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On the one hand, there is no shortage of exposition about real numbers within the constructible hierarchy $L$, for example this thread. On the other hand you have the model $L(\mathbb{R})$, which would seem redundant if $L$ already contained the real numbers (or I guess to be precise, a set order-isomorphic to the real numbers).

After puzzling over this a bit, what I think I understand is: $L$ doesn't actually contain all the real numbers (the entire set $\mathbb{R}$ as it's normally thought of in analysis, etc.), but rather $L$ contains only what one might call the "constructible" real numbers.

Is this right, or is there some other subtlety that I'm missing?

NikS
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Well, it is consistent that $L$ contains all the real numbers, e.g. if we assume $V=L$ holds true in our universe. In fact, if we force over $L$ by a forcing that does not add new reals (e.g. adding a branch to a Suslin tree; shooting a club into some stationary subset of $\omega_1$; adding some Cohen subset to a regular uncountable cardinal $\kappa$), then all the reals in that universe will also be in $L$.

But at the same time, if we add a Cohen real to $L$, then we very clearly have new reals that are not in $L$. If we collapse $\omega_1$ to be countable, then the set of reals in $L$ is itself countable, so clearly some new reals have been added.

Large cardinal hypotheses can also tell us something about the real numbers and $L$. If $0^\#$ exists, which follows from the existence of a measurable cardinal, then we can actually show that the reals in $L$ only form a countable set. In fact, $0^\#$ is a kind of tipping point in "how close the universe is to being $L$". So sufficiently large large cardinal axioms imply that the universe is quite far from $L$, and this starts already with the real numbers.

We often work in models like $L(\Bbb R)$ when studying descriptive set theory, and these are taken in models that are either generic extensions of $L$, e.g. the Solovay model, or under the assumption of large cardinals, e.g. infinitely many Woodin cardinals and then some.

On the other hand, we can prove that if there is a perfect set of reals which are all in $L$, then all the reals are in $L$.

So, which is it? Are all the reals in $L$ or not? Well, that depends on your universe.

Asaf Karagila
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  • Thanks Asaf. To flesh out one specific case: Let's say that my set universe is the "standard" cumulative hierarchy (a.k.a. the von Neuman universe) $V$ (so that $V$ is definitely not equal to $L$). In this scenario the "true" set of "all" the reals is (presumably) the set of reals defined in $V$. If I understand you correctly, this would be a scenario where $L$ would be said to not contain "all" the reals? – NikS Oct 27 '22 at 00:37
  • You're misunderstanding things. The von Neumann universe is just an internal construction to any model of ZF which captures the universe (or, if you prefer, it is the largest transitive class model of ZF inside a universe of ZF-Foundation). Indeed, working in L, the von Neumann hierarchy "grows faster" in a sense, but it will not produce sets that are outside of L itself, simply because it's all done internal to your universe which happens to be L. – Asaf Karagila Oct 27 '22 at 09:29
  • To be more specific, when I speak of the "standard" von Neumann universe, what I mean is specifically the definition described here and characterized in Hrbacek & Jech as containing "precisely all the well-founded sets". And I'm reserving the symbol "$V$" to refer exclusively to the universe defined in exactly this way (so, I'm specifically excluding any scenario where we assert "axiomatically" that $V$ equals a constructible hierarchy like $L$, or equals a forcing extension of it). – NikS Oct 31 '22 at 07:32
  • The question at hand is: based on the specific $V$ just defined, does every subset of $\omega$ that appears $V$ also appear in $L$? The one thing I see mentioned so far that might lead the answer to be "yes" is the condition that "there is a perfect set of reals which are all in $L$". So, what can be said about whether $L$ does or does not contain a perfect set of reals? Does it depend on whether one assumes some large cardinal axiom, or something like that? – NikS Oct 31 '22 at 07:32
  • @NikS: But you are not telling me the information that is needed. Are you assuming $V=L$? Are you assuming the existence of large cardinals, if so, which ones? Do you assume that CH fails? All you're saying is that you want to know if the von Neumann hierarchy provides reals that are not in $L$, to which I responded. The von Neumann hierarchy is internal to any model of ZF. So $L$ has one, as will any other model. So your question just makes no sense. – Asaf Karagila Oct 31 '22 at 07:42
  • OK, I think I see what you're getting at. I think clarifying this merits separate sub-question, which I have created here – NikS Nov 03 '22 at 06:44