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I have a curiosity: is it true that every countable ordinal $\lambda$, there is a monotone function $F : \mathcal{P}(\mathbb{N}) \to \mathcal{P}(\mathbb{N})$, whose closure ordinal is $\lambda$?

I've been trying to prove this on my own but no luck. I'm starting to think it might be false.

Here $\mathcal{P}(X)$ means the power set of the set $X$. Also, monotone and closure ordinal represent the usual definitions.

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The power set is a common concept taught at the undergraduate level, but what is a "closure ordinal"? –  Henning Makholm Mar 21 '13 at 0:09
    
@Henning: See the first two paragraphs of Section $2$ of this paper for the definitions that I think Danx is using. –  Brian M. Scott Mar 21 '13 at 0:25
    
@Brian: Thanks, that was clearer than what I could google up immediately. –  Henning Makholm Mar 21 '13 at 0:29

1 Answer 1

If "closure ordinal" means what Brian M. Scott suggested in his comment, you should be able to take $\Gamma:\mathcal P(\lambda)\to\mathcal P(\lambda)$ defined by $$ \Gamma(A)=\begin{cases} A\cup\{\min(\lambda\setminus A)\} & \text{when }A\ne \lambda \\ A & \text{when } A =\lambda \end{cases} $$

Then, in the notation of Brian's reference, $\Gamma^\alpha = \alpha$ for all $\alpha\le\lambda$, and the closure ordinal of $\Gamma$ is exactly $\lambda$.

Since $\lambda$ is assumed countable, you can cojugate $\Gamma$ with the assumed bijection between $\lambda$ and $\mathbb N$ to get a $F:\mathcal P(\mathbb N)\to\mathcal P(\mathbb N)$ with the same structure as $\Gamma$, and therefore the same closure ordinal.

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