What is the minimal theory for subset relation to be connective over those kinds of ordinals?

Question

If we define ordinal as being a set of all transitive proper subsets of it. Formally

$ord(X) \iff X=\{Y \subsetneq X \mid trs(Y) \}$

Where: $trs(X) \iff \forall x \in X \,(x \subseteq X)$

What is the minimal set of axioms of ZFC needed to prove that all ordinals are connected by $\subseteq$? That is: $$\forall \alpha \, \forall \beta: ord(\alpha) \land ord(\beta) \implies \alpha \subseteq \beta \lor \beta \subseteq \alpha$$

Is it the axiom of intersection, i.e. the assertion that $A \cap B$ exists for any sets $A,B$?

Answer 1

Zuhair commented a nice proof on this answer that runs as follows:

Assume towards a contradiction that $\alpha$ and $\beta$ are $\subseteq$-incomparable ordinals, then consider $\alpha\cap\beta$. This set is an ordinal, since it consists of the transitive proper subsets of $\alpha$ that are also transitive proper subsets of $\beta$. Since $\alpha\cap\beta$ is a transitive proper subset of $\alpha$, it must be an element of $\alpha$, and the same logic applies to $\beta$, so we get $\alpha\cap\beta\in\alpha\cap\beta$. But $\alpha\cap\beta$ isn't a transitive proper subset of $\alpha\cap\beta$, so $\alpha\cap\beta$ can't be a member of $\alpha\cap\beta$, which is a contradiction.

The rudimentary functions are a class of very weak set functions that are even weaker than the primitive recursive set functions, and $f(x,y)=x\cap y=x-(x-y)$ is rudimentary. Rudimentary set theory is a very weak set theory that's Kripke-Platek set theory without the collection schema, and it proves the rudimentary functions total, so it proves $\alpha\cap\beta$ exists. Even though it doesn't have powerset, $ord$ is still absolute between transitive models of rud. set theory, since we can formalize $ord(X)$ as the statement $\forall y((y\subsetneq X\land trs(y))\leftrightarrow y\in X)$, and since we only consider transitive models, $trs$ and $y\subsetneq X$ are absolute between them, and the $y$ such that $y\subsetneq X$ are in the intersection of all such models containing $X$.

But rud. set theory also has foundation, so even a theory as weak as rudimentary set theory is overkill for proving this theorem. I believe we can also remove the axiom of union and the theorem is still provable, since no supersets of $\textrm{max}\{\alpha,\beta\}$ are considered.

I'm not sure if we can remove much more than union (e.g. removing so much that it no longer follows that all rud. functions are total) and still prove the theorem. As of this edit I have attempted several times and failed to construct a model of "empty set+extensionality+union" in which the statement of the theorem is false, but I wasn't able to construct such a model.

Answer 2

I wrongly thought that the axiom of intersection over the logical rules of the language of ZFC (i.e. first order logic with equality and membership), would be enough to prove that those kinds of ordinals are connected by the inclusion relation $\subseteq$. To see that, examine the following:

Example: Let $Q=\{0, Q\}; a=\{0, Q,a\}; b=\{0, Q,b\}; a^*=\{0, Q,a^*\}; b^*=\{0, Q,b^*\}$, where $Q,a,a^*,b,b^*$ are all distinct from each other. Take the sets $A=\{0, Q,a,a^*\}; B=\{0, Q,b,b^*\}$. Now $A,B$ would be ordinals, yet they are incomparable by subsethood.

This can be remedied by adding an extra-condition on the definition of ordinals given here, as to have all of their elements be nonself membered, that is:

Define: $ord(X) \iff X=\{Y \subsetneq X \mid trs(Y) \land Y \not \in Y\}$.

The axiom of intersection is precisely the following:

$\forall A \, \forall B \, \exists C \forall x (x \in C \iff x \in A \land x \in B)$

This will assure the existence of an empty set if disjoint sets exist. Accordingly no nonempty ordinals can be disjoint.

Proof: Let $\alpha,\beta$ be ordinals along the above definition. For a proof by negation, suppose that they are not $\subseteq $-comparable, take $\alpha \cap \beta$, if it is in itself, then it is in both ordinals, so directly violating non-self membership of elements of ordinals; if it is not in itself then since it is a proper transitive subset of both ordinals, so it would be an element of both ordinals and of course of itself by definition of intersection, violating the assumption of it being not in itself.

So rules of first order logic together with axiom of intersections (at least for ordinals) is enough for those modified ordinals to be connected by $\subseteq$. I don't know if there is even weaker condition than that.