Proof: (related to "Transitive Recursion") 

• By hypothesis, $X$ is a set and $\Omega$ is the proper class of all ordinals and $\omega\in\Omega$ is a given ordinal number.
• It is sufficient to show that the function $f:\mathbb \Omega\to X$ defined by specifying the value of $f(\alpha)$ for all $\alpha\subseteq \omega$ and for all $\beta\supset\omega$ with a recursion formula $f(\beta)=\mathcal R(f(\alpha)\mid \alpha \subset\beta)$ is well-defined (i.e. unique) for all ordinal numbers.
• Assume, $f$ defined like this is not well-defined for some ordinal numbers.
  • Since ordinal numbers are well-ordered, there is the smallest ordinal number $\omega_0\in\mathbb \Omega$ for which $f$ is not well-defined.
  • But then for $\alpha \subset \omega_0$ the values $f(\alpha)$ are well-defined and $f(\omega_0)=\mathcal R(f(\alpha)\mid \alpha \subset \omega_0).$
  • This is a contradiction.
• Therefore, $f$ is well-defined for all ordinal numbers.

| | | | created: 2020-07-12 10:22:37 | modified: 2020-07-12 10:22:39 | by: | references: [8635]