2.1 Non-Archimedean fields

Here, we recall the basic definitions and some facts regarding non-Archimedean fields. In the following, ${\mathbb {K}}$ will denote an ordered field. We recall that such a field contains (a copy of) the rational numbers. Its elements will be called numbers.

Infinitesimal numbers can be used to formalize the notion of “infinitely close”:

Clearly, the relation “ $\sim $ ” of infinite closeness is an equivalence relation.

The standard part can be regarded as a function:

(3) $$ \begin{align} st:\left\{ x\in \mathbb{K}\mid x\ \text{is finite}\right\} \rightarrow \mathbb{R}. \end{align} $$

Moreover, with some abuse of notation, we can extend $st$ to all $\mathbb {K} $ by setting

$$ \begin{align*} st\left( \xi \right) =\left\{ \begin{array}{cc} +\infty, & \text{if}\ \xi \ \text{is a positive infinite number,} \\ -\infty, & \text{if}\ \xi \ \text{is a negative infinite number.} \end{array} \right. \end{align*} $$

2.2 Construction of the Euclidean numbers

Given any set E we let $\mathbb {V}(E)$ be the superstructure on E, namely the family of sets which is inductively defined as follows:

$$ \begin{align*} \mathbb{V}_{0}(E) &=E, \\ \mathbb{V}_{n+1}(E) &=\mathbb{V}_{n}(E)\cup \mathbf{\wp }\left( \mathbb{V} _{n}(E)\right)\!, \\ \mathbb{V}(E) &=\bigcup\limits_{n=0}^{\infty }\mathbb{V}_{n}(E). \end{align*} $$

If an object $x\in \mathbb {V}_{n+1}(E)\setminus \mathbb {V}_{n}(E)$ we say that its rank is $n+1$ , and we write $rank(x)=n+1$ . With the usual identifications of pairs with Kuratowski pairs and functions and relations with their graphs, we have that $\mathbb {V}(E)$ contains all the usual mathematical objects that can be constructed from E. Moreover, notice that if E is finite then also each finite level $V_{n}(E)$ of the superstructure on E is finite.

Now we let $\mathbb {A}$ be a set of atoms whose cardinality $\kappa $ is the first strongly uncountable inaccessible cardinal number, and we assume that $ \mathbb {R}\subset \mathbb {A}$ . The mathematical universe we will consider in this paper is

$$ \begin{align*} \Lambda =\left\{ E\in \mathbb{V}(\mathbb{A})\ |\ E\text{ is an atom or a set such that}\ |E|<\kappa \right\}, \end{align*} $$

where $|E|$ denotes the cardinality of $E.$

We let $\mathfrak {L}$ be the family of finite subsets of $\Lambda $ :

$$ \begin{align*} \mathfrak{L}=\mathbf{\wp }_{fin}(\Lambda ). \end{align*} $$

$\mathfrak {L}$ , ordered by the inclusion relation $\subseteq $ , is a directed set; if E is any set, we call net (with values in E) any function

$$ \begin{align*} \varphi :\mathfrak{L}\rightarrow E. \end{align*} $$

From now on, we will denote by $\sqsubseteq $ a partial order relation over $ \Lambda $ that extends the inclusion, namely such that $\forall \lambda ,\mu \in \mathfrak {L,}$

$$ \begin{align*} \lambda \subseteq \mu \Rightarrow \lambda \sqsubseteq \mu. \end{align*} $$

We assume that also $\left ( \mathfrak {L},\sqsubseteq \right ) $ is a directed set; for the moment we will not make any other assumption on E. One of the main tasks of this paper is to define $\sqsubseteq $ in such a way to get a numerosity theory which satisfies the requests described in the introduction.

Let

$$ \begin{align*} \mathfrak{F}\left( \mathfrak{L},\mathbb{R}\right) =\left\{ \varphi \in \mathbb{R}^{\mathfrak{L}}\ |\ \exists A\in \Lambda \ \forall\lambda\in\mathfrak{L} \ \varphi \left( \lambda \cap A\right) =\varphi \left( \lambda \right) \right\}^{3} \end{align*} $$

be endowed with the natural operations

$$ \begin{align*} \left( \varphi +\psi \right) (\lambda ) &=\varphi (\lambda )+\psi (\lambda ), \\ \left( \varphi \cdot \psi \right) (\lambda ) &=\varphi (\lambda )\cdot \psi (\lambda ), \end{align*} $$

and the partial ordering

$$ \begin{align*} \varphi \geq \psi \Leftrightarrow \forall \lambda \in \mathfrak{L,\ }\varphi (\lambda )\geq \psi (\lambda ). \end{align*} $$

The field of Euclidean numbers is defined as follows:Footnote ⁴

Definition 2.5. The field of Euclidean numbers $\mathbb {E\supset R}$ is a field so that there exists a surjective map

$$ \begin{align*} J:\mathfrak{F}\left( \mathfrak{L},\mathbb{R}\right) \rightarrow \mathbb{E} \end{align*} $$

with the following properties:

1. (i) Ring homomorphism: J is a ring homomorphism, namely for all $\varphi ,\psi \in \mathfrak {F}\left ( \mathfrak {L},\mathbb {R} \right ) $

   • • $J\left ( \varphi +\psi \right ) =J\left ( \varphi \right ) +J\left ( \psi \right ) $ and

   • • $J\left ( \varphi \cdot \psi \right ) =J\left ( \varphi \right ) \cdot J\left ( \psi \right)\!.$

2. (ii) Monotonicity: For all $\varphi \in \mathfrak {F}\left ( \mathfrak {L},\mathbb {R}\right ) $ , for all $r\in \mathbb {R} $ , if eventually $\varphi (\lambda )\geq r ($ namely there exists $\lambda _{0}\in \mathfrak {L}$ such that $\forall \lambda \sqsupseteq \lambda _{0},\ \varphi (\lambda )\geq r)$ , then

   $$ \begin{align*} J\left( \varphi \right) \geq r. \end{align*} $$

Let us show that such a field exists.Footnote ⁵

Proof. Let $\mathcal {U}$ be a fine ultrafilter on $\mathfrak {L}$ , namely a filter of sets such that

• • Maximality: $Q\in \mathcal {U}\Leftrightarrow \mathfrak {L} \backslash Q\notin \mathcal {U}$ .

• • Fineness: $\forall \lambda \in \mathfrak {L},\ Q\left [ \lambda \right ] \in \mathcal {U}$ , where

  (4) $$ \begin{align} Q\left[ \lambda \right] :=\left\{ \mu \in \mathfrak{L}\ |\ \mu \sqsupseteq \lambda \right\}\!. \end{align} $$

The existence of $\mathcal {U}$ is a well-known and easy consequence of Zorn’s Lemma. We use $\mathcal {U}$ to introduce an equivalence relation on nets, by letting for all $\psi ,\varphi \in \mathfrak {F}\left ( \mathfrak {B}, \mathbb {R}\right ) $

$$ \begin{align*} \varphi \approx _{\mathcal{U}}\psi \Longleftrightarrow \exists Q\in \mathcal{ U}\ \forall \lambda \in Q,\ \varphi \left( \lambda \right) =\psi \left( \lambda \right)\!. \end{align*} $$

We set

$$ \begin{align*} \widetilde{\mathbb{E}}:=\mathfrak{F}\left( \mathfrak{L},\mathbb{R}\right)/{\approx _{\mathcal{U},}} \end{align*} $$

and we denote by $\left [ \varphi \right ] _{\mathcal {U}}$ the equivalence classes. Now we take an injective map

$$ \begin{align*} \Phi :\widetilde{\mathbb{E}}\rightarrow \mathbb{A} \end{align*} $$

such that $\forall r\in \mathbb {R}$ ,

$$ \begin{align*} \Phi \left( \left[ c_{r}\right] _{\mathcal{U}}\right) =r, \end{align*} $$

where $c_{r}$ is the net identically equal to $r.$ Finally we set

$$ \begin{align*} \mathbb{E}=\Phi \left( \widetilde{\mathbb{E}}\right)\!. \end{align*} $$

The operations on $\mathbb {E}$ can be easily defined by letting

$$ \begin{align*} \Phi \left( \left[ \varphi \right] _{\mathcal{U}}\right) +\Phi \left( \left[ \psi \right] _{\mathcal{U}}\right) =\Phi \left( \left[ \varphi +\psi \right] _{\mathcal{U}}\right) ;\ \ \ \Phi \left( \left[ \varphi \right] _{\mathcal{U} }\right) \cdot \Phi \left( \left[ \psi \right] _{\mathcal{U}}\right) =\Phi \left( \left[ \varphi \cdot \psi \right] _{\mathcal{U}}\right)\!. \end{align*} $$

It is very well known (see, e.g., [Reference Keisler19]) and simple to show that, thanks to $\mathcal {U}$ being an ultrafilter, $\mathbb {E}$ endowed with the above operations is a field; moreover, it can be made an ordered field by endowing it with the following ordering:

$$ \begin{align*} \forall \varphi ,\psi \in \mathfrak{F}\left( \mathfrak{L},\mathbb{R}\right)\!,\ \Phi \left( \left[ \varphi \right] _{\mathcal{U}}\right) \geq \Phi \left( \left[ \psi \right] _{\mathcal{U}}\right) :\Longleftrightarrow \exists Q\in \mathcal{U},\ \forall \lambda \in Q\ \ \varphi \left( \lambda \right) \geq \psi \left( \lambda \right)\!. \\[-36pt] \end{align*} $$

⊣

The number $J\left ( \varphi \right ) $ is called the $\Lambda $ -limit of the net $\varphi $ and will be denoted by

$$ \begin{align*} J\left( \varphi \right) =\lim_{\lambda \uparrow \Lambda }\varphi (\lambda ). \end{align*} $$

The reason of this name and notation is that the operation

$$ \begin{align*} \varphi \mapsto \lim_{\lambda \uparrow \Lambda }\varphi (\lambda ) \end{align*} $$

satisfies many of the properties of the usual Cauchy limit, but with the stronger property of existing for every net. More exactly, it satisfies the following properties:

• • Existence: Every net $\varphi :\mathfrak {L}\rightarrow \mathbb {R}$ has a unique limit $L\in \mathbb {E}$ .

• • Monotonicity: For all $r\in \mathbb {R}$ if eventually $ \varphi (\lambda )\geq r$ , then

  $$ \begin{align*} \lim_{\lambda \uparrow \Lambda }\varphi (\lambda )\geq r. \end{align*} $$

• • Sum and product: For all $\varphi ,\psi :\mathfrak {L} \rightarrow \mathbb {R}$

  $$ \begin{align*} \lim_{\lambda \uparrow \Lambda }\varphi (\lambda )+\lim_{\lambda \uparrow \Lambda }\psi (\lambda ) &=\lim_{\lambda \uparrow \Lambda }\left( \varphi (\lambda )+\psi (\lambda )\right)\!, \\ \lim_{\lambda \uparrow \Lambda }\varphi (\lambda )\cdot \lim_{\lambda \uparrow \Lambda }\psi (\lambda ) &=\lim_{\lambda \uparrow \Lambda }\left( \varphi (\lambda )\cdot \psi (\lambda )\right)\!. \end{align*} $$

Notice that, if $\lim _{\lambda \rightarrow \Lambda }\varphi (\lambda )$ denotes the usual Cauchy limit of $\varphi $ , the relationship between the Cauchy limit and the $\Lambda $ -limit is

$$ \begin{align*} \lim_{\lambda\rightarrow\Lambda}\varphi(\lambda)=st\left(\lim_{\lambda \uparrow\Lambda}\varphi(\lambda)\right)\!. \end{align*} $$

Remark 2.7. The notion of Euclidean field defined by Definition 2.5 has been used in several papers with “ $\subseteq $ ” instead of “ $\sqsubseteq $ ” (e.g., [Reference Benci2, Reference Benci and Luperi Baglini7, Reference Benci, Luperi Baglini and Squassina8]). Now, we will explain the main technical reason for using a Euclidean field rather than a “generic” hyperreal field. A set $F\subset \Lambda ^{\ast }$ is called hyperfinite if

$$ \begin{align*} F=\lim_{\lambda \uparrow \Lambda }F_{\lambda }=\left\{ \lim_{\lambda \uparrow \Lambda }\ \varphi (\lambda )\ |\ \varphi (\lambda )\in F_{\lambda }\right\}, \end{align*} $$

where the sets $F_{\lambda }\in \Lambda $ are finite. Hyperfinite sets play a crucial role in many applications of nonstandard analysis. If we use a Euclidean field, we can associate with every set $E\in \Lambda $ a unique hyperfinite set $E^{\circledast }$ defined as follows:

$$ \begin{align*} E^{\circledast }=\lim_{\lambda \uparrow \Lambda }E\cap \lambda =\left\{ \lim_{\lambda \uparrow \Lambda }\ \varphi (\lambda )\ |\ \varphi (\lambda )\in E\cap \lambda \right\}\!. \end{align*} $$

The set $E^{\circledast }$ satisfies the propertyFootnote ⁶

$$ \begin{align*} E^{\sigma }\subset E^{\circledast }\subset E^{\ast }, \end{align*} $$

which is very useful in the applications. Moreover, using a Euclidean field we can easily define the numerosity function over any set $E\in \Lambda $ by setting (see Section 3)

(5) $$ \begin{align} \mathfrak{num}\left( E\right) :=\lim_{\lambda \uparrow \Lambda }\ \left\vert E\cap \lambda \right\vert\!. \end{align} $$

In this paper, replacing “ $\subseteq $ ” with a suitable “ $\sqsubseteq , $ ” the numerosity theory given by Equation (5) is consistent with the main features of the numerosity theories present in the literature (e.g., [Reference Benci1, Reference Benci, Bottazzi and Di Nasso3–Reference Benci and Forti5, Reference Benci and Di Nasso10–Reference Benci, Di Nasso and Forti12, Reference Blass, Di Nasso and Forti14, Reference Di Nasso and Forti16, Reference Forti and Morana17, Reference Mancosu21]).

2.3 Labelled sets

The notion of labelled set has been introduced in [Reference Benci1, Reference Benci and Di Nasso10] to construct a numerosity theory for countable sets. Here we extend this notion to adapt it to the study of numerosity theories for larger sets.

Requirement (i) gives to $\mathfrak {B}$ a lattice structure, whilst requirement (ii) entails that the elements of a label are either atoms or infinite sets.

Having fixed the notion of “labels,” we can now introduce the notion of “labelling”:

Definition 2.9. Let $\mathfrak {B}$ be a set of labels. We call $\mathfrak {B}$ -labelling the map

$$ \begin{align*} \ell :\Lambda \rightarrow \mathfrak{B} \end{align*} $$

defined as follows:

$$ \begin{align*} \ell (a)=\bigcap_{\mu \in I_{a}}\mu , \end{align*} $$

where $I_{a}=\{\mu \in \mathfrak {B}\mid a\in \mathbb {V}(\mu )\}$ . For every $ a\in \mathfrak {L}$ we call $\ell (a)$ the label of a.

Roughly speaking, the label of an object $a\in \Lambda $ is a finite set whose elements allow us to define a by the fundamental finitistic set operations.

There are plenty of sets of labels: just set

(6) $$ \begin{align} \mathfrak{B}_{\max }=\left\{ \mathfrak{t}\in \mathfrak{L}\mid \ \mathfrak{t} \cap \mathfrak{L}=\emptyset \right\}\!. \end{align} $$

Obviously, $\mathfrak {B}_{\max }$ is a label set; more importantly, every label set $\mathfrak {B}$ is a subset of $\mathfrak {B}_{\max }$ .

Example 2.10. Let $a=\mathbb {N}$ . Then, using the $\mathfrak {B}_{\max }$ -labelling,

$$ \begin{align*} \ell \left( \mathbb{N}\right) =\{\mathbb{N}\}. \end{align*} $$

Now we will describe some properties of a set of labels $\mathfrak {B}$ and the corresponding $\mathfrak {B}$ -labelling:

Proof. (i) $\mathfrak {s}$ is finite, and hence by induction it trivially holds that $\mathbb {V}_{n}\left ( \mathfrak {s}\right ) $ is finite for every $n\in \mathbb {N}$ . Therefore $\mathbb {V}\left ( \mathfrak {s}\right ) $ is countable.

(ii) As $\mathfrak {s}$ is finite, by induction it is immediate to prove that $ \mathbb {V}_{n+1}(\mathfrak {s})\setminus \mathfrak {s}$ consists only of finite sets, and hence the thesis follows straightforwardly.

(iii) The implication $\Leftarrow $ is trivial. Let us prove the other implication. If $\mathbb {V}_{m}(\mathfrak {s})\subseteq \mathbb {V}_{m}( \mathfrak {t})$ then, in particular, $\mathfrak {s}\in \mathbb {V}_{m}( \mathfrak {t})$ . Now, if $\mathfrak {s}=\{a_{1},\dots ,a_{n}\}$ , all $ a_{1},\dots ,a_{n}$ are either atoms or infinite sets, and hence by (i), $ a_{1},\dots ,a_{n}\in \mathfrak {t}$ .

(iv) The inclusion $\subseteq $ is trivial. For the reverse inclusion, let $\eta \in \mathbb {V}(\mathfrak {s})\cap \mathbb {V}(\mathfrak {t})$ . In particular, $\eta \in \mathbb {V}_{n}(\mathfrak {s})\cap \mathbb {V}_{m}( \mathfrak {t})$ and so, if $l=\max \{n,m\}$ , we have that $\eta \in \mathbb {V} _{l}(\mathfrak {s})\cap \mathbb {V}_{l}(\mathfrak {t})$ . We proceed by induction on l to show that $\mathbb {V}_{l}(\mathfrak {s})\cap \mathbb {V} _{l}(\mathfrak {t})\subseteq \mathbb {V}(\mathfrak {s}\cap \mathfrak {t})$ .

If $l=0$ , then $\eta \in \mathbb {V}_{0}(\mathfrak {s})\cap \mathbb {V}_{0}( \mathfrak {t})$ if and only if $\eta =\mathfrak {s}=\mathfrak {t}$ , and the desired inclusion trivially holds.

Now let us suppose the inclusion to hold for $l\in \mathbb {N}$ , and let $ \eta \in \mathbb {V}_{l+1}(\mathfrak {s})\cap \mathbb {V}_{l+1}(\mathfrak {t})$ . If $\eta \in \mathbb {V}_{l}(\mathfrak {s})\cap \mathbb {V}_{l}(\mathfrak {t})$ we are done by inductive hypothesis; if not, there are $A\in \mathbb {V}_{l}( \mathfrak {s}),B\in \mathbb {V}_{l}(\mathfrak {t})$ such that $\eta \in \wp (A)\cap \wp (B)$ . In particular, $\eta \in \wp (A\cap B)$ . But $A\cap B\in \mathbb {V}_{l}(\mathfrak {s})\cap \mathbb {V}_{l}(\mathfrak {t})$ , so by induction $A\cap B\in \mathbb {V}(\mathfrak {s}\cap \mathfrak {t})$ , and hence $ \eta \in \mathbb {V}(\mathfrak {s}\cap \mathfrak {t})$ as desired.

(v) The implication $\Leftarrow $ is trivial. For the reverse implication, let

$$ \begin{align*} l=\min \{n\in \mathbb{N}\mid a\in \mathbb{V}_{n}(\mathfrak{t})\}. \end{align*} $$

In particular, we have that $l\geq 1$ . In fact, if $l=0$ then $\{a\}= \mathfrak {t}$ , and this cannot happen as $\mathfrak {t}\cap \mathfrak {L} =\emptyset $ . Hence $a\in \mathbb {V}_{l-1}(\mathfrak {t})$ and we are done.⊣

As we will see in Section 5, the freedom of choosing a particular set of labels allows us to impose certain additional arithmetical properties on numerosities.

Proof. (i) If $a\subseteq b,$ then trivially $a\in \mathbb {V}( \mathfrak {t})$ whenever $b\in \mathbb {V}(\mathfrak {t})$ and hence

$$ \begin{align*} \ell (a)=\bigcap \left\{ \mathbb{V}(\mathfrak{t})\ |\ \mathfrak{t}\in \mathfrak{B},\ a\in \mathbb{V}(\mathfrak{t})\right\} \subseteq \bigcap \left\{ \mathbb{V}(\mathfrak{t})\ |\ \mathfrak{t}\in \mathfrak{B},\ b\in \mathbb{V}(\mathfrak{t})\right\} =\ell (b). \end{align*} $$

(ii) Trivially, for all $\mathfrak {t}\in \mathfrak {B}$ , if $a\in \mathbb {V}(\mathfrak {t})$ then also $\{a\}\in \mathbb {V}(\mathfrak {t})$ , and hence $\ell (\{a\})\subseteq \ell (\{a\}$ . The second claim follows from the fact that, by Proposition 2.11(v), $a\in \mathbb {V}(\mathfrak {t})\Leftrightarrow \{a\}\in \mathbb {V}( \mathfrak {t})$ .

(iii) If $a\in b,$ then $\left \{ a\right \} \subseteq b$ and by (i) and (ii), we have that

$$ \begin{align*} \ell (a)=\ell (\left\{ a\right\} )\subseteq \ell (b). \end{align*} $$

(iv) By definition, $\forall \mu \in I_{a}, a\in \mathbb {V}(\mu );$ hence

$$ \begin{align*} a\in \bigcap_{\mu \in I_{a}}\mathbb{V}(\mu )=\mathbb{V}\left( \bigcap_{\mu \in I_{a}}\mu \right) =\mathbb{V}\left( \ell (a)\right)\!. \end{align*} $$

(v) We have that $\mathfrak {s}=\mathbb {V}_{0}(\mathfrak {s} )\in \mathbb {V}(\mathfrak {s});$ hence $\mathfrak {s}\in I_{\mathfrak {s}}$ and so $\ell (\mathfrak {s})\subseteq \mathfrak {s}.$ Moreover, if $\mathfrak {t} \in I_{\mathfrak {s}}$ , $\mathfrak {s}\in \mathbb {V}(\mathfrak {t})$ and since $ \mathfrak {s}$ is a label, $\mathfrak {s}\subset \mathfrak {t}$ and so

$$ \begin{align*} \mathfrak{s}\subseteq \bigcap_{\mathfrak{t}\in I_{\mathfrak{s}}}\mathfrak{t} =\ell (\mathfrak{s}). \end{align*} $$

(vi) If $\mathfrak {s}\in \mathfrak {B},$ then $\forall \mathfrak {t}\in \mathfrak {B},$ and by Proposition 2.11(iv) we have that

$$ \begin{align*} \mathbb{V}_{m}(\mathfrak{s})\in \mathbb{V}(\mathfrak{t})\Leftrightarrow \mathfrak{s}\subseteq \mathfrak{t}\Leftrightarrow \mathfrak{s}\in \mathbb{V}( \mathfrak{t}). \end{align*} $$

Then

(7) $$ \begin{align} I_{\mathbb{V}_{m}(\mathfrak{s})} &=\{\mathfrak{t}\in \mathfrak{B}\mid \mathbb{V}_{m}(\mathfrak{s})\in \mathbb{V}(\mathfrak{t})\}=\{\mathfrak{t}\in \mathfrak{B}\mid \mathfrak{s}\subseteq \mathfrak{t}\} \\ &=\{\mathfrak{t}\in \mathfrak{B}\mid \mathfrak{s}\in \mathbb{V}(\mathfrak{t} )\}=I_{\mathfrak{s}}; \notag \end{align} $$

hence the thesis follows by the definition of $\mathfrak {B}$ -labeling.

(vii) ( $\Rightarrow $ ) If $\ell \left ( a\right ) \subseteq \mathfrak {s}$ , then, by (i), $\mathbb {V}\left ( \ell \left ( a\right ) \right ) \subseteq \mathbb {V}\left ( \mathfrak {s}\right)\!;$ therefore, by (iv), $\ a\in \mathbb {V}\left ( \mathfrak {s}\right)\!.$

( $\Leftarrow $ ) If $a\in \mathbb {V}(\mathfrak {s}),$ then $\mathfrak {s}\in I_{a}$ and so $\ell (a)\subseteq \mathfrak {s}$ .

(viii) For $\mathfrak {t}\in I_{\left \{ a,b\right \} },$ since $ \left \{ a,b\right \} \in \mathbb {V}(\mathfrak {t})$ , but $\left \{ a,b\right \} \notin \mathfrak {t}$ , we have that $a\in \mathbb {V}(\mathfrak {t})$ and $b\in \mathbb {V}(\mathfrak {t});$ then

$$ \begin{align*} I_{\left\{ a,b\right\} } &=\{\mathfrak{t}\in \mathfrak{B}\mid a\in \mathbb{V }(\mathfrak{t})\ \text{and}\ b\in \mathbb{V}(\mathfrak{t})\} \\ &=\{\mathfrak{t}\in \mathfrak{B}\mid a\in \mathbb{V}(\mathfrak{t})\}\cap \{ \mathfrak{t}\in \mathfrak{B}\mid b\in \mathbb{V}(\mathfrak{t})\}=I_{a}\cap I_{b}. \end{align*} $$

Hence

$$ \begin{align*} \ell (\left\{ a,b\right\} ) &=\bigcap_{\mathfrak{t}\in I_{\left\{ a,b\right\} }}\mathfrak{t}=\bigcap_{\mathfrak{t}\in I_{a}\cap I_{b}} \mathfrak{t}=\{x\in \Lambda \mid x\in \mathfrak{t}\ \ \text{and}\ \ \mathfrak{t}\in I_{a}\cap I_{b}\} \\ &=\{x\in \Lambda \mid (x\in \mathfrak{t}\ \text{and}\ \mathfrak{t}\in I_{a})\ \text{or}\ (x\in \mathfrak{t}\ \text{and}\ \mathfrak{t}\in I_{b})\} \\ &=\{x\in \Lambda \mid (x\in \mathfrak{t}\ \text{and}\ \mathfrak{t}\in I_{a})\}\cup \{x\in \Lambda \mid (x\in \mathfrak{t}\ \text{and}\ \mathfrak{t} \in I_{b})\} \\ &=\left( \bigcap_{\mathfrak{t}\in I_{a}}\mathfrak{t}\right) \cup \left( \bigcap_{\mathfrak{t}\in I_{b}}\mathfrak{t}\right) =\ell (a)\cup \ell (b). \end{align*} $$

(ix) We have that

$$ \begin{align*} \ell \left( \left( a,b\right) \right) =\ell \left( \{a,\{a,b\}\}\right) =\ell \left( a\right) \cup \ell \left( \{a,b\}\right) =\ell \left( b\right) \ell \left( a\right) \cup \ell \left( a\right) \cup \ell \left( b\right) =\ell \left( a\right) \cup \ell \left( b\right)\!. \end{align*} $$

(x) First set $\mathfrak {s=\ell }\left ( \lambda \right)\!.$ Let us first prove the inclusion $\subseteq $ . Let $x\in E$ be such that $\ell (x)\subseteq \mathfrak {s}$ . By the definition of labelling then $x\in \mathbb {V}(\mathfrak {s})$ , and the inclusion is proven. For the reverse inclusion, let $x\in E\cap \mathbb {V}(\mathfrak {s})$ . In particular, it must be $\ell (x)\subseteq \mathfrak {s}$ , and we are done.

(xi) If $E\in \Lambda $ then E has a finite rank n, which means that $E\cap \mathbb {V}(\mathfrak {s})=E\cap \mathbb {V}_{n}(\mathfrak {s})$ , and the conclusion follows by (x) as, by construction, $\mathbb {V} _{n}(\mathfrak {s})$ is finite.⊣

The notion of $\mathfrak {B}$ -labelling allows us to equip $\mathfrak {L}$ with a partial order structure $\sqsubseteq $ :

Definition 2.13. We set

$$ \begin{align*} \mathfrak{L}_{0}\left(\mathfrak{B}\right):=\{\mathbb{V}_{m}(\mathfrak{t})\ |\ \ m\in \mathbb{N}_{0},\ \mathfrak{t}\in \mathfrak{B}\}, \end{align*} $$

and for every $\lambda ,\mu \in \mathfrak {L}$ , we set

$$ \begin{align*} \lambda \sqsubseteq \mu \Leftrightarrow \lambda \subseteq {\mathop{\displaystyle \bigcap }} \left\{ \tau \in \mathfrak{L}_{0}\mathfrak{\ }|\ \mu \subseteq \tau \right\}\!. \end{align*} $$

Notice that, by definition

$$ \begin{align*} \lambda \sqsubseteq \mu \Leftrightarrow I_{\lambda}\subseteq I_{\mu}, \end{align*} $$

where $I_{\lambda }$ has been introduced in Definition 2.9, and that

$$ \begin{align*} \mathfrak{L}_{0}\left(\mathfrak{L_{0}}\left(\mathfrak{B}\right)\right)= \mathfrak{L}_{0}\left(\mathfrak{B}\right)\!. \end{align*} $$

Clearly $\sqsubseteq $ induces a lattice structure on $\mathfrak {L}_{0}\left ( \mathfrak {B}\right )$ , since

$$ \begin{align*} \lambda \vee \mu :={\mathop{\displaystyle \bigcap }} \left\{ \tau \in \mathfrak{L}_{0}\mathfrak{\ }|\ \lambda \cup \mu \subseteq \tau \right\} , \end{align*} $$

$$ \begin{align*} \lambda \wedge \mu :=\mathop{\displaystyle \bigcup} \left\{ \tau \in \mathfrak{L}_{0}\mathfrak{\ } |\ \tau \subseteq \lambda \cap \mu \right\}\!. \end{align*} $$

Since $\lambda \subseteq \mu \Rightarrow \lambda \sqsubseteq \mu ,$ we can use the directed set $\left ( \mathfrak {L}_{0}\left (\mathfrak {B} \right ),\sqsubseteq \right ) $ to define a field of Euclidean numbers as in Definition 2.5. From now on, $\mathbb {E}$ will denote such a field.

3.1 Definition and first properties

Definition 3.1. Let E be a set in $\Lambda $ . We call numerosity of E the Euclidean number

$$ \begin{align*} \mathfrak{num}\left( E\right) :=\lim_{\lambda \uparrow \Lambda }\ \left\vert E\cap \lambda \right\vert\!. \end{align*} $$

The set of numerosities will be denoted by $\mathbf {Num}$ .

The notion of numerosity allows us to “give a name” to some hyperreal number. We set

(8) $$ \begin{align} \alpha =\mathfrak{num}\left( \mathbb{N}\right) ;\ \beta =\mathfrak{num} \left( \left[ 0,1\right) \right)\!. \end{align} $$

The numerosity of a set depends on the choice of the set of labels $ \mathfrak {B}$ , as well as on the ultrafilter $\mathcal {U}$ on $\mathfrak {B}$ chosen to construct $\mathbb {E}$ . However the properties which will be listed below are independent of any choice.

Proof. (i) If $|E|=n<\infty $ , then for every $\lambda \in \mathfrak {L}$ , we have $\left \vert E\cap \lambda \right \vert =n$ , and the thesis then follows by taking the $\lambda $ -limit.

(ii) If $E\subset F$ , eventually $\left \vert E\cap \lambda \right \vert <\left \vert F\cap \lambda \right \vert $ , so $\lim _{\lambda \uparrow \Lambda }\left \vert E\cap \lambda \right \vert <\lim _{\lambda \uparrow \Lambda }\left \vert E\cap \lambda \right \vert $ .

(iii) Take $\lambda =\mathbb {V}_{m}(\mathfrak {s})$ with $m\geq rank(E)$ and $\mathfrak {s\in B}$ . Then, by Proposition 2.12.(x)

$$ \begin{align*} E\cap \lambda =E\cap \mathbb{V}_{m}(\mathfrak{s})=E\cap \mathbb{V}(\mathfrak{ s})=E_{\lambda }. \end{align*} $$

(iv) By hypothesis we have that for all $\lambda \in \mathfrak {L} |E_{\lambda }|=|F_{\lambda }|$ , and so by the labels principle

(9) $$ \begin{align} \mathfrak{num}\left( E\right) =\lim_{\lambda \uparrow \Lambda }\ \left\vert E\cap \lambda \right\vert =\lim_{\lambda \uparrow \Lambda }\ \left\vert F\cap \lambda \right\vert =\mathfrak{num}\left( F\right)\!. \end{align} $$

(v) Just notice that $|E\cup F|_{\lambda }=|E_{\lambda }|+|F_{\lambda }|$ for every $\lambda \in \mathfrak {L}$ ; hence the thesis follows by Definition 2.5(2) and by the labels principle.

(vi) Let $\lambda \in \mathfrak {L}$ . By property (ix) in Proposition 2.12, we have that $\left ( E\times F\right ) _{\lambda }=E_{\lambda }\times F_{\lambda }$ , and hence $|\left ( E\times F\right ) _{\lambda }|=|E_{\lambda }\times F_{\lambda }|=|E_{\lambda }|\cdot |F_{\lambda }|$ , and the thesis then follows immediately, again by the labels principle.

(vii) Let $\lambda =\mathbb {V}_{m}(\mathfrak {s})\in \mathfrak {L}_{0}$ ( $ m>rank(E))$ , and let $a\in \mathbf {\wp }_{fin}\left ( E\right ) \cap \mathbb {V}( \mathfrak {s})$ . Then by Proposition 2.12(x) we have that it must be $ a\in \mathbf {\wp }_{fin}\left ( E_{\lambda }\right ) $ . Conversely, if $a\in \mathbf {\wp }_{fin}\left ( E_{\lambda }\right ) $ it is immediate to see that $ a\in \mathbf {\wp }_{fin}\left ( E\right ) \cap \mathbb {V}(\mathfrak {s})$ . Hence, by Proposition 2.12 we have

$$ \begin{align*} \left\vert \left[ \mathbf{\wp }_{fin}\left( E\right) \right] _{\lambda }\right\vert =\left\vert \mathbf{\wp }_{fin}\left( E\right) \cap \mathbb{V}( \mathfrak{s})\right\vert =\left\vert \mathbf{\wp }_{fin}\left( E_{\lambda }\right) \right\vert =2^{\left\vert E_{\lambda }\right\vert }, \end{align*} $$

and so by the labels principle

$$ \begin{align*} \mathfrak{num}\left( \mathbf{\wp }_{fin}\left( E\right) \right) =\lim_{\lambda \uparrow \Lambda }\ 2^{\left\vert E_{\lambda }\right\vert }=2^{\mathfrak{num}\left( E\right)}\!. \end{align*} $$

(viii) We set $\lambda =\mathbb {V}_{m}(\mathfrak {s})\in \mathfrak {L}_{0}$ , $m>rank(f).$ Let $f\in \mathfrak {F}_{fin}\left ( X,E\backslash \{a\}\right ) \cap \mathbb {V}(\mathfrak {s})$ , and let D be the domain of f. By identifying functions with Kuratowski pairs, and by our definition of labellings on pairs, it is immediate to see that $f\in \mathfrak {F} _{fin}\left ( X,E\backslash \{a\}\right ) \cap \mathbb {V}(\mathfrak {s})$ if and only if $D\left ( f\right ) \subset X\cap \mathbb {V}(\mathfrak {s} )=X_{\lambda }$ and $Im\left ( f\right ) \subset \left ( E\backslash \{a\}\right ) \cap \mathbb {V}(\mathfrak {s})=E_{\lambda }\backslash \{a\}.$ Therefore

$$ \begin{align*} \mathfrak{F}_{fin}\left( X,E\backslash \{a\}\right) \cap \mathbb{V}( \mathfrak{s})=\mathfrak{F}_{fin}\left( X_{\lambda },E_{\lambda }\backslash \{a\}\right)\!. \end{align*} $$

Notice that

$$ \begin{align*} \left\vert \mathfrak{F}_{fin}\left( X_{\lambda },E_{\lambda }\backslash \{a\}\right) \right\vert =\left\vert \mathfrak{F}\left( X_{\lambda },E_{\lambda }\right) \right\vert\!. \end{align*} $$

In fact, the association $g\in \mathfrak {F}_{fin}\left ( X_{\lambda },E_{\lambda }\backslash \{a\}\right ) \rightarrow \widetilde {g}\in \mathfrak { F}\left ( X_{\lambda },E_{\lambda }\right ) $ , with

$$ \begin{align*} \widetilde{g}(x)= \begin{cases} g(x), & \text{if}\ x\in X_{\lambda }, \\ a, & \text{otherwise,} \end{cases} \end{align*} $$

is a bijection. Hence, again by the labels principle,

$$ \begin{align*} \mathfrak{num}\left( \mathfrak{F}_{fin}\left( X,E\backslash \{a\}\right) \right) &=\lim_{\lambda \uparrow \Lambda }\ \left\vert \mathfrak{F} _{fin}\left( X,E\backslash \{a\}\right) \cap \mathbb{V}(\lambda )\right\vert =\lim_{\lambda \uparrow \Lambda }\ \left\vert \mathfrak{F}\left( X_{\lambda },E_{\lambda }\right) \right\vert \\ &=\lim_{\lambda \uparrow \Lambda }\ \left\vert E_{\lambda }\right\vert ^{\left\vert X_{\lambda }\right\vert }=\mathfrak{num}\left( E\right) ^{ \mathfrak{num}\left( X\right)}\!. \\[-38pt] \end{align*} $$

⊣

4.1 The ordinal numerosities

Let $\mathbf {Num}$ be the set of numerosities.

Definition 4.1. The set $\mathbf {Ord}\subset \mathbf {Num}$ of ordinal numerosities is defined as follows: $\tau \in \mathbf {Ord}$ if and only if

$$ \begin{align*} \tau =\mathfrak{num}\left( \Omega _{\tau }\right), \end{align*} $$

where

$$ \begin{align*} \Omega _{\tau }=\left\{ x\in \mathbf{Ord\ |\ }x<\tau \right\}\!. \end{align*} $$

It is easy to see by transfinite induction that this is a good definition. In fact, it is immediate to check that

• • $0\in \mathbf {Ord}$ and

• • if $\tau \in \mathbf {Ord,}$ then $\tau +1=\mathfrak {num}\left ( \Omega _{\tau }\cup \left \{ \tau \right \} \right ) \in \mathbf {Ord}$ (and hence $ \mathbb {N}\subset \mathbf {Ord}$ ).

Moreover, if $\tau _{k}=\mathfrak {num}\left ( \Omega _{k}\right)\!, k\in K,$ ( $\left \vert K\right \vert <\kappa $ ) are ordinal numerosities, then

$$ \begin{align*} \tau :=\mathfrak{num}\left( \mathop{\displaystyle \bigcup}\limits_{k\in K}\Omega _{k}\right) \in \mathbf{Ord.} \end{align*} $$

In fact, this holds as $\mathop {\displaystyle \bigcup }\limits _{k\in K}\Omega _{k}=\{x\in \mathbf {Ord }\mid x<\tau \}$ : the inclusion $\mathop {\displaystyle \bigcup }\limits _{k\in K}\Omega _{k}\subseteq \{x\in \mathbf {Ord}\mid x<\tau \}$ holds trivially, as if $x\in \mathop {\displaystyle \bigcup } \limits _{k\in K}\Omega _{k}$ then $x\in \mathbf {Ord}$ and $x\in \Omega _{k}$ for some k, and so $x<\tau _{k}<\tau $ ; conversely, if $x\in \mathbf {Ord}$ is such that $x<\tau $ , if $x\notin \mathop {\displaystyle \bigcup }\limits _{k\in K}\Omega _{k}$ we would have that $\Omega _{x}\supseteq \mathop {\displaystyle \bigcup }\limits _{k\in K}\Omega _{k}$ , and so by taking numerosities we would get $x\geq \tau $ , which is absurd.

Definition 4.2. If $\tau _{k}, k\in K, (\left \vert K\right \vert <\kappa )$ are ordinals, we set

$$ \begin{align*} \underset{k\in K}{\sup }\tau _{k}=\mathfrak{num}\left( \mathop{\displaystyle \bigcup}\limits_{k\in K}\Omega _{\tau _{k}}\right), \end{align*} $$

where $\tau _{k}=\mathfrak {num}\left ( \Omega _{\tau _{k}}\right)\!.$

Then $\tau =\sup _{k\in K}\tau _{k}$ is the least element in $\mathbf {Ord}$ equal or greater than every $\tau _{k},$ namely $\tau \in \mathbf {Ord}$ and

(10) $$ \begin{align} \forall k\in K,\ \tau \geq \tau _{k}, \end{align} $$

(11) $$ \begin{align} \forall k\in K,\ \forall \xi \geq \tau _{k}\Rightarrow \xi \geq \tau. \end{align} $$

However $\tau $ is not the least element in $\mathbf {Num}$ greater or equal to every $\tau _{k}\mathbf {.}$ In fact, as we have seen, if $\sup _{k\in K}\tau _{k}$ is not a maximum, there are numerosities $\xi \in \mathbb {E}$ , greater than every $\tau _{k}$ and smaller than $\tau $ , e.g., $ \left (\sup _{k\in K}\tau _{k}\right )-1$ .

Our construction of the ordinal numbers is similar to the construction of Von Neumann. However, whilst a Von Neumann ordinal $\tau $ is the set of all the Von Neumann ordinals contained in $\tau $ , in our construction an ordinal $\tau $ is the numerosity of the set of ordinals smaller than $\tau $ . Hence, here, an ordinal number, as any other numerosity, is an atom.

Obviously, not all numerosities are ordinals: for example, $\mathfrak {num} \left ( \mathbb {N}\right ) $ is not an ordinal. In fact, if $\alpha =\mathfrak { num}\left ( \mathbb {N}\right ) $ were an ordinal then:

$$ \begin{align*} \alpha &=\mathfrak{num(}\left\{ x\in \mathbf{Ord}\ |\ x<\mathfrak{num} \left( \mathbb{N}\right) \right\} )=\mathfrak{num}(\mathbb{N}_{0}) \\ &=\mathfrak{num}(\mathbb{N}\cup \left\{ 0\right\} )=\alpha +1. \end{align*} $$

In a similar way, one can prove that no infinite numerosity smaller than $ \mathfrak {num}\left ( \mathbb {N}\right ) $ is an ordinal. However, $\alpha +1$ is an ordinal:

$$ \begin{align*} \alpha +1=\mathfrak{num}\left( \mathbb{N}_{0}\right) =\mathfrak{num}\left( \{x\in \mathbf{Ord}\ |\ x<\alpha \}\right)\!. \end{align*} $$

Actually $\alpha +1$ is the smallest infinite ordinal. From now on, we will call it $\omega $ .

As we expect, $\mathbf {Ord}$ is a well-ordered set; in fact if $E\subset \mathbf {Ord,}$ the minimum is given by

$$ \begin{align*} \min \ E=\sup \left\{ x\in \mathbf{Ord}\ |\ \forall a\in E,\ x\leq a\right\}\!. \end{align*} $$

4.2 Sums and products of ordinals

In this section we will show that the set of ordinal numerosities is closed under sums and products, and we will show that there is a relationship between sums and products of ordinal numerosities and the natural operations between Cantor ordinals.

First, we start by showing that the operations between numerosities are consistent with the order structure over the ordinals.

Theorem 4.3. For all ordinal numbers $\sigma ,\tau \in \mathbf {Ord}$ we have that

$$ \begin{align*} \mathfrak{num}(\Omega _{\sigma })+\mathfrak{num}(\Omega _{\tau }) &= \mathfrak{num}(\Omega _{\sigma +\tau }), \\[3pt] \mathfrak{num}(\Omega _{\sigma })\cdot \mathfrak{num}(\Omega _{\tau }) &= \mathfrak{num}(\Omega _{\sigma \tau }). \end{align*} $$

In particular, $\sigma +\tau \in \mathbf {Ord}$ and $\sigma \tau \in \mathbf { Ord}$ .

Proof. First let us prove that

$$ \begin{align*} \mathfrak{num}(\Omega _{\sigma +\tau })=\mathfrak{num}(\Omega _{\sigma })+ \mathfrak{num}(\Omega _{\tau }) \end{align*} $$

acting by induction on $\tau .$ If $\tau =0,$ then this relation is obvious. If $\tau =\gamma +1,$ then

$$ \begin{align*} \mathfrak{num}(\Omega _{\sigma +\tau }) &=\mathfrak{num}(\Omega _{\sigma +\gamma +1})=\mathfrak{num}(\Omega _{\sigma +\gamma }\cup \left\{ \sigma +\gamma +1\right\} )\\[3pt] &=\mathfrak{num}(\Omega _{\sigma +\gamma })+\mathfrak{num} (\left\{ \sigma +\gamma +1\right\} ) =\mathfrak{num}(\Omega _{\sigma })+\mathfrak{num}(\Omega _{\gamma })+1\\[3pt] &= \mathfrak{num}(\Omega _{\sigma })+\mathfrak{num}(\Omega _{\gamma }\cup \left\{ \gamma +1\right\} )=\mathfrak{num}(\Omega _{\sigma })+\mathfrak{num}(\Omega _{\tau }). \end{align*} $$

If $\tau =\ \sup _{k\in K}\tau _{k}$ (where $\tau _{k}=\mathfrak {num}\left ( \Omega _{k}\right ) $ ) is a limit ordinal, then

$$ \begin{align*} \mathfrak{num}(\Omega _{\sigma +\tau })=\sup_{k\in K}\ \mathfrak{num}(\Omega _{\sigma +\tau _{k}})=\sup_{k\in K}\ \left[ \mathfrak{num}(\Omega _{\sigma })+\mathfrak{num}(\Omega _{\tau _{k}})\right]. \end{align*} $$

Since $\sigma +\tau _{k}=\mathfrak {num}(\Omega _{\sigma })+\mathfrak {num} (\Omega _{\tau _{k}})$ is an ordinal number, $\tau $ satisfies (10) and (11) and hence

$$ \begin{align*} \forall k\in K,\ \sigma +\tau \geq \sigma +\tau _{k}, \end{align*} $$

$$ \begin{align*} \forall k\in K,\ \forall \xi\in\mathbf{Ord} \ \sigma +\xi \geq \sigma +\tau _{k}\Rightarrow \sigma +\xi \geq \sigma +\tau. \end{align*} $$

Then,

$$ \begin{align*} \sup_{k\in K}\ \left( \sigma +\tau _{k}\right) =\sigma +\sup_{k\in K}\ \tau _{k}, \end{align*} $$

and so

$$ \begin{align*} \mathfrak{num}(\Omega _{\sigma +\tau })=\sigma +\sup_{k\in K}\ \tau _{k}= \mathfrak{num}(\Omega _{\sigma })+\sup_{k\in K}\ \left[ \mathfrak{num} (\Omega _{\tau _{k}}\right] =\mathfrak{num}(\Omega _{\sigma })+\mathfrak{num} (\Omega _{\tau }). \end{align*} $$

Similarly we act with the product. If $\tau =0,$ then this relation is obvious. If $\tau =\gamma +1,$ then

$$ \begin{align*} \mathfrak{num}(\Omega _{\sigma \tau }) &=\mathfrak{num}(\Omega _{\sigma \left( \gamma +1\right) })=\mathfrak{num}(\Omega _{\sigma \gamma +\sigma })= \mathfrak{num}(\Omega _{\sigma \gamma })+\mathfrak{num}(\Omega _{\sigma }) \\[3pt] &=\mathfrak{num}(\Omega _{\sigma })\cdot \mathfrak{num}(\Omega _{\gamma })+ \mathfrak{num}(\Omega _{\sigma })=\mathfrak{num}(\Omega _{\sigma })\left[ \mathfrak{num}(\Omega _{\gamma })+1\right] \\[3pt] &=\mathfrak{num}(\Omega _{\sigma })\cdot \mathfrak{num}(\Omega _{\tau }). \end{align*} $$

If $\tau =\ \sup _{k\in K}\tau _{k}$ (where $\tau _{k}=\mathfrak {num}\left ( \Omega _{k}\right ) $ ) is a limit ordinal, then

$$ \begin{align*} \mathfrak{num}(\Omega _{\sigma \tau })=\sup_{k\in K}\ \mathfrak{num}(\Omega _{\sigma \tau _{k}})=\sup_{k\in K}\ \left[ \mathfrak{num}(\Omega _{\sigma })\cdot \mathfrak{num}(\Omega _{\tau _{k}})\right]. \end{align*} $$

Since $\tau $ satisfies (10) and (11),

$$ \begin{align*} \forall k\in K,\ \sigma \tau \geq \sigma \tau _{k}, \end{align*} $$

$$ \begin{align*} \forall k\in K,\ \forall\xi\in\mathbf{Ord} \ \sigma \xi \geq \sigma \tau _{k}\Rightarrow \sigma \xi \geq \sigma \tau. \end{align*} $$

Then

$$ \begin{align*} \sup_{k\in K}\ \left( \sigma \tau _{k}\right) =\sigma \cdot \sup_{k\in K}\ \tau _{k}; \end{align*} $$

hence

$$ \begin{align*} \mathfrak{num}(\Omega _{\sigma \tau })=\sigma \cdot \sup_{k\in K}\ \tau _{k}= \mathfrak{num}(\Omega _{\sigma })\cdot \mathfrak{num}(\Omega _{\tau }).\\[-38pt] \end{align*} $$

⊣

4.3 Numerosities and Cantor ordinals

The relation with the Cantor definition of ordinal is the following: if $ \tau \in \mathbf {Ord,} \Omega _{\tau }$ is a well-ordered set and hence $ ot(\Omega _{\tau })$ (the order type of $\Omega _{\tau })$ is a Cantor ordinal. From now on, to avoid confusion, we will denote the Cantor ordinals by $\bar {\tau }$ and their set by $\mathbf {COrd}$ . Whilst a Cantor ordinal is an equivalence class of well-ordered sets, in our definition an ordinal is the numerosity of a suitable well-ordered set; in particular, if we let $\omega $ be the smallest infinite ordinal, then $\omega =\mathfrak { num}\left ( \mathbb {N}_{0}\right ) $ and $\bar {\omega }=ot(\mathbb {N}_{0})$ .

Now, let us consider the map

(12) $$ \begin{align} \Phi :\mathbf{Ord}\rightarrow \mathbf{COrd:\ \ }\Phi \left( \tau \right) =ot(\Omega _{\tau }):=\bar{\tau}, \end{align} $$

which identifies the “numerosity ordinals” with the “Cantor ordinals.” So, by construction $\Phi $ is an isomorphism between the ordered sets $\left ( \mathbf {Ord,<}\right ) $ and $\left ( \mathbf {COrd,<}\right)\!.$

In general, the map does not preserve the operations $+,\cdot $ , as $+$ and $ \cdot $ are commutative on $\mathbf {Num}\subset \mathbb {N}^{\ast }$ but not on $\mathbf {COrd}$ . However, the situation is more interesting if we consider the natural operations $\oplus ,\otimes $ between ordinals. We recall that each ordinal $\bar {\sigma }$ has a unique normal form

$$ \begin{align*} \bar{\sigma}=\sum_{n=0}^{m}\bar{\omega}^{j_{n}}a_{n}, \end{align*} $$

where $a_{n}\in \mathbb {N}$ and $n_{1}<n_{2}\Rightarrow j_{n_{1}}>j_{n_{2}}$ .

By using the normal form, the natural ordinal operations can be defined as follows: given

(13) $$ \begin{align} \bar{\sigma}=\sum_{n=0}^{m}\bar{\omega}^{j_{n}}a_{n}\,\,\ \mathrm{and}\,\,\ { \bar{\tau}}=\sum_{n=0}^{m}\bar{\omega}^{j_{n}}b_{n}, \end{align} $$

we let

(14) $$ \begin{align} \bar{\sigma}\oplus {\bar{\tau}}=\sum_{n=0}^{m}\bar{\omega}^{j_{n}}\left( a_{n}+b_{n}\right) \ \ \mathrm{and}\,\,\,\bar{\sigma}{\otimes \bar{\tau}} =\bigoplus_{n,h=0}^{m}a_{n}b_{h}\bar{\omega}^{j_{n}\oplus j_{h}}, \end{align} $$

where $a_{n}+b_{n}$ and $a_{l}b_{m}$ are the usual operations on natural numbers.

In order to compare the operations between numerosities and the natural ordinal operations, we extend a notion used for the Cantor ordinals to the numerosities.

Definition 4.4. An ordinal $\theta>0$ is called irreducible if

$$ \begin{align*} \sigma ,\tau ,\gamma <\theta \Rightarrow \sigma \tau +\gamma <\theta. \end{align*} $$

If $\theta $ is irreducible then

$$ \begin{align*} \sigma ,\tau \in \Omega _{\theta }\Rightarrow \sigma +\gamma <\theta \ \text{ and}\ \sigma \tau <\theta; \end{align*} $$

we need to prove that $\sigma +\gamma$ and $\ \sigma \tau \in \Omega _{\theta }$ .

We denote by $\theta _{j}$ , $j\in \mathbf {Ord}$ the sequence of irreducible ordinals, namely

• • $\theta _{0}=\omega ,$

• • $\theta _{j}=\min \left \{ x\in \mathbf {Ord}\ \mathbf {|\ }\forall m\in \mathbb {N}_{0},\ \forall k<j,\ x>\theta _{k}^{m}\right \} .$

Proposition 4.5. If $\tau \in \mathbf {Ord}$ , we have that

$$ \begin{align*} \tau <\theta _{j+1}\Leftrightarrow \tau =\sum_{k=0}^{m}b_{k}\theta _{j}^{k} \end{align*} $$

with $b_{k}\in \Omega _{\theta _{j}}.$

Proof. This proof is based only on the order structure of $\mathbf { Ord}$ and hence it could be considered well known. However we will report it for completeness and for the sake of the reader.

$\left ( \Leftarrow \right ) $ Trivial.

$\left ( \Rightarrow \right ) $ If $\tau <\theta _{j+1},$ we take

$$ \begin{align*} n=\max \left\{ m\in \mathbb{N}_{0}\ |\ \theta _{j}^{m}\leq \tau \right\}\!. \end{align*} $$

Such an m exists by the definition of $\theta _{j+1}.$ Then we set

$$ \begin{align*} b_{m}=\sup \left\{ x\in \Omega _{\theta _{j}}\ |\ x\theta _{j}^{m}\leq \tau \right\} \end{align*} $$

and

$$ \begin{align*} y_{j,m}=\tau -b_{m}\theta _{j}^{m}. \end{align*} $$

Then,

(15) $$ \begin{align} \forall z\in \Omega _{\theta _{j}},\ y_{j,m}\leq z. \end{align} $$

Now, by induction over $k=m-1,\ldots ,0$ , we set

$$ \begin{align*} b_{k}=\sup \left\{ x\in \Omega _{\theta _{j}}\ |\ \sum_{l=k+1}^{m}b_{l}\theta _{j}^{l}+x\theta _{j}^{k}\leq \tau \right\} \end{align*} $$

and

$$ \begin{align*} y_{j,k}=\tau -\sum_{l=k}^{m}b_{l}\theta _{j}^{l}, \end{align*} $$

so we have that

(16) $$ \begin{align} \forall z\in \Omega _{\theta _{j}},\ y_{j,m}\leq z. \end{align} $$

Now we claim that

(17) $$ \begin{align} \tau -\sum_{k=0}^{m}b_{k}\theta _{j}^{k}=0. \end{align} $$

In order to prove this we argue by induction over $j\in \mathbf {Ord}\cup \left \{ -1\right \} $ by proving that

(18) $$ \begin{align} y_{jk}=0. \end{align} $$

If $j=-1$ , $\tau \in \Omega _{\theta _{0}}=\mathbb {N}_{0},$ then $\forall n\in \mathbb {N}_{0},\ y_{00}\leq 0$ and hence $y_{00}=0$ . If (18) holds $\forall \tau \in \Omega _{\theta _{j}},$ then by (15) and (16), equality (18) holds also for $\tau \in \Omega _{\theta _{j+1}}.$ ⊣

Proof. By Proposition 4.5,

$$ \begin{align*} \sigma =\sum_{k=0}^{n}a_{k}\theta _{j}^{k},\ \tau =\sum_{k=0}^{n}b_{k}\theta _{j}^{k}, \end{align*} $$

for some $j\in \mathbf {Ord}$ and hence

$$ \begin{align*} \sigma +\tau =\sum_{k=0}^{n}\left( a_{k}+b_{k}\right) \theta _{j}^{k};\ \sigma \tau =\sum_{h,k=0}^{n}\left( a_{h}b_{k}\right) \theta _{j}^{h+k}.\\[-42pt] \end{align*} $$

⊣

Now we describe the sequence of the irreducible ordinal numerosities: we set

• • $\theta _{0}=\bar {\omega }$ ,

• • $\theta _{\bar {j}}=\sup \left \{ \theta _{\bar {k}}^{n}\ |\ n\in \mathbb { N},\ \bar {k}<\bar {j}\right \} $ .

So we have that

$$ \begin{align*} \theta _{0} &=\omega , \\\theta _{1} &=\omega ^{\omega }, \\\theta _{2} &=\omega ^{\omega ^{\omega }}, \\& \kern12pt\dots \\\theta _{j+1} &=\theta _{j}^{\omega } ,\\& \kern12pt\dots \\\theta _{\omega } &=\varepsilon _{0}, \\& \kern12pt\dots \end{align*} $$

and so on. Since the definition of $\theta _{j}$ depends only on the order structure of $\left ( \mathbf {Ord,<}\right ) $ , then

$$ \begin{align*} \Phi \left( \theta _{j}\right) =\bar{\theta}_{j}. \end{align*} $$

It is well known and easy to check that any ordinal number ${\bar {\tau }}\in \mathbf {COrd,}\bar {\tau }<\bar {\theta }_{j+1},$ can be written as follows:

$$ \begin{align*} {\bar{\tau}}=\mathop{\displaystyle \bigoplus }\limits_{n=0}^{m}\bar{a}_{n}\otimes \bar{\theta} _{j}^{n},\ \ \bar{a}_{n}<\bar{\theta}_{j}, \end{align*} $$

and the natural operations $\oplus ,\otimes $ take the following form:

$$ \begin{align*} \left( \mathop{\displaystyle \bigoplus }\limits_{n=0}^{m}\bar{a}_{n}\otimes \bar{\theta} _{j}^{n}\right) \oplus \left( \mathop{\displaystyle \bigoplus }\limits_{n=0}^{m}\bar{b}_{n}\otimes \bar{\theta}_{j}^{n}\right) =\mathop{\displaystyle \bigoplus }\limits_{n=0}^{m}\left( \bar{a} _{n}\oplus \bar{b}_{n}\right) \otimes \bar{\theta}_{j}^{n}, \end{align*} $$

$$ \begin{align*} \,\bar{\sigma}{\otimes \bar{\tau}}=\mathop{\displaystyle \bigoplus }\limits_{n,h=0}^{m}\left( \bar{a }_{n}\otimes \bar{b}_{h}\right) \otimes \bar{\theta}_{j}^{n+h}. \end{align*} $$

Theorem 4.7. The map (12) is an isomorphism between the semirings $( \mathbf {Ord,}+,\cdot )$ and $(\mathbf {COrd,}\oplus ,\otimes ),$ namely

$$ \begin{align*} \Phi \left( \sigma +\tau \right) &=\bar{\sigma}\oplus \bar{\tau}, \\ \Phi \left( \sigma \tau \right) &=\bar{\sigma}\otimes \bar{\tau}. \end{align*} $$

Proof. Let $\tau =\sum _{k=0}^{m}b_{k}\theta _{j}^{k}$ be an ordinal numerosity. Then

$$ \begin{align*} \sum\limits_{n=0}^{m}\bar{b}_{n}\otimes \bar{\theta} _{j}^{n}&=ot\left( \left\{ \mathop{\displaystyle \bigoplus }\limits_{n=0}^{m}\bar{a}_{n}\otimes \bar{\theta }_{j}^{n}\in \mathbf{COrd}\ |\ \mathop{\displaystyle \bigoplus }\limits_{n=0}^{m}\bar{a}_{n}\otimes \bar{\theta}_{j}^{n}<\mathop{\displaystyle \bigoplus }\limits_{n=0}^{m}\bar{b}_{n}\otimes \bar{ \theta}_{j}^{n}\right\} \right) \\ &=ot\left( \left\{ \sum_{n=0}^{m}a_{n}\theta _{j}^{n}\in \mathbf{Ord}\ |\ \sum_{n=0}^{m}a_{n}\theta _{j}^{n}<\sum_{n=0}^{m}b_{n}\theta _{j}^{n}\right\} \right) =ot\left(\Omega_{\tau}\right)=\bar{\tau}, \end{align*} $$

namely

$$ \begin{align*} \Phi \left( \tau \right) =\Phi \left( \sum_{n=0}^{m}b_{n}\theta _{j}^{n}\right) =\mathop{\displaystyle \bigoplus }\limits_{n=0}^{m}\bar{b}_{n}\otimes \bar{\theta} _{j}^{n}=\bar{\tau}. \end{align*} $$

Hence $\Phi $ is an isomorphism.⊣

Remark 4.8. Theorems 4.7 and 4.3 provide a new interpretation for the natural operations $\oplus $ and $\otimes $ namely

$$ \begin{align*} \bar{\sigma}\oplus {\tau }=ot(\Omega _{\sigma +\tau })\ \ \mathrm{and}\,\,\, \bar{\sigma}\otimes \bar{\tau}=ot(\Omega _{\sigma \tau }). \end{align*} $$

This fact is somewhat surprising since the operations $+$ and $\cdot $ between numerosities have been introduced in a natural way for the numerosity theory and, a priori, they should not have any relation with the natural operations between ordinal numbers.

Notice, however, that not all operations are the same between numerosity ordinals and Cantor ordinals: for example, let $\bar {\varepsilon }_{0}=\bar { \theta }_{\bar {\omega }}$ be the Cantor ordinal that corresponds to the numerosity ordinal $\theta _{\omega }$ . If we use the ordinal exponentiation, we have that

$$ \begin{align*} \bar{\omega}^{\bar{\varepsilon}_{0}}=\bar{\varepsilon}_{0}, \end{align*} $$

whilst on the contrary, if we use the Euclidean exponentiation, we get that

$$ \begin{align*} \omega ^{\varepsilon _{0}}>\theta _{\omega }. \end{align*} $$

In particular the equation

$$ \begin{align*} \bar{\omega}^{x}=\bar{\varepsilon}_{0} \end{align*} $$

in the world of Cantor ordinals has the solution $x=\bar {\varepsilon }_{0}$ while the equation

$$ \begin{align*} \omega ^{x}=\varepsilon _{0} \end{align*} $$

in the world of Euclidean numbers has the solution $\xi =\log _{\omega }\varepsilon _{0}. \xi $ is a well-defined Euclidean number, but it is not an ordinal number since

$$ \begin{align*} \xi <\varepsilon _{0}=\mathfrak{num}\left( \mathop{\displaystyle \bigcup}\limits_{k<\omega }\Omega _{\theta _{k}}\right)\!. \end{align*} $$

However it is easy to prove that the ordinal exponentiation agrees with the Euclidean exponentiation for numbers in $\Omega _{\theta _{\omega }}$ .

5.1 The general strategy

Theorem 3.2 describes the fundamental properties of numerosities, which are satisfied for all choices of the label set $\mathfrak {B}$ (and of the ultrafilter $\mathcal {U}$ ). However, certain additional properties are satisfied only for some choices of $\mathfrak {B}$ : in fact, they depend on the ultrafilter $\mathcal {U}$ over $\mathbf {\wp }_{fin}\left ( \mathfrak {B} \right ) $ , whose existence depends on Zorn’s lemma which cannot be explicit and hence it is impossible to prove or disprove some of them. However, if we choose a suitable label set $\mathfrak {B}$ (and, consequently we restrict the choice of $\mathcal {U}$ ), it is possible to show that some properties, as the ones mentioned in the Introduction, are satisfied independently of $ \mathcal {U}$ . The goal of Section 5 is to show how a suitable choice of $\mathfrak {B}$ allows the numerosity function to satisfy interesting properties in many specific cases.

The smaller the set $\mathfrak {B}$ is, the more properties are satisfied by the numerosity function. So the idea is to begin with a set $\mathfrak {B} _{\max }=\left \{ \mathfrak {t}\in \mathfrak {L\ }|\ \mathfrak {t}\cap \mathfrak { L}=\varnothing \right \} $ and to construct smaller label sets $\mathfrak {B} _{\max }\supset \mathfrak {B}_{1}\supset \mathfrak {B}_{2}\supset \cdots $ which provide a richer and richer structure to the theory. In this paper we are interested in the numerosity of some specific subsets of $\mathbb {N}_{0}$ , $\mathbb {Z}$ , $\mathbb {Q}$ , $\mathbb {R}$ and so we will construct sets of labels $\mathfrak {B}_{\max }\supset \mathfrak {B}\left ( \mathbb {N}_{0}\right ) \supset \mathfrak {B}\left ( \mathbb {Z}\right )\supset \mathfrak {B}\left ( \mathbb {Q}\right )$ . Depending on the desired properties, other label sets could be added to the list. Each set of labels allows us to enrich the theory with new theorems; all these theorems are independent of the ultrafilter employed in the sense that every ultrafilter which satisfies the fineness propertyFootnote ⁷ does the job.Footnote ⁸

The construction which we will present in the next sections is based on the following definition:

Definition 5.1. If $\mathfrak {D}\subset \mathfrak {B}_{\max }$ is a directed set (with respect to $\subseteq $ ), we define

$$ \begin{align*} \overline{\mathfrak{D}}=\mathfrak{G}\left(\{ \mathfrak{s\in B}_{\max }\mid \exists \mathfrak{t}\in \mathfrak{D},\ \mathfrak{s}\sqsupseteq \mathfrak{t} \}\right), \end{align*} $$

where $\mathfrak {G}(F)$ denotes the smallest lattice containing $F.$

Notice that, by definition, if $\mathfrak {D}\subset \mathfrak {B}_{\max }$ is a directed set then

$$ \begin{align*} \overline{\overline{\mathfrak{D}}}=\overline{\mathfrak{D}}. \end{align*} $$

The numerosity of a set depends on the set of labels $\mathfrak {B}$ and an ultrafilter $\mathcal {U}$ consistent with $\mathfrak {B.}$ As in this section we will discuss also coherence properties between different label sets, we will use the notation $\mathfrak {num}_{\mathfrak {B}}^{\mathcal {U}}$ to denote the numerosity function obtained using labels in $\mathfrak {B}$ and the ultrafilter $\mathcal {U}$ and similarly we denote by $\ell _{\mathfrak {B} }(x)$ the label relative to $\mathfrak {B}$ (see Definition 2.9). This notation will be used only when there is danger of confusion, as multiple sets of labels are used at once. We will keep to the simpler notation $\mathfrak {num}$ whenever there is no danger of such confusion.

By definition, the $\overline {\mathfrak {D}}$ -labelling of $a\in \Lambda $ is given by

$$ \begin{align*} \ell _{\overline{\mathfrak{D}}}(a)=\bigcap \{\mathfrak{s}\in \overline{ \mathfrak{D}}\mid a\in \mathbb{V}\left( \mathfrak{s}\right)\}=\bigcap \{ \mathfrak{s}\in \mathfrak{B}_{\max}\mid a\in \mathbb{V}\left( \mathfrak{s} \right) \ \text{and} \ \exists \mathfrak{t}\in \mathfrak{D},\ \mathfrak{s} \supseteq \mathfrak{t}\}; \end{align*} $$

in particular, we have that

(19) $$ \begin{align} \mathfrak{s}\in \mathfrak{D}\Rightarrow \ell _{\overline{\mathfrak{D}}}( \mathfrak{s})=\mathfrak{s}. \end{align} $$

Proposition 5.3. If $\mathfrak {D}_{1}\subset \mathfrak {D}_{2}$ then $\overline { \mathfrak {D}}_{1}\subset \overline {\mathfrak {D}}_{2};$ hence for all ultrafilter $\mathcal {U}$ consistent with $\overline {\mathfrak {D}}_{1}$ , for every set A in $\Lambda $

$$ \begin{align*} \mathfrak{num}_{\overline{\mathfrak{D}}_{1}}^{\mathcal{U}}(A)=\mathfrak{num} _{\overline{\mathfrak{D}}_{2}}^{\mathcal{U}}(A). \end{align*} $$

Lemma 5.4. If $\lambda \in \overline {\mathfrak {D}},$ then $\lambda $ can be split as follows:

$$ \begin{align*} \lambda =\mathfrak{s}\cup \mathfrak{t}, \end{align*} $$

where $\mathfrak {s}\in \mathfrak {D}$ and $\mathfrak {t}$ is such that

$$ \begin{align*} \forall \sigma \in \mathfrak{D},\ \mathfrak{t}\cap \sigma =\varnothing. \end{align*} $$

Proof. Given $\lambda \in \overline {\mathfrak {D}},$ we set

$$ \begin{align*} \mathfrak{s}:=\mathop{\displaystyle \bigcup} \{\mathfrak{u}\in \mathfrak{D}\mid \mathfrak{u} \subset \lambda \} \end{align*} $$

and

$$ \begin{align*} \mathfrak{t}:=\lambda \backslash \mathfrak{s}. \end{align*} $$

Then $\mathfrak {s}\in \overline {\mathfrak {D}}$ , as $\overline {\mathfrak {D}}$ is a lattice and the union defining $\mathfrak {s}$ is finite, and $\forall \sigma \in \mathfrak {D},\ \mathfrak {t}\cap \sigma =\varnothing .$ In fact, if we set $\mathfrak {u}=\mathfrak {t}\cap \sigma $ then, as $\mathfrak {s}\supseteq \sigma $ , we have

$$ \begin{align*} \varnothing =\mathfrak{s}\cap \mathfrak{t}\supseteq \sigma \cap \mathfrak{t}=\mathfrak{u}; \end{align*} $$

hence $\mathfrak {u}=\varnothing $ .⊣

Remark 5.5. We can look at the splitting given by Lemma 5.4 thinking of $\mathfrak {B\ }$ as a vector space over $\mathbb {Z}_{2};$ in this case we can write

$$ \begin{align*} \mathfrak{B}=\mathfrak{D}\oplus \mathfrak{D}^{\perp }, \end{align*} $$

and the splitting $\lambda =\mathfrak {s}\cup \mathfrak {t\ }$ implies that

$$ \begin{align*} \mathfrak{s\in D\ }\text{and}\ \mathfrak{t}\in \mathfrak{D}^{\perp }. \end{align*} $$

5.2 Numerosity of the natural numbers

In what follows, we set $\mathbb {N}:=\{1,2,3,\dots \}$ and $\mathbb {N}_{0}:= \mathbb {N}\cup \{0\}$ , and we let $\alpha $ denote the numerosity of $ \mathbb {N}$ . We will consider numbers in $\mathbb {N}_{0}$ , as well as more generally in $\mathbb {R}$ , as atoms. Our goal is to find a label set $ \mathfrak {B}(\mathbb {N}_{0})\subset \mathfrak {B}_{\max }$ so that we can prove some properties of $\alpha $ and to describe the numerosity of some subsets of $\mathbb {N}_{0}$ by functions of $\alpha $ .

We define $\mathfrak {D}\left (\mathbb {N}_{0}\right )$ as follows:

$$ \begin{align*} \lambda \in \mathfrak{D}(\mathbb{N}_{0})\Leftrightarrow \exists m\in \mathbb{ N}\ \text{such that}\ \lambda =\left\{ 0,\dots ,m!^{m!}\right\}, \end{align*} $$

and we set

$$ \begin{align*} \mathfrak{B}(\mathbb{N}_{0}):=\overline{\mathfrak{D}(\mathbb{N}_{0})}\!. \end{align*} $$

which, by Lemma 5.2, is a label set. By Definition, we have that for every $n\in \mathbb {N}_{0} $ ,

$$ \begin{align*} \ell (n)=\left\{ 0,\dots ,f(n)\right\} , \end{align*} $$

where

$$ \begin{align*} f(n):=\min \left\{ m!^{m!}\ |\ m\in \mathbb{N},\ m!^{m!}\geq n\right\}\!. \end{align*} $$

The main reason for such a peculiar labelling is to ensure the following algebraical properties for $\alpha $ :

Proof. (i) For $i=0,\dots ,n-1$ let

$$ \begin{align*} A_{i}=\{m\in \mathbb{N}_{0}\mid m\equiv i\mod n\}. \end{align*} $$

Then for every $\lambda \sqsupseteq \{0,1,\dots ,n!^{n!}\}$ , with $\lambda \in \mathfrak {B}\left ( \mathbb {N}_{0}\right ) $ , for every $0\leq i,j<n$ we have

$$ \begin{align*} |A_{i}\cap \lambda |=|A_{j}\cap \lambda |, \end{align*} $$

as $\lambda \cap \mathbb {N}=\{0,1,\dots ,f(m)\}$ for some $m\geq n$ , and n divides $f(m)$ for every such m. In particular, this shows that $\mathfrak { num}\left ( A_{i}\right ) =\mathfrak {num}\left ( A_{j}\right ) $ for every $ 0\leq i,j<n$ ; hence

$$ \begin{align*} \alpha =\mathfrak{num}\left( \mathbb{N}_{0}\right) =\sum_{i=0}^{n-1} \mathfrak{num}\left( A_{i}\right) =n\cdot \mathfrak{num}\left( A_{0}\right)\!. \end{align*} $$

(ii) Let $\lambda \sqsupseteq \{1,\dots ,n!^{n!}\}$ , with $\lambda \in \mathfrak {B}\left ( \mathbb {N}_{0}\right ) $ . As noticed in (i) above, it must be $ \lambda \cap \mathbb {N}=\{1,\dots ,m!^{m!}\}$ for some $m\geq n$ . If $a=m!^{ \frac {m!}{n}}$ , we can rewrite $\{1,\dots ,m!^{m!}\}$ as $\{1,\dots ,a^{n}\}$ . Hence $|\left \{ m^{n}\ |\ m\in \mathbb {N}\right \} \cap \lambda |\ =a=| \mathbb {N}\cap \lambda |^{\frac {1}{n}}$ . The thesis is reached by taking the $\Lambda $ -limit on the above equality.⊣

Remark 5.7. Of course, the choice of $\ \mathfrak {D}\left (\mathbb {N}_{0}\right )$ is not intrinsic, and has been done so to make it possible to have the properties listed in Proposition 5.6. Some additional motivations for this choice of $\mathfrak {D}\left (\mathbb {N}_{0}\right )$ can be found in [Reference Benci and Di Nasso11]; different motivations have led the authors of [Reference Benci and Forti5] to make the following different choice:

$$ \begin{align*} \lambda \in \mathfrak{D}_{1}(\mathbb{N}_{0})\Leftrightarrow \exists m\in \mathbb{N}\ \text{such that}\ \lambda =\left\{ 0,\dots ,2^{m}-1\right\}\!. \end{align*} $$

This can be seen as a feature of this approach: different algebraic properties of the numerosity can be rather easily obtained by changing the label set.

5.3 Numerosity of the integers

We proceed as in the case of the natural numbers. We define $\mathfrak {D}( \mathbb {Z})$ as follows:

$$ \begin{align*} \lambda \in \mathfrak{D}\left( \mathbb{Z}\right) \Leftrightarrow \exists m\in \mathbb{N}\ \text{such that}\ \lambda =\left\{ -m!^{m!},\dots ,m!^{m!}\right\}\!. \end{align*} $$

Clearly $\mathfrak {D}\left ( \mathbb {Z}\right ) \subset \mathfrak {B}(\mathbb {N} _{0})$ and hence, by Lemma 5.2,

$$ \begin{align*} \mathfrak{B}(\mathbb{Z}):=\overline{\mathfrak{B}\left(\mathbb{N} _{0}\right)\cap \mathfrak{D}(\mathbb{Z})} \end{align*} $$

is a label set. Using this label basis for every $z\in \mathbb {Z}$ ,

$$ \begin{align*} \ell (z)\cap \mathbb{Z}=\{-n(z),\dots ,n(z)\}, \end{align*} $$

where

$$ \begin{align*} n(z):=\min \left\{ m!^{m!}\ |\ m\in \mathbb{N},\ m!^{m!}\geq |z|\right\}\!. \end{align*} $$

Moreover, as $\mathfrak {B}(\mathbb {Z})\subseteq \mathfrak {B}(\mathbb {N})$ , by Proposition 5.3 the numerosities constructed with $\mathfrak {B}( \mathbb {Z})$ are coherent with those constructed with $\mathfrak {B}(\mathbb {N })$ .

With this choice of $\mathfrak {D}\left ( \mathbb {Z}\right)\!, \mathfrak {num} \left ( \mathbb {Z}\right ) =2\alpha +1$ and we have that

(20) $$ \begin{align} \mathfrak{num}\left( \mathbb{Z}_{<0}\right) =\mathfrak{num}\left( \mathbb{Z} _{>0}\right) =\alpha; \end{align} $$

this equality agrees with the intuition that the positive numbers are as many as the negative numbers.

Just as an example of a possible application, let us prove the following result for subgroups of $\mathbb {Z}$ , which reminds us of Lagrange’s Theorem for finite groups:

Theorem 5.8. Let $S:=m\mathbb {Z}$ be a subgroup of $(\mathbb {Z},+)$ . Then

(21) $$ \begin{align} \frac{\mathfrak{num}\left( \mathbb{Z}\right) }{\mathfrak{num}\left( S\right) }\sim m=\mathfrak{num}\left( \mathbb{Z}_{m}\right)\!. \end{align} $$

Proof. By definition, $S=\{mn\mid n\in \mathbb {Z}\}$ . We write $ S=S_{+}\cup S_{-}\cup \{0\}$ , where

$$ \begin{align*} S_{+}=\{a\in S\mid a>0\},S_{-}=\{a\in S\mid a<0\}. \end{align*} $$

By Proposition 5.6 we know that $\mathfrak {num}\left ( S_{+}\right ) =\frac {\alpha }{m}$ , and it is trivial to show that $\mathfrak {num}\left ( S_{-}\right ) =\mathfrak {num}\left ( S_{+}\right ) $ . Hence $\mathfrak {num} \left ( S\right ) =\mathfrak {num}\left ( S_{+}\right ) +\mathfrak {num}\left ( S_{-}\right ) +1=2\frac {\alpha }{m}+1$ . As $\mathfrak {num}\left ( \mathbb {Z} \right ) =2\alpha +1,$ we have

$$ \begin{align*} \frac{\mathfrak{num}\left( \mathbb{Z}\right) }{\mathfrak{num}\left( S\right) }=\frac{2\alpha +1}{\frac{2\alpha }{m}+1}=\frac{2\alpha +1}{\frac{1}{m} \left( 2\alpha +m\right) }\sim m, \end{align*} $$

as $\alpha $ is infinite.⊣

Remark 5.9. Let us notice that, with our labelling, in the above proposition we do not have the equality

(22) $$ \begin{align} \frac{\mathfrak{num}\left( \mathbb{Z}\right) }{\mathfrak{num}(S)}=m, \end{align} $$

because not all lateral classes $\left [ k\right ] $ in the quotient have the same numerosity:

$$ \begin{align*} \mathfrak{num}\left( \left[ k\right] \right) =\frac{2\alpha }{m}\ \ \text{if} \ \ k\neq 0;\ \ \mathfrak{num}\left( \left[ 0\right] \right) =\frac{2\alpha }{m}+1. \end{align*} $$

If we want the equality in (21), then we can replace $\mathfrak {D} \left ( \mathbb {Z}\right ) $ with $\mathfrak {D}_{1}\left ( \mathbb {Z}\right ) $ defined as follows:

$$ \begin{align*} \lambda \in \mathfrak{D}_{1}\left( \mathbb{Z}\right) \Leftrightarrow \exists m\in \mathbb{N}\ \text{such that}\ \lambda =\left\{ -m!^{m!}+1,\dots ,m!^{m!}\right\}\!. \end{align*} $$

In this case, we get (22), but $\mathfrak {num}\left ( \mathbb {Z} \right ) =2\alpha $ and the equality (20) is violated.

5.4 Numerosity of the rationals

The labelling of $\mathbb {Z}$ given in Section 5.3 can be extended in several ways to the rationals. A natural one is obtained by setting

$$ \begin{align*} \mathfrak{D}\left( \mathbb{Q}\right) :=\left\{ \ \mathbb{H}_{n}\mid \exists m\in \mathbb{N},\ n=m!^{m!}\right\}, \end{align*} $$

where

$$ \begin{align*} \mathbb{H}_{n}:=\left\{\frac{a}{n}\mid a\in\mathbb{Z}, \ -n^{2}<a<n^{2}\right\}\!. \end{align*} $$

By Lemma 5.2,

$$ \begin{align*} \mathfrak{B}(\mathbb{Q}):=\overline{\mathfrak{D}(\mathbb{Q})} \end{align*} $$

is a label set.

As $\mathfrak {B}\left ( \mathbb {Q}\right ) \subset \mathfrak {B}(\mathbb {Z})$ , by Proposition 5.3 the numerosities constructed with $\mathfrak {B}( \mathbb {Q})$ are coherent with those constructed with $\mathfrak {B}(\mathbb {Z })$ . Using the label basis $\mathfrak {B}(\mathbb {Q})$ we have that, for every $q\in \mathbb {Q}$ ,

$$ \begin{align*} \ell (q)\cap \mathbb{Q}=\mathbb{H}_{n(q)}\!, \end{align*} $$

where

$$ \begin{align*} n(q):=\min \left\{ m!^{m!}\ |\ m\in \mathbb{N},\ m!^{m!}\geq |q|\right\}\!. \end{align*} $$

This labelling has been chosen in order to have the following results:

Proof. (i) Take $H_{m}\in \mathfrak {B}_{\mathbb {Q}}$ with m larger than $n+1$ . Then $|\left ( \mathbb {Q}\cap \left [ n,n+1\right ) \right ) \cap \mathbb {H}_{m}|\ =m$ , and hence eventually $|\left ( \mathbb {Q}\cap \left [ n,n+1\right ) \right ) \cap \mathbb {H}_{m}|\ =|\mathbb {N}\cap \mathbb {H}_{m}|$ , and the thesis follows by taking the $\Lambda $ -limit.

(ii) Take $\mathbb {H}_{m}\in \mathfrak {B}_{\mathbb {Q}}$ with m larger than $|p|,|q|$ . Then $\left ( \mathbb {Q}\cap \left [ p,q\right ) \right ) =(p-q)m$ if $p\in \mathbb {H}_{m}$ and $\left ( \mathbb {Q}\cap \left [ p,q\right ) \right ) =(p-q)m-1$ if $p\notin \mathbb {H}_{m}$ . By taking the $\Lambda $ -limit we have that either $\mathfrak {num}\left ( \mathbb {Q}\cap \left [ p,q\right ) \right ) =(p-q)\alpha -1$ or $\mathfrak {num}\left ( \mathbb {Q}\cap \left [ p,q\right ) \right ) =(p-q)\alpha $ , and the thesis follows as, by (i), $ \mathfrak {num}\left ( \mathbb {Q}\cap \left [ 0,1\right ) \right ) =\alpha $ .

(iii) Let us first compute $\mathfrak {num}\left ( \mathbb {Q}_{>0}\right ) $ . Let $\lambda =\mathbb {H}_{n}\in \mathfrak {B}_{\mathbb {Q}}$ . Then $|\mathbb {H} _{n}\cap \mathbb {Q}_{>0}|=n^{2}$ , and hence if $\varphi $ is the enumeration of $ \mathbb {Q}_{>0}$ and $\psi $ is the enumeration of $\mathbb {N}$ , we have that $\varphi (\lambda )=\psi (\lambda )^{2}$ , so

$$ \begin{align*} \mathfrak{num}\left( \mathbb{Q}_{>0}\right) =\lim_{\lambda \uparrow \Lambda }\varphi (\lambda )=\lim_{\lambda \uparrow \Lambda }\psi ^{2}(\lambda )=\left( \lim_{\lambda \uparrow \Lambda }\psi (\lambda )\right) ^{2}=\alpha ^{2}. \end{align*} $$

Therefore, as each $\mathbb {H}_{n}$ is symmetrical with respect to $0$ , we also have that $\mathfrak {num}\left ( \mathbb {Q}_{<0}\right ) =\alpha ^{2}$ , and so

$$ \begin{align*} \mathfrak{num}\left( \mathbb{Q}\right) =\mathfrak{num}\left( \mathbb{Q} _{<0}\right) +\mathfrak{num}\left( \mathbb{Q}_{>0}\right) +1=2\alpha ^{2}+1.\\[-34pt] \end{align*} $$

⊣

As an example of a possible application, let us prove the following result:

Theorem 5.11. Let $m_{PJ}$ denote the Peano–Jordan measure of an $m_{PJ}$ -measurable set $ E. $ Then

(23) $$ \begin{align} m_{PJ}\left( E\right) =st\left( \frac{\mathfrak{num}\left( E\cap \mathbb{Q} \right) }{\mathfrak{num}\left( \left[ 0,1\right) \cap \mathbb{Q}\right) } \right) =st\left( \frac{1}{\mathfrak{\alpha }}\cdot \mathfrak{num}\left( E\cap \mathbb{Q}\right) \right)\!. \end{align} $$

Proof. If E is an interval then the result follows from Proposition 5.10. We can extend this result to a plurinterval $ E=\mathop {\displaystyle \bigcup } E_{i}$ by the Sum Principle (see Theorem 3.2(v)). In general, if E is $m_{PJ}$ -measurable, $\forall \varepsilon \in \mathbb {R}_{>0}$ there are two plurintervals A and B such that

$$ \begin{align*} A\subseteq E\subseteq B, \end{align*} $$

(24) $$ \begin{align} \left\vert m_{PJ}(B)-m_{PJ}(E)\right\vert <\varepsilon \ \text{and}\ \left\vert m_{PJ}(E)-m_{PJ}(A)\right\vert <\varepsilon. \end{align} $$

By the Euclid Principle (see Theorem 3.2(ii)), we have that

$$ \begin{align*} \mathfrak{num}\left( A\cap \mathbb{Q}\right) \subseteq \mathfrak{num}\left( E\cap \mathbb{Q}\right) \subseteq \mathfrak{num}\left( E\cap \mathbb{Q} \right), \end{align*} $$

then

$$ \begin{align*} m_{PJ}(A)\leq st\left( \frac{1}{\mathfrak{\alpha }}\cdot \mathfrak{num} \left( E\cap \mathbb{Q}\right) \right) \leq m_{PJ}(B). \end{align*} $$

The conclusion follows by the inequality above, Equation (24), and the arbitrariness of $\varepsilon $ .⊣

6.1 A suitable labelling

Let $\mathbb {\hat {R}}^{N}, N\in \mathbb {N}$ , $\mathbb {\hat {R}}^{N}\subset \mathbb {A}$ , be a family of sets such that

$$ \begin{align*} \mathbb{\hat{R}}^{0} &=\mathbb{R}, \\ \mathbb{\hat{R}}^{N} &\subset \mathbb{\hat{R}}^{N+1}, \end{align*} $$

and each $\mathbb {\hat {R}}^{N+1}$ is isomorphic to $\mathbb {R}^{N}$ . This awkward distinction between $\mathbb {\hat {R}}^{N}$ and $\mathbb {R}^{N}$ is useful since, in this context, it is easier to deal with atoms and the points of $\mathbb {R}^{N}$ are N-ples. Moreover, we need to assume that the isomorphism

$$ \begin{align*} \Psi :\mathbb{R}^{N}\rightarrow \mathbb{\hat{R}}^{N} \end{align*} $$

preserves also the labels, namely, if $\left ( x_{1},\dots ,x_{N}\right ) \in \mathbb {R}^{N},$ then

(25) $$ \begin{align} \ell \left[ \Psi \left( x_{1},\dots,x_{N}\right) \right] =\max \left\{ \ell \left( x_{1}\right)\!,\dots,\ell \left( x_{N}\right) \right\}\!. \end{align} $$

If $A\in \wp \left ( \mathbb {\hat {R}}^{N}\right ) $ for some N, we denote by $m_{L}\left ( A\right )$ its N-dimensional Lebesgue measure. We introduce on $\wp \left ( \mathbb {A}\right ) $ the following order relation: given $A,B\in \wp \left ( \mathbb {A}\right )$ ,

• • if $A,B$ are Lebesgue measurable subsets of $\hat {\mathbb {R}}^{N}$ for some $N\in \mathbb {N}$ , we let

  $$ \begin{align*} A\sqsubseteq B\Leftrightarrow \ m_{L}(A)\leq m_{L}\left( B\right)\!; \end{align*} $$

• • otherwise, we let $A\sqsubseteq B\Leftrightarrow \left \vert A\right \vert \leq \left \vert B\right \vert $ .

If $A\sqsubseteq B$ and $B\sqsubseteq A,$ we will write $A\equiv B.$

We define $\mathfrak {D}\left ( \mathbb {A}\right ) $ as follows: $\lambda \in \mathfrak {D}\left ( \mathbb {A}\right ) $ if and only if

$$ \begin{align*} \lambda =\Xi \cup \mathfrak{A}, \end{align*} $$

where

• • $\Xi $ is a finite set,

• • $\mathfrak {A}\in \wp _{fin}\left ( \wp \left ( \mathbb {A}\right ) \right ), $

• • for all $A,B\in \mathfrak {A}$ , the following property holds:

  (26) $$ \begin{align} A\sqsubset B\Rightarrow \left\vert A\cap \Xi \right\vert <\left\vert B\cap \Xi \right\vert\!. \end{align} $$

The definition of $\sqsubseteq $ and $\mathfrak {D}\left ( \mathbb {A}\right ) $ is rather peculiar; it has been done in such a way that the numerosity theory that will be constructed later will respect cardinalities, in the sense that larger sets in the sense of cardinality theory will get larger numerosities, and it will also respect the Lebesgue measure, in a sense that we make precise in Section 7.

Lemma 6.1. If $A_{1},\dots ,A_{m}\subset \mathbb {A}$ and $F\subset \mathbb { A}$ is a finite set, there exists $\Xi \in \wp _{fin}\left ( \mathbb {A} \right ) $ such that $F\subseteq \Xi $ and

(27) $$ \begin{align} \Xi \cup \left\{ A_{1},\dots ,A_{m}\right\} \in \mathfrak{D}\left( \mathbb{A} \right)\!. \end{align} $$

Proof. Let $A_{1},\dots ,A_{m}\subset \mathbb {A}$ and $F\subset \mathbb {A}$ be given. We order the $A_{j}$ ’s so that

$$ \begin{align*} j<k\Rightarrow A_{j}\sqsubseteq A_{k}, \end{align*} $$

in such a way that those sets of cardinality $\mathfrak {c}$ are ordered starting with the non-Lebesgue measurable ones first. We construct a sequence of labels

$$ \begin{align*} \lambda _{k}=\Xi _{k}\cup \left\{ A_{1},\dots ,A_{k}\right\} ,\ k\leq m, \end{align*} $$

such that $\lambda _{k}\in \mathfrak {D}\left ( \mathbb {A}\right)\!,\ \lambda _{k}\subset \lambda _{k+1}$ , and $\Xi _{k}\supseteq F$ .

We do it by induction. If $k=1$ , we just set

$$ \begin{align*}\lambda _{1}=F \cup \left\{ A_{1}\right\},\end{align*} $$

so that $\lambda _{1}\in \mathfrak {D}\left ( \mathbb {A}\right )$ has the desired property by construction.

If $m>2$ , assume we built $\lambda _{k}=\Xi _{k}\cup \{A_{1},\dots ,A_{k}\}$ for $k<m$ . Take $A_{k+1}$ . We consider three cases.

Case 1: $A_{k+1}$ is finite.

In this case, for all $i\leq k A_{i}$ is finite. We just set $\Xi _{k+1}=\Xi _{k}\cup A_{1}\cup \dots \cup A_{k+1}$ , so that $|A_{i}\cap \Xi _{k+1}|=|A_{i}|$ for all $i\leq k+1$ , which shows that $\lambda _{k+1}=\Xi _{k+1}\cup \{A_{1},\dots ,A_{k+1}\}$ satisfies Condition (26), and hence $\lambda _{k}\in \mathfrak {D}\left (\mathbb {A}\right )$ .

Case 2: $|A_{k+1}|$ is not a Lebesgue measurable subset of $\hat {R}^{N}$ for some N or $A_{k+1}$ is Lebesgue measurable but $m_{L}\left (A_{k+1}\right )=0$ .

We let $I=\{i\leq k\mid \ |A_{i}|<|A_{k+1}|\}$ and $S=\bigcup _{i\in I} A_{i}$ . As S is a finite union of sets with a cardinality smaller than that of $A_{k+1}$ , $|S|< |A_{k+1}|$ . Therefore, we can take $a_{1},\dots ,a_{s+1}\in A_{k+1}\setminus S$ and we set $\Xi _{k+1}=\Xi _{k}\cup \{a_{1},\dots ,a_{s+1}\}$ . Then $\lambda _{k+1}=\Xi _{k+1}\cup \{A_{1},\dots ,A_{k+1}\}$ satisfies Condition (26): in fact, for $i<j\leq k+1$ with $A_{i}\sqsubset A_{j}$ :

• • if both $i,j\in I$ , we just observe that $|A_{i}\cap \Xi _{k+1}|=|A_{i}\cap \Xi _{k}|<|A_{j}\cap \Xi _{k}|=|A_{j}\cap \Xi _{k+1}|$ by inductive hypothesis;

• • if $i\in I,j\notin I$ , and $j\neq k+1$ , just observe that $|A_{i}\cap \Xi _{k+1}|=|A_{i}\cap \Xi _{k}|<|A_{j}\cap \Xi _{k}|\leq |A_{j}\cap \Xi _{k+1}|$ by inductive hypothesis and construction;

• • if $i\in I,j=k+1$ then $|A_{i}\cap \Xi _{k+1}|=|A_{i}\cap \Xi _{k}|\leq s<|A_{k+1}\cap \Xi _{k+1}|$ by construction; and

• • the case $i\notin I$ cannot happen, as in this case $|A_{i}|=|A_{k+1}|$ so $A_{i}\equiv A_{k+1}$ .

Therefore $\lambda _{k+1}\in \mathfrak {D}\left (\mathbb {A}\right )$ .

Case 3: $|A_{k+1}|$ is a Lebesgue measurable subset of $\hat {\mathbb {R}}^{N}$ for some N with $m_{L}\left (A_{k+1}\right )>0$ .

Let

$$ \begin{align*}I=\{i\leq k\mid A_{i} \ \text{is not Lebesgue measurable or it has null Lebesgue measure}\}.\end{align*} $$

Let $i=(\max I)+1$ . For all $j=i,\dots ,k+1$ , $A_{j}$ is Lebesgue measurable, $m_{L}(A_{j})>0$ , and (if $i\leq k$ ) $m_{L}(A_{j})\leq m_{L}(A_{j+1})$ .

Set

$$ \begin{align*}S=\bigcup_{j<i}A_{j} \ \text{and} \ s=|\Xi_{k}\cap S|.\end{align*} $$

For $i\leq j_{1},j_{2}\leq k+1$ we set $A_{j_{1}}\sim A_{j_{2}}$ if and only if $m_{L} \left (A_{j_{1}}\Delta A_{j_{2}}\right )=0$ . For $i\leq j\leq k+1$ we set

$$ \begin{align*}I_{j}=\{b\mid i\leq b \leq k+1 \ \text{and} \ A_{b}\sim A_{j}\}\end{align*} $$

andFootnote ⁹

$$ \begin{align*}\widetilde{A_{j}}=\left(\bigcap_{t\in I_{j}}A_{t}\right)\setminus S.\end{align*} $$

For $i\leq j\leq k+1$ we let

$$ \begin{align*}r_{j}=\left|\left(A_{j}\setminus \left(\widetilde{A_{j}}\cup S\right)\right)\cap \Xi_{k}\right|\end{align*} $$

and $r=\max _{i\leq j\leq k+1} r_{j}$ .

Observe that $m_{L} \left (\widetilde {A_{j}}\right )=m_{L}\left (A_{j}\right )$ , and hence $\widetilde {A_{1}}\sqsubseteq \widetilde {A_{2}}\sqsubseteq \dots \sqsubseteq \widetilde {A_{k+1}}$ . In this chain, some of the sets $\widetilde {A_{j}}$ are repeated (this happens for all indices $j_{1},j_{2}$ with $j_{1}\sim j_{2}$ ). We let $H\subseteq \{i,\dots ,k+1\}$ be a minimal set of indices (with respect to inclusion) so that for all $i\leq j \leq k+1$ there is $h\in H$ with $\widetilde {A_{j}}=\widetilde {A_{h}}$ .

For all $G\subseteq H$ we let

$$ \begin{align*}B_{G}=\bigcap_{i\in G}\widetilde{A_{i}}.\end{align*} $$

Let

$$ \begin{align*}X=\{B_{G}\mid G\subseteq H\}\setminus\{\emptyset\},\end{align*} $$

and let

$$ \begin{align*}Y=\{B_{G}\in X\mid B_{G} \ \text{is minimal with respect to inclusion}\}.\end{align*} $$

Let $Y=\{B_{G_{1}},\dots ,B_{G_{l}}\}$ . Notice that the sets in Y are nonempty, disjoint, and that, for all $j=i,\dots ,k+1$ there exists $L_{j}\subseteq \{1,\dots ,l\}$ such that

$$ \begin{align*}\bigcup_{t\in L_{j}} B_{G_{t}}=\widetilde{A_{j}}.\end{align*} $$

Now let $\alpha \in \mathbb {R}_{>1}$ be such that

1. 1. For all $i\leq j_{1},j_{2}\leq k+1$ , if $m_{L}\left (\widetilde {A_{j_{1}}}\right )-m_{L}\left (\widetilde {A_{j_{2}}}\right )>0$ then

   (28) $$ \begin{align}\alpha\left(\mu\left(\widetilde{A_{j_{1}}}\right)- m_{L}\left(\widetilde{A_{j_{2}}}\right)\right)>l+s+r+1.\end{align} $$

2. 2. For allFootnote ¹⁰ $j\leq l$

   (29) $$ \begin{align}\alpha m_{L}\left(B_{G_{j}}\right)>m(l+s+r)>l+s.\end{align} $$

For all $j=1,\dots ,l$ let $C_{j}$ be a subset of $B_{G_{j}}$ with

(30) $$ \begin{align}|C_{j}|=c_{j}=\left\lfloor \alpha\beta m_{L}\left(B_{j}\right)\right\rfloor.\end{align} $$

We let

$$ \begin{align*}\Xi_{k+1}=\Xi_{k}\cup\bigcup_{j=1}^{l} C_{j}.\end{align*} $$

Claim. For all $i\leq j\leq k+1$

$$ \begin{align*} \alpha m_{L}\left(\widetilde{A_{j}}\right)-l\leq |A_{j}\cap\Xi_{k+1}| \leq \alpha m_{L}\left(\widetilde{A_{j}}\right)+r+s. \end{align*} $$

First, let us show that the claim entails that $\lambda _{k+1}=\Xi _{k+1}\cup \{A_{1},\dots , A_{k+1}\}\in \mathfrak {D}\left (\mathbb {A}\right )$ . In fact, let $1\leq j_{1} < j_{2} \leq k+1$ with $A_{j_{1}}\sqsubset A_{j_{2}}$ .

• • If $j_{1}<i$ and $j_{2}<k+1$ , then simply $|A_{j_{1}}\cap \Xi _{k+1}|=|A_{j_{1}}\cap \Xi _{k}|< |A_{j_{2}}\cap \Xi _{k}|\leq |A_{j_{2}}\cap \Xi _{k+1}|$ by inductive hypothesis.

• • If $j_{1}<i$ and $j_{2}=k+1$ , then $|A_{j_{1}}\cap \Xi _{k+1}|=|A_{j_{1}}\cap \Xi _{k}|\leq s$ whilst $|A_{k+1}\cap \Xi _{k}|\geq \alpha m_{L}\left (A_{j}\right )-l>s$ by Condition (30).

• • If $j_{1}\geq i$ then the hypothesis $A_{j_{1}}\sqsubset A_{j_{2}}$ entails that $ m_{L}\left (A_{j_{1}}\right )< m_{L}\left (A_{j_{2}}\right )$ . So

  $$ \begin{align*}|A_{j_{2}}\cap\Xi_{k+1}|-|A_{j_{1}}\cap\Xi_{k+1}|\geq \alpha m_{L}\left(\widetilde{A_{j_{2}}}\right)-l-\alpha m_{L}\left(\widetilde{A_{j_{2}}}\right)-r-s>0\end{align*} $$
  by Condition (28) and the fact that $ m_{L}\left (\widetilde {A_{j}}\right )= m_{L}\left (A_{j}\right )$ for all $i\leq j \leq k+1$ .

It remains to prove the claim. Let $i\leq j\leq k+1$ . Then

$$ \begin{align*} |A_{j}\cap\Xi_{k+1}| &= |A_{j}\cap S \cap \Xi_{k+1}|+\left|\left(A_{j}\setminus \left(S\cup \widetilde{A_{j}}\right)\right)\cap \Xi_{k+1}\right|+|\widetilde{A_{j}}\cap \Xi_{k+1}|\\ & = |A_{j}\cap S \cap \Xi_{k+1}|+r_{j}+\left|\bigcup_{t\in L_{j}} B_{G_{t}}\cap \Xi_{k+1}\right|\\ & = |A_{j}\cap S \cap \Xi_{k+1}|+r_{j}+\sum_{t\in L_{j}}c_{t}, \end{align*} $$

and we conclude as, by construction, $0\leq |A_{j}\cap S \cap \Xi _{k+1}|\leq s$ , $0\leq r_{j}\leq r$ and $\alpha m_{L}\left (\widetilde {A_{j}}\right )-l\leq \sum _{t\in L_{j}}c_{t}\leq \alpha m_{L}\left (\widetilde {A_{j}}\right )$ by Condition (30).⊣

Proof. For $i=1,2$ let

$$ \begin{align*} \lambda _{i}=\Xi ^{i}\cup \left\{ A_{1}^{i},\dots ,A_{l_{i}}^{i}\right\} \in \mathfrak{D}_{0}\left( \mathbb{A}\right)\!. \end{align*} $$

We set

$$ \begin{align*} F=\Xi ^{1}\cup \Xi ^{2}; \end{align*} $$

then, by Lemma 6.1, we can add points to F and get a set $\Xi \supset \Xi ^{1}\cup \Xi ^{2}$ so that Condition (26) is satisfied.⊣

Using Lemma 5.2, we define the label set

(31) $$ \begin{align} \mathfrak{B}(\mathbb{A}):=\overline{\mathfrak{D}\left( \mathbb{A}\right)}\!. \end{align} $$

A key observation for what follows is that, by Condition (29), if $A,B$ are two Lebesgue measurable sets with a positive measure in $\hat {\mathbb {R}}^{N}$ with $m_{L}(A)=m_{L}(B)$ then

(32) $$ \begin{align} st\left(\frac{\mathfrak{num}(A)}{\mathfrak{num}(B)}\right)=1. \end{align} $$

In fact, given $A,B$ let $\lambda =\Xi \cup \{A,B\}$ be given as in Lemma 6.1. By the claim we showed in the proof of Lemma 6.1, we have that (in the notations of the proof of Lemma 6.1, with an index $\lambda $ to denote the dependence from the construction)

$$ \begin{gather*} \alpha_{\lambda} m_{L}\left(A\right)-l_{\lambda}\leq |A\cap\Xi_{\lambda}| \leq \alpha_{\lambda} m_{L}\left(A\right)+r_{\lambda}+s_{\lambda}, \\ \alpha_{\lambda} m_{L}\left(B\right)-l_{\lambda}\leq |B\cap\Xi_{\lambda}| \leq \alpha_{\lambda} m_{L}\left(B\right)+r_{\lambda}+s_{\lambda}. \end{gather*} $$

Hence, as $m_{L}(A)=m_{L}(B)$ ,

$$ \begin{align*}\frac{\alpha_{\lambda} m_{L}\left(A\right)-l_{\lambda}}{\alpha_{\lambda} m_{L}\left(B\right)+r_{\lambda}+s_{\lambda}}\leq \frac{|A\cap\Xi_{\lambda}|}{|B\cap\Xi_{\lambda}|}\leq \frac{\alpha_{\lambda} m_{L}\left(A\right)+r_{\lambda}+s_{\lambda}}{\alpha_{\lambda} m_{L}\left(B\right)-l_{\lambda}},\end{align*} $$

and we conclude as, by Condition (29),

$$ \begin{align*}\alpha_{\lambda} m_{L}\left(A\right)\geq m_{\lambda}(l_{\lambda}+s_{\lambda}+r_{\lambda})-l_{\lambda};\end{align*} $$

hence $\lim _{\lambda \uparrow \Lambda } \alpha _{\lambda } m_{L}\left (A\right )$ is infinitely larger than $\lim _{\lambda \uparrow \Lambda }l_{\lambda }+s_{\lambda }+r_{\lambda }$ , as $\lim _{\lambda \uparrow \Lambda } m_{\lambda }$ is infinite. So

$$ \begin{align*}\lim_{\lambda\uparrow\Lambda}\frac{\alpha_{\lambda} m_{L}\left(A\right)-l_{\lambda}}{\alpha_{\lambda} m_{L}\left(A\right)+r_{\lambda}+s_{\lambda}}=\lim_{\lambda\uparrow\Lambda}\frac{\alpha_{\lambda} m_{L}\left(A\right)-l_{\lambda}}{\alpha_{\lambda} m_{L}(A)}\cdot \lim_{\lambda\uparrow\Lambda}\frac{\alpha_{\lambda} m_{L}\left(A\right)}{\alpha_{\lambda} m_{L}\left(A\right)+r_{\lambda}+s_{\lambda}}\sim 1,\end{align*} $$

and similar for $\lim _{\lambda \uparrow \Lambda }\frac {\alpha _{\lambda } m_{L}\left (A\right )+r_{\lambda }+s_{\lambda }}{\alpha _{\lambda } m_{L}\left (A\right )-l_{\lambda }}$ .

6.2 Cardinal numbers and numerosities

A property that is natural to expect, when one has a numerosity theory for all sets in $\Lambda $ , is that it must be coherent with cardinalities, namely it must satisfy the following property:

Cantor property: If $A,B\subset \Lambda \backslash \mathbb {A}$ then

(33) $$ \begin{align} \left\vert A\right\vert <\left\vert B\right\vert \Rightarrow \mathfrak{num} \left( A\right) <\mathfrak{num}\left( B\right)\!. \end{align} $$

Using the labelling $\mathfrak {B}(\mathbb {A})$ defined by (31), the following result holds:

Theorem 6.3. If $A,B\subset \mathbb {A}$ then

$$ \begin{align*} \left\vert A\right\vert <\left\vert B\right\vert \Rightarrow \mathfrak{num} \left( A\right) <\mathfrak{num}\left( B\right)\!. \end{align*} $$

Proof. Given two sets $A,B\subset \mathbb {A}$ with $\left \vert A\right \vert <\left \vert B\right \vert $ , we take a label $\lambda \supseteq \lambda _{0}:=\Xi \cup \left \{ A,B\right \} \in \mathfrak {B}(\mathbb {A}).$ Then, by (26)

$$ \begin{align*} \left\vert A\cap \lambda \right\vert =\left\vert A\cap \Xi \right\vert <\left\vert B\cap \Xi \right\vert =\left\vert B\cap \lambda \right\vert\!. \end{align*} $$

The conclusion follows taking the $\Lambda $ -limit.⊣

By Theorem 3.2, it follows that the numerosity function is well defined for every set belonging to the family

$$ \begin{align*} K=\left\{ E\in \mathbb{V}\left( F\right) \ |\ F\in \wp _{fin}\left( \mathbb{A }\right) \right\}, \end{align*} $$

since $\mathfrak {D}(\mathbb {A})$ provides a label to the elements of $K.$ In particular, by the Comparison Principle (Theorem 3.2(iv)) and (25), we have that for every set $E\subset \mathbb {R}^{N},$

$$ \begin{align*} \mathfrak{num}\left( E\right) =\mathfrak{num}\left[ \Psi \left( E\right) \right]. \end{align*} $$

Now, we want to extend the notion of numerosity to any set A in $\Lambda $ in such a way that the Cantor property (33) is satisfied. The simplest way to realize this task is to consider the family of infinite sets

$$ \begin{align*} \mathbb{S}:=\Lambda \backslash \left( \mathfrak{L}\cup \mathbb{A}\right) \end{align*} $$

and to assign a label to each of them. We can take an injective map

$$ \begin{align*} \Phi :\mathbb{S\rightarrow A} \end{align*} $$

and set

$$ \begin{align*} \ell (A)=\ell (\Phi \left( A\right) ). \end{align*} $$

Then, every set in $\Lambda \backslash \mathbb {A}$ has a label in $\mathfrak { B}(\mathbb {A})$ . By the Comparison Principle (Theorem 3.2(iv)), we get our desired final result:

Theorem 6.4. If $A,B\in \Lambda \backslash \mathbb {A}$ , then

$$ \begin{align*} \left\vert A\right\vert <\left\vert B\right\vert \Rightarrow \mathfrak{num} \left( A\right) <\mathfrak{num}\left( B\right)\!. \end{align*} $$

7.1 The general theory

Given a numerosity theory and a set $E\in \Lambda $ , we put

$$ \begin{align*} \mu _{\gamma }(E)=st\left( \frac{\mathfrak{num}\left( E\right) }{\mathfrak{ \gamma }}\right), \end{align*} $$

where $\gamma \in \mathbb {N}^{\ast }$ . $\mu _{\gamma }$ is called numerosity measure. As we will see, an interesting case occurs if you take $\mathfrak { \gamma =num}\left ( \left [ 0,1\right ) \right ) ^{d}$ with $d\in \mathbb {R} _{\geq 0}$ . In this case we will say that $\mu _{\gamma }$ is the canonical $ d$ -dimensional numerosity measure.

Proof. (i) This is a trivial consequence of the additivity of the numerosity.

(ii) By Theorem 3.2, we have that for all $N\in \mathbb {N}$ ,

$$ \begin{align*} \mathfrak{num}\left( A\right) \geq \mathfrak{num}\left( \mathop{\displaystyle \bigcup}\limits_{n=0}^{N}A_{n}\right) =\sum_{n=0}^{N}\mathfrak{num}\left( A_{n}\right)\!; \end{align*} $$

hence

$$ \begin{align*} st\left( \frac{\mathfrak{num}\left( A\right) }{\gamma }\right) \geq st\left( \sum_{n=0}^{N}\frac{\mathfrak{num}\left( A_{n}\right) }{\gamma }\right) =\sum_{n=0}^{N}st\left( \frac{\mathfrak{num}\left( A_{n}\right) }{\gamma } \right)\!; \end{align*} $$

therefore,

$$ \begin{align*} \mu _{\gamma }\left( A\right) \geq \sum_{n=0}^{N}\mu _{\gamma }\left( A_{n}\right)\!. \end{align*} $$

The conclusion follows taking the Cauchy limit in the above inequality for $ N\rightarrow \infty $ .⊣

7.2 Numerosity of the subsets of $\mathbb {R}$

In this section, we will show that $\mu _{\beta }$ agrees with the Lebesgue measure, namely, if E is a Lebesgue measurable set, then

(34) $$ \begin{align} m_{L}\left( E\right) =\mu _{\beta }\left( E\right) =st\left( \frac{\mathfrak{ num}\left( E\right) }{\beta }\right), \end{align} $$

where

(35) $$ \begin{align} \beta :=\mathfrak{num}\left( \left[ 0,1\right) \right)\!. \end{align} $$

First, let us show that this holds for intervals:

Moreover, we have the following property:

Proof. Let $E\in \wp (\mathbb {R})$ ; wlog, we assume $E\in \wp \left (\mathbb {R}_{\geq 0}\right )$ , as the result for a generic E will then follow easily by splitting $E=E^{+}\cup E^{-}$ . Let

$$ \begin{align*} E=\bigcup_{j\in\mathbb{N}} E_{j} \end{align*} $$

be a partition of E, with all $E_{j}$ ’s Lebesgue measurable. Let $\varepsilon =\frac {a}{b}\in [0,1)$ ; for N large enough we have

$$ \begin{align*} m_{L}\left( E\right) \leq m_{L}\left( \bigcup_{j=1}^{N}E_{j}\right) +\varepsilon. \end{align*} $$

Now $E\cap \left [ -1,-1+\varepsilon \right ] =\varnothing $ , so

$$ \begin{align*} m_{L}\left( E\right) \leq m_{L}\left( \bigcup_{j=1}^{N}E_{j}\cup \left[ -1,-1+\varepsilon \right] \right)\!. \end{align*} $$

By Property 27 of our labelling, we have

$$ \begin{align*} \mathfrak{num}\left( E\right) \leq \mathfrak{num}\left( \bigcup_{j=1}^{N}E_{j}\cup \left[ -1,-1+\varepsilon \right] \right)\!. \end{align*} $$

By Theorem 7.2

$$ \begin{align*}&\mathfrak{num}\left(\bigcup_{j=1}^{N}E_{j}\cup \left[ -1,-1+\varepsilon \right] \right)\\ & \quad =\sum_{j=1}^{N}\mathfrak{num}\left( E_{j}\right)+\mathfrak{num}\left([-1,-1+\varepsilon]\right)\sim \sum_{j=1}^{N}\mathfrak{num}\left( E_{j}\right) +\varepsilon \beta;\end{align*} $$

hence

$$ \begin{align*} \mu _{\beta }\left( E\right) \leq \sum_{j=1}^{N}\mu _{\beta }\left( E_{j}\right) +\varepsilon \leq \sum_{j=1}^{\infty }\mu _{\beta }\left( E_{j}\right) +\varepsilon. \end{align*} $$

The arbitrariness of $\varepsilon $ gives the desired inequality

$$ \begin{align*} \mu _{\beta }\left( E\right) \leq \sum_{j=1}^{\infty }\mu _{\beta }\left( E_{j}\right)\!. \\[-38pt] \end{align*} $$

⊣

We can now prove our desired final result:

The results of this section could be generalized to prove that, for any measurable set $A\subset \mathbb {R}^{N}$ ,

$$ \begin{align*} m_{L}(A)=st\left( \frac{\mathfrak{num}\left( A\right) }{\beta ^{N}}\right)\!. \end{align*} $$

Similarly, modifying the ordering $\sqsubseteq $ to handle Hausdorff measures instead of Lebesgue’s, one could consider the “fractal measure” of any fractal set $A\subset \mathbb {R}^{N}$ , defined as follows:

$$ \begin{align*} \mathfrak{m}_{d}(A)=st\left( \frac{\mathfrak{num}\left( A\right) }{\beta ^{d} }\right)\!,\ \ \ d\in \left[ 0,N\right]. \end{align*} $$

We are not going to study this fractal measure in detail here but, analogously to what has been done here with the Lebesgue measure, it would not be too difficult to check that $\mathfrak {m}_{d}(A)$ is equal to the normalized Hausdorff measure $H_{d}$ .

7.3 Numerosity and nonstandard measures

It is well known that the Lebesgue measure can be realized using a counting procedure based on hyperfinite sets: this is, e.g., at the core of the construction of Loeb measures, which is the most known and used of such constructions. Loeb measures were introduced in mid-70s, see [Reference Loeb20]; see also [Reference Ross22] for an overview of Loeb methods and applications, and [Reference Ross, Cutland, Di Nasso and Ross23] for an overview of other applications of nonstandard analysis in measure theory. To confront Loeb construction with our approach, here we shortly recall Loeb construction following Goldblatt’s presentation (see [Reference Goldblatt18, Section 16.8]).

Let N be an infinite hypernatural number, and let $S=\{\frac {k}{N}\mid -N^{2}\leq k\leq N^{2}, k$ hyperinteger $\}$ . Let $\wp _{I}(S)$ be the set of internal subsets of S, and for every $A\in \wp _{I}(S)$ let

$$ \begin{align*} m(A):=st\left( \frac{|A|}{N}\right)\!, \end{align*} $$

where $|A|$ denotes the internal cardinality of A. Then $m:\wp _{I}\left ( S\right ) \rightarrow \lbrack 0,+\infty ]$ is a finitely additive measure on $ \wp _{I}\left ( S\right ) $ . The Loeb measure is obtained by means of the usual Carathéodory extension procedure applied to m (we will denote also the Loeb measure by m). What Loeb proved is that the Lebesgue measure can be seen as a restriction of m, in the sense that for every Lebesgue measurable set X the Lebesgue measure $m_{L}(X)$ is equal to the Loeb measure of the so-called pre-shadow $st^{-1}(X)$ of X, namely

$$ \begin{align*} m_{L}(X)=st\left( m\left( st^{-1}\left( X\right) \right) \right)\!, \end{align*} $$

where $st^{-1}(X)=\{\xi \in S\mid st(\xi )\in X\}$ .

The similarity between our approach is that we have that, actually, $\mu _{\beta }$ is obtained as the standard part of a quotient similar to Loeb’s. In fact, $\mu _{\beta }\left ( A\right ) =st\left ( \frac {|A^{\ast }\cap \Gamma |}{|[0,1)^{\ast }\cap \Gamma |}\right ) $ , where:

1. 1. $|\cdot |$ denotes the internal cardinality of a set and

2. 2. $\Gamma $ is the hyperfinite set obtained by taking $ \lim _{\lambda \uparrow \Lambda }\lambda \cap \mathbb {R}$ .

However, in our approach the use of Carathéodory extension procedure, as well as of pre-shadows, is substituted with the choice of a particular labelling set, which can be equivalently seen as a particular choice of the hyperfinite set used in the quotient. A similar result in a general nonstandard setting was first obtained by Bernstein and Wattenberg (see [Reference Bernstein, Wattenberg and Luxemburg13]; see also [Reference Cutland15, Section 2] for a comparison of Bernstein–Wattenberg’s result and Loeb measures), who in fact proved that there exists hyperfinite subsets $S\subseteq \lbrack 0,1]^{\ast }$ such that for all Lebesgue measurable $A\subseteq \lbrack 0,1]$

$$ \begin{align*} m_{L}(A)=st\left( \frac{|A^{\ast }\cap S|}{|S|}\right)\!. \end{align*} $$

As we said before, Theorem 7.4 provides a new proof of the above result by taking

$$ \begin{align*} S=\lim_{\lambda \uparrow \Lambda }\ \left( \lambda \cap \mathbb{R}\right)\!. \end{align*} $$

Finally, the problem of the relationship between numerosities and Lebesgue measure in general has been addressed in [Reference Benci, Bottazzi and Di Nasso3, Reference Benci, Bottazzi and Di Nasso4]. In these papers, the authors introduced the notion of “elementary numerosity” (see [Reference Benci, Bottazzi and Di Nasso3, Definition 1.1]), that we recall:

The main connection between the “elementary numerosity” and Lebesgue measure is given by the following result, which is one of the instances of Theorem 3.1 in [Reference Benci, Bottazzi and Di Nasso4]:

Once again, Theorem 7.4 provides another proof of the above result, as $\mathfrak {num}$ , when restricted to $\wp \left ( \mathbb {R} \right ) $ is, in fact, an elementary numerosity on $\mathbb {R}$ .

The interest of Theorem 7.4 lies on the fact that it is based on a numerosity theory which satisfies many other additional properties.