Fix(S^{\prime})$ and $x\in V_{1}$. Thus, there is $l\in\mathbb{N}$ such that $x\in Y_{2l+1}$. We obtain $\pi\cdot x\in\pi\star Y_{2l+1}=Y_{2l+1}$, and so $\pi\cdot x\in V_{1}$. Thus, $V_{1}$ is supported by $S^{\prime}$. Analogously, $V_{2}$ is supported by $S^{\prime}$, and so $X$ is FSM non-amorphous.

Conversely, the set $A+A=\{0,1\}\times A$ (the disjoint union of $A$ and $A$) is obviously non-amorphous because $\{(0,a)\,|\,a\in A\}$ is equivariant, infinite and coinfinite. One can define the equivariant bijection $f:\wp_{fs}(A)\times\wp_{fs}(A)\to\wp_{fs}(\{0,1\}\times A)$ by $f(U,V)=\{(0,x)\,|\,x\in U\}\cup\{(1,y)\,|\,y\in V\}$ for all $U,V\in\wp_{fs}(A)$. Clearly $f$ is equivariant because for each $\pi\in S_{A}$ we have $f(\pi\star U,\pi\star V)=\pi\star f(U,V)$. However, $\wp_{fs}(A)\times\wp_{fs}(A)$ is not FSM Dedekind infinite according to Corollary 4.7(2) and Theorem 4.3(6).

It is also worth noting that non-uniformly amorphous FSM sets are non-amorphous FSM sets since uniformly supported sets are obviously finitely supported. The converse however is not valid since $\wp_{fin}(A)$ is non-amorphous but it has no infinite uniformly supported subset (the only finite subsets of atoms supported by a finite set $S$ of atoms being the subsets of $S$), and so it cannot be non-uniformly amorphous.

Corollary 4.15

Let $X$ be a finitely supported amorphous subset of an invariant set (i.e. any finitely supported subset of $X$ is either finite or cofinite). Then each finitely supported surjective mapping $f:X\to X$ should be injective.

Proof. Since any finitely supported subset of $X$ is either finite or cofinite, then any uniformly supported subset of $X$ is either finite or cofinite. From Proposition 4.14, $\wp_{fin}(X)$ is not FSM Dedekind infinite. For the rest of the proof we follow step-by-step the proof of Proposition 4.13 (and of Lemma 3.7). If $X$ is finite, we are done, so assume $X$ is infinite. If $Y\in\wp_{fin}(X)$, then $f^{-1}(Y)\in\wp_{fs}(X)$ (supported by $supp(f)\cup supp(X)\cup supp(Y)$). Since $X$ is amorphous, it follows that $f^{-1}(Y)$ is either finite or cofinite. If $f^{-1}(Y)$ is cofinite, then its complementary $\{x\in X\;|\;f(x)\notin Y\}$ is finite. This means that all but finitely many elements in $X$ would have their image under $f$ belonging to the finite set $Y$. Therefore, $Im(f)$ would be a finite subset of $X$, which contradicts the surjectivity of $f$. Thus, $f^{-1}(Y)$ is a finite subset of $X$. In this sense we can well-define define the function $g:\wp_{fin}(X)\rightarrow\wp_{fin}(X)$ by $g(Y)=f^{-1}(Y)$ which is supported by $supp(f)\cup supp(X)$ and injective. Since $\wp_{fin}(X)$ is not FSM Dedekind infinite, it follow that $g$ is surjective, and so $f$ is injective exactly as in the last paragraph of the proof of Proposition 4.13.

Proposition 4.16

Let $X$ be an FSM Dedekind infinite set. Then there exists a finitely supported surjection $j:X\to\mathbb{N}$. The reverse implication is not valid. 2. 2.

If $X$ is a finitely supported subset of an invariant set such that there exists a finitely supported surjection $j:X\to\mathbb{N}$, then $\wp_{fs}(X)$ is FSM Dedekind infinite. The reverse implication is also valid.

Proof. 1. Let $X$ be an FSM Dedekind infinite set. According to Theorem 4.3(1), there is a finitely supported injection $i:\mathbb{N}\to X$. Let us fix $n_{0}\in\mathbb{N}$. We define the function $j:X\to\mathbb{N}$ by

[TABLE]

Since $Im(i)$ is supported by $supp(i)$ and $n_{0}$ is empty supported, by verifying the condition in Proposition 2.11 we have that $j$ is supported by $supp(i)\cup supp(X)$. Indeed, when $\pi\in Fix(supp(i)\cup supp(X))$, then $x\in Im(i)\Leftrightarrow\pi\cdot x\in Im(i)$, and $n=i^{-1}(\pi\cdot x)\Leftrightarrow i(n)=\pi\cdot x\Leftrightarrow\pi^{-1}\cdot i(n)=x\Leftrightarrow i(\pi^{-1}\diamond n)=x\Leftrightarrow i(n)=x\Leftrightarrow n=i^{-1}(x)$, where $\diamond$ is the trivial action on $\mathbb{N}$; similarly $y\notin Im(i)\Leftrightarrow\pi\cdot y\notin Im(i)$ and $j(\pi\cdot y)=n_{0}=\pi\diamond n_{0}=\pi\diamond j(y)$. Clearly, $j$ is surjective. However, the reverse implication is not valid because the mapping $f:\wp_{fin}(A)\to\mathbb{N}$ defined by $f(X)=|X|$ for all $X\in\wp_{fin}(A)$ is equivariant and surjective, but $\wp_{fin}(A)$ is not FSM Dedekind infinite.

2. Suppose now there exists a finitely supported surjection $j:X\to\mathbb{N}$. Clearly, for any $n\in\mathbb{N}$, the set $j^{-1}(\{n\})$ is non-empty and supported by $supp(j)$. Define $f:\mathbb{N}\to\wp_{fs}(X)$ by $f(n)=j^{-1}(\{n\})$. For $\pi\in Fix(supp(j))$ and an arbitrary $n\in\mathbb{N}$ we have $j(x)=n\Leftrightarrow j(\pi^{-1}\cdot x)=n$, and so $x\in j^{-1}(\{n\})\Leftrightarrow\pi^{-1}\cdot x\in j^{-1}(\{n\})$, which means $f(n)=\pi\star f(n)$ for all $n\in\mathbb{N}$, and so $f$ is supported by $supp(j)$. Since $f$ is also injective, by Theorem 4.3(1) we have that $\wp_{fs}(X)$ is FSM Dedekind infinite.

Conversely, assume that $\wp_{fs}(X)$ is FSM Dedekind infinite. As in the proof of Theorem 4.3(8), we can define a uniformly supported, countable family $(Y_{n})_{n\in\mathbb{N}}$ of subsets of $X$ that are non-empty and pairwise disjoint. The mapping $f$ can be defined by f(x)=\left\{\begin{array}[]{ll}n,&\text{if}\;\exists n.x\in Y_{n};\\ 0,&\text{otherwise}\end{array}\right., and, obviously, $f$ is supported by $supp(n\mapsto Y_{n})$.

Proposition 4.17

Let $X$ be an infinite finitely supported subset of an invariant set. Then there exists a finitely supported surjection $f:\wp_{fs}(X)\to\mathbb{N}$.

Proof. Let $X_{i}$ be the set of all $i$-sized subsets from $X$, i.e. $X_{i}=\{Z\subseteq X\,|\,|Z|=i\}$. The family $(X_{i})_{i\in\mathbb{N}}$ is uniformly supported by $supp(X)$ and all $X_{i}$ are non-empty and pairwise disjoint. Define the mapping $f$ by f(Y)=\left\{\begin{array}[]{ll}n,&\text{if}\>Y\in X_{n};\\ 0,&\text{if}\;Y\;\text{is infinite}.\end{array}\right. According to Proposition 2.11, $f$ is supported by $supp(X)$ (since any $X_{n}$ is supported by $supp(X)$) and it is surjective. We actually proved the existence of a finitely supported surjection from $\wp_{fin}(X)$ onto $\mathbb{N}$.

The sets $A$ and $\wp_{fin}(A)$ are both FSM usual infinite and none of them is FSM Dedekind infinite. We prove below that $A$ is not FSM ascending infinite, while $\wp_{fin}(A)$ is FSM ascending infinite.

Proposition 4.18

• •

The set $A$ is not FSM ascending infinite.

• •

Let $X$ be a finitely supported subset of an invariant set $U$. If $X$ is FSM usual infinite, then the set $\wp_{fin}(X)$ is FSM ascending infinite.

Proof. In order to prove that $A$ is not FSM ascending infinite, we prove firstly that each finitely supported increasing countable chain of finitely supported subsets of $A$ must be stationary. Indeed, if there exists an increasing countable chain $X_{0}\subseteq X_{1}\subseteq\ldots\subseteq A$ such that $n\mapsto X_{n}$ is finitely supported, then, according to Proposition 2.11 and because $\mathbb{N}$ is a trivial invariant set, each element $X_{i}$ of the chain must be supported by the same $S=supp(n\mapsto X_{n})$. However, there are only finitely many such subsets of $A$ namely the subsets of $S$ and the supersets of $A\setminus S$. Therefore the chain is finite, and, because it is ascending, there exists $n_{0}\in\mathbb{N}$ such that $X_{n}=X_{n_{0}},\forall n\geq n_{0}$. Now, let $Y_{0}\subseteq Y_{1}\subseteq\ldots\subseteq Y_{n}\subseteq\ldots$ be a finitely supported countable chain with $A\subseteq\underset{n\in\mathbb{N}}{\cup}Y_{n}$. Then $A\cap Y_{0}\subseteq A\cap Y_{1}\subseteq\ldots\subseteq A\cap Y_{n}\subseteq\ldots\subseteq A$ is a finitely supported countable chain of subsets of $A$ (supported by $supp(n\mapsto Y_{n})$) which should be stationary (finite). Furthermore, since $\underset{i\in\mathbb{N}}{\cup}(A\cap Y_{i})=A\cap(\underset{i\in\mathbb{N}}{\cup}Y_{i})=A$, there is some $k_{0}$ such that $A\cap Y_{k_{0}}=A$, and so $A\subseteq Y_{k_{0}}$. Thus, $A$ is not FSM ascending infinite.

We know that $\wp_{fin}(X)$ is a subset of the invariant set $\wp_{fin}(U)$ supported by $supp(X)$. Let us consider $X_{n}\!=\!\{Z\!\in\!\wp_{fin}(X)\,|\,|Z|\!\leq\!n\}$. Clearly, $X_{0}\subseteq X_{1}\subseteq\ldots\subseteq X_{n}\subseteq\ldots$. Furthermore, because permutations of atoms are bijective, we have that for an arbitrary $k\in\mathbb{N}$, $|\pi\star Y|=|Y|$ for all $\pi\in S_{A}$ and all $Y\in X_{k}$, and so $\pi\star Y\in X_{k}$ for all $\pi\in Fix(supp(X))$ and all $Y\in X_{k}$. Thus, each $X_{k}$ is a subset of $\wp_{fin}(X)$ finitely supported by $supp(X)$, and so $(X_{n})_{n\in\mathbb{N}}$ is finitely (uniformly) supported by $supp(X)$. Obviously, $\wp_{fin}(X)=\underset{n\in\mathbb{N}}{\cup}X_{n}$. However, there exists no $n\in\mathbb{N}$ such that $\wp_{fin}(X)=X_{n}$. Thus, $(\wp_{fin}(X),\star)$ is FSM ascending infinite.

Theorem 4.19

Let $X$ be a finitely supported subset of an invariant set $(Z,\cdot)$.

If $X$ is FSM Dedekind infinite, then $X$ is FSM Mostowski infinite. 2. 2.

If $X$ is FSM Mostowski infinite, then $X$ is FSM Tarski II infinite. The reverse implication is not valid.

Proof. 1. Suppose $X$ is FSM Dedekind infinite. According to Theorem 4.3(1) there exists an uniformly supported infinite injective sequence $T=(x_{n})_{n\in\mathbb{N}}$ of elements from $X$. Thus, each element of $T$ is supported by $supp(T)$ and there is a bijective correspondence between $\mathbb{N}$ and $T$ defined as $n\mapsto x_{n}$ which is supported by $supp(T)$. If we define the relation $\sqsubset$ on $T$ by: $x_{i}\sqsubset x_{j}$ if and only if $i<j$, we have that $\sqsubset$ is a (strict) total order relation supported by $supp(T)$. Thus $T$ is an infinite finitely supported (strictly) totally ordered subset of $X$, and so $X$ is FSM Mostowski infinite since any strict total order can be extended to a total order.

2. Suppose that $X$ is not FSM Tarski II infinite. Then every non-empty finitely supported family of finitely supported subsets of $X$ which is totally ordered by inclusion has a maximal element under inclusion. Let $(U,<)$ be a finitely supported strictly totally ordered subset of $X$ (any total order relation induces a strict total order relation). We prove that $U$ is finite, and so $X$ is not FSM Mostowski infinite. In this sense it is sufficient to prove that $<$ and $>$ are well-orderings. Since both of them are (strict) total orderings, we need to prove that any finitely supported subset of $U$ has a least and a greatest element wrt $<$, i.e. a minimal and a maximal element (because $<$ is total). Let $Y$ be a finitely supported subset of $U$. The set $\downarrow z=\{y\in Y\,|\,y<z\}$ is supported by $supp(z)\cup supp(Y)\cup supp(<)$ for all $z\in Y$. The family $T=\{\downarrow z\,|\,z\in Y\}$ is itself finitely supported by $supp(Y)\cup supp(<)$ because for all $\pi\in Fix(supp(Y)\cup supp(<))$ we have $\pi\cdot\downarrow z=\downarrow\pi\cdot z$. Since $<$ is transitive, we have that $T$ is (strictly) totally ordered by inclusion, and so it has a maximal element, which means $Y$ has a maximal element. Analogously, the set $\uparrow z=\{y\in Y\,|\,z<y\}$ is supported by $supp(z)\cup supp(Y)\cup supp(<)$ for all $z\in Y$ and the family $T^{\prime}=\{\uparrow z\,|\,z\in Y\}$ is itself finitely supported by $supp(Y)\cup supp(<)$ because for all $\pi\in Fix(supp(Y)\cup supp(<))$ we have $\pi\cdot\uparrow z=\uparrow\pi\cdot z$. The family $T^{\prime}$ is (strictly) totally ordered by inclusion, and so it has a maximal element, from which $Y$ has a minimal element. We used the obvious properties $z<t$ if and only if $\downarrow z\subset\downarrow t$, and $z<t$ if and only if $\uparrow t\subset\uparrow z$.

Conversely, according to Proposition 4.18, $\wp_{fin}(A)$ is FSM ascending infinite, and so it is FSM Tarski II infinite. However, $\wp_{fin}(A)$ is not FSM Mostowski infinite, according to Corollary 4.24.

Proposition 4.20

Let $X$ be a finitely supported subset of an invariant set $(Z,\cdot)$. If $X$ is FSM Mostowski infinite, then $X$ is non-amorphous meaning that $X$ can be expressed as a disjoint union of two infinite finitely supported subsets. The reverse implication is not valid.

Proof. Suppose that there is an infinite finitely supported totally ordered subset $(Y,\leq)$ of $X$. Assume, by contradiction, that $Y$ is amorphous, meaning that any finitely supported subset of $Y$ is either finite or cofinite. As in the proof of Theorem 4.19 (without making the requirement that $\leq$ is strict, which anyway would not essentially change the proof), for $z\in Y$ we define the finitely supported subsets $\downarrow z=\{y\in Y\,|\,y\leq z\}$ and $\uparrow z=\{y\in Y\,|\,z\leq y\}$ for all $z\in Y$. We have that the mapping $z\mapsto\downarrow z$ from $Y$ to $T=\{\downarrow z\,|\,z\in Y\}$ is itself finitely supported by $supp(Y)\cup supp(\leq)$. Furthermore it is bijective, and so $T$ is amorphous. Thus, any subset $Z$ of $T$ is either finite or cofinite, and obviously any subset $Z$ of $T$ is finitely supported. Analogously, the mapping $z\mapsto\uparrow z$ from $Y$ to $T^{\prime}=\{\uparrow z\,|\,z\in Y\}$ is finitely supported and bijective, which means that any subset of $T^{\prime}$ is either finite or cofinite, and clearly any subset of $T^{\prime}$ is finitely supported.

We distinguish the following two cases:

1. There are only finitely many elements $x_{1},\ldots,x_{n}\in Y$ such that $\downarrow x_{1},\ldots,\downarrow x_{n}$ are finite. Thus, for $y\in U=Y\setminus\{x_{1},\ldots,x_{n}\}$ we have $\downarrow y$ infinite. Since $\downarrow y$ is a subset of $Y$, it should be cofinite, and so $\uparrow y$ is finite (because $\leq$ is a total order relation). Let $M=\{\uparrow y\,|\,y\in U\}$. As in Theorem 4.19 we have that $M$ is totally ordered with respect to sets inclusion. Furthermore, for an arbitrary $y\in U$ we cannot have $y\leq x_{k}$ for some $k\in\{1,\ldots,n\}$ because $\downarrow y$ is infinite, while $\downarrow x_{k}$ is finite, and so $\uparrow y$ is a subset of $U$. Thus, $M$ is an infinite, finitely supported (by $supp(U)\cup supp(\leq)$), totally ordered family formed by finite subsets of $U$. Since $M$ is finitely supported, for each $y\in U$ and each $\pi\in Fix(supp(M))$ we have $\pi\cdot\uparrow y\in M$. Since $\uparrow y$ is finite, we have that $\pi\cdot\uparrow y$ is finite having the same number of elements as $\uparrow y$. Since $\pi\cdot\uparrow y$ and $\uparrow y$ are comparable via inclusion, they should be equal. Thus, $M$ is uniformly supported. Since $\leq$ is a total order, for $\pi\in Fix(supp(\uparrow y))$ we have $\uparrow\pi\cdot y=\pi\cdot\uparrow y=\uparrow y$, and so $\pi\cdot y=y$, from which $supp(y)\subseteq supp(\uparrow y)$. Thus, $U$ is uniformly supported. Since any element of $U$ has only a finite number of successors (leading to the conclusion that $\geq$ is an well-ordering on $U$ uniformly supported by $supp(U)$) and $U$ is uniformly supported, we can define an order monomorphism between $\mathbb{N}$ and $U$ which is supported by $supp(U)$. For example, choose $u_{0}\neq u_{1}\in U$, then let $u_{2}$ be the greatest element (w.r.t. $\leq$) in $U\setminus\{u_{0},u_{1}\}$, $u_{3}$ be the greatest element in $U\setminus\{u_{0},u_{1},u_{2}\}$ (no choice principle is used since $\geq$ is an well-ordering, and so such a greatest element is precisely defined), and so on, and find an infinite, uniformly supported countable sequence $u_{0},u_{1},u_{2},\ldots$. Since $\mathbb{N}$ is non-amorphous (being expressed as the union between the even elements and the odd elements), we conclude that $U$ is non-(uniformly) amorphous containing two infinite uniformly supported disjoint subsets.

2. We have cofinitely many elements $z$ such that $\downarrow z$ is finite. Thus, there are only finitely many elements $y_{1},\ldots,y_{m}\in Y$ such that $\downarrow y_{1},\ldots,\downarrow y_{m}$ are infinite. Since every infinite subset of $Y$ is cofinite, only $\uparrow y_{1},\ldots,\uparrow y_{m}$ are finite. Let $z\in Y\setminus\{y_{1},\ldots,y_{m}\}$ which means $\uparrow z$ infinite. Since $\uparrow z$ is a subset of $Y$ it should be cofinite, and so $\downarrow z$ is finite. As in the above item, the set $M^{\prime}=\{\downarrow z\,|\,z\in Y\setminus\{y_{1},\ldots,y_{m}\}\}$ is an infinite, finitely supported, totally ordered (by inclusion) family of finite sets, and so it has to be uniformly supported, from which $Y\setminus\{y_{1},\ldots,y_{m}\}$ is uniformly supported, and so $\leq$ is an FSM well ordering on $Y\setminus\{y_{1},\ldots,y_{m}\}$. Therefore, $Y\setminus\{y_{1},\ldots,y_{m}\}$ has an infinite, uniformly supported, countable subset, and so $Y\setminus\{y_{1},\ldots,y_{m}\}$ is non-(uniformly) amorphous containing two infinite uniformly supported disjoint subsets. Thus $Y$ is non-amorphous, and so $X$ is non-amorphous.

Conversely, the set $A+A$ (the disjoint union of $A$ and $A$) is obviously non-amorhpous because because $\{(0,a)\,|\,a\in A\}$ is equivariant, infinite and coinfinite. However, if we assume there exists a finitely supported total order relation on an infinite subset of $A+A$, then there should exist an infinite, finitely supported, total order on at least one of the sets $\{(0,a)\,|\,a\in A\}$ or $\{(1,a)\,|\,a\in A\}$, which leads to an infinite finitely supported total order relation on $A$. However $A$ is not FSM Mostowski infinite by Corollary 4.24.

Theorem 4.21

Let $X$ be a finitely supported subset of an invariant set $(Z,\cdot)$. If $X$ contains no infinite uniformly supported subset, then $X$ is not FSM Mostowski infinite.

Proof. Assume, by contradiction, that $X$ is FSM Mostowski infinite, meaning that $X$ contains an infinite, finitely supported, totally ordered subset $(Y,\leq)$. We claim that $Y$ is uniformly supported by $supp(\leq)\cup supp(Y)$. Let $\pi\in Fix(supp(\leq)\cup supp(Y))$ and let $y\in Y$ an arbitrary element. Since $\pi$ fixes $supp(Y)$ pointwise and $supp(Y)$ supports $Y$, we obtain that $\pi\cdot y\in Y$, and so we should have either $y<\pi\cdot y$, or $y=\pi\cdot y$, or $\pi\cdot y<y$. If $y<\pi\cdot y$, then, because $\pi$ fixes $supp(\leq)$ pointwise and because the mapping $z\mapsto\pi\cdot z$ is bijective from $Y$ to $\pi\star Y$, we get $y<\pi\cdot y<\pi^{2}\cdot y<\ldots<\pi^{n}\cdot y$ for all $n\in\mathbb{N}$. However, since any permutation of atoms interchanges only finitely many atoms, it has a finite order in the group $S_{A}$, and so there is $m\in\mathbb{N}$ such that $\pi^{m}=Id$. This means $\pi^{m}\cdot y=y$, and so we get $y<y$ which is a contradiction. Analogously, the assumption $\pi\cdot y<y$, leads to the relation $\pi^{n}\cdot y<\ldots<\pi\cdot y<y$ for all $n\in\mathbb{N}$ which is also a contradiction since $\pi$ has finite order. Therefore, $\pi\cdot y=y$, and because $y$ was arbitrary chosen form $Y$, $Y$ should be a uniformly supported infinite subset of $X$.

Looking to the proof of Proposition 4.20, the following result follows directly.

Corollary 4.22

Let $X$ be a finitely supported subset of an invariant set $(Z,\cdot)$. If $X$ is FSM Mostowski infinite, then $X$ is non-uniformly amorphous meaning that $X$ has two disjoint, infinite, uniformly supported subsets.

Remark 4.23

In a permutation model of set theory with atoms, a set can be well-ordered if and only if there is a one-to-one mapping of the related set into the kernel of the model. Also it is noted that axiom of choice is valid in the kernel of the model [5]. Although FSM/nominal is somehow related to (has connections with) permutation models of set theory with atoms, it is independently developed over ZF without being necessary to relax the axioms of extensionality or foundation. FSM sets are ZF sets together with group actions, and such a theory makes sense over ZF without being necessary to require the validity of the axiom of choice on ZF sets. Thus, FSM is the entire ZF together with atomic sets with finite support (where the set of atoms is a fixed ZF formed by element whose internal structure is ignored and which are basic in the higher order construction). There may exist infinite ZF sets that do not contain infinite countable subsets, and as well there may exist infinite uniformly supported FSM sets (particularly such ZF sets) that do not contain infinite countable, uniformly supported, subsets.

Corollary 4.24

The sets $A$, $A+A$ and $A\times A$ are FSM usually infinite, but there are not FSM Mostowski infinite, nor FSM Tarski II infinite. 2. 2.

None of the sets $\wp_{fin}(A)$, $\wp_{cofin}(A)$, $\wp_{fs}(A)$ and $\wp_{fin}(\wp_{fs}(A))$ is Mostowski infinite in FSM. 3. 3.

None of the sets $A^{A}_{fs}$, $T_{fin}(A)^{A}_{fs}$ and $\wp_{fs}(A)^{A}_{fs}$ is FSM Mostowksi infinite.

Proof. In the view of Theorem 4.21 it is sufficient to prove that none of the sets $A$, $\wp_{fin}(A)$, $\wp_{cofin}(A)$, $\wp_{fs}(A)$, $A+A$, $A\times A$, $A^{A}_{fs}$,$T_{fin}(A)^{A}_{fs}$ and $\wp_{fs}(A)^{A}_{fs}$ contain infinite uniformly supported subsets. For $A$, $\wp_{fin}(A)$, $\wp_{cofin}(A)$ and $\wp_{fs}(A)$ this is obvious since for any finite set $S$ of atoms there are at most finitely many subsets of $A$ supported by $S$, namely the subsets of $S$ and the supersets of $A\setminus S$. Moreover, $\wp_{fin}(\wp_{fs}(A))$ does not contain an infinite uniformly supported subset according to Lemma 4.4 since $\wp_{fs}(A)$ does not contain an infinite uniformly supported subset.

Regarding $A^{A}_{fs}$, the things are also similar with Corollary 4.7(4). According to Lemma 4.10, any $S$-supported function $f:A\to A$ should have the property that either $f|_{A\setminus S}=Id$ or $f|_{A\setminus S}$ is an one-element subset of $S$. For each possible definition of such an $f$ on $S$ we have at most $|S|+1$ possible ways to define $f$ on $A\setminus S$, and so at most $|S|+1$ possible ways to completely define $f$ on $A$. If there was an infinite uniformly $S$-supported sequence of finitely supported functions from $A$ to $A$, there should exist infinitely many finitely supported functions from $S$ to $A$ supported by the same finite set $S$. But this contradicts the fact that $A^{|S|}$ does not contain an infinite uniformly supported subset (this follows by applying finitely many times the result that $X\times X$ does not contain an infinite uniformly supported subset whenever $X$ does not contain an infinite uniformly supported subset). Analyzing the proofs of Corollary 4.7(6) and (7), we also conclude that $T_{fin}(A)^{A}_{fs}$ and $\wp_{fs}(A)^{A}_{fs}$ do not contain infinite uniformly supported subsets.

We also have that $A$ is not FSM Tarski II infinite because $\wp_{fs}(A)$ contains no infinite uniformly supported subsets, and so every totally ordered subset (particularly via inclusion) of $\wp_{fs}(A)$ should be finite meaning that it should have a maximal element. Furthermore, we have that there is an equivariant bijection between $\wp_{fs}(A+A)$ and $\wp_{fs}(A)\times\wp_{fs}(A)$. Since $\wp_{fs}(A)$ does not contain an infinite uniformly supported subset, we have that $\wp_{fs}(A)\times\wp_{fs}(A)$ does not contain an infinite uniformly supported subset (the proof is quasi-identical to the one of Theorem 4.3(6) without taking count on the countability of the related infinite uniformly supported family). Therefore, any infinite totally ordered (via inclusion) uniformly supported family of $\wp_{fs}(A+A)$ should be finite containing a maximal element. There is an equivariant bijection between $\wp_{fs}(A)^{A}_{fs}$ and $\wp_{fs}(A\times A)$. Therefore any uniformly supported totally ordered subset of $\wp_{fs}(A\times A)$ should be finite containing a maximal element.

Corollary 4.25

Let $X$ be a finitely supported subset of an invariant set $Y$ such that $X$ does not contain an infinite uniformly supported subset. Then the set $\wp_{fin}(X)$ is not FSM Mostowski infinite.

Proof. According to Lemma 4.4, $\wp_{fin}(X)$ does not contain an infinite uniformly supported subset. Thus, by Theorem 4.21, $\wp_{fin}(X)$ is not FSM Mostowski infinite.

Theorem 4.26

Let $X$ be a finitely supported subset of an invariant set $(Y,\cdot)$.

If $X$ is FSM Tarski I infinite, then $X$ is FSM Tarski III infinite. The converse does not hold. However if $X$ is FSM Tarski III infinite, then $\wp_{fs}(X)$ is FSM Tarski I infinite. 2. 2.

If $X$ is FSM Tarski III infinite, then $X$ is FSM Dedekind infinite. The converse does not hold. However if $X$ is FSM Dedekind infinite, then $\wp_{fs}(X)$ is FSM Tarski III infinite.

Proof.

1. We consider the case when $X$ has at least two elements (otherwise the theorem is trivial). Let $X$ be FSM Tarski I infinite. Then $|X\times X|=|X|$. Fix two elements $x_{1},x_{2}\in X$ with $x_{1}\neq x_{2}$. We can define an injection $f:X\times\{0,1\}\to X\times X$ by f(u)=\left\{\begin{array}[]{ll}(x,x_{1})&\text{for}\>u=(x,0)\\ (x,x_{2})&\text{for}\>u=(x,1)\end{array}\right.. Clearly, by checking the condition in Proposition 2.11 and using Proposition 2.5, we have that $f$ is supported by $supp(X)\cup supp(x_{1})\cup supp(x_{2})$ (since $\{0,1\}$ is necessarily a trivial invariant set), and so $|X\times\{0,1\}|\leq|X\times X|$. Thus, $|X\times\{0,1\}|\leq|X|$. Obviously, there is an injection $i:X\to X\times\{0,1\}$ defined by $i(x)=(x,0)$ for all $x\in X$ which is supported by $supp(X)$. According to Lemma 3.4, we get $2|X|=|X\times\{0,1\}|=|X|$.

Let us consider $X=\mathbb{N}\times A$. We make the remark that $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$ by considering the equivariant injection $h:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ defined by $h(m,n)=2^{m}3^{n}$ and using Lemma 3.4. Similarly, $|\{0,1\}\times\mathbb{N}|=|\mathbb{N}|$ by considering the equivariant injection $h^{\prime}:\mathbb{N}\times\{0,1\}\to\mathbb{N}$ defined by $h^{\prime}(n,0)=2^{n}$ and $h^{\prime}(n,1)=3^{n}$ and using Lemma 3.4. We have $2|X|=2|\mathbb{N}||A|=|\mathbb{N}||A|=|X|$. However, we prove that $|X\times X|\neq|X|$. Assume the contrary, and so we have $|\mathbb{N}\times(A\times A)|=|\mathbb{N}\times A\times\mathbb{N}\times A|=|\mathbb{N}\times A|$. Thus, there is a finitely supported injection $g:A\times A\to\mathbb{N}\times A$, and by Proposition 3.10 there is a finitely supported surjection $f:\mathbb{N}\times A\to A\times A$. Let us consider three different atoms $a,b,c\notin supp(f)$. There exists $(i,x)\in\mathbb{N}\times A$ such that $f(i,x)=(a,b)$. Since $(a\,b)\in Fix(supp(f))$ and $\mathbb{N}$ is trivial invariant set, we have $f(i,(a\,b)(x))=(a\,b)f(i,x)=(a\,b)(a,b)=((a\,b)(a),(a\,b)(b))=(b,a)$. We should have $x=a$ or $x=b$, otherwise $f$ is not a function. Assume without losing the generality that $x=a$, which means $f(i,a)=(a,b)$. Therefore $f(i,b)=f(i,(a\,b)(a))=(a\,b)f(i,a)=(a\,b)(a,b)=(b,a)$. Similarly, since $(a\,c),(b\,c)\in Fix(supp(f))$, we have $f(i,c)=f(i,(a\,c)(a))=(a\,c)f(i,a)=(a\,c)(a,b)=(c,b)$ and $f(i,b)=f(i,(b\,c)(c))=(b\,c)f(i,c)=(b\,c)(c,b)=(b,c)$. But $f(i,b)=(b,a)$ contradicting the functionality of $f$. Therefore, $X$ is FSM Tarski III infinite, but it is not FSM Tarski I infinite.

Now, suppose that $X$ is FSM Tarski III infinite, which means $|\{0,1\}\times X|=|X|$. We define the mapping $\psi:\wp_{fs}(X)\times\wp_{fs}(X)\to\wp_{fs}(\{0,1\}\times X)$ by $f(U,V)=\{(0,x)\,|\,x\in U\}\cup\{(1,y)\,|\,y\in V\}$ for all $U,V\in\wp_{fs}(X)$. Clearly $\psi$ is well defined and bijective, and for each $\pi\in Fix(supp(X))$ we have $\psi(\pi\star U,\pi\star V)=\pi\star\psi(U,V)$ which means $\psi$ is finitely supported. Therefore, $|\wp_{fs}(X)\times\wp_{fs}(X)|=|\wp_{fs}(\{0,1\}\times X)|=|\wp_{fs}(X)|$. The last equality follows by applying twice Lemma 3.7 (using the fact that there is a finitely supported surjection from $X$ onto $X\times\{0,1\}$ and a finitely supported surjection from $X\times\{0,1\}$ onto $X$, we obtain there is a finitely supported injection from $\wp_{fs}(X\times\{0,1\})$ into $\wp_{fs}(X)$, and a finitely supported injection from $\wp_{fs}(X)$ into $\wp_{fs}(X\times\{0,1\})$) and Lemma 3.5.

2. Let us assume that $X$ is FSM Tarski III infinite. Let us consider an element $y_{1}$ belonging to an invariant set (whose action is also denoted by $\cdot$) with $y_{1}\notin X$ (such an element can be, for example, a non-empty element in $\wp_{fs}(X)\setminus X$). Fix $y_{2}\in X$. One can define a mapping $f:X\cup\{y_{1}\}\to X\times\{0,1\}$ by f(x)=\left\{\begin{array}[]{ll}(x,0)&\text{for}\>x\in X\\ (y_{2},1)&\text{for}\>x=y_{1}\end{array}\right.. Clearly $f$ is injective and it is supported by $S=supp(X)\cup supp(y_{1})\cup supp(y_{2})$ because for all $\pi$ fixing $S$ pointwise we have $f(\pi\cdot x)=\pi\cdot f(x)$ for all $x\in X\cup\{y_{1}\}$. Therefore, $|X\cup\{y_{1}\}|\leq|X\times\{0,1\}|=|X|$, and so there is a finitely supported injection $g:X\cup\{y_{1}\}\to X$. The mapping $h:X\to X$ defined by $h(x)=g(x)$ is injective, supported by $supp(g)\cup supp(X)$, and $g(y_{1})\in X\setminus h(X)$, which means $h$ is not surjective. It follows that $X$ is FSM Dedekind infinite.

Let us consider $X=A\cup\mathbb{N}$. Since $A$ and $\mathbb{N}$ are disjoint, we have that $X$ is an invariant set (similarly as in Proposition 2.5). Clearly $X$ is FSM Dedekind infinite. Assume, by contradiction, that $|X|=2|X|$, that is $|A\cup\mathbb{N}|=|A+A+\mathbb{N}|=|(\{0,1\}\times A)\cup\mathbb{N}|$. Thus, there is a finitely supported injection $f:(\{0,1\}\times A)\cup\mathbb{N}\to A\cup\mathbb{N}$, and so there exists a finitely supported injection $f:(\{0,1\}\times A)\to A\cup\mathbb{N}$. We prove that whenever $\varphi:A\to A\cup\mathbb{N}$ is finitely supported and injective, for $a\notin supp(\varphi)$ we have $\varphi(a)\in A$. Assume, by contradiction, that there is $a\notin supp(\varphi)$ such that $\varphi(a)\in\mathbb{N}$. Since $supp(\varphi)$ is finite, there exists $b\notin supp(\varphi)$, $b\neq a$. Thus, $(a\,b)\in Fix(supp(\varphi))$, and so $\varphi(b)=\varphi((a\,b)(a))=(a\,b)\diamond\varphi(a)=\varphi(a)$ since $(\mathbb{N},\diamond)$ is a trivial invariant set. This contradicts the injectivity of $\varphi$. We can consider the mappings $\varphi_{1},\varphi_{2}:A\to A\cup\mathbb{N}$ defined by $\varphi_{1}(a)=f(0,a)$ for all $a\in A$ and $\varphi_{2}(a)=f(1,a)$ for all $a\in A$, that are injective and supported by $supp(f)$. Therefore, $f(\{0\}\times A)=\varphi_{1}(A)$ contains at most finitely many element from $\mathbb{N}$, and $f(\{1\}\times A)=\varphi_{2}(A)$ also contains at most finitely many element from $\mathbb{N}$. Thus, $f$ is an injection from $(\{0,1\}\times A)$ to $A\cup Z$ where $Z$ is a finite subset of $\mathbb{N}$. It follows that $f(\{0\}\times A)$ contains an infinite subset of atoms $U$, and $f(\{1\}\times A)$ contains an infinite subset of atoms $V$. Since $f$ is injective, it follows that $U$ and $V$ are infinite disjoint subsets of $A$, which contradicts Proposition 2.9 stating that $A$ is amorphous.

Now, if $X$ is FSM Dedekind infinite, we have that there is a finitely supported injection $h$ from $X$ onto a finitely supported proper subset $Z$ of $X$. Consider an element $y_{1}$ belonging to an invariant set with $y_{1}\notin X$. We can define an injection $h^{\prime}:X\cup\{y_{1}\}\to X$ by taking $h^{\prime}(x)=h(x)$ for all $x\in X$ and $h^{\prime}(y_{1})=b$ with $b\in X\setminus Z$. Clearly $h^{\prime}$ is supported by $supp(h)\cup supp(y_{1})\cup supp(b)$. Since there also exists an $supp(X)$-supported injection from $X$ to $X\cup\{y_{1}\}$, according to Lemma 3.5, one can define a finitely supported bijection $\psi$ from $X$ to $X\cup\{y_{1}\}$. According to Lemma 3.7 the mapping $g:\wp_{fs}(X\cup\{y_{1}\})\to\wp_{fs}(X)$ defined by $g(V)=f^{-1}(V)$ for all $V\in\wp_{fs}(X\cup\{y_{1}\})$ is finitely supported and injective. Therefore, $2^{|X|}\geq 2^{|X|+1}=2\cdot 2^{|X|}$ which in the view of Lemma 3.5 leads to the conclusion that $\wp_{fs}(X)$ is FSM Tarski III infinite.

Corollary 4.27

The following sets are FSM usual infinite, but they are not FSM Tarski I infinite, nor FSM Tarski III infinite.

The invariant set $A$. 2. 2.

The invariant set $\wp_{fs}(A)$. 3. 3.

The invariant sets $\wp_{fin}(A)$ and $\wp_{cofin}(A)$. 4. 4.

The set $\wp_{fin}(X)$ where $X$ is a finitely supported subset of an invariant set containing no infinite uniformly supported subset.

Proof. The result follows directly because the related sets are not FSM Dedekind infinite, according to Theorem 4.3 and Corollary 4.7.

Corollary 4.28

Let $X$ be an infinite finitely supported subset of an invariant set. Then $\wp_{fs}(\wp_{fs}(\wp_{fs}(X)))$ is FSM Tarski III infinite and, consequently, $\wp_{fs}(\wp_{fs}(\wp_{fs}(\wp_{fs}(X))))$ is FSM Tarski I infinite.

Proof. Since $\wp_{fs}(\wp_{fs}(X))$ is FSM Dedekind infinite, as in the proof of Theorem 4.26(2) one can prove $|\wp_{fs}(\wp_{fs}(X))|+1=|\wp_{fs}(\wp_{fs}(X))|$. The result now follows directly using arithmetic properties of FSM cardinalities proved above.

In a future work we intend to prove an even stronger result claiming that $\wp_{fs}(\wp_{fs}(X))$ is FSM Tarski III infinite and, consequently, $\wp_{fs}(\wp_{fs}(\wp_{fs}(X)))$ is FSM Tarski I infinite, whenever $X$ is an infinite finitely supported subset of an invariant set.

Corollary 4.29

The sets $A^{\mathbb{N}}_{fs}$ and $\mathbb{N}^{A}_{fs}$ are FSM Tarski I infinite, and so they are also Tarski III infinite.

Proof. There is an equivariant bijection $\psi$ between $(A^{\mathbb{N}})^{2}_{fs}$ and $A^{\mathbb{N}\times\{0,1\}}_{fs}$ that associates to each Cartesian pair $(f,g)$ of mappings from $\mathbb{N}$ to $A$ a mapping $h:\mathbb{N}\times\{0,1\}\to A$ defined as follows.

[TABLE]

The equivariance of $\psi$ follows from Proposition 2.11 because if $\pi\in S_{A}$ we have $\psi(\pi\widetilde{\star}f,\pi\widetilde{\star}g)=h^{\prime}$ where $h^{\prime}(n,0)=(\pi\widetilde{\star}f)(n)=\pi(f(n))$ and $h^{\prime}(n,1)=(\pi\widetilde{\star}g)(n)=\pi(g(n))$. Thus, $h^{\prime}(u)=\pi(h(u))$ for all $u\in\mathbb{N}\times\{0,1\}$ which means $h^{\prime}=\pi\widetilde{\star}h=\pi\widetilde{\star}\psi(f,g)$.

There also exists an equivariant bijection $\varphi$ between $(\mathbb{N}^{A})^{2}_{fs}$ and $(\mathbb{N}\times\mathbb{N})^{A}_{fs}$ that associates to each Cartesian pair $(f,g)$ of mappings from $A$ to $\mathbb{N}$ a mapping $h:A\to\mathbb{N}\times\mathbb{N}$ defined by $h(a)=(f(a),g(a))$ for all $a\in A$. The equivariance of $\varphi$ follows from Proposition 2.11 because if $\pi\in S_{A}$ we have $\varphi(\pi\widetilde{\star}f,\pi\widetilde{\star}g)=h^{\prime}$ where $h^{\prime}(a)=((\pi\widetilde{\star}f)(a),(\pi\widetilde{\star}g)(a))=(f(\pi^{-1}(a)),g(\pi^{-1}(a)))=h(\pi^{-1}(a))=(\pi\widetilde{\star}h)(a)$ for all $a\in A$, and so $h^{\prime}=\pi\widetilde{\star}h=\pi\widetilde{\star}\varphi(f,g)$. Therefore $|(A^{\mathbb{N}})^{2}_{fs}|=|A^{\mathbb{N}\times\{0,1\}}_{fs}|=|A^{\mathbb{N}}_{fs}|$. Therefore, $|(\mathbb{N}^{A})^{2}_{fs}|=|(\mathbb{N}\times\mathbb{N})^{A}_{fs}|=|\mathbb{N}^{A}_{fs}|$ according to Proposition 3.11(3) and Lemma 3.4 (we used $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$).

Theorem 4.30

Let $X$ be a finitely supported subset of an invariant set $(Y,\cdot)$. If $\wp_{fs}(X)$ is FSM Tarski I infinite, then $\wp_{fs}(X)$ is FSM Tarski III infinite. The converse does not hold.

Proof. The direct implication is a consequence of Theorem 4.26(1). Thus, we focus on the proof of the invalidity of the reverse implication.

Firstly we make the remark that whenever $U,V$ are finitely supported subsets of an invariant set with $U\cap V=\emptyset$, we have that there is a finitely supported (by $supp(U)\cup supp(V)$) bijection from $\wp_{fs}(U\cup V)$ into $\wp_{fs}(U)\times\wp_{fs}(V)$ that maps each $X\in\wp_{fs}(U\cup V)$ into the pair $(X\cap U,X\cap V)$. Analogously, whenever $B,C$ are invariant sets there is an equivariant bijection from $\wp_{fs}(B)\times\wp_{fs}(C)$ into $\wp_{fs}(B+C)$ that maps each pair $(B_{1},C_{1})\in\wp_{fs}(B)\times\wp_{fs}(C)$ into the set $\{(0,b)\,|\,b\in B_{1}\}\cup\{(1,c)\,|\,c\in C_{1}\}$. This follows directly by verifying the conditions in Proposition 2.11.

Let us consider the set $A\cup\mathbb{N}$ which is FSM Dedekind infinite. According to Theorem 4.26(2), we have that $\wp_{fs}(A\cup\mathbb{N})$ is FSM Tarski III infinite. We prove that it is not FSM Tarski I infinite. Assume, by contradiction that $|\wp_{fs}(A\cup\mathbb{N})\times\wp_{fs}(A\cup\mathbb{N})|=|\wp_{fs}(A\cup\mathbb{N})|$ which means $|\wp_{fs}(A+\mathbb{N}+A+\mathbb{N})|=|\wp_{fs}(A\cup\mathbb{N})|$, and so $|\wp_{fs}(A+A+\mathbb{N})|=|\wp_{fs}(A\cup\mathbb{N})|$. Thus, according to Proposition 3.11(4), there is a finitely supported injection from $\wp_{fs}(A+A)$ to $\wp_{fs}(A\cup\mathbb{N})$, which means there is a finitely supported injection from $\wp_{fs}(A)\times\wp_{fs}(A)$ to $\wp_{fs}(A)\times\wp_{fs}(\mathbb{N})$, and so there is a finitely supported injection from $A\times A$ to $\wp_{fs}(A)\times\wp_{fs}(\mathbb{N})$. According to Proposition 3.10, there should exist a finitely supported surjection $f:\wp_{fs}(A)\times\wp_{fs}(\mathbb{N})\to A\times A$. Let us consider two atoms $a,b\notin supp(f)$ with $a\neq b$. It follows that $(a\,b)\in Fix(supp(f))$. Since $f$ is surjective, there exists $(X,M)\in\wp_{fs}(A)\times\wp_{fs}(\mathbb{N})$ such that $f(X,M)=(a,b)$. According to Proposition 2.11 and because $\mathbb{N}$ is a trivial invariant set meaning that $(a\,b)\star M=M$, we have $f((a\,b)\star X,M)=f((a\,b)\otimes(X,M))=(a\,b)\otimes f(X,M)=(a\,b)\otimes(a,b)=((a\,b)(a),(a\,b)(b))=(b,a)$. Due to the functionality of $f$ we should have $((a\,b)\star X,M)\neq(X,M)$, which means $(a\,b)\star X\neq X$.

We prove that if both $a,b\in supp(X)$, then $(a\,b)\star X=X$. Indeed, suppose $a,b\in supp(X)$. Since $X\in\wp_{fs}(A)$, from Proposition 2.9 we have that $X$ is either finite or cofinite. If $X$ is finite, then $supp(X)=X$, and so $a,b\in X$. Therefore, $(a\,b)\star X=\{(a\,b)(x)\,|\,x\in X\}=\{(a\,b)(a)\}\cup\{(a\,b)(b)\}\cup\{(a\,b)(c)\,|\,c\in X\setminus\{a,b\}\}=\{b\}\cup\{a\}\cup(X\setminus\{a,b\})=X$. Now, if $X$ is cofinite, then $supp(X)=A\setminus X$, and so $a,b\in A\setminus X$. Since $a,b\notin X$, we have $a,b\neq x$ for all $x\in X$, which means $(a\,b)(x)=x$ for all $x\in X$, and again $(a\,b)\star X=X$.

Thus, one of $a$ or $b$ does not belong to $supp(X)$. Assume $b\notin supp(X)$. Let us consider $c\neq a,b$, $c\notin supp(f)$, $c\notin supp(X)$. Then $(b\,c)\in Fix(supp(X))$, and so $(b\,c)\star X=X$. Moreover, $(b\,c)\in Fix(supp(f))$, and by Proposition 2.11 we have $(a,b)=f(X,M)=f((b\,c)\star X,M)=f((b\,c)\otimes(X,M))=(b\,c)\otimes f(X,M)=(b\,c)\otimes(a,b)=(a,c)$ which is a contradiction because $b\neq c$.

Proposition 4.31

Let $X$ be a finitely supported subset of an invariant set $(Y,\cdot)$. If $X$ is FSM Tarski III infinite, then there exists a finitely supported bijection $g:\mathbb{N}\times X\to X$. The reverse implication is also valid.

Proof. By hypothesis, there is a finitely supported bijection $\varphi:\{0,1\}\times X\to X$. Let us consider the mappings $f_{1},f_{2}:X\to X$ defined by $f_{1}(x)=\varphi(0,x)$ for all $x\in X$ and $f_{2}(x)=\varphi(1,x)$ for all $x\in X$, that are injective and supported by $supp(\varphi)$ according to Proposition 2.11. Since $\varphi$ is injective we also have $Im(f_{1})\cap Im(f_{2})=\emptyset$, and because $\varphi$ is surjective we get $Im(f_{1})\cup Im(f_{2})=X$. We prove by induction that the $n$-times auto-composition of $f_{2}$, denoted by $f_{2}^{n}$, is supported by $supp(f_{2})$ for all $n\in\mathbb{N}$. For $n=1$ this is obvious. So assume that $f_{2}^{n-1}$ is supported by $supp(f_{2})$. By Proposition 2.11 we must have $f_{2}^{n-1}(\sigma\cdot x)=\sigma\cdot f_{2}^{n-1}(x)$ for all $\sigma\in Fix(supp(f_{2}))$ and $x\in X$. Let us fix $\pi\in Fix(supp(f_{2}))$. According to Proposition 2.11, we have $f_{2}^{n}(\pi\cdot x)=f_{2}(f_{2}^{n-1}(\pi\cdot x))=f_{2}(\pi\cdot f_{2}^{n-1}(x))=\pi\cdot f_{2}(f_{2}^{n-1}(x))=\pi\cdot f_{2}^{n}(x)$ for all $x\in X$, and so $f_{2}^{n}$ is finitely supported from Proposition 2.11. Define $f:\mathbb{N}\times X\to X$ by $f((n,x))=f_{2}^{n}(f_{1}(x))$. Let $\pi\in Fix(supp(f_{1})\cup supp(f_{2}))$. According to Proposition 2.11 and because $(\mathbb{N},\diamond)$ is a trivial invariant set we get $f(\pi\otimes(n,x))=f((n,\pi\cdot x))=f_{2}^{n}(f_{1}(\pi\cdot x))=f_{2}^{n}(\pi\cdot f_{1}(x))=\pi\cdot f_{2}^{n}(f_{1}(x))=\pi\cdot f((n,x))$ for all $(n,x)\in\mathbb{N}\times X$, which means $f$ is supported by $supp(f_{1})\cup supp(f_{2})$. We prove the injectivity of $f$. Assume $f((n,x))=f((m,y))$ which means $f_{2}^{n}(f_{1}(x))=f_{2}^{m}(f_{1}(y))$. If $n>m$ this leads to $f_{2}^{n-m}(f_{1}(x))=f_{1}(y)$ (since $f_{2}$ is injective) which is in contradiction with the relation $Im(f_{1})\cap Im(f_{2})=\emptyset$. Analogously we cannot have $n<m$. Thus, $n=m$ which leads to $f_{1}(x)=f_{1}(y)$, and so $x=y$ due to the injectivity of $f_{1}$. Therefore, $f$ is injective. Since we obviously have a finitely supported injection from $X$ into $\mathbb{N}\times X$ (e.g $x\mapsto(0,x)$ which is supported by $supp(X)$), in the view of Lemma 3.4 we can find a finitely supported bijection between $X$ and $\mathbb{N}\times X$.

The reverse implication is almost trivial. There is a finitely supported injection from $\{0,1\}\times X$ into $\mathbb{N}\times X$. If there is a finitely supported injection from $\mathbb{N}\times X$ into $X$, then there is a finitely supported injection from $\{0,1\}\times X$ into $X$. The desired result follows from Lemma 3.4.

5 Countability

Definition 5.1

Let $Y$ be a finitely supported subset of an invariant set $X$. Then $Y$ is countable in FSM (or FSM countable) if there exists a finitely supported onto mapping $f:\mathbb{N}\to Y$.

Proposition 5.2

Let $Y$ be a finitely supported countable subset of an invariant set $(X,\cdot)$. Then $Y$ is uniformly supported.

Proof. There exists a finitely supported onto mapping $f:\mathbb{N}\to Y$. Thus, for each arbitrary $y\in Y$, there exists $n\in\mathbb{N}$ such that $f(n)=y$. According to Proposition 2.11, for each $\pi\in Fix(supp(f))$ we have $\pi\cdot y=\pi\cdot f(n)=f(\pi\diamond n)=f(n)=y$, where $\diamond$ is the necessarily trivial action on $\mathbb{N}$. Thus, $Y$ is uniformly supported by $supp(f)$.

Proposition 5.3

Let $Y$ be a finitely supported subset of an invariant set $X$. Then $Y$ is countable in FSM if and only if there exists a finitely supported one-to-one mapping $g:Y\to\mathbb{N}$.

Proof. Suppose that $Y$ is countable in FSM. Then there exists a finitely supported onto mapping $f:\mathbb{N}\to Y$. We define $g:Y\to\mathbb{N}$ by $g(y)=min[f^{-1}(\{y\})]$, for all $y\in Y$. According to Proposition 2.11, $g$ is supported by $supp(f)\cup supp(Y)$. Obviously, $g$ is one-to-one. Conversely, if there exists a finitely supported one-to-one mapping $g:Y\to\mathbb{N}$, then $g(Y)$ is supported is equivariant as a subset of the trivial invariant set $\mathbb{N}$. Thus, there exists a finitely supported bijection $g:Y\to g(Y)$, where $g(Y)\subseteq\mathbb{N}$. We define $f:\mathbb{N}\to Y$ by

[TABLE]

where $t$ is a fixed element of $Y$. According to Proposition 2.11, we have that $f$ is supported by $supp(g)\cup supp(Y)\cup supp(t)$. Moreover, $f$ is onto.

Proposition 5.4

Let $Y$ be an infinite, finitely supported, countable subset of an invariant set $X$. Then there exists a finitely supported bijective mapping $g:Y\to\mathbb{N}$.

Proof. First we prove that for any infinite subset $B$ of $\mathbb{N}$, there is an injection from $\mathbb{N}$ into $B$. Fix such a $B$. It follows that $B$ is well ordered. Define $f:\mathbb{N}\to B$ by: $f(1)=min(B)$, $f(2)=min(B\setminus f(1))$, and recursively $f(m)=min(B\setminus\{f(1),f(2),...,f(m-1)\})$ for all $m\in\mathbb{N}$ (since $B$ is infinite). Since $\mathbb{N}$ is well ordered, choice is not involved. Obviously since both $B$ and $\mathbb{N}$ are trivial invariant sets, we have that $f$ is equivariant. Since $B$ is a subset of $\mathbb{N}$ we also have an equivariant injective mapping $h:B\to\mathbb{N}$. According to Lemma 3.3, there is an equivariant bijection between $B$ and $\mathbb{N}$ (we can even prove that $f$ is bijective).

Since $Y$ is countable, there exists a finitely supported one-to-one mapping $u:Y\to\mathbb{N}$. Thus, the mapping $u:Y\to u(Y)$ is finitely supported and bijective. Since $u(Y)\subseteq N$, we have that there is an equivariant bijection $v$ between $u(Y)$ and $\mathbb{N}$, and so there exists a finitely supported bijective mapping $g:Y\to\mathbb{N}$ defined by $g=v\circ u$.

From [1] we know that the (in)consistency of the choice principle CC(fin) in FSM is an open problem, meaning that we do not know whether this principle is consistent or not in respect of the FSM axioms. A relationship between countable union principles and countable choice principles is presented in ZF in [4]. Below we prove that such a relationship is preserved in FSM.

Definition 5.5

The Countable Choice Principle for finite sets in FSM CC(fin) has the form “Given any invariant set $X$, and any countable family $\mathcal{F}=(X_{n})_{n}$ of finite subsets of $X$ such that the mapping $n\mapsto X_{n}$ is finitely supported, there exists a finitely supported choice function on $\mathcal{F}$." 2. 2.

The Countable Union Theorem for finite sets in FSM, CUT(fin), has the form “Given any invariant set $X$ and any countable family $\mathcal{F}=(X_{n})_{n}$ of finite subsets of $X$ such that the mapping $n\mapsto X_{n}$ is finitely supported, then there exists a finitely supported onto mapping $f:\mathbb{N}\to\underset{n}{\cup}X_{n}$" 3. 3.

The Countable Union Theorem for $k$-element sets in FSM, CUT(k), has the form “Given any invariant set $X$ and any countable family $\mathcal{F}=(X_{n})_{n}$ of $k$-element subsets of $X$ such that the mapping $n\mapsto X_{n}$ is finitely supported, then there exists a finitely supported onto mapping $f:\mathbb{N}\to\underset{n}{\cup}X_{n}$" 4. 4.

The Countable Choice Principle for sets of $k$-element sets in FSM, CC(k) has the form “Given any invariant set $X$ and any countable family $\mathcal{F}=(X_{n})_{n}$ of $k$-element subsets of $X$ in FSM such that the mapping $n\mapsto X_{n}$ is finitely supported, there exists a finitely supported choice function on $\mathcal{F}$."

Proposition 5.6

In FSM, the following equivalences hold.

CUT(fin)* $\Leftrightarrow$ CC(fin);* 2. 2.

CUT(2)* $\Leftrightarrow$ CC(2);* 3. 3.

CUT(n)* $\Leftrightarrow$ CC(i) for all $i\leq n$.*

Proof. 1. Let us assume that CUT(fin) is valid in FSM. We consider the finitely supported countable family $\mathcal{F}=(X_{n})_{n}$ in FSM, where each $X_{n}$ is a non-empty finite subset of an invariant set $X$ in FSM.

From CUT(fin), there exists a finitely supported onto mapping $f:\mathbb{N}\to\underset{n}{\cup}X_{n}$. Since $f$ is onto and each $X_{n}$ is non-empty, we have that $f^{-1}(X_{n})$ is a non-empty subset of $\mathbb{N}$ for each $n\in\mathbb{N}$. Consider the function $g:\mathcal{F}\to\cup\mathcal{F}$, defined by $g(X_{n})=f(min[f^{-1}(X_{n})])$. We claim that $supp(f)\cup supp(n\mapsto X_{n})$ supports $g$. Let $\pi\in Fix(supp(f)\cup supp(n\mapsto X_{n}))$. According to Proposition 2.11, and because $\mathbb{N}$ is a trivial invariant set and each element $X_{n}$ is supported by $supp(n\mapsto X_{n})$, we have $\pi\cdot g(X_{n})=\pi\cdot f(min[f^{-1}(X_{n})])=f(\pi\diamond min[f^{-1}(X_{n})])=f(min[f^{-1}(X_{n})])=g(X_{n})=g(\pi\star X_{n})$, where by $\star$ we denoted the $S_{A}$-action on $\mathcal{F}$, by $\cdot$ we denoted the $S_{A}$-action on $\cup\mathcal{F}$ and by $\diamond$ we denoted the trivial action on $\mathbb{N}$. Therefore, $g$ is finitely supported. Moreover, $g(X_{n})\in X_{n}$, and so $g$ is a choice function on $\mathcal{F}$.

Conversely, let $\mathcal{F}=(X_{n})_{n}$ be a countable family of finite subsets of $X$ such that the mapping $n\mapsto X_{n}$ is finitely supported. Thus, each $X_{n}$ is supported by the same set $S=supp(n\mapsto X_{n})$. Since each $X_{n}$ is finite (and the support of a finite set coincides with the union of the supports of its elements), as in the proof of Lemma 4.4, we have that $Y=\underset{n\in\mathbb{N}}{\cup}X_{n}$ is uniformly supported by $S$. Moreover, the countable sequence $(Y_{n})_{n\in\mathbb{N}}$ defined by $Y_{n}=X_{n}\setminus\underset{m<n}{\cup}X_{m}$ is a uniformly supported (by $S$) sequence of pairwise disjoint uniformly supported sets with $Y=\underset{n\in\mathbb{N}}{\cup}Y_{n}$. Consider the infinite family $M\subseteq\mathbb{N}$ such that all the terms of $(Y_{n})_{n\in M}$ are non-empty.

For each $n\in M$, the set $T_{n}$ of total orders on $Y_{n}$ is finite, non-empty, and uniformly supported by $S$. Thus, by applying CC(fin) to $(T_{n})_{n\in M}$, there is a choice function $f$ on $(T_{n})_{n\in M}$ which is also supported by $S$. Furthermore, $f(T_{n})$ is supported by $supp(f)\cup supp(T_{n})=S$ for all $n\in M$. One can define a uniformly supported (by $S$) total order relation on $Y$ (which is also a well order relation on $Y$) as follows

$x\leq y$ if and only if \left\{\begin{array}[]{ll}x\in Y_{n}\;\text{and}\;y\in Y_{m}\;\text{with}\;n<m\\ \text{or}\\ x,y\in Y_{n}\;\text{and}\;xf(T_{n})y\end{array}\right..

Clearly, if $Y$ is infinite, then there is an $S$-supported order isomorphism between $(Y,\leq)$ and $M$ with the natural order, which means, in the view of Proposition 5.4, that $Y$ is countable.

2. As in the above item $\textbf{CUT(2)}\Rightarrow\textbf{CC(2)}$.

For proving $\textbf{CC(2)}\Rightarrow\textbf{CUT(2)}$, let $\mathcal{F}=(X_{n})_{n}$ be a countable family of 2-element subsets of $X$ such that the mapping $n\mapsto X_{n}$ is finitely supported. According to CC(2) we have that there exists a finitely supported choice function $g$ on $(X_{n})_{n}$. Let $x_{n}=g(X_{n})\in X_{n}$. As in the above item, we have that $supp(n\mapsto X_{n})$ supports $x_{n}$ for all $n\in\mathbb{N}$.

For each $n$, let $y_{n}$ be the unique element of $X_{n}\setminus\{x_{n}\}$. Since for any $n$ both $x_{n}$ and $X_{n}$ are supported by the same set $supp(n\mapsto X_{n})$, it follows that $y_{n}$ is also supported by $supp(n\mapsto X_{n})$ for all $n\in\mathbb{N}$.

Define $f:\mathbb{N}\to\underset{n}{\cup}X_{n}$ by f(n)=\left\{\begin{array}[]{ll}x_{\frac{n}{2}}&\text{if}\>n\;\text{is even}\\ \\ y_{\frac{n-1}{2}}&\text{if}\>n\;\text{is odd}\end{array}\right.. We can equivalently describe $f$ as being defined by $f(2k)=x_{k}$ and $f(2k+1)=y_{k}$. Clearly, $f$ is onto. Furthermore, because all $x_{n}$ and all $y_{n}$ are uniformly supported by $supp(n\mapsto X_{n})$, we have that $f(n)=\pi\cdot f(n)$, for all $\pi\in Fix(supp(n\mapsto X_{n}))$ and all $n\in\mathbb{N}$. Thus, according to Proposition 2.11, we obtain that $f$ is also supported by $supp(n\mapsto X_{n})$, and so $\underset{n}{\cup}X_{n}$ is FSM countable.

3. As in the proof of the first item.

We can easily remark that under CC(fin) a finitely supported subset $X$ of an invariant set is FSM Dedekind infinite if and only if $\wp_{fin}(X)$ is FSM Dedekind infinite.

Proposition 5.7

Let $Y$ be a finitely supported countable subset of an invariant set $X$. Then the set $\underset{n\in\mathbb{N}}{\cup}Y^{n}$ is countable, where $Y^{n}$ is defined as the n-time Cartesian product of $Y$.

Proof. Since $Y$ is countable, we can order it as a sequence $Y=\{x_{1},\ldots,x_{n},\ldots\}$. The other sets of form $Y^{k}$ are uniquely represented in respect of the previous enumeration of the elements of $Y$. Since $Y$ is finitely supported and countable, all the elements of $Y$ are supported by the same set $S$ of atoms. Thus, in the view of Proposition 2.5, for each $k\in\mathbb{N}$, all the elements of $Y^{k}$ are supported by $S$. Fix $n\in\mathbb{N}$. On $Y^{n}$ define the S-supported strict well order relation $\sqsubset$ by: $(x_{i_{1}},x_{i_{2}},\ldots,x_{i_{n}})\sqsubset(x_{j_{1}},x_{j_{2}},\ldots,x_{j_{n}})$ if and only if \left\{\begin{array}[]{ll}i_{1}<j_{1}\\ \text{or}\\ i_{1}=j_{1}\;\text{and}\;i_{2}<j_{2}\\ \text{or}\\ \ldots\\ \text{or}\\ i_{1}=j_{1},\ldots,i_{n-1}=j_{n-1}\;\text{and}\;i_{n}<j_{n}\end{array}\right..

Now, define an $S$-supported strict well order relation $\prec$ on $\underset{n\in\mathbb{N}}{\cup}Y^{n}$ by

$u\prec v$ if and only if \left\{\begin{array}[]{ll}u\in Y^{n}\;\text{and}\;v\in Y^{m}\;\text{with}\;n<m\\ \text{or}\\ u,v\in Y_{n}\;\text{and}\;u\sqsubset v\end{array}\right..

Therefore, there exists an $S$-supported order isomorphism between $(\underset{n\in\mathbb{N}}{\cup}Y^{n},\prec)$ and $(\mathbb{N},<)$.

6 Conclusion

It is known that, when an infinite family of elements having no internal structure is considered by weakening some axioms of the ZF set theory, the results in ZF may lose their validity. According to Theorem 5.4 in [3], multiple choice principle and Kurepa’s antichain principle are both equivalent to the axiom of choice in ZF. However, in Theorem 9.2 of [5] it is proved that multiple choice principle is valid in the Second Fraenkel Model, while the axiom of choice fails in this model. Furthermore, Kurepa’s maximal antichain principle is valid in the Basic Fraenkel Model, while multiple choice principle fails in this model. This means that the following two statements (that are valid in ZF) ‘Kurepa’s principle implies axiom of choice’ and ‘Multiple choice principle implies axiom of choice’ fail in Zermelo Fraenkel set theory with atoms.

FSM is related to set theory with atoms, however in our approach $A$ is considered as a ZF set (without being necessary to modify the axioms of foundation or of extensionality), and invariant sets are defined as sets with group actions. Additionally, FSM involves an axiom of finite support which states that only atomic finitely supported structures (under a canonical hierarchical set-theoretical construction) are allowed in the theory. Therefore, there is indeed a similarity between the development of permutation models of set theory with atoms and FSM, but this framework is developed over the standard ZF in the form ‘usual sets together with actions of permutation groups’ without being necessary to consider an alternative set theory. The goal of this paper is to answer to a natural question whether the theorems involving the usual/non-atomic ZF sets remain valid in the framework of atomic sets with finite supports modulo canonical permutation actions. It is already known that there exist results that are consistent with ZF, but the are invalid when replacing ‘non-atomic structure’ with ‘atomic finitely supported structure’. The ZF results are not valid in FSM unless we are able to reformulate them with respect to the finite support requirement. The proofs of the FSM results should not brake the principle that any structure has to be finitely supported, which means that the related proofs should be internally consistent in FSM and not retrieved from ZF. The methodology for moving from ZF into FSM is based on the formalization of FSM into higher order logic (and this is not a simple task due to some important limitations) or on the hierarchical construction of supports using the $S$-finite support reasoning that actually represents an hierarchical method for defining the support of a structure using the supports of the sub-structures of the related structure. Since any structure has to be finitely supported in FSM, specific results (that are not derived from ZF) can also be obtained.

In this paper we study infinite cardinalities of finitely supported structures. The preorder relation $\leq$ on FSM cardinalities defined by involving finitely supported injective mappings is antisymmetric, but not total. The preorder relation $\leq^{*}$ on FSM cardinalities defined by involving finitely supported surjective mappings is not antisymmetric, nor total. Thus, Cantor-Schröder-Bernstein theorem (in which cardinalities are ordered by involving finitely supported injective mappings) is consistent with the finite support requirement of FSM. However, the dual of Cantor-Schröder-Bernstein theorem (in which cardinalities are ordered by involving finitely supported surjective mappings) is not valid for finitely supported structures. Several other specific properties of cardinalities are presented in Theorem 3.13.

The idea of presenting various approaches regarding ‘infinite’ belongs to Tarski who formulates several definitions of infinite in [8]. The independence of these definitions was later proved in set theory with atoms in [6]. Such independence results can be transferred into classical ZF set theory by employing Jech-Sochor’s embedding theorem stating that permutation models of set theory with atoms can be embedded into symmetric models of ZF, and so a statement which holds in a given permutation model of set theory with atoms and whose validity depend only on a certain fragment of that model, also holds in some well-founded model of ZF. In this paper we reformulate the definitions of (in)finiteness from [8] internally into FSM, in terms of finitely supported structures. The related definitions for ‘FSM infinite’ are introduced in Section 4. We particularly mention FSM usual infinite, FSM Tarski (of three types) infinite, FSM Dedekind infinite, FSM Mostowski infinite, FSM Kuratowski infinite, or FSM ascending infinite. We were able to establish comparison results between them and to present relevant examples of FSM sets that satisfy certain specific infinity properties. These comparison results are proved internally in FSM, by employing only finitely supported constructions. Some of the results are obtained by using the classical translation technique from ZF into FSM involving the $S$-finite support principle, while many other properties (especially those revealing uniform supports) are specific to FSM. We also provide connections with FSM (uniformly) amorphous sets. We particularly have focused on the notion of FSM Dedekind infinity, and we proved a full characterization of FSM Dedekind infinite sets. For example, we were able to prove that $T_{fin}(A)$, $\wp_{fin}(\wp_{fs}(A))$, $A^{A}_{fs}$, $\wp_{fin}(A^{A}_{fs})$, $(A^{n})^{A}_{fs}$ (for a fixed $n\in\mathbb{N}$), $T_{fin}(A)^{A}_{fs}$ and $\wp_{fs}(A)^{A}_{fs}$ are not FSM Dedekind infinite (nor FSM Mostowski infinite), while $\wp_{fs}(\wp_{fin}(A))$ and $T^{\delta}_{fin}(A)$ are FSM Dedekind infinite. The notion of ‘countability’ is described in FSM in Section 5, where we present connections between countable choice principles and countable union theorems within finitely supported sets.

In Figure 1 we point out some of the relationships between the FSM definitions of infinite. The ‘red arrows’ symbolize strict implications (of from $p$ implies $q$, but $q$ does not imply $p$), while ‘black arrows’ symbolize implications for which we have not proved yet if they may be strict or not (analyze this in respect of Remark 4.23). Blue arrows represent equivalences.

[TABLE]

In this final table we present the forms of infinity satisfied by the classical FSM sets.