Let $ T$ be a theory written in Mono-sorted first order logic with equality, with extralogical primitives: $<, \in$.
Define: $x \leq y \iff x < y \lor x=y$
Axioms:
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$\textbf{Well ordering: }\\\textit{Transitive:} \ x < y \land y < z \to x < z \\ \textit{Connective:} \ x \neq y \leftrightarrow (x < y \lor y < x) \\ \textit{Well founded:} \ \exists n \in x \to \exists n \in x \forall m \in x (n \leq m)$
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$\textbf{Finiteness: }\exists n \in x \to \exists n \in x \forall m \in x (m \leq n)$
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$\textbf{Sets: } \forall n \exists! x \forall m (m \in x \leftrightarrow m \leq n \land \phi)$, if $x$ is not free in formula $\phi$.
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$\textbf{Closure: } x \in y \to x < y$
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$\textbf{Emergence: }\max ( x \setminus y) < \max (y \setminus x) \to x < y $
Where, $\max x$ is the maximal element of $x$ with respect to relation $<$; and $x \setminus y = \{m \in x \mid m \not \in y\}$.
Theory $T$ is an example of an interwoven theme of order and set theory. It is synonymous with $\sf PA$. [see here]
If we remove Finiteness, and add the existence of a limit number, that is:
- $\textbf{Infinity: } \exists l \neq \emptyset: \forall x < l \exists y: x < y < l$
, call the resultant theory $“T – {\sf Fin + I}" $.
What would be the set theory that is synonymous to $T – {\sf Fin + I}$?
My guess is that it would be $\sf Z + Ranks$ plus adding a bijection $j$ to the language of $\sf Z$ from sets to von Neumann ordinals in a manner such that sets are assigned higher ordinals than their elements. The Ranks axiom is $(\forall \alpha \exists V_\alpha)$ where $V_\alpha = \{x \in j^{-1}(\beta) \mid \beta < \alpha \}$. So, the theory would prove Global Choice, Transitive closures, Ranks, etc…
[EDIT:] The above theory had been proved inconsistent as shown in the accepted answer. A possible salvage along the original intentions is to restrict axiom of Emergence to apply to natural numbers (i.e. objects prior to the first infinite number). We call $T$ with this restriction as $T^r$. We add the following axiom:
- $\textbf {Respective: } \forall n \in x (m > n) \land m \in y \to x < y$.
So, the question is re-stated in connection with theory $T^r – {\sf Fin + I + Respective}$.
Best Answer
We can define $0$ and $S(n)$ as in my previous answer - $0$ is the smallest number, $S(n)$ is the smallest number greater than $n$. My proof that these exist did not use Finiteness.
Define $n$ to be a natural number if, for any set $x$, if $0 \in x$, and for all $k \in x$, $S(k) \in x$, then $n \in x$. For a limit number $l$, all natural numbers are less than $l$: $0$ is clearly less than $l$, and for any $n < l$, there exists $y$ such that $n < y < l$, and therefore $S(n) \leq y < l$. So the set of natural numbers exists, and I will write it as $\mathbb{N}$.
If $n, m$ are natural numbers and $n < m$, then $\mathbb{N} \setminus \{m\} < \mathbb{N} \setminus \{n\}$, by Emergence: $(\mathbb{N}\setminus\{m\})\setminus(\mathbb{N}\setminus\{n\}) = \{n\}$, and similarly $(\mathbb{N}\setminus\{n\})\setminus(\mathbb{N}\setminus\{m\}) = \{m\}$, and $\max\{n\} < \max\{m\}$.
For all $n \in \mathbb{N}$, $\mathbb{N} \setminus \{n\} \leq \mathbb{N} \setminus \{0\}$, so the set $\{\mathbb{N} \setminus \{n\} : n \in \mathbb{N}\}$ exists. For any element $\mathbb{N} \setminus \{n\}$ of this set, $\mathbb{N} \setminus \{S(n)\}$ is a smaller element, so this set has no minimum. This contradicts well-foundedness, so $T - {\sf Fin + I}$ is inconsistent.