3
$\begingroup$

A monotone operator $\Gamma: \mathcal{P}(A) \to \mathcal{P}(A)$ is an operator such that, if $X \subseteq Y \subseteq A$, then $\Gamma(X) \subseteq \Gamma(Y)$. A monotone operator is $\kappa$-based if, whenever $x \in \Gamma(Y)$, $x \in \Gamma(X)$ for some $X \subseteq Y$ of cardinality less than $\kappa$. Define by transfinite recursion $\Gamma^\lambda$ such that $\Gamma^\lambda = \bigcup\limits_{\mu<\lambda}\Gamma^\mu \cup \Gamma(\bigcup\limits_{\mu < \lambda} \Gamma^\mu)$ (i.e. $\Gamma^\lambda$ is basically the result of iterating $\Gamma \lambda$ times); $\Gamma^ {<\lambda}$ is an abbreviation for $\bigcup\limits_{\mu<\lambda}\Gamma^\mu$. A fixed point of $\Gamma$ is a set $K \subseteq A$ such that $\Gamma(K) = K$; the set $I(\Gamma) = \bigcap K \{K \subseteq A \; | \; K \text{ is a fixed point}\}$ is the least fixed point of $\Gamma$. The closure ordinal for $\Gamma$ is the least $\lambda$ such that $I(\Gamma) = \Gamma^\lambda$.

Using the above, it's now possible to prove the theorem in the title of this question (the proof here is due to Aczel, in his article on inductive definitions in Barwise's Handbook of Mathematical Logic):

Let $\Gamma$ be a $\kappa$-based monotone operator where $\kappa$ is regular. Then the closure ordinal of $\Gamma$ is $\kappa$.

Proof: By monotonicity, it's clear that $\Gamma^{< \kappa} \subseteq \Gamma^\kappa$, so it suffices to show that $\Gamma^\kappa \subseteq \Gamma^{<\kappa}$. Let $\Gamma$ be as in the hypothesis of the theorem and consider an arbitrary $x \in \Gamma^\kappa = \Gamma(\Gamma^{<\kappa})$. By the hypothesis, since $\Gamma$ is $\kappa$-based, $x \in \Gamma(X)$ for some $X \subseteq \Gamma^{\kappa}$ of cardinality less than $\kappa$. By the regularity of $\kappa$, $X \subseteq \Gamma^{< \lambda}$ for some $\lambda < \kappa$, so $x \in \Gamma(\Gamma^{<\lambda}) \subseteq \Gamma^{<\kappa}$, which is what we wanted to prove.

Question: what is the work being done by the regularity of $\kappa$?

I'm still getting used to working with these ordinal properties, so if someone could help me understand how this work, preferably explaining why the theorem fails for $\kappa$ singular, it'd be much appreciated.

$\endgroup$
2
$\begingroup$

You have $x\in\Gamma^\kappa=\Gamma(\Gamma^{<\kappa})$, so $x\in\Gamma(X)$ for some $X\subseteq\Gamma^{<\kappa}$ (not $\Gamma^\kappa$) of cardinality less than $\kappa$. Thus, for each $y\in X$ there is a $\beta_y<\kappa$ such that $y\in\Gamma^{\beta_y}$. Let $B=\{\beta_y:y\in X\}$; then $B\subseteq\kappa$, and $|B|<\kappa$ Finally, $\kappa$ is regular, so it's not the limit of fewer than $\kappa$ smaller ordinals: if $\lambda=\sup\{\beta+1:\beta\in B\}$, then $\lambda<\kappa$. This is where regularity of $\kappa$ is used. (Clearly $X\in\Gamma^{<\lambda}$.)

$\endgroup$
  • $\begingroup$ Thanks. The idea is that, if $\kappa$ is singular, then it could turn out that $\lambda > \kappa$? $\endgroup$ – Nagase Aug 22 '15 at 16:50
  • 1
    $\begingroup$ @Nagase: Not quite: it could turn out that $\lambda=\kappa$, in which case the fact that $X\in\Gamma^{<\lambda}$ does you no good: it just says that $x\in\Gamma(\Gamma^{<\kappa})=\Gamma^\kappa$, which you already knew. In order to get $x\in\Gamma^{<\kappa}$, you need to get $X\subseteq\Gamma^{<\lambda}$ for some $\lambda<\kappa$, so that $x\in\Gamma(\Gamma^{<\lambda})=\Gamma^\lambda\subseteq\Gamma^{<\kappa}$. $\endgroup$ – Brian M. Scott Aug 22 '15 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.