Is there a notion similar to cardinality that respects Euclid's axiom?

Question

Is there a notion similar to cardinality that respects Euclid's axiom of "The whole is greater than the part"? I know that a set can have the same cardinality as a proper subset of that set. Basically, what I am looking for is a preorder $\preceq$ on the universe of sets that satisfies at least the following axioms:

1. The empty set $\emptyset$ is the smallest element of the preorder.
2. If two sets $A$ and $B$ are finite and also have the same number $n$ of elements, then $A$ and $B$ precede each other.
3. If $A$ is a proper subset of $B$, then $A$ strictly precedes $B$.
4. For any two sets $A$ and $B$, either $A$ precedes $B$, or $B$ precedes $A$. In other words, the preorder is a linear preorder.
5. If $A$ is a proper subset of $B$, where $B$ has exactly one more element than $A$, then there is nothing in the preorder strictly between $A$ and $B$.

Has such a notion been defined and studied in the mathematical research literature?

Answer 1

This can be done, though we’ll need to assume the axiom of global choice and work in NBG for this proof.

First, recall that $A \Delta B$ is defined to be $(A \setminus B) \cup (B \setminus A)$. This is the “symmetric difference” operator, and it is associative and commutative.

Note that the relation “$A \Delta B$ is finite” is an equivalence relation. We write this as $A \cong B$. Let $[A]$ be the class of $A$ in $V / \cong$ (which we can construct using Scott’s Trick).

Notation: if $A \preceq B$ and $B \preceq A$, we write $A \approx B$. Furthermore, we write “strictly precedes” as $\prec$.

Let’s explore some properties $\preceq$ should have if it satisfies (1) through (5). Note that if we have $x, y \notin A$, then $A \cup \{x\} \approx A \cup \{y\}$. For by (4), we may suppose WLOG that $A \cup \{x\} \preceq A \cup \{y\}$. Then by (3), note that $A$ strictly precedes $A \cup \{x\}$. Therefore, by (5), $A \cup \{x\}$ does not strictly precede $A \cup \{y\}$; therefore, $A \cup \{x\} \approx A \cup \{y\}$.

We can generalise this result. If $C, D$ are both disjoint with $A$, and if $|C| = |D|$ is finite, then $A \cup C \approx A \cup D$. This follows by applying the above result and induction on $|C|$.

In slightly more generality, if we have $C, D$ both disjoint from $A$ and $|C| < |D|$ where both are finite, then $A \cup C \prec A \cup D$. In full generality, if $C$ and $D$ are finite sets both disjoint from $A$, then $A \cup C \preceq A \cup D$ iff $|C| \leq |D|$.

Now we have established what to do when we are comparing two sets $A, B$ in the same $\cong$ class. But what restrictions are there for when $A$ and $B$ are not in the same class?

For starters, suppose $A \subseteq B$ and $B \setminus A$ is infinite. Then it’s easy to see that for all $C \in A$, $C \prec B$. This is because we can write $C \subseteq A \cup D$ for some finite $D$. Take some $E \subseteq B \setminus A$ such that $|E| = |D|$. Then $C \subseteq A \cup D \approx A \cup E \subsetneq B$, so $C \prec B$.

In fact, we can slightly generalise this. If $A \subseteq B$, $B \setminus A$ is not empty, $A \cong C$, and $B \cong D$, then $C \prec D$. For in particular, consider $A’ = A \cap D = A \setminus (B \setminus D)$. Then $A’ \cong A \cong C$, $A’ \subseteq D$, and $D \setminus A’$ is infinite. By the previous paragraph, $C \prec D$.

This leads us to the following definition. We say $[A] < [B]$ if and only if $A \setminus B$ is finite and $B \setminus A$ is infinite. Then we have defined a strict partial order $<$ on $V / \cong$. And if $[A] < [B]$ then $A \prec B$.

We have discovered some critical properties of $\preceq$. Now, using global choice, we construct a preorder $\preceq$ satisfying these properties.

First, we can use the axiom of global choice to extend $<$ to a total order $<‘$ on $V / \cong$.

Then, we define $A \preceq B$ to mean $[A] <‘ [B] \lor ([A] = [B] \land |A \setminus B| \leq |B \setminus A|)$. Note that under this definition, $A \approx B$ if and only if $|A \setminus B|$ and $|B \setminus A|$ are equal and finite.

We see that this $\preceq$ satisfies axiom (3). For if $A \subseteq B$, then either $B \setminus A$ is infinite, in which case $[A] < [B]$ and consequently $A \prec B$, or it is finite, in which case $[A] = [B]$ and $A \approx B$ iff $B = A$.

Clearly, $\preceq$ also satisfies (4). This is straightforward.

$\preceq$ also satisfies (5). Given $x \notin A$, anything $C$ such that $A \preceq C \prec A \cup \{x\}$ must satisfy $[C] = [A]$. A careful analysis of this case shows that either $C \approx A$ or $C \approx A \cup \{x\}$.

And conditions (1) and (2) are redundant.