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I have three three questions, the first two of which probably have the same answer and the third of which is more vague.

For a set $A$ let $L_\alpha(A)$ be the constructible universe up to $\alpha$, built from $A$ as a set (and not a predicate). Further let $X = (B, f)$ where $B$ is a transitive set and $f$ is a bijection from $\omega$ to $B$.

Also assume that the background universe has whatever large cardinals you would like (or that would be helpful). In particular though there is at least one inaccessible cardinal in $L$.

(1) Suppose $L_\alpha\models ZFC$. Is it the case that $\omega_1^L = \omega_1^{L_\alpha}$?

(2) Suppose $L_\alpha(X)\models ZFC$. Is it the case that $\omega_1^{L(X)} = \omega_1^{L_\alpha(X)}$?

(3) If the answer to (1), (2) is yes, is there any simpler way for $L_\alpha$ to know that $\omega_1^{L_\alpha} = \omega_1^L$ (other than $L_\alpha\models ZFC$)?

Finally I will just make one observation to highlight why this question isn't trivial. If you replace $ZFC$ with $KP$ then there are many countable admissible sets $L_\alpha\models KP$ with countable (in $L$) ordinals $\beta\in L_\alpha$ such that $L_\alpha \models \omega_1 = \beta$.

Thanks

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What is $\kappa$? Just an ordinal? A cardinal? An $L$-cardinal? The smallest ordinal $\alpha$ with $L_\alpha\models ZFC$ is countable in $L$: –  Goldstern Oct 28 '12 at 6:12
    
@ Goldstern: I changed $L_\kappa$ to $L_\alpha$. Thanks –  Nate Ackerman Oct 28 '12 at 13:17
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2 Answers

up vote 7 down vote accepted

The answer is no. If there is a transitive set model $M$ of set theory (and this is all you need), then if $\alpha$ is its height (that is, if $\alpha=\mathsf{ORD}^M$), then $L_\alpha$ is a model of $\mathsf{ZFC}+V=L$. Note that the assumption is strictly stronger than the existence of an $\omega$-model of $\mathsf{ZFC}$, which in turn is strictly stronger than the mere consistency of $\mathsf{ZFC}$, but it is strictly weaker than the existence of inaccessible cardinals.

Now work in $L$, and consider a countable elementary substructure $Y\preceq L_\alpha$. Note that $Y$ is well-founded and satisfies $V=L$. Its collapse is then $L_\beta$ for some $\beta$ that, by necessity, is countable in $L$.

(Note that this also addresses (2), by taking $X=\emptyset$.)


By the way, this highlights some of the difficulties a challenger of "$V=L$" must face. Even if there are no inaccessibles, the model $M$ we began with may perfectly well be constructible, and satisfy that there are supercompact cardinals, or whatever. About this, you may also want to look at this post by Joel.

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Andres Caicedo has answered Questions 1 and 2 as stated, but, just in case you intended $\kappa>\omega_1^L$ in Question 1 (since that's obviously necessary for the proposed conclusion), let me add that this is also sufficient for an affirmative answer to Question 1. The point is that, if an ordinal $\alpha$ is countable in $L$ then a bijection between $\alpha$ and $\omega$ appears in the construction of $L$ well before stage $\omega_1^L$. This is essentially contained in Gödel's proof of GCH in $L$. It uses much less than the full strength of ZFC in $L_\kappa$; you just need enough set theory to talk sensibly about countability of ordinals.

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