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I have an issue with a proof given "Théorie axiomatique des ensembles" by J.L Krivine.

Ordinal definition given of $\alpha$:

  1. $"x \in y"$ defines a strict, total, well-foundedness relation R in $\alpha$
  2. $x \in \alpha \Rightarrow x \subset \alpha$

We are trying to prove:

Given two ordinal sets $\alpha$ and $\beta$, one of these cases is true: $\alpha \in \beta$ or $\beta \in \alpha$ or $\alpha = \beta$

The proof starts by considering $\xi = \alpha \cap \beta$

The sketch of the proof is:

$\xi \: initial \: segment \: of \: \alpha \Rightarrow \xi = \alpha \: or \: \xi \in \alpha$

$\xi \: initial \: segment \: of \: \beta \Rightarrow \xi = \beta \: or \: \xi \in \beta$

4 cases are studied. The last case $\xi \in \alpha$ and $\xi \in \beta$ is excluded because it will lead to $\xi \in \xi$ (impossible for ordinals)

My question is: how do we know $\xi \ne \emptyset$ ? Why two random ordinal sets have a common element ? Or does the proof holds if $\xi = \emptyset$ ?

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    $\begingroup$ Where does the proof uses that $\xi\neq\varnothing$? $\endgroup$ – Asaf Karagila Jun 13 at 8:12
  • $\begingroup$ If $\xi = \emptyset$ then would it still be an initial segment of $\alpha$ ? $\endgroup$ – g.lahlou Jun 13 at 8:17
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    $\begingroup$ Is the empty set a subset of $\alpha$? $\endgroup$ – Asaf Karagila Jun 13 at 8:17
  • $\begingroup$ The empty set is a subset of $\alpha$ but it does not prove it is an element of $\alpha$, does it ? Every initial segment of $\alpha$ should be an element of $\alpha$, but is $\emptyset$ an element of every ordinal? $\endgroup$ – g.lahlou Jun 13 at 8:22
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    $\begingroup$ Is $\varnothing$ a transitive set? Is $\in$ a well-order of it? $\endgroup$ – Asaf Karagila Jun 13 at 8:23
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Okay, I think I may have an answer to my question. Thanks @Asaf for pointing me to the right direction. According to a comment in this thread, if an ordinal is not the emptyset, it contains the emptyset. So, if $\xi$ is the emptyset, it is still an initial segment of $\alpha$ and $\beta$ considering none of them is an emptyset. If $\alpha$ or $\beta$ is the emptyset, the theorem we are trying to prove is trivial.

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