# $\wp^{(\omega)}(A)=?$

Let $A$ be a set,

$$\wp^{(0)}(A)=A$$ $$\wp^{(n+1)}(A)=\wp(\wp^{(n)}(A))$$ But what sense does $\wp^{(\alpha)}(A)$ make where $\alpha$ is a limit ordinal number? The most natural way is let $$\wp^{(\alpha)}(A)=\lim_{\xi \uparrow \alpha}\wp^{(\xi)}(A),$$ but what is this 'limit' means? Note that $\wp^{(n)}(A)$ probably not the subset of $\wp^{(n+1)}(A)$.

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Is that symbol $\wp$ often used for power set? – GEdgar Jun 14 '12 at 14:28
@GEdgar Yes, $\mathcal{P}$ and $\wp$ are both familiar. Wiki_powerset – Popopo Jun 14 '12 at 14:59

If you’re going to define it at all, the only reasonable definition is that $$\wp^{(\eta)}(A)=\bigcup_{\xi<\eta}\wp^{(\xi)}(A)$$ when $\eta$ is a limit ordinal.

For $n\in\omega$, elements of $\wp^{(n)}(A)$ are (intuitively speaking) sets that have members of $A$ buried inside $n$ layers of curly braces. You can’t have things buried exactly $\omega$ layers deep, because you need an outer layer of braces, but you can have things buried at most $\omega$ layers deep, though of course that just turns out to be things buried less than $\omega$ layers deep.

It’s more natural to look at the operation $A\mapsto A\cup\wp(A)$: then it’s clear that at limit stages you just want to take the union of the earlier stages.

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This seems to be related to cumulative hierarchy in the sense that $V_\alpha=\wp^{\alpha}(\emptyset)$. – Martin Sleziak Jun 14 '12 at 14:03
Yes, it is a definition. But as Mr.Scott noticed, it will turns out to be things buried less than $\omega$ layers deep. And that is what I worried about. If $A=\{apple\}$ is a box which contains an apple, than for $\wp^{(n)}(A)$ we have to open at least $n+1$ boxes to get an apple. However, if we packed them more and more, suddenly an apple occurred at the outermost box, it seems incredible. – Popopo Jun 14 '12 at 14:52
So I consider about whether we can accept $\omega$ layers. If we see a set as a transitive tree with a valuation, for example $\{a\}$ can be seen as $$((v_0 \rightarrow v_1),V)$$ s.t. $V(v_0)=\emptyset$ and $V(v_1)=a$. Then can we see $\{\{\ldots a \ldots\}\}$($\omega+1$ many layers) as $$((v_0 \to v_1 \to \ldots v_{\omega}),V)$$ s.t. $V(v_n)=\emptyset$ whereas $V(v_\omega)=a$? – Popopo Jun 14 '12 at 14:52

Another possibility would be to take $\mathscr P^{(\alpha)}(A)$ to be the set of sequences $(X_\beta)_{\beta<\alpha}$ such that $X_\beta \subseteq \mathscr P^{(\beta)}(A)$ for all $\beta$ and $X_\beta = \bigcup X_{\beta+1}$ whenever $\beta+1<\alpha$. That is isomorphic to the usual iterated power set when $\alpha$ is finite, but generalizes more smoothly to $\omega$.

(It looks like it will need a bit more work to stay smooth above $\omega$, though).

(In fact it needs a bit more work in order to work at all, since as it stands it doesn't depend on $A$ at all).

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It looks nice for me, let us sleep on it. – Popopo Jun 14 '12 at 17:53