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Can we add a primitive binary relation $<$ to the language of ZFC, and add the following axioms on top of axioms of ZFC?

  1. Well ordering: $<$ is a well ordering on the universe.
  2. Membership: $x \in y \to x < y$

Where 1. is the following schema:

$x < y \to y \not < x \\ x < y < z \to x < z \\ x \neq y \leftrightarrow [x < y \lor y < x] \\ \exists x \phi(x) \to \exists x \phi(x) \land \forall y (\phi(y) \to x \leq y)$

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    $\begingroup$ What about $V=L$ with this ordering? $\endgroup$
    – Nate Eldredge
    Apr 17, 2020 at 13:27
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    $\begingroup$ Simply well-order each $V_{\alpha+1} \setminus V_\alpha$ individually and then put these orders on top of each other. $\endgroup$
    – Jonathan Schilhan
    Apr 17, 2020 at 13:40

1 Answer 1

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Whenever $E$ is a well-founded and set-like relation on a class $A$, and there is some (class) well-order $\triangleleft$ on $A$ then there is a (class) well-order on $A$ containing $E$.

Namely consider the rank function $\rho \colon A \to ON$ associated with $E$. This is definable. Then we define $x <' y$ iff $\rho(x) < \rho(y)$ or $\rho(x) = \rho(y)$ and $x \triangleleft y$, which is a well-order on $A$ containing $E$, since if $x E y$, then $\rho(x) < \rho(y)$.

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  • $\begingroup$ let me try fathom what you are saying, lets take for example $ZF + V=L$, now this supports the existence of even a definable well order $\triangleleft$ on the whole universe $L$, now lets take the usual ranking function $r$, then I can well order $L$ by taking $ x R y \iff r(x) < r(y) \lor (r(x)=r(y) \land x \triangleleft y)$, so $R$ would have $\in$ as a subclass of it. right? $\endgroup$
    – Zuhair
    Apr 17, 2020 at 15:21
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    $\begingroup$ Exactly. In fact, the cannonical well-order of $L$ that you can find in many textbooks already is of the form that you require. Moreover for any model of ZFC there is an NGB model with the same sets as the original model and a global well order (which is a class). $\endgroup$
    – Jonathan Schilhan
    Apr 17, 2020 at 15:37
  • $\begingroup$ Thanks Jonathan. $\endgroup$
    – Zuhair
    Apr 17, 2020 at 15:38

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