Let there be an omega sequence of ordinals such that the first is the least $\Sigma_1$-admissible ordinal and the $n+1$st is the least $\Sigma_{n+1}$-admissible ordinal. What is the name, if any, of the union of all of these? Incidentally, $L$ at the level of this ordinal would be the minimal model of ZFC minus the power set axiom.
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asked Jul 21, 2013 at 20:51
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2$\begingroup$ I don't see why the last sentence is true. $\endgroup$– François G. DoraisJul 21, 2013 at 21:49
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This ordinal is not as large as you claim. In particular, the constructible universe up to this ordinal does not satisfy ZFC minus power set, and the ordinal is not even admissible.
To see this, let $\gamma_n$ be the least $\Sigma_n$-admissible ordinal (which I understand you to mean that $L_{\gamma_n}$ satisfies KP plus $\Sigma_n$-replacement, or equivalently, $\Sigma_n$-collection); and then the ordinal in question is $\gamma=\sup_n\gamma_n$ and the corresponding structure $L_\gamma$. But this structure does not satisfy KP, since the map $n\mapsto\gamma_n$ on domain $n\in \omega$ is $\Sigma_1$-definable in $L_\gamma$. The point is that whether some $L_\beta$ satisfies a certain sentence is a $\Delta_0$ espressible assertion about $L_\beta$, since all quantifiers are bounded by $L_\beta$, and whether $\beta=\gamma_n$ is a $\Sigma_1$-property of $\beta$ inside $L_\gamma$. Namely, $\beta=\gamma_n$ if and only if there is a transitive set $X$ that thinks it is $L_\beta$ and there is a satisfaction predicate on this set fulfilling Tarski's recursive definition of truth and according to this satisfaction predicate, the set $X$ satisfies KP and every instance of $\Sigma_n$-collection, and finally no smaller $\beta'\lt\beta$ satisfies all those.
answered Jul 21, 2013 at 21:51
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$\begingroup$ It seems that one can get it down to $\Delta_1$, since the structure $L_\beta$ is unique and so is its satisfaction predicate. So the relation $\beta=\gamma_n$ is a $\Delta_1$ property of $\beta$ and $n$ inside $L_\gamma$. $\endgroup$ Jul 21, 2013 at 22:06
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$\begingroup$ Would the least admissible ordinal above be at the level where ZFC minus power set has a model? $\endgroup$ Jul 21, 2013 at 22:26
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2$\begingroup$ @FrodeBjørdal Unless I'm very confused, the next admissible ordinal strictly above any given $\alpha$ is never $\Sigma_2$-admissible. $\endgroup$ Jul 21, 2013 at 22:55
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$\begingroup$ Let S be a set that does not contain zero such that S is not definable by ZFC minus the power set, and let f be a function from N onto S. Is the union of the sets in the omega sequence s(0)=the f(0)'th Sigma (0)-admissible ordinal and s(n+1)=the f(n+1)'th Sigma(n+1) admissible ordinal and ordinal where L models ZFC minus power set? $\endgroup$ Jul 21, 2013 at 23:25
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$\begingroup$ I would expect that the answer is negative. Note that your $f$ need not grow very rapidly. If $A$ is any set of natural numbers that isn't definable in ZFC minus power set, then you could take $S$ to be $\{2n:n\in A\}\cup\{2n+1:n\notin A\}$. $\endgroup$ Jul 22, 2013 at 23:00