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Proof

(related to Theorem: Trichotomy of Ordinals)

• Let \(\alpha, \beta\) be fixed ordinals and \(\gamma=\alpha\cap \beta\).
• By the lemma about equivalence of set inclusion and element inclusion of ordinals, it follows from \(\gamma\subseteq\alpha\) that \(\gamma\in\alpha\) or \(\gamma=\alpha\), and from \(\gamma\subseteq\beta\) that \(\gamma\in\beta\) or \(\gamma=\beta\).
• Altogether, we have the following cases:
  • \(\gamma\in\alpha\) and \(\gamma=\beta\). From this case it follows that \(\beta\in\alpha\).
  • \(\gamma\in\beta\) and \(\gamma=\alpha\). From this case it follows that \(\alpha\in\beta\).
  • \(\gamma=\beta\) and \(\gamma=\alpha\). From this case it follows that \(\alpha=\beta\), and by the axiom of foundation, neither \(\alpha\in \beta\) nor \(\beta\in \alpha\) is possible.
  • \(\gamma\in\beta\) and \(\gamma\in\alpha\). However, this case is not possible, since otherwise we would have \(\gamma\in\alpha\cap\beta=\gamma\), contradicting the axiom of foundation.
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References

Bibliography

1. Hoffmann, Dirk W.: "Grenzen der Mathematik - Eine Reise durch die Kerngebiete der mathematischen Logik", Spektrum Akademischer Verlag, 2011
2. Hoffmann, D.: "Forcing, Eine Einführung in die Mathematik der Unabhängigkeitsbeweise", Hoffmann, D., 2018