Consider a first-order language $L$ and the infinitary language $L(\omega_1,\omega)$ obtained from $L$ by allowing infinite conjonctions and disjonctions. Let $X$ be an infinite set of sentences, including (possibly infinitely many) infinitary formulae and infinitely many finite formulae (let $\Delta$ be an infinite set of finite formulas, $\Delta \subset X$). Let $\mathcal{A}$ be the least admissible set containing $X$ ($X \subset \mathcal{A}$). Is it possible that $\mathcal{A}$ does not contain $\Delta$ as an element? ($\Delta \notin \mathcal{A}$).
1 Answer
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I suspect you didn't mean literally what you asked, but if you did, then the answer is yes. You required only that $X$ be a subset of $\mathcal A$, so even if $X$ consists entirely of finitary sentences and $\Delta$ is all of $X$, it wouldn't follow that it's an element of $\mathcal A$.
If you require $X$ to be an element of $\mathcal A$, to avoid the situation in the preceding paragraph, the answer is still yes, since you require $\Delta$ just to be some subset of $X$ consisting of finitary formulas. So, for example, $X$ could be the set of all finitary formulas of $L$, in which case $\mathcal A$ would contain just the hyperarithmetical subsets of $X$, while $\Delta$ could be some far more complicated subset of $X$.
If you require $X$ to be an element of $\mathcal A$ and require $\Delta$ to be the set of all the finitary formulas in $X$, then I believe the answer is no, because the notion of "finitary formula" will be $\Delta_1$-definable.
answered Aug 25, 2012 at 15:35