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 $a = b$.


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.

  • $\begingroup$ You can just add in an empty element and the equivalence relation that identifies $0$ and $1$ is still a counterexample. $\endgroup$ – Eric Wofsey Mar 18 at 4:58
  • $\begingroup$ And if I assume that $\epsilon$ is well-founded? $\endgroup$ – Popov Florino Mar 18 at 5:02
  • $\begingroup$ Then both axioms are equivalent to ordinary extensionality (for Axiom 2, prove that $[a]_{\sim}=\{a\}$ by $\epsilon$-induction on $a$). $\endgroup$ – Eric Wofsey Mar 18 at 5:05
  • 1
    $\begingroup$ It's obviously not equivalent to well-foundedness. I would suggest you actually try and think about some examples before you ask more questions. $\endgroup$ – Eric Wofsey Mar 18 at 5:12

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