# Are these two extensionality-axioms equivalent?

Let $$\epsilon$$ be a binary relation on a set $$U$$.

A subset $$A \subseteq U$$ is called $$\epsilon$$-transitive iff $$a \mathrel{\epsilon} b \wedge b \in A \Rightarrow a \in A$$ for all $$a,b \in U$$. For $$a \in U$$ we define the transitive closure $$T_\epsilon(a) \subseteq U$$ of $$a$$ as the smallest $$\epsilon$$-transitive superset of $$\{x \in U; x \mathrel{\epsilon} a\}$$. Further we say that $$a,b \in U$$ are $$\epsilon$$-isomorphic iff there exists an isomorphism $$\varphi : (T_\epsilon(a) \cup \{a\},\epsilon) \to (T_\epsilon(b) \cup \{b\},\epsilon)$$ with $$\varphi(a) = b$$.

Axiom 1:

If $$a,b \in U$$ are $$\epsilon$$-isomorphic, then $$a = b$$.

and

Axiom 2:

If $$\sim$$ is an equivalence relation on $$U$$ such that $$[a]_\sim = [b]_\sim \quad\Rightarrow\quad \{[x]_\sim; x \in U, x \mathrel{\epsilon} a\} = \{[y]_\sim; y \in U, y \mathrel{\epsilon} b\}$$ for all $$a,b \in U$$, then $$\sim$$ has to be the equality $$=$$.

Question: Are these axioms equivalent?

No, Axiom 2 is much stronger than Axiom 1. For instance, consider $$U=\{0,1\}$$ with $$\epsilon={\leq}$$. This satisfies Axiom 1 but fails Axiom 2 because the equivalence relation with one equivalence class satisfies the condition in Axiom 2. Indeed, the equivalence relation with one equivalence class will be a counterexample to Axiom 2 for any $$(U,\epsilon)$$ that has no "empty" element.
• You can just add in an empty element and the equivalence relation that identifies $0$ and $1$ is still a counterexample. – Eric Wofsey Mar 18 at 4:58
• And if I assume that $\epsilon$ is well-founded? – Popov Florino Mar 18 at 5:02
• Then both axioms are equivalent to ordinary extensionality (for Axiom 2, prove that $[a]_{\sim}=\{a\}$ by $\epsilon$-induction on $a$). – Eric Wofsey Mar 18 at 5:05