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In the Constructible Universe $L$, the power set operation $P(X)$ becomes relativized to the definable power set operation $P^L(X)$ of all definable subsets of $X$. But if $X\subset\omega$, then there are only countably many definable subsets of $X$, so we must have $P^L(X)$ countable!?

Something has clearly gone wrong here. We know that $|L_\alpha|=|\alpha|$ for every ordinal, therefore in order for there to be uncountably many subsets of $\omega$ (which we know to be true!) we must have some subsets of $\omega$ with rank at least $\omega_1$. However I don't see how this is possible, as when we apply the relativized power set operation to a set, its rank only increases by one in the Universe $L$. Sorry for being silly, I know I must be overlooking something fundamental here.

Elie Bergman
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    The definitions allow parameters, and definability is local, meaning that at each stage we only local at sets definable over the model indexed by that stage. – Andrés E. Caicedo Aug 11 '18 at 16:17
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    The notation $\mathcal P(X)^M$ or $\mathcal P^M(X)$ usually means "the interpretation of the power set operation inside the class $M$", in the case that $M$ is transitive this is just $M\cap\mathcal P(X)$. – Asaf Karagila Aug 11 '18 at 16:20
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    Incidentally, the sense in which every real in $L$ is definable is: for every real $r\in L$, there is a finite tuple $\alpha_1,...,\alpha_n$ of $L$-countable ordinals such that $r$ is definable in $L$ with parameters $\alpha_1,...,\alpha_n$. In fact, we could use just one parameter, and replace "definable in $L$" with "definable in $L_{\omega_1^L}$;" also this generalizes to arbitrary sets in the sense that all sets in $L$ are definable in $L$ (or a long enough initial segment of $L$) from an ordinal parameter. Put another way: there is no undefinability in $L$ except coming from ordinals. – Noah Schweber Aug 11 '18 at 19:28

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Your confusion is rooted by the mistake that being constructible does not mean being definable.

It is true that the constructible hierarchy constructs each step by taking definable subsets. But just because a set is definable over one structure doesn't mean it is definable over a different one. Not to mention, that the definable subset are taken with parameters anyway.

The point, in summary, is that being constructible means that you are definable with parameters over some initial segment of the hierarchy. But since there are many initial segments in which subsets of $\omega$ (or any ordinal) are added, there are uncountably many sets and all is fine.

 

Also, note that for $\omega$ to have uncountably many subsets, you don't need subsets of $\omega$ with rank at least $\omega_1$, just that for unboundedly many ordinals below $\omega_1$, we add new subsets to $\omega$. This is indeed what happens.

Asaf Karagila
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  • To be specific, are you saying that in $L$ the power set of $\omega$ is $P(\omega)\cap L$ rather than $P^L(\omega)$? – Elie Bergman Aug 11 '18 at 16:17
  • Those are the same things. I am saying that $\mathcal P(\omega)^L\neq\operatorname{Def}(\omega)$. – Asaf Karagila Aug 11 '18 at 16:17
  • I see. That was the error. I don't know if you're familiar with this, but as the ranks of the subsets of $\omega$ are cofinal in $\omega_1$, how could we in practice calculate the ranks of particular sets like say, the even numbers, or the primes, etc... I'm interested in what types of sets end up falling in each particular rank. – Elie Bergman Aug 11 '18 at 16:20
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    Recursion theory, or computability theory in its modern name, often deals with these kind of questions. Note that since $L_\omega=V_\omega$ has a natural interpretation of Peano axioms, every arithmetic subset of $\omega$ already appears in $L_{\omega+1}$. You can search this site for a question from not "too long ago" (let's say past year?) about gaps, which is the key word you want to search for. – Asaf Karagila Aug 11 '18 at 16:23
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    Here it is. – Andrés E. Caicedo Aug 11 '18 at 16:38
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    @ElieBergman From a computability-theoretic perspective, each "step up" in the $L$-hierarchy is intuitively like the $\omega$th Turing jump. A finer-grained hierarchy can be developed, closer to the "iterate the jump" picture (and in particular for all computable ordinals $\alpha$ the sets of natural numbers appearing by level $\omega+\alpha$ are those computable from $0^{(\alpha)}$). As we push further and further up $\omega_1^L$ (aside: the ranks of reals are cofinal in $\omega^L$, not necessarily $\omega_1$), things get a bit odd however and we have to be careful. (cont'd) – Noah Schweber Aug 11 '18 at 19:23
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    These "set-theoretic iterated jumps" are the mastercodes. They can be generalized somewhat beyond $L$. I think the text to look at here is Hodes' Jumping through the transfinite, which is a bit technical but does a good job giving the overall picture. – Noah Schweber Aug 11 '18 at 19:25
  • Thanks for these responses. I've rewritten a question here dealing more specifically with the 'fine structure' of $L$; https://math.stackexchange.com/questions/2879792/ranks-of-reals-in-the-constructible-universe-l

    I should preface my question by saying that I have a very very limited knowledge of computer science, so if you could give a brief description of any technical terms like $0^{\alpha}$, that would be very much appreciated!

     – Elie Bergman Aug 11 '18 at 21:30